my dear students before starting this chapter there are few basic terminologies which you must know and those basic terminologies are from the chapter redox reactions so I'll try to cover all the terms which are from the chapter redox reactions but we use those terms in this particular chapter so let's try to cover those Basics first and then I'll be starting this particular chapter electrochemistry right because redox Electro they are enter linked let's get to know all the terminologies of redox which are frequently used in this particular chapter right and the first terminology with which we
are going to start that is something which you call as oxidation that is something which you call as oxidation I'm sure majority of you guys will be already knowing what oxidation is all about oxidation as per modern definition is defined as loss of electrons my dear students any specie which is going to lose the electrons we say that particular species under go oxidation any specie which under go loss of electrons basically which are which is going to lose the electrons we are going to say that particular specie is undergoing oxidation what it means exactly before
that I'll tell you one more thing whenever in a chemical reaction you need to check whether a substance is undergoing oxidation or reduction what you guys are going to do at that time just remember one simple Point increase in the oxidation state increase in the oxidation state of an element is termed as oxidation whenever you see an element is undergoing increase in the oxidation state you'll directly say that particular element is undergoing oxidation now let's get to know what it exactly means have a look people for example for example I have got an element M
for example this is the element M I'm assuming this m is losing let's say one electron and it is getting converted into M positive right so I'm assuming that this m it has lost one electron it is getting converted into M positive now my dear students just try to analyze one simple thing if I ask you what is the charge present on this metal here what is the charge present on this element here it is zero what is the charge present on this m here it is + one tell me one thing from 0 to
+1 what it means has the oxidation state of M increased or decreased I would say the oxidation state of M is increasing from 0 to +1 and that particular element whose oxidation state increases during the reaction we say that particular specie is undergoing oxidation so I would say this m is undergoing oxidation in this particular case right as simple as that so I taken the element M it is losing one electron getting converted into M positive the initial the initial oxidation of this m the initial oxidation state of this that is zero final oxidation state
is plus + one so 0 to+ one means inre increas in the oxidation state and increase in the oxidation state is something which we call as oxidation and what is oxidation loss of electrons so you can categorically say this m has lost one electron and got converted into M positive right similarly there's a term called as reduction I'm sure all of you must be familiar with this particular term as well so dear students how do you define the term reduction if oxidation involves loss of electrons then I would say reduction is going to involve the
gain of electrons reduction is going to involve the gain of electrons number one number two decrease in the oxidation state of an element decrease in the oxidation state of an element is termed as reduction as well whenever you see the oxidation state of any element decreasing with time do remember that particular spey will be undergoing reduction for example let's say I have got an element a and I'm giving one electron to this element this a after gaining one electron let's assume that it's getting converted into a negative right so first of all is this process
involving gain of electrons absolutely this particular process is involving gain of electrons so this particular process is something which I'll be calling as reduction right now at the same time if I ask you what is the oxidation state of a here what is the charge on a here it's zero what is the charge on this particular a it is minus one check it out whether the oxidation state of a is increasing or decreasing 0 to minus1 0 to minus1 means decrease in the oxidation state and decrease in the oxidation state is something which you call
as reduction so I must say this a is undergoing reduction over here and reduction as you already know that involves gain of electrons so I must say this a is gaining one electron and getting converted into a negative I hope this particular basic scenario is clear to everyone right so you got to know what oxidation is all about you got to know what reduction is all about now my dear students you might have heard about something called as oxidizing agent and reducing agent let's get to know what oxidizing agent is what reducing agent is and
how do we identify them in a chemical reaction let's get to know about that this again completely Basics only okay perfect guys there is a term called as oxidizing agent there's a term called as oxidizing agent oxidizing agent is also called as oxidant it is also called as oxidant now how do you define the term oxidizing agent my dear students oxidizing agent is the one or I'll say that specie that specie which oxidizes others that spey which oxidizes others but itself under goes a reduction but itself under goes reduction that particular specie which is going
to oxidize others and itself will undergo reduction we'll be calling that particular spey as oxidizing agent what it means you'll get the idea in some time before making you understand that I'll be writing the definition of reducing agent as well reducing agent or you'll be calling it as reductant as well you'll be calling this as reductant as well now how do you define the reducing agent and how do you define the reducing agent let's get to know about that reducing agent it is that specie it is that species which which reduces others that speci which
reduces others but itself under goes oxidation so that particular specie which is going to reduce others but itself is undergoing oxidation I'll be calling that as the reducing agent let's try to understand these two definitions over here try to understand what exactly I'm going to say this is something important which I'm going to tell you now see guys for example I have got a reaction and the reaction is like this zinc solid plus copper di positive Aquis let's say it gives zinc di positive Aquis plus copper solid let's assume this is the reaction okay now
if I ask you what is the charge present on this zinc over here it is zero that means its oxidation state is zero the charge present on copper right here is plus two that means its oxidation state is plus two the charge present on this zinc over here is plus two it's oxidation state plus two the charge present on copper is zero its oxidation state is zero now my dear students I want you to check exactly whether the oxidation state of zinc is increasing or decreasing what do you think I would say the initial oxidation
state of zinc is zero the final oxidation state of zinc is plus2 so I would say the oxidation state of zinc is increasing with time and any such specie whose oxidation state increases with time we say that particular specie is undergoing oxidation so you got to know the zinc since its oxidation state is increasing so it is undergoing oxidation if it is undergoing oxidation definitely it will be losing the electrons right it will be losing the electrons and let me tell you that particular specie which itself under go oxidation that is something which you call
as a reducing agent or reductant so you identified in this particular reaction your zinc solid that is behaving like the reducing agent right now at the same time check it out whether the oxidation state of copper is increasing or decreasing I would say the oxidation state of copper is changing from plus 2 to 0 what is it increase or decrease it is decrease in the oxidation state plus 2 to 0 decrease in the oxidation state and decrease in the oxidation state is something which you call as a reduction so I would say this copper diosi
is undergoing a reduction yeah and my dear students that particular species which under go reduction that will be involving gain of electrons so I would say this copper di positive will be will be gaining certain electrons right now you must be thinking from where those electrons are going to come those electrons are going to come from the zinc which is undergoing oxidation zinc is losing electrons and copper dios will be definitely gaining electrons right and is students that particular specie which under goes reduction that's something which you call as oxidizing agent I hope this particular
thing is clear to everyone now you should be able to identify which reactant in a particular reaction behaves like the oxidizing agent and which reactant in a chemical reaction it behaves like the reducing agent is this clear to everyone quickly in the chats let me know quickly in the charts this was example number one let's have a look on example number two as well see guys for example I'm writing a reaction like this H2 gas plus H2 gas plus 2 * AG positive let's say it gives these are some reactions which I'm writing as such
right so let's say it gives 2 * AG positive Aquis and 2 * AG solid let's say this this is the reaction which I've taken into consideration and in this particular reaction I need to check which particular species is the oxidizing agent and which one is the reducing agent let's getting about that if I ask you what's the charge on this hydrogen over here you'll say it's zero charge on Silver over here it is+ one charge on this hydrogen over here it's plus one charge on this silver over here it's zero right now check it
out whether the oxidation state of hydrogen is increasing or decreasing from 0 to + one increase in the oxidation State 02 Plus for increase in the oxidation state and increase in the oxidation state is something which you call as oxidation so this H2 is undergoing oxidation here and the one which under goes oxidation that's something which you'll be calling as reducing agent in the similar way take about the check the oxidation state of this particular silver + one + 1 to 0 + 1 to 0 means decrease in the oxidation state and decrease in the
oxidation state is something which you call as reduction decrease in the oxidation state something which you call as reduction and that particular species which under goes reduction that's something which you will be calling as oxidizing agent right so let me know quickly in the chats if you got to know how to identify oxidizing and reducing agent quickly my dear student in the chats quickly in the charts everyone yeah quickly quickly n watch the the redox reaction chapter first right and mole concept and equilibrium right I think you're here for the first time quickly let me
know in the chats all the things are clear perfect one more example and with with this we'll start the chapter exactly I'm writing the reaction for example M4 Nega plus f d positive let's say it's giving MN di positive and with this you are writing f a tri positive as well now let me know in the chats which specie will be undergoing oxidation and which specie will be undergoing reduction quickly quickly people in the chats quickly people in the chats see guys have a look first of all if I ask you what is the oxidation
state of this magnes over here you will have to calculate it it's going to be X oxygen shows min-2 oxygen State there are four oxygen atoms net charge over here is minus one so the value of x will come out to be + 7 I would say this magnes oxidation state is+ 7 in this case what is the charge present on this iron over here it's plus two charge present on the magnes it's plus two and charge present on iron right here is + three right now check exactly whether the ox oxidation state of magnes
is increasing or decreasing + 7 to +2 decrease in the oxidation state decrease in the oxidation state is something which we call as reduction so I would say this M4 negative is undergoing reduction in this particular case and that particular spey which under go reduction that is something which you'll be calling as oxidizing agent as simple as that in the similar way take the oxidation state of iron it is changing from +2 to + 3 plus2 to+ 3 means in increase in the oxidation state increase in the oxidation state is something which you call as
oxid oxidation and that particular specie which under go oxidation that's something which I'll be calling as the reducing agent quickly let me know in the chats if all these things are clear quickly my dear students do remember one thing no doubt the oxidation state of magnes is decreasing but you will never say that magnes is undergoing reduction you will say this whole compound is undergoing reduction you will say this whole compound is undergoing reduction right okay perfect is it clear people one last example one last example let me give you so that I'll get that
satisfaction you got to know how to identify the oxidizing and reducing agent one last example with this let's say I'm writing the reaction like this cr207 D Nega cr207 D Nega Plus s n di positive let's say it gives chromium Tri positive plus SN + 4 this is the reaction which I mentioned over here now my dear students I would want you guys to I would want you guys to check which reactant among the two is going to be the oxidizing agent and which one is going to be the reducing agent how exactly you guys
are going to check first of all I'll calculate the oxidation state of this particular chromium right it has to be 2x -4 equal -2 so the value of x will come out to be plus 6 right what is the oxidation St of this 10 over here it's plus2 oxidation St of chromium it's + three oxidation state of 10 it is +4 right now check whether the oxidation state of chromium is increasing or decreasing plus 6 to + 3 plus 6 to + 3 decrease in the oxidation state decrease in the oxidation state is something which
you call as reduction so I would say this particular species this cr207 D negative it is undergoing reduction and that specie which under goes reduction that is going to behave like the oxidizing agent similarly plus 2 to +4 plus 2 to +4 again it is increase in the oxidation state increase in the oxidation state is something which you call as oxidation and that particular speci which undergoes oxidation that's something which I'll be calling as a reducing agent perfect yeah right people manprit this is class 12th class 12th I think this is chapter number four I
believe perfect people all right now if all these basic things are clear to everyone let's try to go into the first topic of the electrochemistry chapter are you ready for that are you all ready for that are you all ready for that perfect guys so let's get going let's get started the first topic which I'm mentioning on the screen that is electrochemical cell electrochemical cell now what this electrochemical cell exactly is all about what is electrochemical cell let me tell you electrochemical cell it is the device basically the device which converts the device which converts
which converts either chemical energy into electrical energy or electrical energy into chemical energy so electrochemical cell it is basically a device which is going to convert either chemical energy into electrical or electrical into chemical how exactly I'll let you know in some time right how exactly this is going to happen how chemical energy is going to get converted into electrical energy how electrical energy will be converting into chemical energy that's something which we shall be discussing in detail in this particular chapter right perfect now this electrochemical cell it's actually of two types this electrochemical cell
it's actually of two types my dear students one is called as galvanic cell or you call it as the volt cell as well galvanic or volt cell number two that is something which you call as electrolytic cell that's what you call as electrolytic cell so what is the galvanic cell and what is the electrolytic cell how do we Define them right now I'll be just giving you the definition and some basic points which you need to remember first of all before going into the details of this particular chapter so basically in this chapter we have
to see exactly how Daniel cell works right how Daniel cell works how electrolytic cell works these are two wor topics which we have to cover in this particular chapter only but right now I'll just give you the definition of these two cells then we'll go into the details one by one yeah okay my dear students when I talk about the galvanic cell galvanic cell it is a device it is a device which converts it is a device which converts chemical energy into what into electrical energy it is a device which converts chemical energy into electrical
energy point number one how exactly we shall be doing we shall be discussing that in some time number one number two in case of Gonic cells what happens exactly let me tell you in case of Gonic cells spontaneous cell reactions take place spontaneous cell reactions will take place in case of galonic cell how exactly again you'll get to in sometime number three number three for any process to be spontaneous you must be knowing for any process to be spontaneous Delta G for the system at constant pressure and temperature that has to be negative so I
would say if spontaneous cell reactions will be taking place in the galmic cell so I would say it's Delta G for the reaction will be less than zero and my dear students in this galonic cell I'll be using two electrodes one I'll be calling as anode and another one I'll be calling as cathode one I'll be calling as anode another one I'll be calling as cathode dear students remember one simple thing the rod of the anode the rod of the anode in galonic cell it will carry the negative charge and the rod of the cathode
in the galvanic cell it will be carrying the positive charge what is the logic behind that why the rod of anode will carry negative Rod of cathode will carry positive there's a logic behind that again that's something we shall be discussing in some time right and do remember one simple thing at anode always oxidation takes place which involves loss of electrons and at cathode always reduction takes place which involves gain of electrons so these are few basic pointers which you need to know right which we shall be discussing one by one in detail in this
particular chapter related to what related to galmic cell okay similarly my dear students electrolytic cell this is one more topic which we have to cover in detail but right now let me give you certain basic things about this electrolytic cell as as well let me tell you this electrolytic cell it is just the reverse it converts electrical energy into chemical energy it converts electrical energy into chemical energy and in this particular cell the cell reactions the cell reactions are carried out are carried out by supplying cell reactions are carried out by supplying external voltage external
voltage again what that means keep it as such we shall be discussing in some time at the same time if Cell reactions are carried out by supplying external voltage I shall be saying that cell reactions are non-s spontanous in this particular case so Delta G has to be positive right similarly in this particular cell I'll be using using two electrodes one will be calling as anode and one we shall be calling as cathode but over here there's only one difference the rod of the anode hair carries the positive charge and Rod of the cathode hair
carries the negative charge why it happens again there's a logic we shall be discussing in some time but one thing that is certain that at anode always oxidation is going to take place and at cathode always reduction is going to take please so these are few points which you should remember first of all okay right these are some points which you should remember first of all right people now now our actual topic in the today's session that will be exactly how this galonic cell works how exactly it converts chemical energy into electrical energy that will
be our first topic actually in the today's discussion so let me know first of all whatever I discussed till now is every single thing clear is every single thing clear quickly yeah I just gave you the overview of the things I did not teach anything in detail yet now it's the time to see things in detail but before that I would want you guys to let me know if every single thing is clear quickly people everyone everyone all right now people now people let me tell you one important thing the cell which we have to
discuss that is something which you call as galvanic cell right the typical example of the galvanic cell the most common example of the galvanic cell that is Daniel cell So Daniel cell what is it it is the most typical example it's the most common example of the GIC cell what this Daniel cell exactly is going to do this Daniel cell since it is the most common example of the gallic cell this Daniel cell it is going to convert chemical energy into what it is going to convert chemical energy into electrical energy now the point is
how how this Daniel say which is the most common example which is the most common example of the galic cell how it's going to convert chemical energy into electrical energy that is the discussion here right but again before that few points which you must know see guys first of all before going into the details of the D if I ask you how electrodes are made you should be exactly knowing how electrodes exactly are made for example people let's say I have to make make a zinc electrode how do I make a zinc electrode how do
I make it similarly if I ask you how do we make the copper electrode how copper electrodes exactly are made let's get to know about this how electrodes are made first of all then only you can understand the concept of the Daniel so try to understand what exactly I'm going to say my dear students since we are going to make we are planning to make the zinc electrode so what I'll be doing I'll be taking a container let's assume this is the container and in this container I'm going to keep a solution I'm going to
keep a solution containing containing the salt of zinc for example I have kept zinc sulfate in the container let's say I kept zinc sulfate in the container my dear students this zinc sulfate since it is an electrolyte so it would have got dissociated into its ions zinc die positive and S so4 D negative right so basically in this particular container you have got zinc die positives as well as s so4 d negatives so these are the two types of ions which are present in the container now at the same time since I have to make
a zinc electrode so what I'll be doing now now my dear students I'll be placing a zinc Rod I'll be placing I'll be dipping a zinc Rod this is the zinc Rod this is the zinc Rod so what exactly I have done I have placed zinc Rod into a solution which contains its own ions on ions means which contains zinc di positive ions right whenever you need to make an electrode of certain metal what exactly you need to do you need to take the rod of that particular metal and you have to insert that Rod
into the solution which contains the ions of the same matter that's what I've done over here I've taken zinc Rod I inserted a zinc Rod into a solution which contains the ions of zinc my dear students this whole setup over here I'll be calling as zinc electrode this whole setup I'll be calling as the zinc electrode right in the similar way let's say you have to make the copper electrode how exactly you'll be making the copper electrode again you'll be doing the same procedure you'll be taking a container let's assume this is the container and
my dear students in this particular container I'm going to keep I'm going to keep a solution I'm going to keep a solution containing an electrolyte right and that electrolyte should be the the salt of copper let's say that electrolyte is copper sulfate let's assume that electrolyte is copper sulfate so my students this copper sulfate it would have got dissociated completely into its ions for example its ions are copper di positive and S so4 DGA so in this particular solution we have got copper di positive ions we have got s so4 d negative ions as well
now what exactly I should be doing since I'm going to make the electrode of copper so what I'll be doing I'll be taking a copper rod and I'll be inserting that copper Rod into a solution I'll be inserting this copper Rod into a solution containing its own ions so this whole setup I'll be calling as electrode isn't it simple isn't it simple people right so for example whenever you need to make the electrode of some metal what exactly you'll be doing you'll be taking Rod of that particular metal and that Rod of a particular metal
that has to be introduced into a solution containing ions of the same metal the whole setup you shall be calling as the electrode right now if I ask you can you make can you make electrode of hydrogen can you make electrode of hydrogen say yes or no see I showed you how to make the electrode of zinc I showed you how to make the electrode of copper can you make the electrode of hydrogen right so that means you have to take the hydrogen Rod now how come it's possible to take the hydrogen Rod hydrogen is
a gas right perfect that's something which we have to discuss in detail how to make the hydrogen electrode but right here I'll just give you the idea of how do we make the hydrogen electrode just the quick glimpse of how do we make the hydrogen electrod see guys what exactly you should be doing you'll be taking a rod which is going to be either platinum or graphite you'll be taking a rod which is going to be platinum or graphite let's say I have taken Platinum my dear students there is property I mean this Platinum it
has got a property associated with it and what is that property if by chance if by chance you'll keep hydrogen gas molecules if by chance you'll keep hydrogen gas molecules in the vicinity of this Platinum Rod let's assume that you're going to keep the hydrogen gas molecules in the vicinity of this uh this Platinum Rod over here what exactly is going to happen this these hydrogen molecules these are going to be Ed absorbed these are going to get absorbed on the surface of this Rod they'll get absorbed on the surface of the rod right now
if these hydrogen gas molecules are absorbed on the surface of the rod what is going to happen can I say after some time the whole surface of platinum the whole surface of the Platinum would have got covered by hydrogen gas yes I would say the whole surface of platinum would have got covered by the hydrogen gas now my dear students if I ask you whether this rod from outside is it looking like a platinum Rod now see imagine this was the Platinum Rod now H2 gas has been absorbed on the surface of this Platinum right
this entire surface the entire outer covering of this Rod is covered by hydrogen gas now from outside is it looking like Platinum no it looks like hydrogen Rod to me right it looks like a hydrogen Rod to me because I can see the entire surface covered by what H2 gas now what you'll be doing you'll be introducing this H2 Rod you'll be introd ucing this H2 Rod into a solution containing the ions of hydrogen for example contains H positives so I introduced the rod into the solution which contains its own ions so again I shall
be calling this as the electrode this is the hydrogen electrode right just the quick idea right this is not something which is detailed in the detailed way I'll show you how to make the hydrogen electrode after some time but right now understand this so I hope you got to know I hope you got to know exactly how do we make the electrodes yeah take the rod of the metal introduce that into a solution containing the ions of the same metal right the entire the complete setup is called as the electrode okay now guys let's get
into the details of what let's get into the details of the dial cell now okay now if I ask you what is the Daniel cell Daniel cell it is the common example of the GIC cell and what does a galic cell do it converts chemical energy into electrical energy so this danial cell is going to convert chemical energy into electrical energy now how exactly let's see about let's talk about that see guys in case of Daniel cell you will be using two electrodes in case of Daniel cell you'll be using two electrodes one is going
to be your zinc electrode and one is going to be your copper electrode in case of your Daniel cell you'll be using two electrodes one is going to be your zinc electrode one is going to be your copper electrode now my dear students if I compare zinc and copper if I compare zinc and copper let me tell you zinc is comparatively zinc is comparatively more electropositive zinc is comparatively more electropositive than that of what than that of copper if I compare zinc with copper let me tell you zinc is comparatively more electropositive than copper if
zinc is more electropositive what it's going to do it's going to lose electrons electropositive metal electropositive element what it does it has got the tendency to lose electrons so I would say the cell which I'm going to make in which the zinc is going to behave like the I mean in which the zinc and copper these are the two electrodes right in that particular cell this particular zinc electrode since zinc is more electropositive so it has got more tendency to lose electrons so I would say the zinc it will undergo oxidation and if zinc under
goes oxidation I would say this copper will be undergoing reduction this copper will be undergoing reduction and already you know oxidation it takes place at anode and reduction it takes place at cathode so my dear students since you are going to make the Daniel cell and in the danial cell you are going to use two electrodes zinc and copper and you got to know easily which one is going to behave like the anod and which one is going to behave like the cathode right so do remember in the danial cell your zinc electrode that will
be behaving like the anode and copper electrode that is going to behave like the cathode right okay how exactly you got to know that zinc is considered to be more electropositive than copper so more tendency to lose electrons loss of electrons is something which you call as oxidation and oxidation always takes place at anode right this is how you can remember it okay now it's time to make the Daniel cell and see exactly how it works okay it is time to make the Daniel S and see exactly how it works try to understand everything in
detail now because this is the most important point over here which I'm going to discuss see guys first of all I'm going to take how many electrodes here two electrodes one is zinc one is copper let's try to make the zinc electrode first of all for example my dear students this is the container which I've taken and in this particular container let's say I have taken zinc sulfate the zinc sulfate let's assume it has got completely dissociated into its ions and its ions are zinc di positive Aquis plus s so4 d negative okay so in
the container I have taken zinc sulfate this zinc sulfate has got completely dissociated into its ions so I would say the container which I have taken in this particular container you have got equal number of cats and anions right in this particular container you have got equal number of cats and anions right if there are equal number of cat and ANS present in this container so I would say the solution is right now electrically neutral right since there are equal number of cats anions present so this particular solution it has to be electrically neutral and
it is electrically neutral now people what I'll be doing I will be introducing a rod which Rod I'm going to introduce I'm going to introduce a zinc Rod over here I'm going to introduce a zinc Rod over here so try to understand how introduced zinc zinc Rod into a solution containing its own ions I've introduced zinc Rod into a solution containing its own ions so this whole setup I'll be calling as zinc electrode so this is your zinc electrode over here right similarly I would need one more electrode as well and you already know what
that electrode is that is going to be copper electrode so again I have taken one more container and in this container what exactly I have done I taken copper sulfate for example and I'm assuming this copper sulfate has got completely dis assciated into its ions and its ions are copper D positive and S so4 D negative so I would say this particular solution right now again it contains equal number of cats and anions so this solution again is electrically neutral it is again electrically neutral now I'm going to introduce a rod I'm going to introduce
a rod which Rod I introduced copper Rod into a solution containing its own I I've introduced copper Rod into a solution containing its own ions so this setup again I'll be calling as what I'll be calling it as copper electrode right so how many electrodes have I taken how many electrodes have I taken quickly how many electrodes have I taken I have taken two electrodes one is zinc and one is copper correct now dear students I am going to connect these electrodes with the help of a voltmeter I connected them externally with the help of
voltmeter right these two electrodes are connected with the help of voltmeter and at the same time internally I'll be connecting them internally I'll be connecting them with the help of inverted U type tube this is inverted U type tube over here this is inverted u- type tube Over Here My Dear students this inverted U type tube what does it contain what does it contain let me tell you this inverted U type two which I have used over here which I have used over here this inverted U type tube it contains an inert electrolyte the example
of the inert electrolyte is for example your KCl for example it's k204 let's say it is nh4cl these are the examples of the inert electrolytes and my dear students this inverted U type tube I'm going to fill with inert electrolyte and that in inert electrolyte it is mixed with that inert electrolyte it is mixed with it is mixed with gelatin or aagar this inert electrolyte is mixed with gelatin or aagar and let me tell you when inert electrolyte is mixed with gelatin or agar AAR it forms a jelly like paste it forms a jelly like
paste and let me tell you that jelly like paste is basically introduced in this inverted U type tube so this inverted U type tube it contains a jelly like paste which contains inert electrolyte which was mixed with gelatin or agag right and at the same time the ends of the inert I mean the ends of this inverted U type tube the ends of this inverted U type tube it is sealed with cotton balls so these are the cotton balls which I have used over here right now people if I ask you if I ask you
imagine that you have used k2so4 as the inert electrolyte if I ask you which types of ions are present in this salt in this inverted U type tube you will say it's going to be K positive and S so4 D negative let's say I've used k204 as the inert electrolyte over here so this inverted U type tube it's going to contain K positive and S4 negative and this inverted U type tube containing inert electrolyte this is something which you call a solid bridge this is something which you call a solid Bridge so the first thing
which you I would want everyone of you to remember what is a Sol Bridge what does Sal Bridge contains salt bridge is basically the inverted U type tube which contains iner electrolyte which is mixed with gelatin or agag and when we mix inert electrolyte with gelatin or agag it leads to the formation of the jelly like paste and that jelly like paste is inserted in this inverted type tube and the ends of this inverted U type tube are sealed with the help of the cotton balls right yeah what is the function of the salt bridge
you'll get to know in some time okay tell me one thing since I told you I told you already I told you already that this zinc electrode in the Daniel cell it behaves like the it is going to behave like the anode and this copper electrode it is going to behave like the it is going to behave like the cathode right this is something which I mentioned already I've told you already right right and people tell me one thing at anode what's going to happen at anode oxidation is going to happen and similarly at cathode
what's going to happen at cathode I must say at cathode I must say reduction is going to happen at cathode reduction is going to take place right okay right people someone is saying spelling of jelly is incorrect yes it's incorrect it's J L right is that wow so intelligent people I'm teaching right now yeah people is everything clear to he so intelligent right literature people people just kidding okay so I've used two electrodes one is zinc one is copper right the zinc electrode is going to behave like the anode and copper electrode is it's going
to behave like the cathode yeah and my dear students at anode we know oxidation takes place and at cathode we know reduction takes place now if I ask you whether this particular cell is complete right now or not what do you think what do you think is this complete is the circuit complete or not absolutely the circuit is externally complete as well as it is internally complete it's a closed loop it's a closed loop right the circuit is complete right now okay and if the circuit is complete what is going to happen I told you
at anode oxidation is going to take place so the first thing what is going to happen at anode I told you at anode oxidation is going to take place oxidation means loss of electrons right now people tell me one thing since you have taken a rod over here which is a zinc Rod which is absolutely made up of zinc atoms so this Rod is made up of zinc atoms now the rod since it is made up of zinc atoms those zinc atoms of which Rod is made they'll undergo oxidation those zinc atoms of which Rod
is made they'll undergo oxidation oxidation means loss of electrons so I would say the zinc atom for example zinc solid it is going to lose some electrons and it's going to get converted into zinc di positive I would say the zinc atom of which Rod is made it will undergo oxidation it will get converted into zinc di positive and with that you'll be getting two electrons now the zinc di positive which we got by the oxidation of zinc atom of the rod that zinc die positive is going to enter into the solution the solution initially
was electrically neutral now I would say an extra zinc diosi is going into the left container due to which the left container carries the positive charge now agreed what is happening the zinc Rod is made of zinc atoms and those zinc atoms of which Rod is made those zinc atoms undergoing oxidation and when zinc atom under go oxidation it gets convered convert into zinc D positive and with that it gives two electrons initially the solution was electrically neutral now in the solution an extra zinc dios is entering due to which this left container carries a
positive charge now these two electrons they are going to get accumulated on the rod they are going to get accumulated on the rod similarly one more zinc solid one more zinc atom will undergo oxidation one more zinc diosi is going to enter into the solution and two more electrons are going to accumulate on the rod similarly I would say the electrons are continuously going to be accumulated on the rod and if electrons are accumulated on the rod tell me which charge this Rod is going to carry since electrons are being accumulated on the rod I
would say the rod is going to carry the negative charge that's something which I told you already right the rod of the zinc the rod of the zinc electrod it's going to carry which charge negative charge right perfect yeah right people now at the same time this copper electrode it's going to behave like the cathode and at cathode what happens at cathode reduction takes place reduction means gain of electrons so at cathode I must say reduction is going to take place and you know reduction involves gain of electrons right now try to understand this particular
solution it contains two types of ions copper D positive s so4 d negative and it contains equal number of cats and anions so this particular solution was electrically neutral in the beginning now what's going to happen my dear students if I ask you between copper di positive and S so4 D negative which one is electron deficient which one is electron deficient is it copper diosi absolutely so the copper di positive which is in the solution that is going to collide with this particular Rod the copper diosi which is in the solution it's going to collide
with a rod and this copper diosi which is going to collide with the rod it's going to take take two electrons from the rod so what's going to happen the copper diosi to Aquis which was there in the solution it is going to take two electrons from the rod and when copper di positive will take two electrons from the rod it will get converted into copper solid right now try to understand since copper dios in the solution is taking electrons from the rod right from the rod electrons are being snatched so I must say the
rod is going to carry the positive charge now because electrons are taking electrons are being taken away from from what from this copper dip over correct now we will try to understand one thing try to understand one simple thing at anode oxidation happened at cathode reduction happened now on this Rod electrons are accumulated on this Rod there's a positive charge can I say this particular positive charge is going to attract these electrons towards itself since the path is already created the path is already created now this particular positive charge is going to attract these electrons
right it's going to pull these electrons towards itself perfect now tell me what is happening can I say electrons are basically moving from anode to cathode this positive Char charge Rod it's going to attracting it's attracting the electrons towards itself due to which due to which electrons are moving from anode to cathode from zinc to Copper from zinc to Copper now if I ask you one thing whether it is just the moment of electrons no it is not just the moment of electrons it is moment of electrons in a particular direction it is moment of
charge in a particular direction and when whenever you see the directional moment of electrons whenever you see the directional moment of charge whenever you see electrons moving in a particular direction they say current will be automatically generated here right and you must be knowing the direction of current is always opposite to the direction of electrons so the direction of current is going to be from cathode to this is the first question which can be asked in case of your galvanic cell in case of your Daniel cell what is the direction of the moment of electrons
anode to cathode what is the direction of current is going to be from cathode to anode Point number one point number two my dear students what is happening at anode at anode oxidation is happening due to which zinc solid zinc atoms zinc atoms of which Rod was made up of that's undergoing oxidation getting converted into zinc dios2 that zinc dios is entering into the solution right similarly due to the oxidation of one more zinc atom from the rod one more zinc atom will get oxidized will get converted to zinc di positive that zinc dios will
enter into the solution then one more zinc atom will get oxidized will get converted into zinc diosi will enter into the solution so what is happening to the thickness of this rod with time is the thickness of Rod increasing or decreasing quickly if the thickness of this particular Rod increasing or decreasing I would say the thickness of zinc Rod the thickness of zinc Rod it decrease with time it decreases with time because the rod is made up of zinc atoms and those zinc atoms are continuously undergoing oxidation continuously getting converted into zinc di positives and
those zinc die positives are going into the solution right so this one more conclusion the thickness of zinc Rod decreases with time similarly if you ask me what is happening to the thickness of copper Rod what is happening to the thickness of copper Rod see see Di positives which were there in the solution they collided with the rod they took two electrons from the rod got converted into copper solid and that copper solid will get deposited on the rod my dear students so since copper solid is being deposited on the rod due to which the
thickness of the copper Rod that increase with time the thickness of the copper Rod it increases with time right since this particular cell it is made of two electrodes anode cathode right at anode oxidation is taking place at cathode reduction is is stay in place now if I ask you what is going to be the net reaction what is the net reaction what is the net reaction which is happening in the cell my dear students whenever you are supposed to write the net reaction taking place in the cell first you shall be writing reaction taking
place at anode then you shall be writing reaction taking place at cathode right reaction at anode reaction at cathode now what you should be doing you should be you should be I mean you are going to balance these electrons then write the reaction at anode first then write the reaction at cathode try to balance these electrons and in both the reactions electrons are balanced only now you can directly add up if the electrons were not balanced you were supposed to balance them imagine over here there was one electron then you were supposed to multiply this
reaction by two to make the electrons balanced so first of all reaction at anode reaction at cathode try to balance the electrons after balancing the electrons try to what try to add these two reactions and when you add these two reactions I'll get something like this zinc solid plus copper di positive Aquis it gives zinc di positive Aquis and with this you will be writing copper solid as well so this is your net reaction which is happening in this Daniel cell over here right it is the net reaction which is happening in the Daniel cell
now if I ask you how many electrons got cancelled you'll say two electrons got cancelled what is meant by that it means that it means that number of moles number of moles of electrons it means that number of moles of electrons exchanged exchanged in the net cell reaction exchanged in the net cell reaction which I'm representing with n that is equal to two so basically there's a moment of two moles of electrons from anod to cathode right since electrons are moving from anod to cathode so how many moles of electrons you got to know two
moles of electrons are being exchanged from anode to cathode in this particular galonic cell in this particular danial cell right okay is this clear to everyone people is this clear to everyone right so simply just add the reaction when you add the reactions the number of electrons which gets canceled that gives you the value of n basically okay now people you must be thinking what was the need of this solid Bridge why did we introduce the solt bridge there's a logic for that as well there's a logic for that as well right salt bridge why
do we use the salt bridge the first thing the first thing the first thing it it maintains it maintains electrical neutrality it maintains the electrical neutrality in both the Solutions in both the solutions this is the first point now what that means exactly what what is meant by it maintains electrical neutrality in both the solutions try to understand what exactly I'm going to say by students this particular solution it was electrically neutral in the beginning this particular solution it was also electrically neutral in the beginning right then what happened when oxidation of zinc happened what
happened the rod is made up of zinc atoms those zinc atoms they they they have got they have undergone oxidation those zinc atoms got converted into zinc di positive right and those zinc di positives entered into the cell solution the solution was electrically neutral in the beginning now due to the presence of extra zinc D positive positive charge got accumulated over here right similarly my dear students initially there were equal number of cats and anions in this solution due to which this particular solution was electrically neutral now one of the CU di positive one of
the CU positive collided with the rod took two electrons from the rod got converted into CU solid and that CU solid got accumulated on the rod right but what happened to the concentration of Cu positives what happened to the number of C positives in the solution they decreased initially there were equal number of cat and ANS now one of the cuos is undergoing reduction it is getting converted into CU solid that CU solid is deposited on the rod now what is happening to the number of Cu positives in the solution number of Cu posi are
decreasing and if the number of Cu positives are increasing with time so what will happen to the charge on the solution will the solution get the negative charge absolutely the solution will get negative charge right initially equal number of Cates and anions now one of the ctin is lost it has undergone reduction so now one ctin is less so more anions less cats so net negative charge on the solution right so I would say I would say the left container got the positive charge and right container got the negative charge now this solt bridge it
already contains K positive and S so4 D negative I students as soon as this particular solution this particular solution gets the positive charge as soon soon as the solution gets the positive charge at the same time the s so4 d negative is going to come from the solid bridge and it's going to neutralize this charge right similarly my dear students once this particular solution gets the negative charge at the same time K positive is going to come from the Sol bridge and it's going to neutralize this particular solution so can you say the ions of
the salt bridge they are going to neutralize they are going to maintain the neutrality in both the solutions yes they are going to maintain the neutrality in both the Sol Solutions is this clear that's what that's what I exactly wrote over here it maintains the ions of the solid Bridge they are going to maintain the electrical neutrality in both the solutions Point number one as soon as the left container gets the positive charge at the same time S4 D negative is going to come from the salt F similarly as soon as negative charge is accumulated
in the right container at the same time K positive is going to come will make the solutions electrically neutral again right now there is one more point the solid Bridge it avoids it avoids the liquid it avoids the liquid Junction potential it avoids the liquid Junction potential now what is meant by liquid Junction potential and how solt Bridge exactly is going to avoid that let's try to identify see guys see exactly what I am going to do just a second just a second imagine I'm going to take only one container instead of two containers imagine
that I'm only taking one container instead of two containers right I just taken one container and my dear students this particular container which I've taken I'm going to divide this container into two parts with the help of semi permeable membrane so this is the semi permeable membrane which I used in the middle right so how many chambers which we got we got two Chambers basically left chamber right chamber the left chamber I'm filling with zinc sulfate the left chamber I'm filling with zinc sulfate so imagine this is zinc sulfate present in the left chamber similarly
in the right chamber what do we have my dear students imagine that in the right chamber we have got copper sulfate so basically in the left chamber you have got two types of ions zinc di positives and S S4 D negatives similarly in the right chamber you again have two types of ions copper D positives and S so4 D negatives I would say the left chamber contains equal number of Cates and anions the right chamber also contains equal number of Cates and anions so both the solutions right now they're electrically neutral right now now introduce
the zinc rod in the solution introduce a copper rod in this particular solution now this is a zinc rod and this is a copper Rod so can I say this is the zinc electrode now and this is the copper electrode so we have got two electrodes now zinc electrode copper electrode now dear students what exactly you should be doing you should be connecting them externally with the help of the voltmeter connect them externally with the help of voltmeter if I ask you whether this cell is completed externally right whether the circuit is completed externally absolutely
you have complete the circuit externally and internally I would say they're also connected because the solutions they are in direct contact with the help of semi- peral membrane so internally the cell is connected externally also the circuit is complete internally also the circuit is complete so I would say the cell is complete right now it's complete to work I have not used salt bridge right now I have not used salt bridge by chance if I do not use SOL bridge what's going to happen try to understand see this is your anode right this is your
anode and in the similar way this is your cathode at I know what's going to happen oxidation is going to happen the zinc Rod is made of zinc atoms those zinc atoms will undergo oxidation will get converted into zinc di positive and that extra zinc di positive is going to enter into the solution due to which the solution carries what positive charge similarly copper diosi in the right chamber it's going to collide with the rod it's going to take electrons from the rod and it will get converted to CU solid that Co solid will get
deposited on the rod but what happened to the number of Cu diosi in the solution since CU di positives they are getting reduced they are taking electrons from the rod getting converted to CU solid that CU solid is deposited on the rod so I would say this right chamber will become CU di positive deficient already there were equal number of cat and anions now cats will be less if Ks are less I'll say this particular solution this particular chamber it gets a negative charge now tell me one thing left chamber got positive charge right chamber
got the negative charge let me say this positive and negative is going to attract this is positive and this negative they attracted they got accumulated at the junction similarly one more zinc atom will get will undergo oxidation one more zinc positive will come into the solution and similarly one more negative charge here POS negative will attract again this process will keep on happening continuously the process will happen continuously positive negative right the process happen continuously if I talk about this particular Junction on the left of this particular Junction can you see positive charges on the
right of this particular Junction do you see the negative charge absolutely can I say at this Junction potential difference got created can you see at this particular Junction potential difference got created because on the left side you got a positive charge on the right side you got negative charge a potential difference an extra potential difference is getting created at the junction and yes students this particular potential difference which gets created at the junction you call this potential difference as the liquid Junction potential you call this potential difference as the liquid Junction potential right and people
if you do not use the solid Bridge we'll get this X extra potential difference here which is something you call as liquid Junction potential and by chance if you use the solid bridge what's going to happen if you use the salt bridge what's going to happen as soon as positive charge is accumulated in this container in the left chamber at the same time s so4 d negative would have come from salt bridge neutralized it as soon as negative charge got developed in the right container at the same time K positive would have come from the
solt bridge and neutralized it so there was no positive and negative if there was no positive and negative I would say there was no potential difference which which would have got created at Junction right yeah so can I say this particular solid Bridge it is avoiding the liquid Junction potential absolutely it's avoiding the liquid Junction potential right is it clear to everyone quickly in the chats quickly in the chats with the thumbs UPS people Everyone yeah wonderful now have a look on few more things since I told you the solt bridge it contains inert electrolyte
now people have ask you what inert electrolyte does you saw it function so which type of salt we can use as the inert electrolyte condition for the salt to be used as the inert electrolyte condition for the salt to be used as an inert electrolyte number one number one the ions of the inert electrolyte the ions of inert electrolyte should have should have same ionic Mobility this is point number one the speed with which K positive enters into the solution in the with the same speed s so4 d negative enters into another solution so the
C and anion of the iner electroly they should move with the same same speeds they should have got same ionic Mobility number two number two the ions of number two the ions of the ions of inert electrolyte should not should not participate the ions of the iner electrolyte should not participate in the net cell reaction it should not participate in the net cell reaction my dear students if you look at the net cell reaction look at the net cell reaction in this net cell reaction do you see K positive or s04 D negative there's no
K positive or s04 D negative here so the inert electrolyte which you use in the solt bridge let me tell you it should not participate in the Nell reaction right so you'll be using that type of inert electrolyte whose C and anion speed whose C and anion Mobility will be the same number one number two number two what should be there the ions of the iner electrolyte they should not participate in the net cell reaction I hope this is super super super clear to everyone let me know once in the chats if all these things
are clear let me know quickly in the chats if all these things are clear these are the conclusions any conclusion can be asked in your examination any conclusion can you can be asked in the examination people right zinc electrode anode oxidation loss of electrons copper electrode cathode reduction gain of electrons direction of electrons from anode to cathode direction of current cathode to anode zinc electrode keeps on dissolving its thickness reduces thickness of copper electrode increase with time similarly if I talk about solid Bridge completes the internal circuit it avoids the liquid Junction potential maintains the
electrical neutrality in the solt bridge we use inert electrolyte like nh4 NO3 ko3 this this mixed with gelatin to get the Jelly like P etc etc all these things are clear all these things are clear let me know once in the chats R which example you asking what example this is the most typical example of the Gonic cell you will see millions of GIC in some time just wait for it don't spam the chats once you write something in the chats just once don't spam people let me know in the chats if all the things
are clear to quickly quickly quickly everyone quickly quickly all the things are clear perfect guys now it is that it's time to see exactly how do we represent the Gonic cells right how do we represent Gonic cells let's get to know about that see guys first of all let's try to see how do we represent the Daniel cell Daniel cell is the one of the common example it is one of the common example of the GIC cell let's see exactly how this Daniel cell is represented so when you try to represent a galvanic cell in
the middle you'll be using two lines and these two lines they are going to represent a solid Bridge these two lines are going to represent the solid Bridge okay on the left side of the solt bridge you'll be using anode on the right side of the solt bridge you'll be using cathode and you already know at anode what happens at anode oxidation takes place and you already know at cathode what happens at cathode reduction takes place this is something which you already know perfect the general representation of the gmic cell use the two lines in
the middle Sol bridge on the left side you'll be writing the anode on the right side you'll be writing the cathode now let's exactly see how these galic cells are actually represented try to understand what exactly I'm going to do try to understand what exactly I'm going to do my de students in case of galonic cells what do we need exactly we need two electrodes in case of galonic cell we need two electrodes take two electrodes connect them externally connect them internally connect them externally connect them internally right you'll be getting a complete cell perfect
the most common example was Daniel cell in which the two electrodes were zinc and copper right in case of other galonic cells you can take any electrode hydrogen electrode cadmium electrode any two electrodes connect externally as well as internally you'll be getting a galonic cell now how do you represent the galonic cell with the help of their net cell reaction imagine that there's a GIC cell in which you have used two electrodes right this is the net cell reaction taking place in the GIC cell with the help of this particular net cell reaction you should
know how to represent the galic cell try to understand try to understand see guys first of all since this is the net cell reaction if I ask you what is the charge on zinc zero what is the charge on copper plus two what is the charge on this zinc plus two what is the charge on this copper it is zero right okay perfect now people try to understand is the oxidation state of zinc increasing or decreasing it's increasing from 0 to plus2 Copper die positive plus 2 to 0 decrease in the oxidation state increase in
the oxidation state is something which you call as oxidation so I would say zinc is undergoing oxidation copper di positive is undergoing reduction now remember oxidation takes place at anode reduction takes place at you know it takes place at Cath since you identified which one is going to behave like the anode which one is going to behave like the cathode right now people how do you represent it the two lines they are going to represent the salt bridge which we discussed few minutes back and someone is still spamming what is salt bridge what is salt
bridge nonsense girl she is it's been half an hour I've been teaching you what is solid bridge and you are coming and entering into the session and spamming the chats with what is solid Bridge what is Solage are you out of your mind you can get the hell out of my class please it's been one hour and I've been teaching what is Sol Bridge I've given you functions of the Sol Bridge suddenly you entering into the class you were somewhere else till now suddenly you are coming into the class and spamming the chats that too
I've got the option of blocking as well by the way try to understand now these two lines what do they represent they represent solt bridge and on the left of the solid Bridge what do we write we write anode at anode what happens oxidation oxidation of what is happening oxidation of what is happening oxidation of zinc is happening and this zinc is getting converted into zinc dipole positive so on the left side what is happening zinc solid it is getting converted into zinc die positive and this zinc di postive it goes into the left container
left chamber if you remember imagine that imagine that the concentration of zinc dios to in the left chamber imagine that that is C1 M imagine that that is C1 M similarly on the right side what do we write on the right side we write the cathode and at cathode you know reduction takes place reduction of what is taking place copper di positive and copper di positive is getting converted into copper solt so at cathode you got to know that copper diosi which was there in the right solution right chamber that collided with the rod took
two electrons from the rod and got converted into CU solid I hope you remember that so this copper dios which was there in the right chamber let's assume that its concentration is C2 M right and it got converted into copper solid so this is how you represent your galic cell whose n cell reaction was given to us I hope this is clear right I hope this is clear I hope this is clear similarly let me take one more example with that every single thing will be clear to you imagine my dear students you have got
a gvic cell imagine you have got a galonic cell in which the net cell reaction is this one imagine you have got a Gonic cell in which net cell reaction is this one from the net reaction can we write the galonic cell can we represent the galic cell absolutely we can do that what is the charge on this iron it's zero charge on this copper it's plus two charge on this particular iron it's plus two charge on this copper it's zero identify first of all which one is your anode and which one is your cathode
identify iron is changing its oxidation state from 0 to plus2 0 to plus2 means increase in the oxidation state increase in the oxidation state means oxidation so this particular iron is under goinging oxidation and oxidation takes place at anode right similarly the oxidation St of copper is changing from plus2 to 0 plus2 to 0 means decrease in the oxidation state decrease in the oxidation state is something which you call as reduction reduction always takes place at cathode so first of all you identifi which one is your anode and which one is your cathode now what
I'll be doing I'll be using two lines these two lines represent the solid bridge on the left side of the solid Bridge I'll be writing an weting anode and at anode oxidation takes place and oxidation of what is taking place oxidation of iron is taking place iron is getting converted into f f positive right perfect so what is happening at anode Fe solid is getting converted into Fe dios2 and I must say this Fe dios2 will go into the left container in which the concentration of Fe di positive imagine that assume that that is C1
M on the right right side you will have to write cathode at cathode reduction takes place reduction of what C positive C positive will get converted CU di positive which was there in the right container whose concentration imagine was C2 M it got converted into CU solid right so this is how you represented this particular cell over here yeah is this clear quickly is this clear quickly sh you can check the name of the channel it is unad be neat English okay perfect let's take one more example imagine that you have got a galic cell
imagine that you have got a galic cell in which this is the net cell reaction by looking at the net cell reaction you should be able to write you should be able to represent the cell again what I'll be doing so first of all the charge on zinc is zero the charge on zinc here is plus two 0 to plus2 means increase in the oxidation state increase in the oxidation state is oxidation and oxidation takes place at anode if this is anode that means this has to be your cathode right at cathode reduction takes place
so you identified the anode and cathode perfect now people use the two lines as a solt bridge on the left side you'll be writing the anode at at at anode zinc is getting converted into zinc di positive so zinc solid at anode is getting converted into zinc die positive and this zinc di positive enters into the left chamber in which the concentration of zinc D posi for example C1 M right similarly on the right side of the right side at the right side what's going to happen at the right side you'll be writing the cathode
and at cathode reduction takes place reduction of what H positive and H positive is getting converted into H2 so basically H positive whose concentration in the right chamber is for example C2 M it is getting converted into H2 gas few minutes back only I told you how do you make the hydrogen electrode with the help of platinum solid with the help of platinum solid you make the hydrogen electrode so I'll be writing Platinum solid as well here right I hope this is clear I hope this is clear to everyone okay let me take one more
example and with that we'll be moving on to the New Concept see guys let's assume you have got a Gonic cell in which this is the net cell reaction in which this is the net cell reaction by looking at the net cell reaction try to identify it try to represent the gmic cell how the charge on hydrogen is 0o the charge on copper is plus2 the charge on hydrogen is+ 1 the charge on copper is 0o check it out now 0 to+ one increase in the oxidation state increase in the oxidation state means oxidation oxidation
takes place at anode so you got to know what your anode is copper plus 2 to 0 plus 2 to 0 decrease in the oxidation state decrease in the oxidation state is something which you call as reduction and you know reduction takes place at cathode so you identified your anode and cathode so what you'll be doing again draw the two lines these two lines represent the solid bridge on the left you'll be writing the anode and at anode H2 is getting converted into H positive at anode H2 gas is getting converted into H positive and
this H positive it enters into the left container in which the concentration of H positive imagine that is C1 mol now this is the hydrogen electrode hydrogen electrode is made with the help of what it is made with the help of platinum solid so I've written Platinum solid with it as well right I mentioned Platinum solid here as well now on the right side what do you have to write you'll have to write cathode and at cathode reduction takes place reduction of what copper di positive and this copper di positive this copper di positive it's
undergoing reduction this copper dtive is present in the right chamber right in which the concentration of Cu dii imagine that it is C2 mol and my dear students this copper dtive it is getting converted into copper solid so this is how you represent the cell this is how you represent the cell okay this is how you represent the cell perfect this is how you represent the cell can you represent this one uh I think I've already showed this leave this aside now okay the other one can you represent this particular cell can you represent this
particular cell try to identify what exactly should be done the oxidation St of chromium right here is plus 6 right here is plus two oxidation of chromium here is+ three and here is plus 3 as well perfect you got the oxidation States now tell me one thing is the oxidation state of chromium increasing or decreasing plus 6 to+ 3 decrease decrease in the oxidation state is something which you call as reduction and reduction takes place at cathode so this is your cathode if this is cathode that means this has to be your anode so you
identified your cathode and anode so draw the true lines these two lines will be representing solid bridge on the left side write the anode and at anode you can see see F di positive is getting converted into F Tri positive so I'll be writing like this F di positive is basically getting converted into what F Tri positive similarly on the right side you'll be writing cathode and at cathode cr207 D negative is getting converted into CR Tri positive so cr207 D negative is getting converted into CR Tri positive so this is your cath now first
of all try to understand try to understand one simple thing uh just a second guys just just a second people okay so try to understand one more thing this is Ion this is Ion right this is basically ion ion electrode till now if you remember your zinc electrode we used have a zinc Rod zinc Rod was dipped into zinc die positive here there is no Rod I mean what kind of Rod here is I was taking zinc Rod which was made of zinc atoms right and those zinc atoms were dipped in zinc D positive ions
similarly copper electrode take the copper Rod introduce that in Copper di positive yeah and here you have got ionine here you have got ionine electrode so again in order to make this electrode you'll take the help of platinum solid again over here you have got ion ion ion as well as ion so this particular electrode is also made with the help of solid perfect I hope this is clear I hope this is clear so this is how you represent it for example there is one more cell reaction of some Gonic cell which is given to
me how do I represent it how do I represent it again the simple procedure you'll be following calculate the oxidation set of magnes here it's plus 7 calculate the oxidation of carbon over here it's plus three oxidation of magnes here it's plus two oxidation state of carbon over here it's plus4 agreed now tell me one thing what is happening to the oxidation state of magnes plus 7 to plus2 decrease in the oxidation state decrease in the oxidation state is something which you call as reduction reduction takes place at cathode so this is going to be
your cathode right + three to +4 plus 3 to plus4 increase in the oxidation state means oxidation oxidation takes place at anode so you identified your anode and cathode you identified your anode and cathode now use the two lines as a solt bridge on the left side try to write anode and at anode the oxalate ion it is getting converted into carbon dioxide the oxalate ion c24 D Nega it is getting converted into carbon dioxide similarly on the right you'll be writing the cathode and at cathode reduction takes place M4 negative gets converted to m&
D positive so I would say M4 negative is getting converted into MN di positive here right again have a look this is gas this is Ion so this a new type of electrode so what kind of electrodes we have discussed till now we have discussed metal metal ion electrod metal for example zinc Rod introduced in zinc di positive so metal metal electrode copper Rod introduced in CU di positive so metal metalon electrode this is not metal metal electrode this is not your metal metalon electrode right this is Ion this is Ion so whenever you do
not have the metal metal and Electrode you'll be using Platinum with it right so in order to make this particular electrode you'll take the help of platinum here also you'll take the help of platinum so this is how you're going to represent this particular cell as well yeah perfect I hope this type of these type of things are clear to you now we will try to understand few more things try to understand few more things till now I was teaching you how to represent a cell right how was represent a cell now comes the most
important part if the cell is given to you if the cell is given to you after looking at the cell how do you write the cell reaction after looking at the cell how do you write the cell reactions that is important right so Mark The Heading as how to write the cell reactions this is most important part over here because afterwards I'm going to teach you the nnest equation nnest equation you can only do if you understand this particular topic right how to write how to write the net cell reaction how to write the net
cell reaction okay tell me one thing guys do you remember how to write the equilibrium constant do you do you remember how do we write the equilibrium constant of some reaction do you remember that quickly do you remember that yeah yes or no I hope you do right okay now for example let's say you have got a cell like this zinc solid getting converted into zinc dipositive salt bridge copper die positive getting converted into copper solid imagine this is the cell which is given to us after looking at the cell what all parameters we should
conclude from the cell what all parameters we should get from the cell see guys see Point number one point number one this is the solt bridge on the left side side of the solt bridge you know it's going to be anode always on the right side of the solt bridge it's going to be cathode always right and at anode what happens at anode what happens at anode oxidation takes place your zinc is getting converted into zinc di positive zinc is getting converted to zinc di positive see 0 to plus2 0 to plus2 increase in the
oxidation state means oxidation means loss of electrons so basically at anode oxidation takes place and oxidation means loss of electrons so I would say zinc has to go into zinc diosi so the reaction has to be zinc solid it will be getting converted into zinc di positive Aquis and with this you will be writing two electrons because the reaction is taking place at anode and at anode oxidation takes place oxidation means loss of electrons so zinc is getting converted into zinc di positive 0 to plus two increase in the oxidation state means oxidation means loss
of electrons so when zinc will be getting converted into zinc di positive it will be giving two electrons right perfect s what is happening at cathode what is happening at cathode you know at cathode reduction takes place at cathode reduction takes place look at the cathode plus 2 to 0 plus 2 to 0 means decrease in the oxidation state plus 2 to 0 means decrease in the oxidation state decrease in the oxidation state means reduction reduction means gain of electrons so I would say copper di positive will be gaining some electrons and then only it'll
be getting converted into copper solid so if I ask you how many electrons this copper dios will be gaining it's simple I would say copper D positive will be gaining two electrons and it will be getting converted into copper solid this is the reaction taking place at cathode now you got the reaction at anode you got the reaction at cathode what is the next step you will have to write the net reaction my dear students before writing the net reaction before writing the net reaction what you need to do you need to balance the electrons
are the electrons already balanced yes the electrons are balanced now you can directly add them and when you add them what exactly you'll be writing you should be writing zinc plus copper di positive Aquis what does it give us it gives us zinc diosi Aquis and with zinc dios to Aquas you'll be writing copper solid as well so this is the net cell reaction which is taking place in the galonic cell which is given to us right this is a net cell reaction if I ask you since you got to know the net cell reaction
if I ask you how many moles of electrons got exchanged in the net cell reaction you will directly say two moles of electrons got exchanged in the net cell reaction if I ask you how do you write the reaction quotient QC reaction quotient my dear students if you remember reaction quotient expression is written in the similar way as you write the equilibrium constant expression right you'll be starting with the product you'll be starting with the products right in order to write the QC you'll be starting with the product and that reactant or product which will
be present in aqua State you'll be using the term concentration for that that reactant product which will be present in gas State you'll be using partial pressure pressure term for that that reactant or product which is in solid state or liquid state right you are not going to consider that it's active mass is taken as Unity leave that part aside tell me one thing in order to write the expression for QC you'll be starting with the product this particular product is an aqua state so you'll write concentration of zinc D positive raised to the power
it's toet coent that's one right this one is in solid state it's act2 mass is Unity nothing to do with it divided by this is also in solid state it's AC masses Unity nothing to do with it this copper D positive it's an aquous state so you'll be using the term concentration so concentration of Cu ra^ one so this is how you write the expression of QC for this particular cell that's given to us these are three important points which you have to learn at any cost then only you can Master the electrochemistry first write
the net cell reaction get the value of N and write the QC expression let me take few more examples wait let me take few more examples let me take few more examples for example the second example which I'm taking it is like this understand guys let's say you got a cell like this AG solid AG positive salt bridge H positive H2 gas and here this is platinum solid let's assume this is a cell that's given to me right so after looking at this cell I should know all the parameters first thing what is the first
thing on the left side of the solt bridge what do I write I'll be writing the anode on the right side I'll be writing the cathode correct one is anode one is cathe perfect now people at anode what happens oxidation takes place so the reaction taking place at anode will be the oxidation reaction oxidation means loss of electrons see this is 0o this is + one 0 to+ 1 increase in the oxidation state means oxidation right so AG has to get converted into AG positive then only I can say that oxidation is place so AG
solid it has to get converted into AG positive right then only I'll say oxidation will take place now when it's oxidation loss of electrons takes place loss of electrons now how many electrons are lost I would say when AG is getting converted into AG positive this AG will be losing some electrons how many how do you check that final oxidation State minus initial oxygen State final minus initial will give you one so it is losing one electron here basically in short right similarly at cathode at cathode what is happening at cathode I must say reduction
is taking place your Edge positive is getting converted into H2 so + 1 to 0+ 1 to 0 means decrease in the oxidation state plus 1 to 0 means decrease in the oxidation state decrease in the oxidation state means reduction reduction means gain of electrons so H positive has to get con converted into H2 then only reduction will take place so what has to be the reaction what has to be the reaction so what should we form we have to form H2 gas so to form H2 gas how many H positives do we need we
need 2 H positives we need 2 H positives so H positive should be gaining electrons then only it will be getting converted into H2 correct how many electrons check it out how many electrons how many electrons H positive should how many electrons will be participating in this reaction can you let me know that how many electrons will be participating in this reaction say for example if you have 1 H postive 1 H pos2 will gain one electron will get converted into H but do I have to make H or H2 I have to make H2
to make H2 you should take two H positives 2 H positives will gain two electrons right perfect so plus two electrons now tell me one thing in order to write the net cell reaction do I have to add the reactions directly no I'm not supposed to add the reactions directly so what do I need to do I'll be balancing the electrons first in the first reaction you got one electron in the second reaction you got two electrons so I I'll be multiplying this reaction by two when you multiply this particular reaction by two this becomes
two times this becomes two times even this becomes two times now add them up it is going to be 2 * AG solid plus 2 * H positive Aquis it gives what it gives 2 * AG positive Aquis plus H2 gas this is your net cell reaction this is your net cell reaction and my de students in this particular net cell reaction if I ask you how many moles of electrons got exchanged you'll say two moles of electrons got exchanged in the net cell reaction if I ask you how do we write the QC expression
QC is going to be start with the product this product is an aqua state so use the term concentration concentration of AG positive raise par to isometric that's two this H2 is in gaseous State use the term partial pressure partial pressure of H2 raet that's one divided by this is in solid state so it's active masses Unity nothing to do with this H positive use its concentration ra power what ra power two so this is how you write QC expression as well I hope this is super clear to everyone right okay clear clear people perfect
for example this is the cell that's given to me try to write all the parameters try to write all the parameters related to this particular cell whatever I have taught you till now utilize all the concepts and write out see guys these two two lines they are representing the solid bridge on the left side you always write anode on the right side you always write cathode right and you know at anode oxidation takes place loss of electrons right you can check it out also H2 is 0 here it's + one 0 to plus one is
increase in the oxidation state means oxidation loss of electrons right now tell me the reaction taking place at anode I would say at anode H2 has to get converted into H positive at anode I must say H2 gas it has to get converted into H positive but should I write H positive or 2 * H positive absolutely I should write 2 * H positive right because this H2 perfect now you know when H2 is getting converted into H positive 0 to+ one increase means oxidation loss of electrons so H2 has to lose some electrons then
only it will get converted into H positive now how many electrons will it lose check it out final minus initial final minus initial the value is one but one is the change in the oxidation state for one atom of H hydrogen but I have two right so how many electrons it will be losing it'll be losing two electrons it's simple it's simple guys see when one hydrogen one hydrogen has got one electron right when it loses one electron it gets converted into H positive right but do I have one hydrogen no we have got two
hydrogens so two hydrogens will be giving 2 H positives and with that you'll be getting two electrons as well perfect so you got the reaction at anode now it's time to write the reaction at cathode time to write the reaction at cathode at cathode what happens reduction plus 4 to plus2 plus 4 to plus2 decrease decrease means reduction gain of electrons right so I would say SN plus4 Aquis SN plus4 Aquis at cathode should get converted into what it has to get convert into SN plus2 Aquas now at cathode reduction takes place gain of electrons
so I must say SN plus4 has to gain electrons then only it will get converted into SN plus2 right now how many electrons this SN Plus 4 will be gaining final oxidation state minus initial I would say two electrons it will be gaining tell me one thing whether electrons are balanced in both the reactions or not yes the electrons are balanced electrons are balanced in both the reactions right yeah electrons are balanced in both the reactions so directly you are going to add them up when you add them up what do we get we get
H2 gas plus SN plus 4 Aquis it gives 2 * H positive Aquis plus SN di positive Aquis perfect now people if I ask you how many moles of electrons got exchanged you'll directly say two moles of electrons got exchanged right two moles of electrons got exchanged so I would say n value is equal to two two moles of electrons got exchanged in the net cell reaction now if you ask me what is the QC value start with the product start with the product this H positive is in AC state so use the term concentration
concentration of H positive ra that's too SN posi again Aqua state so concentration of s and D positive ra stric coefficient divided by H2 is in gaseous state so use the term pressure pressure of H2 raise barometric coefficient SN +4 is in again Aqua state so use the term concentration raise par tetric coent this is how you write the expression for Q is this clear is this clear people is this clear quickly for example this is the question you have to write the net cell reaction you have to write the QC you have to get
the N value Guys these are the most important things which I'm teaching you right now because if all these things are not clear definitely electrochemistry can never be clear honestly I'm telling you okay try to understand try to understand see guys these two lines they'll be representing what solid bridge on the left side we'll be writing anode on the right side we'll be writing cathode so you have got anode and cathode right now dear students at anode what happens oxidation takes place at anode oxidation takes place oxidation means loss of electrons see oxidation state here
is zero it's plus4 0 to plus4 means increase increase means oxidation oxidation means loss of elect so I would say in has to get converted into SN +4 I would say SN solid should get converted into SN +4 right then only oxidation will happen now it's evident when when SN gets converted into SN plus4 increase in the oxidation state means oxidation means loss of electrons so SN has to lose certain electrons then only it will get converted into SN plus 4 now how many electrons final minus initial the value is four so I would say
it'll be losing four electrons yeah similarly what is going to happen at cathode I would say at cathode reduction takes place reduction takes place reduction means gain of electrons see the oxidation state here is zero here it's minus1 0 to minus1 means decrease in the oxidation state decreases the reduction so at cathode cl2 has to get converted into CL Nega at cathode cl2 has to get converted into CL Nega so I must say cl2 gas it has to get converted into CL Nega right and you know add cathode reduction takes place gain of electrons so
I would say cl2 will be gaining some electrons then only it will be getting converted into cl2 now this is cl2 make it 2 * CL NE tell me one thing 0 to minus one what is the change one one is the change power of one atom how many we have two so basically it'll be gaining two electrons are the electrons balanced in both no you'll be balancing the electrons so multiply this particular reaction by number two this becomes two times this becomes four times this becomes four times now add these two reactions I would
say four and four got canceled so the net reaction is going to be 10 solid 10 solid plus 2 * cl2 gas plus 2 * cl2 gas just a second yeah 2 * cl2 gas it gives what it gives SN +4 Aquis and with that you'll be writing four * CL AIS as well so this is your net cell reaction if you ask me how many moles of electrons got exchanged how many moles of electrons got exchanged I'll directly say four moles of electrons got exchanged what is the value of QC what is the value
of QC start with the product this is an Aquis state so it has to be concentration of SN plus 4 ra^ 1 right it has to be concentration of Cl negative ra^ 4 divided by this is in solid state it's mass is Unity the cl2 is in gaseous state so use the term pressure so pressure of cl2 rais power two so this is how you are going to write QC for the net reaction as well right is this clear to everyone is this clear to everyone quickly okay one example in which you might do a
mistake one example in which you might do a mistake for example I'm writing like this I'm representing a cell like this Platinum solid this is Fe D positive getting convert into F Tri positive this is salt bridge for example and this is M4 negative getting converted into MN di positive this is one more Platinum solid let's assume this is a cell which is given to me imagine that it is a cell which is given to me now you have to do every single procedure anode reaction cathode reaction net reaction n value and QC here you
can do a mistake try to understand what that mistake is first of all I'll be writing the reaction which should take place at anode at anode oxidation takes place your Fe pos2 is getting converted into Fe Tri postive right plus 2 to plus 3 increase in the oxidation state loss of electrons so basically Fe di positive Aquis will be losing certain electrons and then only getting converted into Fe Tri positive Aquis now how many electrons final minus initial I'll say one electron this is the reaction at anode right now if I ask you what is
going to be the reaction at cathode you know at cathode reduction takes place at cathode reduction Tak takes place gain of electrons now my dear students the oxidation state of magnes here is + 7+ 7 to +2 decrease in the oxidation state plus 7 plus2 decrease decrease means reduction so basically at cathode M4 negative will be finally getting converted into m& di positive right since it's a reduction reaction plus 7 to plus2 means decrease decrease means reduction gain of electrons how many electrons will be gained M4 negative will be gaining certain electrons how many fin
oxidation state minus initial so it'll be gaining five electrons then only it'll be getting converted to m& positive correct now people are the electrons balanced the electrons are not balanced in the reactions so I'll be balancing the electrons multiply this reaction by number five so this has to be five times even this has to be five times even this automatically becomes five times right now you are going to add these two reactions just add them up just add them up this is going to be 5 * f a DI positive Aquis plus what plus M4
negative Aquis and it's going to give me 5 * f a tri positive Aquis plus MN dipositive Aquis this is your net reaction perfect this is your net reaction now tell me one thing how many electrons are being exchanged in the net cell reaction I would say five moles of electrons are exchanged in the net cell reaction right now one thing is there is the reaction balanced or not this is not the net cell reaction yet the reaction is not completely balanced you are supposed to balance the reaction completely you have to write the reaction
in the complete balanced format it is not present in the complete balanced format right so what you shall be doing understand understand people how many oxygen atoms you have on the left side four oxygen atoms right how many oxygen atoms are on the right side no oxygen atoms so which side is oxygen deficient LS or rhs rhs is oxygen deficient how many oxygen deficiency are there on the right side four oxygen deficiency number of oxygen deficiency is equal to number of water molecules added on the oxygen deficient side so on this side I'll be adding
four water molecules I'll be adding four water molecules now tell me one thing how many hydrogen on this side eight how many hydrogen on this side no so which side is hydrogen deficient L's side hydrogen deficiency is eradicated through H positives so on this side we have got eight hydrogen deficiency the number of hydrogen deficiency is equal the number of H positives added on that side which is hydrogen deficient so basically oxygen is balanced with the help of water and hydrogen is balanced with the help of H positives something which I've taught you in redox
reactions as well right yes perfect people a prant work everywhere has got the sign minus we always start with minus sign minus integral V1 to V2 P gas multiplied by DV that's for reversible and for irreversible it is minus P external Delta V V2 minus V1 this negative is in chemistry basically in physics you do not start with negative sign in physics your work expression is simple P Delta V here it is minus P Delta V and minus here indicates that internal energy decreases during expansion Intergy of the system I'm talking about right okay so
this is the net reaction now n value is calculated now it is the time to get QC expression in order to get the QC we have to start with the products we have to start with the product right see this is Aquis so use the term concentration concentration of f Tri positive ra power 5 this is Aquis it is concentration of MN di positive raise part stet that's one this water is in liquid state nothing to do with this divided by this H posit is in aqua state so concentration of H positive ra^ h f
di posi is in aqua state so concentration of f d positive ra power 54 Aqua state so it's concentration of M4 negative ra^ 1 so this is how you write the expression for QC as well perfect in these questions see guys you know why these are not the actual questions the actual question will come in the nnest equation but Nest equation while solving the question based on nnest equation you should know all these parameters basically that's why I'm clearing all the parameters right here that's why I'm clearing all the parameters right here so that when
I teach you an Earnest equation there'll be no issues at all in doing the questions yeah clear people clear clear clear okay let me give one last example look at this particular example guys look at this particular example this is the last example of the today's session okay after this we'll end the session the next session will be the next session will be or should I give you as this as the homework what do you think what do you suggest are you going to solve this or let me solve this the other one I'll be
giving you as the homework okay and let me tell you the next session will be on Tuesday 6 p.m. lecture number two of electrochemistry let me solve this the other question I'll be giving is the homework okay let's try to solve this first of all these two lines they represent the solt brid on the left side you'll be writing the anode right so this is the anodic part this is the cathodic part you know at anode oxidation takes place at anode oxidation takes place so Fe di positive is getting converted into F Tri positive right
so Fe di positive Aquis is getting converted into Fe Tri to Aquis since it is the oxidation reaction so loss of electrons has to happen right so plus 2 to+ 3 final minus initial value is one the value comes out be one here right so one electron is lost in the oxidation reaction similarly the reaction which will be taking place at cathode you know at cathode what happens reduction at cathode what happens reduction try to understand oxidation state here is plus 6 there it is plus three so plus 6 to plus 3 decrease in the
oxidation state reduction gain of electrons so I would say when cr207 DGA when cr27 DGA will be getting converted into CR Tri positive definitely this cr27 D negative will be gaining some electron now how many plus 6 to + three change is three three is the change in the oxidation state for one atom of chromium but I have two atoms so 3 2 are six so it'll be gaining some six electrons and at the end it'll be getting converted into two times CR trios CU you got two chrom right now tell me one thing are
the electrons balanced the electrons are not balanced yet okay six and this is one so multiply this particular reaction by six so this has to be six times this has to be six times this has to be six times now the electrons are balanced after balancing the electrons what you need to do just add the reactions when you add the reactions six and six got cancelled the net reaction has to be 6 * Fe di positive Aquis plus what plus cr27 cr27 D negative Aquis it gives 6 * F Tri positive Aquis and with this
I'll be writing two times CR trios there'll be one thing how many moles of electrons got exchanged I would say six moles of electrons got exchanged in the net cell reaction can we write the expression for QC before writing the expression for QC check it out is the reaction balanced or not the reaction is not balanced yet it is time to balance the reaction first right it is time to balance the reaction first on the left side how many oxygen we have seven on the right side there is no oxygen so right side is oxygen
deficient so add seven water molecules on right side right how many hydrogen 14 so on this side you'll be adding 14 H positives perfect now the reaction is balanced now it's time to write the expression for QC how do we write the expression for QC start with the product it's an Aquis form so use the term concentration so concentration of f Tri positive ra power 6 similarly it's going to be concentration of C Tri positive ra stric that's to it's in liquid form nothing to do with this divide by again use the concentration of Fe
positive ra^ 6 concentration of cr207 D negative rais St that's one and here it's going to be concentration of H positive rais 14 so this is how you are going to write the expression for QC as well right is this clear quickly people one question I'll be giving you guys the homework you are going to let me know the number of moles of electrons exchanged in the net cell reaction okay in the comments in the comments once the session ends okay and tell me let me tell you one more thing on Tuesday 6 p.m. day
after tomorrow there is one more session of the chapter electrochemistry you have to be there on time right on this channel itself the ones who have not subscribe to the channel yet do subscribe to the channel as well okay perfect for example the cell is like this Platinum solid this is SN di positive SN +4 salt bridge cr207 D negative CR Tri positive this is platinum solid this is the cell which I'm giving you try to write every single thing related to this particular cell right the N Net cell reaction n value as well as
QC okay and do let me know the answers in the comment section of this particular video so dear students every single parameter is it clear or not is it clear or not do let me know in the comment section at the end did you get every single thing okay do let me know in the comment section at the end not in the chats the chat box right I'm telling you do write in the comment section at the end whether you got all the concepts which I taught or not perfect Perfecto people Perfecto yeah so the
first session is done and it's okay it's okay it's okay don't spam from now onwards because I get frustrated when somebody spams right either he lives or I live yeah Perfect People sir I need your WhatsApp number you're not supposed to ask the whats numbers publicly right welcome back welcome back people welcome back so my dear students as you all must be knowing we are done with the first session of electrochemistry in the last session if you remember and today it's going to be basically the second session well I would say this is the the
most important session of the chapter electrochemistry right because in this particular session you are going to learn something really really very important right something which is which students mostly find tough basically right but again we'll be starting exactly from the basics as we keep on doing in every session okay so we'll start with electrode potential then we'll see the EMF we'll see how to calculate standard EMF we'll talk about spontaneity right we'll talk about the nnest equation so all all the most important things of this particular chapter will be done in the today's session perfect
but before starting the session just let me know once in the chats if all of you have watched the first session of electrochemistry can you let me know quickly in the chats I just want to know are you guys done with the first session of electrochemistry which I took like 3 4 days back yeah quickly people quickly say yes or no in the chats quickly okay that's great that's great that's great then Perfect all right then so let's get going let's get started with the topic that is electrode potential right electrode potential just give me
a second electrode potential my dear students how do we Define the term electrode potential this is the first topic let me quickly write its definition then I'll make you understand what it exactly means okay this students electrode potential it is basically defined as it is basically defined as the potential difference the potential difference that gets created the potential difference that gets created between between the rod and the electrolytic solution the potential difference that gets created between the rod and the electrolytic solution what does it mean let's try to understand this in detail okay let's try
to understand this in detail my dear students imagine that imagine that you are taking a container let's say this is the container which you are taking right for example in this particular container what exactly I'm going to take I'm taking zinc sulfate into the Container let's say I'm taking zinc sulfate into the Container okay now dear students let's assume that the zinc sulfate it got completely dissociated into its ions and its SS are zinc di positive plus s so4 d Nega I'm assuming this particular electrolyte the zinc sulfate which you have taken in the container
I'm assuming this zinc sulfate has got completely dissociated into its ions so in this particular container right now can I say there will be equal number of cats and anions absolutely in this particular container right now there will be equal number of cats and anions now what exactly I'm going to do since there are equal number of cats and anions in the solution so I must say this particular solution right now is electrically neutral there is no net charge present in the solution okay now dear students just do one thing thing just place a zinc
Rod just place a zinc rod in this particular solution what would you call this particular setup as you have taken the zinc rod and you have introduced the zinc Rod into the solution containing its own ions right so what you should be calling the setup as this particular setup will be called as electrode right as simple as that so I just took a zinc Rod I just introduced the zinc Rod into the solution containing its o NS so I'll be calling this whole setup as the electrode Point number one this is the zinc electrode basically
now my dear students this particular zinc electrode it can behave like the anode it can behave like cathode as well right let's exactly see what are the possibilities what exactly will happen if this particular electrode behaves like the anode if the same electrode behaves like the cathode let's see what exactly is going to happen try to understand I'm assuming that the zinc electrode for example which you took let's say it is behaving like the anode you already know at anode oxidation takes place if this electrode behaves like the anode so oxidation will be taking place
here right and what will happen due to oxidation I must say the zinc atoms of which this Rod is made up of the zinc atoms of which this Rod is made up of those zinc atoms will undergo oxidation right those zinc atoms will undergo oxidation so I would say the zinc atoms of which Rod is made up of right those zinc atoms will undergo oxidation oxidation means loss of electrons so this zinc solid it will lose two electrons and will get converted into zinc die positive the zinc solid right the zinc atoms of which Rod
is made up of they'll undergo oxidation they'll lose two electrons get converted into Z and I positive my dear students what exactly happened try to understand the zinc atoms of which this Rod is made up of the zinc Rod is undergoing oxidation it is losing two electrons and getting converted into zinc di positive tell me one thing initially in the container there were equal number of cat and anions initially there were equal number of cats and anions and the solution initially was electrically neutral now what happened the zinc atoms of which Rod is made up
of those zinc atoms are undergoing oxidation and those zinc atoms are getting converted into zinc diol positives I would say those zinc di positives will enter into the solution right the extra zinc di positives which are produced by the oxidation of zinc atoms of the rod those extra zinc di positives will enter into the solution correct initially the solution was electrically neutral now an extra zinc di positive is entering into the solution due to which this particular solution gets the positive charge right so the electrolytic solution which you took here this electrolytic solution is getting
the positive charge the two electrons which are generated due to the oxidation of the zinc atom right these two electrons are accumulated on the rod Perfect Since electrons are being accumulated on the rod so I must say due to the accumulation of electrons the rod will carry the negative charge perfect so I took one electrode I assumed that this electrode is behaving like the anode if the electrode behaves like the anode oxidation will take place oxidation of what will take place oxidation of the rod oxidation of the zinc atoms of which Rod is made up
of and those zinc atoms of which Rod is made up of they got oxidized got converted into zinc di positive gave certain electrons the extra zinc di positives they entered into the solution due to which solution got positive charge and these electrons they got accumulated on the rod due to which Rod got negative charge now tell me one thing what kind of charge do you see in the solution positive what kind of charge do you see on the rod negative can I say since solution is getting the positive charge Rod is getting the negative charge
can I say polarity get got developed over here can I say potential difference got created between the rod and the solution my dear students this particular potential difference which got created between the rod and the solution this potential difference is something which you call as electrode potential that is a definition which I mentioned over here right the potential difference which gets created between what between the rod and the electrolytic solution the potential difference is something which you call as electrode potential I hope this is clear to everyone right now imagine my dear students the same
electrode was behaving like the cathode imagine that the same electrode was behaving like the cathode right what would have happened then what would have happened imagine the same which I took over here right imagine the same electrode is behaving like what imagine the same electrode is behaving like the cathode if the same electrode behaves like the cathode what will happen I would say reduction will happen reduction involves gain of electrons so what is exactly going to happen I would say the zinc di positives which are there in the solution these zinc di positives will collide
with a rod will take two electrons from the rod and will get converted into zinc solid if by chance reduction happens here reduction of what will happen since in the solution there are equal number of zinc di positives and S4 D negatives I would say the zinc die positives which are there in the solution they are going to collide with the rod they are going to take two electrons from the rod so what will happen the zinc di positives which are there in the solution they are going to take two electrons from the rod will
get converted into zinc solid now tell me one thing initially in the the solution there were equal number of catons and anions equal number of catons and anions now now C is undergoing reduction this zinc die positive it's gaining electrons from the rod it's taking electrons from the rod it is getting converted into zinc solid now tell me one thing electrons are being taken out from the rod if electrons are snatched from the rod I would say right now the rod is going to get the positive charge because electrons are snatched from the rod and
at the same time initially equal number of cats and anions were there in the container now C collided with a rod took two electrons from the rod right got converted into zinc solid that zinc solid gets deposited on the rod right due to which due to which tell me what happened to the number of cat in the solution are the number of cattins in the solution increasing or decreasing quickly are the number of Cates in the solution increasing or decreasing initially the there were equal number of cats and anions due to which there was no
charge no net charge in the solution now cats are decreasing if Cates are decreasing can I say the solution overall will get the negative charge can I say again polarity got developed between the rod and the solution right rod is right now getting the positive charge solution is getting the negative charge can I say again a potential difference is getting created between the rod and the solution absolutely and this particular potential difference between the rod and the solution this is something which you call as electrode potential my dear students right perfect so I hope you
got to know exactly what is actually the concept of electrode potential so electrode potential is just the potential difference which gets created between the rod and the solution right perfect let me know once in the chats if it is clear to everyone let me know once in the chat chats if all these things are super clear to everyone because I'm going to teach you some important stuff now quickly people everyone everyone in the chats quickly yes wonderful now guys try to understand few more things this electrode potential which I'm talking about right now this electrode
potential it is represented by what it is represented by E okay electrode potential it is represented by capital E okay and this electrode potential we further divide it into two categories one is called as oxidation potential one is called called as reduction potential one is called as oxidation potential one is called as reduction potential the electrode potential which is represented by E it's further classified into two types one is called as oxidation potential one is called as reduction potential right my dear students this oxidation potential and reduction potential it gives us the idea of how
easily oxidation and reduction takes place at the electrodes which I'll let you know in some time but before that is there any other way to represent the oxidation potential yes there are many ways my dear students you can represent oxidation potential like this as well e of M change into m n positive M changing into MN positive this is one more way of representing the oxidation potential and have you exactly got to know that this term represents oxidation potential since if you look at the oxidation St of M over here at zero the oxidation state
of M over here is plus n 0 to plus N means increase in the oxidation state increase in the oxidation state is something which you call as oxidation right so this represents your oxidation potential what about reduction potential my dear students reduction potential is represented like this e of e of MN positive change to m e of MN positive change to M this is how you represent the reduction potential now check it out the oxidation state of M over here is plus n the oxidation state of M over here is 0 plus n to 0
decrease in the oxidation state decrease in the oxidation state is what you call as reduction so this particular term it represents the reduction potential so my dear students electrode potential is exactly of two types one is oxidation potential one is reduction potential right I hope this particular thing is clear to everyone and these are the ways by means of which you can represent either oxidation potential or reduction potential which are the types of what which are the types of electrode potential okay perfect what is electrode potential electrod potential is the potential difference which gets generated
between the rod and the solution if the rod gets negative charge solution gets positive charge that means I'm making sure the electrode behaves like the anode that means that potential difference which we calculated between the rod and the solution when the rod got negative solution got positive that potential difference is what you'll be calling as oxidation potential right similarly similarly when the rod gets positive solution gets negative that means I made sure the electrode behaves like the cathode at that point of time the potential difference which got created between the rod between the positively charged
rod and negatively charged solution that potential difference I'll be calling as a reduction potential I hope you got the difference between the two right one is oxidation potential one is reduction potential perfect oxidation potential is the potential difference between negatively charged rod and positively charged solution reduction potential in short that is the potential difference between positively charged rod and negatively charged solution correct I hope this is clear now guys understand one more thing understand one more thing Let Me Tell You electrode potential electrode potential when measured under standard conditions electrode potential when measured under standard
conditions is termed as standard electrode potential whenever this electrode potential whenever this electrode potential which is basically which is basically the potential difference between the rod and the solution when it's measured under standard conditions you call it as standard electrode potential now you must be thinking what this standard conditions are what these standard conditions are my dear students if I talk about standard conditions if I talk about standard conditions what are basically your standard conditions under standard conditions pressure is to be taken as one bar or 1 atm you can say right temperature is taken
as constant generally it is taken as 25° Cen concentration of electrolyte in the solution is taken as 1 mol my dear students these conditions when pressure is kept as one bar temperature is taken as constant generally 25° centigrade and at the same time concentration of electrolyte in the container is taken as one molar or you can say concentration of ion in the solution is one molar at that point of time I'll be calling these particular conditions as standard condition now imagine that you have got the electrode and the electrode is kept at pressure one bar
temperature is kept constant the concentration of ions in the container is kept as one molar that means I have taken electrode under standard conditions and under standard conditions whatever will be the potential difference between the rod and the solution that potential difference I'll be calling as standard electrode potential is this clear to everyone is this clear to everyone so in short electrode potential when measured under standard conditions it's going to give you what it's going to give you it is going to give you the standard electrode potential I hope this is clear to everyone okay
so electrode potential when measured under standard conditions that's called as standard electrode potential and this standard electrode potential my dear students it is represented by E not standard electrode potential it is represented by E not and de students this standard electrode potential it is further of two types one is called as standard oxidation potential one is called as standard reduction potential standard oxidation potential you can call as sop standard reduction potential you can call as SRP right or there is one more way of representing the standard oxidation potential you can represent like this e not
of M changing into MN positive right and standard reduction potential you can represent like this e of MN positive change to M these are the ways of representing the standard oxidation and standard reduction potentials right perfect I hope this is super clear to everyone my dear students there is one more thing there is one more point there is one more point when we talk about the standard electrode potential in general when we talk about the standard electrode potential of an electrode in general be it s o or SRP let me tell you s o or
SRP of an element whose electrode you have made of an element whose electrode you have made is calculated is calculated with the help of is calculated with the help of with the help of a reference electrode and that reference electrode is something which you call as standard hydrogen electrode that reference electrode is something which you call as standard hydrogen electrode so my dear students let me tell you the electrode whatever electrode you have made for example zinc electrode Co oper electrod whatever electrode you have perfect every element has got a particular value of its standard
electrode potential be it sop or SRP every element has got a particular value of standard electrode potential be it sop or SRP and that sop or SRP of an element is measured is measured with the help of one more electrode and that one more electrode is the reference electrode with the help of which we calculate the standard electrode potentials of different elements and that reference electrode is something which you call as standard hydrogen electrode I'll let you know in some time what this standard hydrogen electrode is all about right so why am I introducing the
standard hydrogen electrode here because every element has got a particular value of sop SRP right sop as well as SRP and those sop and SRP values are measured with the help of a reference electrode and that reference electrode is taken as standard hydrogen electrode so standard hydrogen electrode is the one with the help of which we calculate the Sops and srps of different elements okay perfect let me know once in the chats if all the things are clear to here quickly let me know quickly in the chats people let me know quickly in the chats
everyone okay all right now there are few more things which I would want to share with you and what are those things those are again very important things try to understand let me tell you for an element for an element whatever element you have it sop is always equal to minus times its SRP this is important for a particular element for a particular element the SOP value is always equal to minus * its SRP for example if I say that the SOP of zinc is 1.1 volt for example right let's say I'm saying the SOP
of zinc is 1.1 volt so what will be the SRP of same zinc the SRP of same electrode is going to be just reverse the sign it's going to be minus times okay just remember this particular thing perfect now you must be thinking what kind of idea this sop and SRP gives us before understanding what kind of information this sop SRP gives us before understanding that it is high time to understand the standard electrode potential first sorry standard hydrogen electrode first let's get to know how the standard hydrogen electrode is made right which is the
reference electrode which is a standard electrode right with the help of which we calculate the standard electrode potentials of different elements let's get to know how this particular how this particular setup is made exactly try to understand people this is important for example for example I'm taking a container over here imagine this is the container and in this particular container what exactly I'm doing in this particular container I'm for example keeping HCL so I've got HCL in this particular container now imagine that this particular HCL I'm taking its concentration as one mol and this HCL
for example it has got completely dissociated into its ions which are H positive CL negative right so at time T is equal to Z the concentration of HCL was one molar this will be zero this will be zero now this HCL it has got 100% dissociated into itss so this is zero this has to be one M this has to be one M too now people try to understand try to understand I took HCL in the container and I'm telling you the concentration of H positive as well as CL negative in the container is one
mol right the concentration of H POS CL Nega I'm taking as one mol the temperature of the system I have kept as 25° C I'm keeping the temperature of the system as 25° C now after this after this dear students I'm going to take a inverted tube like this I'm taking an inverted tube like this and in this particular inverted tube what exactly am I going to do in this particular inverted tube I'm going to keep a platinum wire imagine this is the Platinum wire imagine this is the Platinum wire which is coated with Platinum
black so it is basically a platinum wire which is co with Platinum black now dear students from this side I'm introducing H2 gas I'm introducing H2 gas at one bar pressure I'm introducing H2 gas at one bar pressure what is going to happen what is going to happen exactly my students since you are introducing H2 gas through this particular hole what will happen the H2 gas will start to absorb on this Platinum surface H2 gas will start to absorb it will get sticked it will get sticked with the surface of platinum perfect now imagine imagine
you're introducing H2 gas you're introducing H2 gas until the whole surface of platinum gets covered with hydrogen now imagine once whole surface of platinum got covered with hydrogen imagine this was the Platinum wire now the whole surface of this Platinum is is covered with hydrogen right now how it looks to me is it looking like a platinum wire now no it looks like a hydrogen Rod to me now it's not a platinum wire now it looks like a hydrogen wire to me now you have got the hydrogen wire which is introduced in a solution containing
its own ions you have got the hydrogen wire which is introduced into a solution containing its own ions you'll be calling this particular setup as the hydrogen electrode and which conditions I have kept here I have kept standard conditions so this Hydro electrode I'll be calling as the standard hydrogen electrode my dear students right so this is the standard hydrogen electrode which I taken perfect now this standard hydrogen electrode it's not necessary that every time you'll make it like this no you can represent it like this as well you will simply say I've taken a
container in which we have got H positive ions and you have introduced one H2 rod and it's understood how this H2 Rod was you was made basically it's understood so H2 Rod into H positives this is the hydrogen electrode right so you can represent it like this as well no issues now this particular hydrogen electrode it can behave like the anode it can behave like the cathode as well and let me tell you when this hydrogen electrode behaves like the anode at anode oxidation takes place when this hydrogen electrode behaves like the anode oxidation will
happen right oxidation means loss of electrons on the surface of this Rod you have got H2 right so I say H2 gas which is there on the surface of the rod it will get converted into 2 * H positives and with that you'll be getting two electrons right if this particular electrode be is like the anode oxidation will happen oxidation means loss of electrons now what is happening basically the surface contains H2 molecules now these H2 molecules will undergo oxidation they lose electrons they lose two electrons and will get converted into tce H positives it's
simple guys see when you got the hydrogen atom if one hydrogen atom loses one electron it gets converted into H positive now you have got two hydrogen atoms here so if two hydrogen atoms loses electrons they lose two electrons and will get converted to 2 * H positive nothing to think a lot about it right okay so this is the reaction which will happen when the hydrogen electrod behaves like the this is the reaction which going to happen when your hydrogen electrode behaves like the anode okay if the same electrode behaves like the cathode if
the same electrode behaves like the cathode what's going to happen then reverse the reaction simple reverse the reaction it becomes 2 * H positives so H positives which are there in the solution they'll collide with the rod they'll take two electrons from the rod will get converted into H2 gas so this is the reaction which will happen when the same electrode will behave like will behave like the cathode because at cathode reduction takes place now my dear students if this hydrogen electrode behaves like the anode or if the hydrogen electrode behaves like the cathode in
both the categories sop as well as SRP of the hydrogen electrode sop as well as SRP of the hydrogen electrode is taken as zero as per convention right as per reference the value of sop as well as SRP of the hydrogen electrone it is taken as zero right it is taken as zero as per reference okay if I ask you how do you represent sop of hydrogen electrode how do we represent it I'll write e okay if I write not that means standard right now for example if I write something like this H2 gives H
positive H2 gives H positive H2 gives H positive 02 + 1 02 + 1 increase oxidation so this is sop this is sop of hydrogen electrode similarly what about SRP of hydrogen electrode just reverse it it becomes H2 and this sorry H positive gives H2 plus 1 to Z decrease decrease means reduction so this SRP sop as well as SRP of hydrogen electrode it's taken as zero as per Convention as per reference I hope this is clear I hope this is clear to everyone yeah clear people quickly is every single thing clear till here I
want you to let me know in the charts are all the things clear if all the things are clear then only I can tell you some important things about this electrode potential which are going to be which are going to have a lot of significance okay perfect wonderful now try to understand the significance of electrode potential what kind of information this electrode potential gives us what kind of information it gives us my dear students do remember more the standard oxidation potential of an element more the standard oxidation potential of an element more is the tendency
of an element to undergo oxidation more the standard oxidation potential of an element more is the tendency of an element to undergo oxidation if more is the tendency of an element to undergo oxidation better will be the reducing agent better I'll be calling it as a reducing agent I'll be calling it as a better reducing agent and do remember the one which will be the better reducing agent will have more reducing power this is the first statement which I would want everyone of you to know you'll be given a lot of questions in which you
will have to compare different elements on the base of their reducing agent capabilities on the basis of the reducing power right just do remember one thing more the SOP more the standard oxidation potential of an element more is its tendency to undergo oxidation better the reducing agent more the reducing power as simple as that okay as simple as that now be one more statement it's application you'll get in sometime second statement more the SRP of an element more the standard reduction potential more is the tendency of an element to undergo reduction more the tendency of
an element to undergo reduction better the oxidizing agent better the oxidizing agent more the oxidizing power more the oxidizing power do remember this particular statement as well do remember this particular statement as well right right more the SRP more the standard reduction potential more the tendency to undergo reduction better oxidizing agent more oxidizing power right right people is it clear is it clear people quickly in the chats perfectly done perfectly let's try to understand where these two statements are utilized okay let's try to understand that okay one simple and basic question simple and basic question
if I ask you people see you are given with certain values or leave the values aside let me give you one more question just a second let me give you one simple question so that you can understand it properly so that just to make you understand okay for example I'm writing certain elements x y and Zed these are few Elements which I have and over here I'm writing certain values e not of M gives MN positive tell me in the chats whether it is sop or SRP tell me the charts quickly e not of M
gives MN positive whether it's sop or SRP quickly sop or SRP 0 to plus n increase in the oxidation state means oxidation that means this is sop let's say this value is 1.1 volt this is 2.1 volt this is 3.1 volt so you are given with sop values right now which one has got more sop this is the maximum sop more the SOP more the tendency to undergo oxidation better than the reducing agent so this is the better reducing agent the one which is the better reducing agent will have more reducing power done understood right
okay for example I'm writing something like this e of M gives MN positive let's say it is minus 1.1 volt it is - 2.1 volt it is - 3.1 volt right now again you given with Sops only which one is maximum sop among all this is maximum sop more the SOP more the tendency to undergo oxid better the reducing agent so this is the better reducing agent it has more reducing power now okay now for example I'm writing something like this e of MN positive gives M so plus n to 0 plus n to 0
decrease decrease means reduction reduction means SRP this is SRP now let's say this is 1.1 volt this is 2.1 volt this is 3.1 volt so which one has got more SRP now this is the more SRP value more SRP more SRP means more tendency to undergo reduction more SRP more tendency to undergo reduction better oxidizing agent so better oxidizing agent more oxidizing power okay similarly for example I'm writing something like this e not of MN positive gives M this is - 1.1 volt this is - 2.1 volt this is - 3.1 volt now check it
out check it out which first of all this is SRP again and this is the maximum SRP here maximum SRP means maximum tendency to undergo reduction better oxidizing agent more oxidizing power is this clear to everyone is this clear to everyone people quickly quickly let me know in the chats try to solve this question I'm giving you one minute of time give it a try give it a try people give it a try everyone what do you think is the correct answer of this question hello okay okay okay okay people are saying option D is
it people are going with option D let's get to know what do we have to check strongest reducing agent strongest reducing agent so we have to check the strongest reducing agent reducing agent is the one which undergo oxidation reducing agent is the one which undergo oxidization we have to check strongest reducing agent so strongest reducing agent will be the one which will undergo oxidation easily now which will undergo oxidation easily the one which will be having maximum sop the one which will be having maximum sop will undergo oxidation easily that will be the better reducing
agent that will be the strongest reducing agent now check which one of the following has got maximum sop guys I want you to understand one thing over here in all the these reactions electrons are gained let's say in the first reaction electron is gained by mg di positive mg di positive is gaining electrons so mg di positive hair is undergoing reduction so this is the this is the standard reduction potential value which is given to us right it's not mentioned whether it's sop or SRP look at the reaction carefully in the reaction gain of electrons
is happening gain of electron involves means reduction so this is standard reduction potential of M G positive when getting converted into mg similarly standard reduction potential of zinc die positive when getting converted into zinc minus 0.76 right similarly these are the SRP values basically these are the SRP values of mg di positive zinc diosi nickel dios and Fe trios respectively right these are the SRP values okay now do I have to check on the base of s SRP or sop I will have to check on the base of sop values my dear students if I
reverse reaction if I reverse the first reaction for example if I reverse the first reaction it becomes like this mg gives mg D positive plus two electrons now it's en not value it sign would have got changed now it is 2.37 volt what should I be calling this value now as is this reaction reduction or oxidation this is the oxidation reaction which is undergoing oxidation magnesium is undergoing oxidation here magnesium is losing electrons right so this is the SOP of magnesium sop of magnesium this value was SRP of mg di positive this value was SRP
because mg di positive was undergoing reduction so this value was SRP of mg di positive but this value is going to be the s o of mg because mg is undergoing oxidation mg di positive is undergoing reduction but mg is undergoing oxidation so this is the SRP of mg di positive which is undergoing reduction but this is s o of mg right now check if you reverse the everywhere if you reverse the sign everywhere where do you see the maximum sop value maximum sop value on reversing the sign will be the first one and that
is the maximum sop of what that is the SOP of what that is sop of magnesium right so more the SOP more the tendency to undergo oxidation better reducing agent so the answer is going to be magnesium is this clear to everyone is this clear to everyone quickly people quickly what do you think about this one what do you think about this one this particular question quickly it's again one simple question people see in every reaction as you see all the reactions in every reactions electrons are gained and you know gain of electrons is called
as reduction gain of electrons is called as a reduction he just give me a second just give me a second thank you okay in the first one gain of electrons is happening gain of electron is what we call as a reduction So Co Tri positive is gaining electron So Co Tri positive is undergoing reduction so this is SRP of Co Tri positive this is SRP of C +4 this is SRP of PB +4 this is SRP of B Tri positive so we have SRP values right now if you reverse the first reaction for example if
you reverse the okay don't reverse the reaction tell me one thing what do we have to check the oxidizing power of order the oxidizing power order tell me people one thing I told you f back only the one which is the better oxidizing agent the one which is the better oxidizing agent will always have more oxidizing power and the oxidizing agent always under goes reduction right so the one which under go reduction easily the one which under goes reduction easily will be the better oxidizing agent will have more oxidizing power now which one will undergo
a reduction easily the one which will be having more SRP more the SRP more the tendency of an element to undergo reduction better oxidizing agent more oxidizing power now what do you think should be the answer of this question these are all the srps of the substances which are undergoing reduction in the first reaction copper tripos Cobalt Tri positive is undergoing reduction so this SRP of cobalt Tri positive getting converted into Cobalt dipositive right now where do you see since we got to know these are SRP values right so where do you see SRP value
maximum which one is the maximum SRP 1.81 this is the SRP of what this is the SRP of cobalt Cobalt rosit right after that 1.67 this is the SRP of PV + 4 right after 1.67 1.61 this is the SRP of C + 4 and then at the end it's going to be B Tri positive right so this is the order of SRP values these are the srps basically right more the SRP more the tendency to undergo reduction better oxidizing agent more oxidizing power so what should be the answer Co Tri positive will have maximum
Co Tri positive will have maximum followed by PB plus 4 followed by so it has to be option b right I hope you can solve these sort of questions from now onwards clear people is it clear to everyone what do you think about this one the reducing power reducing power reducing power the one which will be having more reducing power will be always a better reducing agent and the reducing agent is the one which under goes oxidation so the one which under goes oxidation easily will be better reducing agent more reducing power now which one
has got more more tendency to under go oxidation the one whose sop will be more the one whose sop will be more will be undergoing oxidation easily better reducing agent more reducing power now which one among the following will undergo oxidation easily first of all check one simple thing in all the reactions electrons are gained gain of electrons is reduction so these are basically the srps this is the standard reduction potential of what of the substance which is undergoing reduction zinc dios is gaining electrons so zinc dios is undergoing reduction so this is e not
of zinc di positive gives zinc this is e of calcium D positives gives calcium this is e not of mg D posi gives mg this is e not of n die positive gives n right now tell me one thing since these are the standard reduction potentials but I have to solve question in terms of sop so reverse the sign this will be plus this will be plus this will be plus this will be plus now this plus 0.76 plus 0.76 is it going to be S so or SRP initially it was SRP now it is
going to be sop sop of what of the substance which under goes oxidation it is going to be sop of zinc this is going to be sop of calcium this is going to be sop of this this is going to be sop of nickel simple right okay initially we were given SRP of the ions now we got the SOP of these Metals right initially we were given SRP of ions now we got sop of the metals now check which one has got more sop op it is calcium I believe right calcium has got maximum sop
followed by followed by magnesium followed by followed by zinc followed by nickel right this is the SOP order more the SOP more the tendency to undergo oxidation better reducing agent more reducing power so which answer should be the correct one so calcium has got maximum followed by magnesium followed by zinc followed by nickel so it has to be option b again I hope all these things are clear if all these things are clear let me know once in the chats quickly yeah perfectly done guys I hope you can solve all the questions based on electrode
potential yeah I believe that Perfect People now let's get into one more topic that's EMF have you studied this topic before I'm just asking you let me know in the chats quickly have you studied this topic before EMF quickly guys yeah all right so let's get going then let's get started with the topic EMF what EMF exactly is all about how do you define the term EMF it's a simple topic again right see guys first of all if you talk about AMF it it is basically EMF is the EMF is the maximum maximum potential difference
maximum potential difference between the electrodes between the electrodes means between anode and cathode right I hope you remember your galic cell in which we used to have two electrodes right two electrodes connect them externally and internally we used to get a complete cell right now let me tell you the maximum potential difference between anode and cathode when when no current is drawn from the cell or you'll directly say when the cell is not in use when the cell is not in use you call that maximum potential difference between the electrodes as EMF so imagine that
imagine that I hope you remember your Gonic cell in case of your I mean I hope you remember your Daniel cell in case of your Daniel cell you used to have zinc electrode on one side you used to have copper electrod on one more side right and here you use connect an Amer and here a solt bridge right what this Daniel cell used to do Daniel cell it is a typical example of the galic cell it converts chemical energy into electrical energy right basically this Daniel cell it produces current it produces current now since this
Daniel cell produces current you already know current direction in the danial cell it is from cathode to anode this is the direction of current this is the direction of current and since the Daniel cell is producing current right current is produced due to potential difference can I say this cell will only produce current if there will be potential difference between these two electrodes right this cell will only produce current if there will be potential difference between these two electrodes correct the maximum potential difference between these electrodes between cathode and anode when the cell is not
in use when no current is drawn from the cell right that's something which you call us that's something which you call as that's something which you call as EMF now you must be thinking how it's calculated what is how it's done right let me tell you EMF is calculated with the help of potential meter it is calculated with the help of potential meter now guys what I'm going to teach you now that's important try to understand what exactly I'm going to say okay try to understand for example imagine imagine this is zinc sulfate in this
container and here you have got copper sulfate this is something important guys try to understand this properly right copper sulfate and here you are keeping a zinc rod and here you are keeping a copper Rod one is zinc rod and one is copper Rod okay this is your zinc rod and this is your copper Rod you're connecting them with the help of emiter for example right and what this emiter will do it will show deflection maybe right and when it shows deflection that means the cell produces current if it is not showing any deflection that
means the cell is not producing any current and when the cell produces current when the when the emiter shows deflection that means the cell is producing current current is produced due to potential difference between the electrodes that means at that time there will be potential difference between the electrodes now over here I'm using a solt Bridge as well the circuit is complete now my dear students try to understand what exactly I'm going to say try to understand what exactly I'm going to say I told you EMF is measured with the help of potential meter how
exactly from here you'll get the exact feel of what EMF is all about before that let me tell you one thing EMF when measured under standard conditions and you know what standard conditions are when EMF is measured under standard conditions you do not call it as the EMF of the cell you call it as the standard EMF of the cell so there's a difference between a cell and E not cell what is a cell a cell is simply the maximum potential difference between anode and cathode what is a not cell it is the maximum potential
difference under standard conditions between the anod and cathode now try to understand try to understand people I'm assuming this particular cell I'm assuming this particular cell is kept under standard conditions right it is kept under standard conditions okay now guys understand what exactly I'm going to say this particular zinc Rod as you know this zinc Rod it gets the negative charge something discussed already and this cathode it gets the positive charge right now people what we are planning to do we are planning to use an external battery this is the external battery which I'm going
to take positive D of the external battery negative term of the external battery this is the external battery which I've taken over here this is the external battery which I've taken over okay and this external battery my dear students I'm going to connect this external battery with this danial cell over here okay see the positive terminal of the external battery is connected to the negative recharge Rod it is connected with the negatively charged rod and the positively charged Rod is connected to the negatively charged terminal okay I hope you're understanding what exactly I'm doing I
connected the external battery in reverse polarity over here positively charged positively charged with the negatively charged Rod negatively charged with the positivly charged rod that's what I'm doing right my dear students this external which you are using it will be definitely having some voltage right it will have some voltage perfect even there will be potential difference between these two electrodes as well these two electrodes there will be potential difference correct now tell me one thing tell me imagine and understand if by chance the external battery external battery right if by chance the voltage of external
battery is 0 volts if the voltage of external battery is 0 volts will anything happen to the potential difference between the electrodes I'll say the potential difference between the electrodes I'm representing with Asel for example if external voltage is zero it is not going to affect this particular Daniel cell right the Daniel cell will be normally work working in which oxidation takes place at anode reduction takes place at cathode electrons will move from anode to cathode current will move from cathode to anode right now now try to understand if by chance external voltage if by
chance external voltage is equal to what is equal to EMF of the cell if external voltage is equal to the EMF of the cell so first of all try to understand what I'm going to say since between these two electrodes there will be potential difference right now initially the external voltage given was Zero now imagine slowly you are increasing the external voltage slowly you increasing the external voltage imagine that external voltage slowly you are increasing let me tell you e not cell that means the potential difference between these two rods that is experimentally observed it
is 1.1 volt experimentally it's observed that maximum potential difference between these two electrodes under standard condition that is 1.1 volt so initially the external voltage was Zero now slowly you are increasing the external voltage slowly you are increasing the external voltage slowly you are increasing the external voltage can I say there will be a time there will be a time when external voltage will become equal to e not cell and my dear students since this particular battery it is connected with reverse polarity it is connected with reverse polarity now imagine if external voltage becomes equal
to inot cell at that point of time will the cell will this particular C cell produce any current say yes or no will it produce any current at that time will it produce any current at that time understand it properly since external battery you are connected with reverse polarity positive with negative negative with positive initially there was no external voltage there was no external voltage the cell was working like a normal gmic cell now slowly you started increasing this external voltage and slowly you started increasing the external volage voltage initially in absence of external voltage
this this emiter was showing maximum deflection this emiter initially was showing maximum deflection when there was Zero external voltage now slowly when you start increasing external voltage the deflection in the emiter will decrease the deflection in the Emer will keep on decreasing as the external voltage will keep on increasing the deflection in Emer will keep on decreasing and there will be a point there will be an external voltage at which there will be an external voltage at which there will be no deflection in a meter there will be an external voltage at which there will
be no deflection in Emer and that external voltage at which there is no deflection in Emer that is 1.1 volt in case of danial cell right now since external voltage and E cell have become equalized and both are reversely polarized right polarity is reverse so at this particular point of time there'll be no current in the Cell at that point of time there is no current in the cell perfect right people so if you ask me how this 1.1 volt was calculated 1.1 volt was basically 1.1 volt was basically the voltage which was used here
so initially you did not know what was E not cell what you kept on doing you kept on increasing the voltage of external external battery and there was one voltage which was 1.1 volt at which cells stopped working right and that external voltage has to be equal to the inor cell basically yeah I hope this particular point is clear so what is know what is e cell basically it is the maximum potential difference between the electrodes when the cell is not in use when no current is drawn from the cell I hope this particular point
is clear okay now the actual stuff which we have to understand here which we have to remember the actual stuff which we have to remember let's try to understand that let's try to understand that how do you exactly calculate this how do you exactly calculate this e not cell that's important my dear students there is one simple way of calculating this e not cell how e not cell is calculated as e of cathode minus E of anode this is one way of calculating the e c e of cathode minus E of anode what is e
of cathode and what is e of anode e of cathode is basically standard reduction potential of cathode and this is your standard reduction potential of anode right this is the generalized result which I would want all of you to remember e not cathode will be equal e not sorry e not cell will be equal to e not of cathode minus E of anode what is e of cathode this SRP of cathode this is SRP of anode well you can convert them into sop values as well right you just need to write one thing instead of
SRP of cathode you have to write minus * sop of cathode right instead of SRP of anode you have to write minus * s o p of anode you'll get one more formula no need to remember a lot of formulas just remember this particular one so this one General result of calculating e not cell which is e not of cathode minus E of an now people where do I use this equation that's important try to understand one simple thing imagine I got two electrodes A and B imagine I have got two electrodes A and B
okay imagine I'm writing the value e not of a positive gives a is equal 1.1 volt e not of B positive gives B it is equal 2.1 volt imagine I'm writing something like this how many electrodes I have two one is your a one is B now tell me one thing since I've got two electrodes my dear students under understand if I connect them externally as well as internally if I connect two electrodes externally as well as internally can I say I will be getting one galonic cell I'll be getting an electrochemical cell right yeah
just give me a second guys guys just a second now okay righty to understand what exactly I'm going I'm going to say I to crew electrodes A and B I am connecting these electrodes extern and internally you know externally we connect them with the help of Emer internally we connect them with the help of solt bridge now my dear students we got a cell we got a complete cell which is made up of two electrodes now I'm asking you what will be the standard EMF of that particular cell I'm asking you what will be the
standard EMF of that particular cell first of all try to understand which value this is it is sop or SRP what do you think is it sop or SRP plus 1 to zero decrease reduction this SRP this plus 1 to Z decrease this reduction SRP so both are SRP values now which srp's value is more this SRP is more more the SRP more the tendency to undergo reduction and reduction takes place at cathode so this electrode will be behaving like the cathode and this electrode will be behaving like the anode so first of all you'll
try to identify which is your anode and which is your cathode you identified the cathode and anode now after identifying the cathode and anode it's the time to calculate a not cell which is e of cathode minus E of anode SRP of cathode minus SRP of anode what is SRP of cathode SRP of cathod is 2.1 volt SRP of anode is 1.1 volt the value will be exactly 1 volt done and dusted right what this 1 volt is this one volt is basically going to be your EO cell that's it right perfect if this is
clear let me take few more examples let's say people I'm writing something like this and you need to answer me this okay for example you got the electrode like this I'm writing e not of a positive gives a this is 3.1 volt I'm writing e not of uh B gives B positive it is equal 5.1 volt so I've got two electrodes basically I've got two electrodes I'm connecting them externally as well as internally and you know when we connect these electrodes externally as well as internally do we get one complete cell absolutely we get one
complete cell which is made up of two electrodes now dear students how do we calculate e not cell of the cell which is made by connecting these two how do we Cal e cell first of all which potential this is SRP + 1 to 0 which potential this is 0 to + 1 sop one is SRP one is sop so if this is sop so what has to be SRP then SRP will be Min - 5.1 volt right now compare these SRP values compare these SRP Val Val which SRP is more 3.1 orus 5.1 3.1
the one which will be having more SRP will undergo reduction and reduction takes place at cathode so this has to be cathode and this has to be anode so you got to know which one among the two is going to behave like the anode and which one among the two is going to behave like the cathode so my dear students it is the time to calculate e not cell which is e not of cathode SRP of cathode minus SRP of anode this is the inot cell value I hope this is clear I hope this is
clear there's something wrong with my throat I don't know what about this particular question guys give it a try quickly quickly give it a try standard oxidation potential of the half cells so you are given with sop value this is sop value of zinc zinc is undergoing oxidation here zinc is losing electrons here iron is losing electrons so zinc and iron both are undergoing oxidation so these are the SOP values of zinc and iron EMF of this particular cell EMF of this particular cell the cell which is given to us is like this zinc plus
iron di positive gives zinc di positive plus iron tell me one thing here it zero here is+ two here it is + two and here it is zero now tell me 0 to plus2 increase or decrease increase + two to zero increase or decrease decrease increase in the oxidation state is something which you call as oxidation right so zinc is undergoing oxidation oxidation takes place at anode right iron di positive is under going reduction reduction takes place at cathode perfect reduction takes place at cathode so you understood you got to know that in this net
reaction your zinc electrode will be behaving like the anode and iron electrode will be behaving like the cathode so since you identified calculate e not cell a cell is going to be e of cathode minus a of anode right right people yeah perfect so enod of cathode which is SRP of cathode SRP of cathode cathode is iron is it SRP of sop this sop what we have to write SRP so minus 0.41 minus SRP of anode this is sop of anode but we need SRP so minus 0.76 solve it get the eot cell value is
this clear is this clear to everyone people is this clear to everyone quickly clear I believe everything is clear to everyone right okay now guys now comes one more important thing that is spontaneity and e not cell this is again one important topic spontaneity and E not cell my dear students imagine and understand this is something important okay understand it properly imagine that you have got one electrode this one electrode imagine you have got one more electrode this is one more electrode simply you're taking two electrodes you're connecting them with the help of a meter
with the help of solt bridge it's a complete cell what is a complete cell what is a complete cell complete cell is made up of two half cells two electrodes half cell is also called as electrode a complete cell is made of two half cells so imagine you got two half cells here right which are connected externally as well as internally so I would say I would say this is a complete cell which is made up of two half cells no doubt you made a complete cell you made a complete cell now the point is
whether this complete cell will will be working or not whether this complete cell will be generating current or not right that is one more debate so till now you can make many complete cells you can make many galvanic cells right you can make many complete cells just take two electrodes connect them externally connect them internally you'll be getting a complete cell now whether the complete cell will be working or not on what parameters that will depend on what parameters that will depend upon whether the complete cell will be working or not on what Factor it
will depend upon let's try to check that Factor on which the complete cell depends upon understand guys I hope you got the context of what exactly we are going to discuss we are going to set the parameters we are going to set the criteria which a cell must follow then only we can say the cell is working like a normal galmic cell okay right we are going to set the parameters we are going to set the parameters which a cell has to follow then only I'll say the cell is working like a normal galmic cell
in which electrons will move from anode to cathode current will flow from cathode to anode right now try to understand my dear students let's say I'm taking a Daniel cell here imagine this is the Daniel cell which I'm taking try to understand what I'm going to to talk imagine this is the Daniel cell okay this is one Daniel cell over here there's the zinc electrode there's the copper electrode so two electrodes connected externally and internally it's a Cell It's a complete cell now whether the cell will be working or not we have to discuss that
condition only tell me one simple thing if you remember we have studied in thermodynamics we have studied in thermodynamics system always possesses two types of energy IES this cell is my system right now it is under observation it is under investigation this cell is under investigation right now it is my system so this particular system it will be having two types of energies one is called as available one is called as unavailable energy there will be two types of energies present in the system one is available one is unavailable my dear students if you remember
this unavailable energy this unavailable energy it increases the randomness or disorderness of the particles present in the system this unavailable energy it increases the disorderness present in the system right so whatever is the disorderness of the particles whatever is the randomness of the particles present in the system that is on the cost of unavailable energy so unavailable energy it tries to increase the randomness the disorderness present in the system and that disorderness or Randomness that is measured by the factor called as entropy I hope you remember that disorderness is measured by the factor called as
entropy so what is entropy entropy is a factor which gives you the measure of unavailable energy present in the system right more the unavailable energy more the randomness more the disorderness in the system more the entropy more the entropy of the system you know that right now this available energy this available energy this available energy is used for what my dear students let me tell you this available energy is converted into this available energy is converted into useful work the name of the useful work is also non-pv work non-pv work the example of nonp work
is electrical work is electrical work tell me me one thing do you see any electrical work happening here I would say yes yeah do you see any electrical work happening here you know my dear students in case of this particular cell your electrons they move from zinc to copper right you know that you know electrons move from zinc to Copper that means in case of your Daniel cell if you remember in case of your Daniel cell if you remember this part of the cell in this part what is happening basically electrons are shifted from anod
to cathode in case of your danial cell right in case of your danial cell the electrons are shifted from anode to cathode you know that right so I would say is this the is this just the moment of electrons or it is the moment of electrons in a particular direction it is the moment of electrons in a particular direction due to which the cell produces current you know that you know that now my dear students tell me one simple thing so you know electrical work is happening here since electrons are moving from anod to cathode
you know electrical work is happening here electrical work that is the example of nonp work which is also called as useful work right and what is useful work it is that work it is that work which happens on the cost of what on the cost of available energy what does that mean that simply you can understand like this see I can say the has to go from zinc to Copper okay now electron on its own won't go from zinc to Copper I'll say system will utilize some of its energy system will consume some of its
energy to send electron from anode to cathode to send electron from zinc to copper right this particular system this particular cell it is going to convert it is going to consume some of some of its energy it is going to utilize some of its energy then only it will send electron from anode to cathode for example I'm walking I'm walking on the cost of my energy I'm consuming my energy then I'm walking similarly the electron is going from anode to cathode so the system the cell will be consuming its energy the cell will be consuming
its energy right and that energy consumed by the cell will be utilized to send electrons from anode to cathode now which type of energy the cell utilizes in order to do the electrical work in order to do the useful work it is the available energy which the cell consumes it is the available energy which the cell consumes to do what to do the useful work to do the nonp work to do the electron electrical work to send the electron from anod to cathode right so imagine in the beginning There Was 80 jewels of available energy
in the system imagine there were 80 jewels of available energy present in the system initially after some time you observed that only 60 Jews of available energy is left in the system so what has happened has the available energy increased or decreased the available energy of the system has decreased by how much unit by 20 units the available energy has decreased by 20 Jew my dear students the available energy of the system the available energy of the cell is decreasing by 20 units what does that mean can I say those 20 jewles of energy has
been utilized in order to perform the electrical work right those 20 Jews have been utilized to perform what to perform electrical work can I write a simple statement decrease in the available energy decrease in the available energy is equal is equal to the what useful work and which kind of useful work is happening here it is the electrical work it is the electrical work which is happening okay right people I hope this particular statement is clear to everyone if this is clear to everyone let's try to understand one simple thing my dear students I'm writing
minus Delta G as such is equal electrical work how do you calculate electrical work it is charge multiplied by potential difference charge multiplied potential difference right now I can write something like this minus Delta G is equal see the charge on one electron the charge on one electron is how much - 1.6 the magnitude of charge on electron is 1.6 into 10- 19 right my dear students the charge present on one mole of electrons is equal to 965 kums and this 965 kums is something which you call as farad so one farad is basically volt
965 kums imagine that if one mole of electron is moving from anode to cathode imagine imagine if one mole of electron moves from anode to cathode how much charge is transferred from anod to cathode imagine when one mole of electron moves from anod to cathode that means one fad charge is transferred from anod to cathode now imagine n moles of electrons are moving from to cathode that means n fad charge is moving from anod to cathode so if I'm imagining if I'm assuming n moles of electrons moving from anode to cathode that means n farad
charge is going from anode to cathode multiplied by what multiplied by eel so my dear students Delta G we got to know as minus NF eel this is one very very very important result which every one of you are supposed to remember yeah is it clear Delta G is equal minus n of e cell why did we derive this result what was the importance of it why do we need this there's a logic for that as well my dear students just remember thermodynamics for the process to be spontaneous for any process to be spontaneous Delta
G for the system at constant pressure and temperature that has to be negative then only we say the process spontaneous right okay right people for any process to be spontaneous Delta G for the system at constant pressure and temperature that has been negative you know that now imagine you have got two electrodes you are connecting them externally as well as internally only you got one complete cell now for the cell to be working for the cell to be working for the cell to produce current what should happen once you complete the circuit once you introduce
the salt bridge automatically at anode oxidation should start at cathode reduction should start and once oxidation reduction takes place spontanously then you'll directly say electrons will move from anode to cathode current will move from cathode to anode right so for a complete cell to be working for a complete cell to be working like a normal galmic cell I'm writing it over here for for a cell to be working like a normal galmic cell like a normal gamic cell in which in which current will be from cathode toode right for the cell to be working like
a normal galmic cell once you make the cell once you complete the cell once you take two electrodes connect them exter internally if automatically an oxidation takes place at cathode reduction takes place electrons will automatically move from anode to cathode and will move from cathode to anode right so for the cell to be working like a normal galmic cell the cell reactions in the cell should be spawned aneous and the cell reactions in the cell will be only spontaneous if Delta G reaction will be negative Delta G has to be negative Delta G for the
reactions have to be negative Delta G negative means minus NF cell it has to be negative Delta G negative means minus NF cell has to be negative so what does that mean when is this whole term negative this whole term is only negative if e cell is if e cell is positive if e cell is is positive I would say for the cell to be working like a normal galmic cell Delta G for the reactions have to be negative or you can say e cell has to be positive this is the condition my dear students
this is the condition which a cell should follow then only you can say the cell is working like a normal galmic cell in which electrons are moving from anode to cathode and current is moving from cathode to anode right yes I believe it's clear I believe it's clear Point number one Point number two point number two if by chance Delta G for the reactions is equal to Z what does it indicate it indicates e cell is equal to z e cell is equal to Z and this particular condition is called as equilibrium State equilibrium condition
in the cell what is equilibrium condition this is something which I have not discussed yet I'll be discussing this in the next session not now not now okay and my dear students one more thing if by chance Delta G for the cell comes out be positive or you'll say e cell is negative the cell reactions will be non-spontaneous the cell reactions will be nonspontaneous and if the cell reactions are non-spontaneous cell won't work like a normal galonic cell cell won't work like a normal galmic cell right perfect these are the three conditions which you guys
need to remember so for for any cell to be working like a normal gamic cell it cell reactions should happen on its own they should be spontaneous for that purpose Delta G for the reaction should be negative or E cell should be positive right perfect my dear students one more thing if I write the same equation under standard conditions it becomes Delta G is equal minus NF e cell right this is one more equation which is valid under standard conditions sometimes they ask you calculate the standard gibes free energy of the cell Delta G Delta
G is equal minus NF e not cell that's it that's the difference okay so let me quickly summarize all these things what did I say I said one simple thing for example you you are getting two electrodes you connecting them externally you connecting them internally you are getting a complete cell now whether this complete cell will be working or not whether this complete cell will generate current or not whether this complete cell will behave like a normal galic cell or not once you make this cell if reactions at anode and cathode are happening on its
own that means the cell will work right and if the reactions at anode and cathode are not happening on its own the cell will not work like an Mixel I hope this is clear to everyone okay I hope this is clear to everyone okay and for the cell reactions to happen on its own it's Delta G for the reaction has to be negative Delta G negative means e cell has to be positive then only Delta G will be negative okay my dear students on this particular concept there is one more thing there's one more thing
if you would have studied thermodynamics I would have told you Gibs free energy it is an extensive property and EMF is an intensive property Gibbs free energy is an extensive property and EMF is an intensive property and I have told you if you remember I told you in thermodynamics we can we cannot add or subtract intensive properties directly extensive properties can be added or subtracted directly but intensive properties cannot be added or subtracted directly right okay okay intensive we cannot add or subtract the intensive properties like e Cel like eel e Cel is the intensive
property you cannot add these intensive properties directly right whereas you can add or subtract extensive properties for example there is one question which is frequently Asked from this let's say you have got something like this understand F Tri positive plus three electrons it gives Fe solid Fe Dios + 2 electrons gives Fe solid Fe Tri positive + 1 electron gives Fe die positive E1 value is 1 volt E2 value is 2 volts you are supposed to calculate A3 value how do you solve these sort of questions you would have seen these sort of questions right
I believe you would have seen these sort of questions I believe you would have seen these sort of questions my dear students how do we solve these sort of questions understand first of all let me tell you let me call this as reaction number one let me call this as two let me call this as three right I'm supposed to calculate e value for this particular reaction I'm supposed to calculate e value for this particular reaction just remember one thing the reaction whose e is to be calculated the reaction whose e is to be calculated
your supposed to make that reaction out of the given reactions the reaction whose e value is to be calculated you are suppos to make that reaction out of the a values out of the given reactions out of the given reaction now how do you make this particular reaction my dear students third reaction is made when when you subtract second from first first third reaction is made when you subtract second from first if you subtract second from first so this will be three electrons minus two electrons gives one electron Fe solid F solid zero nothing Fe
trios minus Fe dios that minus Fe dios comes to the right side so this becomes the reaction so third reaction we got when these two are subtracted now if you are thinking that on subtracting these two you are getting third if you are thinking you'll do the same algorithm with their Eva vales if you are thinking you write something like this E3 equal e1- E2 it is absolutely wrong there will be option there will be option like this as well but this is wrong why because I told you your EMF your electrode potential they are
intensive properties they cannot be added or subtracted directly so my dear students if your electrode potential EMF if they are intensive properties if they are not added or subtracted directly that means you'll be solving this particular question with the help of Delta G value instead of three instead of three you'll write Delta G3 is equal instead of one you'll write Delta G1 minus instead of two you'll write Delta G2 okay now you already know Delta G3 what that is Delta G3 it is minus N3 F E3 is equal Delta G1 minus N1 F E1 minus
Delta G2 it is minus N2 F E2 correct f f f are canceled if you multiply throw by minus sign at the end you'll be getting something like this E3 is equal N1 E1 E3 is equal N1 E1 minus N2 E2 divide what divide by N3 now E1 E2 are given to us as for the equation right E1 E2 are given to us as for the equation now tell me what is what is N1 N2 N3 how many electrons are exchanged in the first reaction three so N1 value is three in the second reaction two
electrons are exchanged N2 value is two in the third reaction one electron is exchanged so N3 value is one so E1 E2 N1 N2 N3 everything is given put it here and get the E value right so my dear students one thing you need to remember which you are going to remember directly from now onwards the reaction whose e Valu is to be calculated you are supposed to make that reaction out of the given reactions Point number one point number two your intensive I mean your electrode potential EMF they are intensive properties they cannot be
added or subtracted directly whereas your Delta G that's extensive property you can add or subtract that directly that's why this particular sort of question is solved in terms of Delta G not in terms of E Yeah is it clear to everyone quickly in the chats is it clear to everyone people quickly let me know quickly in the chats if it is super clear to everyone can you give it a try can you give it a try three reactions I've already given the E values have given the fourth reaction its e value has to be calculated
I told you one simple thing I told you one simple thing the one whose e is to be calculated you're supposed to make that reaction now you must be thinking how to make this reaction for that purpose you should complete these reactions for that purpose you should complete these reactions tell me one thing oxidation state of magnes here is + 7 + 7 to + 6 the change is 1 plus 7 to plus 6 decrease in the oxidation state reduction so m42 be gaining one electron then only it will be getting converted into M4 DGA
tell me the oxidation of magnes here it is plus 6 here it is +4 so plus 6 to + 4 decrease in the oxidation state reduction gain of electrons how many two electrons right this is+ 4 to +2 decrease in the oxidation state reduction gain of electrons how many two electrons this is+ 7 to plus2 decrease decrease means reduction gain of five electrons now I believe you can now I believe you can I complete the reactions after completing the reactions the reaction whose e value is to be calculated you are supposed to make that reaction
now how can you make the fourth reaction I would say fourth reaction can be made when first second and third will be added because this term this term will get canceled out this term this term will get canceled out 2 + 2 plus 1 makes it five electrons so M4 NE plus 5 electrons gives m& positive M4 NE plus 5 electrons gives m& positive so fourth reaction you making when you adding first three reactions right now are you going to do the same algorithm with their e values are you going to write E4 equal E1
plus E2 plus E3 no way you're going to solve this question in terms of Delta G so my dear students Delta G for the fourth is equal Delta G for the first reaction is equal Delta G1 plus Delta G2 plus Delta G3 now you know Delta G4 is equal minus NF E4 is equal minus N1 F e1- N2 F E2 - N2 F E3 right multiply throw with minus sign and f f f everywhere gets canceled out so from here you can calculate E4 which has to be equal N1 E1 plus N2 E2 + N3
E3 divide by what divide by N4 E1 E2 E3 are given as per question now N1 is 1 N2 is is 2 N3 is 3 sorry N3 is 2 and N4 is 5 N1 N2 N3 N4 everything is given so easily my dear students you can calculate E4 value from here right I gave you the random values I did not give you the exact values this 1 2 3 4 I'm giving you on my own right these are not the exact values perfectly done perfectly done guys yeah yes is it perfectly done now people till
now I told you till now I told you how to calculate a cell how to calculate standard em of the cell how to calculate maximum potential difference between the electrodes when the cell is not in use under standard condition right under standard conditions that's something which I told you how do we calculate standard EMF of the cell but by chance imagine you have got a cell which is made up of two electrodes but imagine the conditions are not standard the conditions are something different imagine the concentration of electrolyte is different imagine temperature is changed imagine
pressure is changed right that means conditions are different if conditions are different at that point of time if I ask you calculate the maximum potential difference between the electrodes that means I'm not asking you to calculate the standard EMF I'm asking you to calculate the normal EMF MF so there are these are two things one is standard one is normal standard is e not normal is e not sorry normal is e normal is E standard is e not right difference one is measured under standard conditions one is simple the maximum potential difference under normal normal
conditions okay now understand properly what I'm going to say I'm going to give you an equation here the most important topic of the chapter the most I would say nnest equation right nnest equation people this Nest equation it has got many uses but right now I'm not going to tell you about the uses let's first of all derive this equation let's see from where the equation comes exactly there is one equation which I I would say everyone would have studied in thermodynamics Delta G is equal Delta G Delta G Plus RT lwn of QC have
you studied this you should have I believe now what is this Delta G it is equal to minus NF e cell what is Delta G minus NF e cell right plus RT it is a lawn of QC If I multiply through by minus and divide with NF what do I get I'll get get e cell is equal e not cell minus RT / NF it is lwn of QC correct I can write this equation in this format as well a cell is equal a not cell minus this is lawn converted in log it is 2.303
RT / NF and it's log of QC this is something which is what you call as nnest equation why did we write this particular equation my dear students be careful with these terminologies e cell is what you call us EMF of the cell under non-standard conditions under non-standard conditions this is standard EMF of the cell standard EMF of the cell which is equal to e cathode minus E anode this n what is n number of moles of electrons exchanged in the net cell reaction number of moles of electrons exchanged in the net cell reaction what
is f f is farad constant whose value is 965 Kum our values 8.314 JW per Calin per mole right and T stands for temperature and QC stands for reaction quotient UC stands for reaction quotient this equation is what we call as Nest equation now dear students this particular equation you won't be using in the questions in the questions I'll be giving you one simplified version of this imagine you are putting the value of RS 8.314 Jew per Kelvin per mole imagine you are keeping the Fades constant as 965 kums imagine temperature is taken as 298
kin let's say all these three parameters you put into this equation after putting in this equation what do we get a cell is equal a cell - 0.0591 / n and then it's going to be log of QC this is the equation which I will be using in majority of the questions guys this is the equation which I'll be using in majority of the questions right and this equation is basically valid when temperature of the cell is kept as 25° C 298 kin okay so this particular equation it has got many uses basically right we
we use this particular equation to calculate lot of things to calculate EMF of the cell to calculate oxidation potential of the electrode reduction potential of the electrode pH of the cell etc etc a lot of things are I mean a lot of things are calculated with the help of this particular equation or this particular equation and what are I mean how do we calculate them right what are its applications to be more precise that's something which I'll be teaching you in the next session because in electrochemistry what all topics we have discussed till now first
of all we started with something called as galvanic cell if you remember right we discussed how galvanic cell exactly Works how Daniel cell exactly works then I told you how do we represent a galvanic cell that was the second topic the third topic was how to write the net cell reactions from a given cell that was the third topic fourth topic was your electrode potential and its significance right fifth topic was how do you calc calate the standard EMF of the cell e not cell which used to be e of cathode minus E of anode
I hope all those things I hope all those things you exactly remember which we already discussed in the chapter electrochemistry right okay now we will try to understand what exactly I'm going to say now we are going to discuss something called as nnest equation so let's exactly get to know what this nerest equation is all about okay let's have a look people let's have a look tell me in the chats whether the video Clarity is proper is the video Clarity proper say yes or no in the chats say yes or no in the chats quickly
quickly everyone is the video Clarity all okay let me know quickly because somebody wrote video Clarity should be increased I think you should increase your video Clarity in the settings yeah because on my side I believe everything is okay yes suanti you can enroll still you can still enroll into the batch no issues the link is there in the description okay so let's try to get this nnest equation my dear students what this nnest equation is I'll be first of all deriving it if you remember there was one equation which you all must be knowing
from the chapter thermodynamics and the equation was Delta G is equal Delta G Plus RT lwn of QC I hope you remember this particular equation which we have discussed in thermodynamics if you do not remember it do remember it from now onwards Delta G it is always equal to Delta G Plus RT lwn of Q where Q is the reaction quotient you know it now my dear students if I particularly talk about the galvanic cell for a galvanic cell in case of the galic cell can you let me know what Del Delta G exactly is
all about we have discussed this in the last session Delta G for the galvanic cell is equal minus NF cell right and Delta G Delta G is equal minus NF a cell these were the things which we have already discussed right now people understand if I put these two parameters into the main equation what do I get instead of Delta G I'll be writing minus NF cell is equal instead of Delta G I'll be writing minus NF a cell minus n of C it's going to be plus this is RT this is lwn of QC
okay now first of all if I divide this equation if I divide this equation throughout throughout by what if I divide this equation throughout by NF I'm dividing this particular term by NF I'm dividing this particular term by NF and at the same time I'll be dividing this particular term with NF as well right so NF andf canceled so I'll be getting okay you can do one more thing you can multiply throughout with the minus sign when you multiply throughout with a minus sign NF NF canceled minus minus becomes plus so you are getting one
equation here which is eil is equal e is equal a c minus E not cell minus this is RT / NF and it's going to be lawn of QC this particular equation which is made over here my dear students this particular equation which is formed over here which is made over here this particular equation is something which you call as nnest equation this is something which you call as nnest equation okay so e cell is equal e cell minus RT / NF L of QC now let me further simplify it a bit let me further
simplify it a bit have a look so you can write something like like this e cell you are going to write as such e cell you are going to write as such is equal e cell you are going to write as as such minus this is lawn of QC so convert it into log so it has to be 2.303 2.303 RT ided by NF and now it's going to be log of QC right this is one more form of the nnest equation which is something I call as the logarithmic form this is something which I
call as the logarithmic form of the nnest equation now my dear students try to understand the parameters which are there in the equation once you get to know the parameters then I can show you its application form where do we exactly need this particular equation before that what is this e cell over here e cell is something which you call as EMF of the cell e cell is something which you call as EMF of the cell but under which conditions under under non-standard conditions under non-standard conditions I hope you all remember what are standard conditions
and what are non-standard conditions in case of standard conditions in case of standard conditions concentration is taken as one M pressure is taken as one bar right and temperature is generally kept constant right temperature is kept constant generally 25° C perfect so e cell is something which you call as the EMF of the cell under non-standard conditions I hope you remember what EMF is what is EMF EMF is the maximum potential difference between the electrodes in the galvanic cell when the cell is not in use discuss that in detail right already we have discussed that
in detail EMF is basically the maximum potential difference between the electrodes when the cell is not in use when no current is drawn from the cell right and EMF is measured with the help of potential meter I hope you remember that now people what is e not cell what is e not cell e not cell is something which you call as standard EMF of the cell this is something which you call a standard EMF of the cell and I hope you guys remember how do we calculate e not cell it is calculated like this e
not cathode minus E not anode what is e not cathode this is something which you call as standard reduction potential of cathode and this particular term is your standard reduction potential of anode this is how you calculate a not cell as well okay I hope you remember what this n exactly is if I ask you what is this n over here n represents number of moles number of moles of electrons exchanged in the net cell reaction number of moles of electrons exchanged in the net cell reaction and I have taught you in detail how do
we get the N value right what is this F over here f is the farad constant f is the Faraday constant whose value is 965 kums right and and what is this QC over here QC is something which you call as reaction quotient QC is something which you call as reaction quotient and if you remember already I have taught you how to get the reaction quotient of a particular reaction what is the c over here e is the temperature at which the cell is working and R is a constant whose value you already know that's
8.314 Jew per Kelvin per mole correct now my dear students try to understand what exactly I'm going to say something important I'm going to discuss with you see we got the equation the equation is something like this eil is equal to e not C- 2.303 RT ided by NF and then we are going to write log of QC this is something which you call as the nnest equation now my dear students in the questions in the questions I'm not going to use this equation always in the questions I'm going to simp simplify this equation a
bit right and after simplifying we'll be getting one comparatively simple equation and that simple equation we'll be using in the questions and how do we get that equation see guys if I keep the temperature if I keep the temperature as 25° Centigrade which means 298 kin right if I take the farad constant as 965 Kum right if I take the value of r as 8.314 JW per Calin per mole if I put all these parameters in into the equation what do I get finally I'll be getting something like this a c is equal it going
to be a c minus 0.0591 divided n and then it's going to be log of QC this is the equation which I'll be using in majority of the questions in case of the nerest right okay e cell is equal to e not cell minus 0.0591 divid by n then it's going to be log of QC my dear students do you remember how do we calculate e cell I have discussed this with you e cell is equal e cathode minus E anode do you remember how do we get the N value number of moles of electrons
exchanged in the net cell reaction for that purpose we write reaction at anode then we write reaction at cathode then we try to balance the electrons then we try to add the reactions and once we add the reactions there will be the electrons which gets cancelled out right that gives you the N value I hope you remember and I hope you remember how to get the QC value as well this is something which we have discussed too right so basically whatever parameters we need in the nnest equation all those parameters we have already discussed now
what we need to do is we need to put those parameters we need to put the value of those parameters into this equation and get the value of e cell okay so number one this equation is what you call as nnest equation now the point is where all do we use this equation what about its application my dear students let me exactly tell you where all do we use this particular nerest equation okay so first of all nerest equation is used in half cells half cells means electrode right you know a complete cell a complete
galvanic cell that's made of two electrodes right or you can say a complete Gonic cell it's made up of two half cells half cell is equal electrode right half cell is equal electrode you can use this particular equation for the half cells why my dear students in order to calculate in order to calculate the oxidation potential of any electrode under non-standard conditions or in order to calculate the reduction potential of any electrode under non-standard conditions I'll be using this particular equation in order to calculate the oxidation potential of the half cell the oxidation potential of
the electrode or the reduction potential of the electrode under non-standard conditions at that point of time I'll be using this particular equation and how exactly I'm going to write this imagine that imagine that I need to calculate reduction potential of some electrode under non standard conditions how do I write this equation you'll be writing this equation in a different format you'll write e r which represents reduction potential of the half cell is equal e not red e not red means standard reduction potential of the same electrode right minus 0.0591 / n and then it's going
to be log of what log of QC so this particular equation this particular equation is used to calculate the reduction potential of the half cell under non-standard conditions okay my dear students in the similar way whenever you need to calculate the oxidation potential of the half cell oxidation potential of the half cell under non-standard conditions again you'll be using this particular equation and how do you write this equation at that time you'll write eox instead of E a cell you'll be writing aox which represents oxidation potential of the half cell under non-standard conditions is equal
e not Ox a ox represents standard oxidation potential of the half cell minus 0.0591 divided n then it's going to be log of QC where do we use this particular equation my dear students whenever you'll be given with the electrode whenever you'll be given with a half cell and you are supposed to calculate you will be supposed to calculate the oxidation potential or reduction potential of that half cell under non-standard conditions you'll be using this Nest equation in these two forms right if you if you are supposed to calculate the reduction potential of the half
cell use this equation if you are supposed to calculate oxidation potential of the half cell use this particular equation that's all right okay is this point clear to everyone now one by one in detail we are going to discuss how this nerest equation exactly is used okay so first of all I'll show you how nerest equation is used in case of electrodes or electrode is something which you call as half cell okay try to understand what exactly I'm going to say imagine that imagine that I taken a container let's say this is the container and
my dear students in this particular container in this particular container imagine that imagine that there is an electrolyte there is an electrolyte is it fine now is it fine now quickly let me know in the chats it should be okay now I I believe yeah guys just refresh once just refresh once just refresh once okay now it's fine all right perfect perfect all right guys try to understand what exactly I'm going to say imagine that this is the container and in this container you have kept the ions which ions MN positive I right and in
this particular container you have dropped a rod there's a rod which is made up of element M there's a rod which is made up of element M and this particular Rod is introduced into a solution containing its own ions right I've taken a rod right and this Rod is introduced into a solution which contains its own eyes perfect so I'll be calling this whole setup as electrode this whole setup should be called as electrode now my dear students this particular particular electrode which are taken over here it can work like cathode it can work like
anode as well you know it already imagine that this particular electrode is working like cathode imagine that the electrode is behaving like the cathode okay now tell me one thing when this particular electrode behaves like the cathode what is going to happen oxidation or reduction reduction will happen because at cathode always reduction takes place a reduction means gain of elect so what exactly is going to happen my dear students these MN positive ions which are there in the solution they are going to collide with the rod they are going to take n electrons from the
rod so the reaction has to be something like this MN positives which were there in the solution they are going to collide with a rod they are going to take n electrons out from the rod and it will get converted into M solid this has to be the reaction when this particular electrode behaves like the cathode right now tell me one thing in this particular reaction if I ask you how many moles of electrons are exchanged in this particular reaction I must say n moles of electrons are exchanged now dear students if I ask you
to write the reaction quotient for this particular reaction you'll start with the product which in solid form it's active Mass's Unity divided by it's going to be concentration of MN positive rais tetric coent that's one okay perfect now my dear students I want you guys to calculate the reduction potential of this particular electrode imagine imagine for example imagine for example the concentration of MN positive is anything anything anything three M four M 5 mol right imagine the concentration of MN positive in the container is anything three M 5 M 6 M anything right so basically
is this electrode present under standard conditions or non-standard conditions I'm not keeping the the concentration of electrolyte here is one mol I'm keeping it different so this particular electrode is present under non-standard conditions now under non-standard conditions I want you to calculate the reduction potential of this half cell the reduction potential of this electrode and reduction potential of the electrode under non-standard conditions is calculated with the help of nnest equation it is calculated with the help of nerest equation right my dear students basically what am I asking you to calculate try to understand one thing
if you remember we have discussed this in detail see basically initially in the container initially in the container there were equal number of cats and anions initially in the container there were equal number of cats and anions now now the C in the container is colliding with the rod it is taking electrons from the rod getting converted into what getting converted into M solid and that M solid will be deposited on the rod that M solid will be deposited on the rod tell me is the amount of ctin in the container increasing or decreasing is
the amount of CES in the container increasing or decreasing I'll say amount of CES in the container is decreasing because MN positives they are undergoing reduction getting converted to molid and that's molid is deposited on the rod basically right so the amount of Cates in the container are decreasing if the CES are decreasing if the cats are decreasing anions in the container are remaining the same but C are decreasing so this solution will effectively get which charge this solution will get effectively negative charge and my dear students these MN positive ions they have collided with
the rod they have taken electrons from the rod if electrons are taken from the rod then the rod gets with charge positive charge right so tell me one thing the rod got positive charge solution got negative charge can I say between the rod and the solution a potential difference is getting created a potential difference differ is getting created between the rod and the solution and that potential difference is what you call as electrode potential basically right that's what you call as electrode potential imagine imagine the conditions are not standard so I'll be calling this electrode
potential under non-standard conditions right perfect so what I simply did leave all this aside we have discussed this in detail I just need one thing here I told you I told you that I have got an electrode and I want to calculate its reduction potential under non-standard conditions okay so till here I'm done now in order to calculate reduction potential under non-standard conditions I'll be using the nnest equation nnest equation says that e red reduction potential of the same electrode under non-standard conditions is equal to a red which is standard reduction potential of the same
electrode minus 0.0591 divided by n and it's going to be log of QC what is QC QC is 1 / concentration of MN positive right this is how you guys are going to calculate this is how you guys are going to calculate the reduction potential of the half cell under non-standard conditions this particular term represents what it represents the reduction potential of the half cell under which conditions under non-standard conditions under non-standard conditions this particular term represents standard reduction potential of this electrode standard reduction potential of the half cell right and N already you know
it represents moles of electrons exchanged now tell me one thing tell me one thing guys one thing I want you guys to learn from here see imagine that imagine that I'm increasing the concentration of electrolyte in the container imagine that I'm increasing the concentration of electrolyte in the container if I increase the concentration of electrolyte in the container tell me what will happen to the concentration of MN positives these M positives how they are coming into the Container when the electrolyte is undergoing dissociation if I'm increasing the amount of electrolyte in the container can I
say the concentration of MN positives in the container will increase absolutely the concentration of MN positives in the container will increase now use Simple mathematics here you are increasing the concentration of MN positives in the container by adding more electrolyte you increasing the electrolyte in the container so you are increasing the concentration of M POS POS that means you increasing the denominator here if you increase the denominator if you increase the denominator what will happen to this particular value this value will decrease you're increase the denominator so this whole value is decreasing if this value
is decreasing I would say this whole value will be decreasing if this whole value is decreasing what will happen to the difference of these two terms if this value is decreasing I must say the difference between these two terms it will increase if the difference between these two terms increases that actually means your reduction potential is increasing that actually means reduction potential of the half cell that is increasing so from here we can generalize one very very very important statement and what is the statement all about statement is very simple on increasing the concentration of
electrolyte in the container on increasing the concentration of electrolyte in the container the reduction potential of the half cell that increases okay what is meant by this let's try to understand this a bit more in detail see guys imagine that this is the zinc electrode I mean this is the zinc rod and you have introduced the zinc Rod into zinc sulfate right you have got one more zinc electrode here you have got one more zinc electrode here this is your zinc Rod which is introduced into zinc sulfate right electrodes are same both are zinc electrodes
right now tell me one thing imagine the concentration of zinc sulfate here is is 2 m and the concentration of zinc sulfate here I have kept as 5 mol will the reduction potential of both the electrodes be same or different since the concentration of electrolyte is kept different that electrode in which the concentration of electrolyte is more that electrode in which the concentration of electrolyte is more will have more reduction potential is it simple so I would say this particular electrode will have more reduction potential than this particular electrode yes agreed so do remember this
is one very important thing with the increase in the concentration of electrolyte reduction potential of the electrode it automatically increases okay now people so this is how you calculate the reduction potential of the half cell okay now now people now the point is how do we calculate oxidation potential of the half cell how do you calculate the oxidation potential of the half cell tell me one thing imagine I'm taking the same electrode which I took few minutes back let's say this is your container and in this container what do we have we have got MN
positive ions and here you have introduced a metal rod here you have introduced a metal rod so this is an electrode now my dear students I want to calculate the oxidation potential of this half cell under non-standard conditions in order to calculate oxidation potential I'll make sure this electrode behaves like the anode why because at anode only oxidation takes place so I'll make sure that this electrode behaves like the anode when this electrode behaves like the anode what will happen oxidation right so this Rod is basically made up of metal atoms and those metal atoms
will undergo oxidation they lose electrons and those metal atoms those metal atoms they'll get converted into MN positives and MN positives will enter into the solution and you'll be getting some n electrons here perfect this is the reaction which will have happen when this particular electrode behaves like the anode behaves like the anode right now try to understand this represents number of moles of electrons exchanged right now I'm asking you to get the QC value start with the product it's going to be concentration of MN positive raise parri that's one this in solid state it's
AC masses Unity now my dear students what exactly am I going to do I made sure that electrode behaves like the anode at anode oxidation takes place so basically I'm asking you to calculate the oxidation potential of this particular half cell oxidation potential of this electrode as for nest equation it has to be equal e Ox e Ox standard oxidation potential of the same electrode - 0.0591 divided by n and then it's going to be log of QC and QC value is nothing that's concentration of MN positive clear so with the help of this particular
equation with the help of this particular equation I will be calculating the oxidation potential of this particular electrode under non-standard condition so this is oxidation potential of what of the half cell under what under non-standard conditions under non-standard conditions what is this ax this is your standard oxidation potential of the same house okay now try to understand one thing Here My Dear students if I increase the concentration of electrolyte in the container if I increase the concentration of electrolyte in the container what will happen due to that due to that I would say the concentration
of MN positives will increase the concentration of M and positives will increase now if the concentration of M and positive is increasing if this term is increasing that means this term is increasing if this term is increasing that means this term is increasing if this particular term is increasing that means the difference between the two will decrease if the difference between the two is decreasing what does that mean that means indirectly oxidation potential is decreasing indirectly oxidation potential is decreasing so do remember this particular scenario as well upon increasing the concentration upon increasing the concentration
of electrolyte in the container upon increasing the concentration of electrolyte in the container the oxidation potential of the half cell the oxidation potential of the electrode it decreases is this point clear to everyone say yes or no in the chats quickly with a thumbs UPS is this point clear to everyone so with the increase in the concentration of electrolyte reduction potential of the electrode increases and it's evident if reduction potential of the electrode increases automatically oxidation potential of the same electrode that will decrease right let me know once in the chats quickly everyone everyone people
everyone yeah it's clear wonderful all right so I hope you got to know how do we calculate the oxidation potential and reduction potential of the half cells of the electrodes under non-standard conditions right and my dear students after some time I'll be giving a lot of questions on the same equations so that you'll understand it in a better way right but right now this much is enough okay after some time I'll give give you the questions before solving the questions there are few more things which I want to share with you which I want to
share with you what are those things nerest equation for hydrogen electrode how do you exactly write the nnest equation for hydrogen electrode try to understand I hope you guys remember how we exactly make the hydrogen electrode I hope you remember this is the container in this particular container what are we keeping in this particular container we are keeping hedge positives right this this concentration of H posi can be anything I'm keeping H positives in the container and my dear students over here this is the inverted U type tube which I'm keeping over here right and
if you remember there's a platinum wire which is coated with Platinum black I hope you remember this as well this is the Platinum wire which is coated with Platinum black right and over here on this side I'm introducing H2 gas I'm introducing H2 gas what is going to happen this H2 gas will absorb on the surface of platinum and I'll make sure this H2 gas which I'm introducing it gets absorbed on the whole surface of platinum right I'll introduce H2 gas still the whole surface of platinum is covered by H2 gas can I say after
continuously the introduction of IO gas can I say there'll be a scenario when the plattinum wi is completely covered with IO gas right so it looks like a hydrogen Rod to me it does not look like a platinum Rod now it looks like a hydrogen Rod because this entire surface is covered with hydrogen gas now put this hydrogen rod and introduce in h positives right so hydrogen Rod introduced into its own ions so this whole setup I'll be calling as electrode now which electrode this is this is a hydrogen electrode this is the hydrogen electrode
right this is the hydrogen electrode now tell me one thing imagine that imagine the pressure at which I'm introducing H2 gas imagine that is for example one bar the pressure at which I'm introducing hydrogen gas imagine that's one bar right now guys this particular hydrogen electrode which we have this hydrogen electrode it can behave like anode or it can behave like cathode right imagine that this particular hydrogen electrode is behaving like the anode when the hydrogen electrode behaves like the anode what's going to happen oxidation is going to happen oxidation is going to happen the
surface is covered with H2 gas right so H2 gas on the the surface it will undergo oxidation and if you remember when H2 under goes oxidation it gets converted into 2 * H positive Aquis and with that you'll be getting two electrons this is something which is also discussed when I taught you the hydrogen electrode formation long back so when this particular hydrogen electrode when it behaves like the anode this is the reaction which going to happen which I call as the oxidation reaction now my dear students in this particular reaction if I ask you
how many moles of electrons are exchanged I would say two moles of electrons are exchanged two moles of electrons are exchanged in this reaction okay tell me what is the value of QC QC value start with the product it is concentration of H positive rais power 2 divided by this is partial pressure of H2 and partial pressure of H2 I kept as one bar so leave that aside okay so this is your QC value now my dear students since since this electrode this hydrogen electrode is behaving like the anode and at anode oxidation takes place
right imagine I want to calculate the oxidation potential of this hydrogen electrode under non-standard conditions so I'll be using the nest equation I'll say eox oxidation potential of this hydrogen electrone is equal ax standard oxidation potential of the same hydrogen electrod minus 0.0591 divided by n that is going to be log of QC this is how you are going to write nus equation to calculate what to calculate oxidation potential of this particular hydrogen electrode under non-standard conditions now my dear students you guys are going to tell me one thing you guys are going to tell
me one thing what is e notx that is standard oxidation potential if you remember s o SRP of hydrogen electrode as per convention is taken as zero do you remember that right sop as well this is sop of hydrogen electrode and you know sop as well as SRP of hydrogen electrone that's taken as zero so this particular term is already zero now- 0.05 9 9 1 divide by what divide by n log of QC so it's going to be log of QC is nothing that's concentration of H posi Square correct now guys tell me one
thing log of M ra n log of M ra n is n log n so I can write it like this e x is equal - 0.0591 divide by n value is 2 so this is 2 log of H positive log of H positive concentration as simple as that right as simple as that now after this what am I going to do try to understand after this what exactly am I going to do try to understand I'll write eox is equal eox is equal this two and two gets canceled out I'll just write 0.0591 and
this minus I'll take inside so it's minus log of H positive minus log of H positive is something which you call us minus log of H positive something which you call us pH my de students this is one General result which I got calculate the oxidation potential of the hydrogen electrode in which the partial pressure of H2 is one bar right in which the partial pressure of H2 is one bar if the partial pressure of H2 is changed if it's not one bar then you cannot use this relation you cannot use this relation this relation
is only valid when the partial pressure of H2 is one bar now guys tell me one thing can I say oxidation potential of the hydrogen electrode depends on the pH of this solution it depends on the pH of this particular solution right okay so can you say on increasing the pH of the solution the oxidation potential of the hydrogen electrode does it increase or decrease will absolutely increase right oxidation potential is directly proportional to pH if you increase the pH of the solution I would say it's oxidation potential will automatically increase right people is it
clear to everyone quickly in the chats quickly in the chats let me know quickly in the chats if all the things are absolutely clear right okay so what I was telling you this was the just this is just a general equation by means of which you can calculate the oxidation potential of the hydrogen electrode under non-standard conditions now guys tell me one thing can't the same hydrogen electrode behaves like behave like the cathode what do you think can't the same hydrogen electrode behave like the cathode absolutely the same hydrogen electrode it can behave like the
cathode as well yeah the same hydrogen electrode it can behave like it can behave like cathode as well and tell me just one thing when the same hydrogen electrode behaves like the cathode when the same hydrogen electrode behaves like the cathode what will happen reduction will happen reduction will happen just reverse this reaction just reverse this reaction so the reaction now is going to be 2 * H positive Aquis plus 2 electrons it gives what it gives H2 gas so this is the reaction when the hydrogen electrode behaves like the cathode now guys how many
moles of electrons are exchanged two moles of electrons are exchanged in the in this particular reaction write the expression for QC QC is going to be start with the product it is in gaseous form so partial pressure of H2 I'm going to take as one bar divided by concentration of what concentration of H positive rais power 2 and it has to be concentration of H positive raised -2 simple now I made sure that hydrogen electrode behaves like the cathode at cathode reduction takes place so indirectly what am I trying to calculate I'm trying to calculate
the reduction potential of the same hydrogen electrode under non-standard conditions and I'll be using the nest equation reduction potential of the hydrogen ELR is equal standard reduction potential of the hydrogen electrode minus 0.0591 / n and it's going to be log of QC right now my dear students in case of hydrogen electrode this particular term is going to be zero this is SRP of hydrogen Zer so it's going to be e red is equal - 0.0591 / n value is just 2 and log of QC QC is concentration of H POS ra^ minus 2 perfect
Now log of m^ n log of M ra for n which has to be n log M right so I would say e r is equal - 0.0591 / by 2 log of M ra n is n log n so min-2 comes to the front then it's going to be log of H positive here perfect simple maths I'm using nothing else so two and two got canceled out when two and two gets canceled out what I'll be left with what I'll be left with see guys try to understand I'll be writing the reduction potential of
the hydrogen electrode is equal - 0.0591 oh this is minus log of H postive and minus log of H positive something which you call us pH minus log of H positive is something which you call us pH so I got one more important result over here the reduction potential of hydrogen electrode does it depend on ph yes it does depend upon the ph and you got to know you got to know on increasing the pH of the solution on increasing the pH of the solution reduction potential of the hydrogen electrode it will decrease because the
sign here is minus because the sign here is minus right perfect and just try to understand one thing look at this equation Eed is equal - 0.0591 multipli pH and look at this equation aox is plus 0.0591 into pH it is evident only this is something which we already knew this is something which I've already told you eox of an electrode is equal minus times it's ered right that's something which was already discussed which was already understood I hope hope all these things are clear I hope all these things are clear to you is that
is that people right so this is one more important result you need to remember this so when you increase the pH when you increase the pH of this particular solution when you increase the pH of this particular solution oxidation potential of hydrogen electrod will increase and reduction potential of the same electrode it will decrease right perfect perfect guys so I hope you got to know how exactly we are going to use nnest equation for half okay so now my dear students let's try to solve few basic basic questions so that you can understand all these
things properly look at the first question look at the first question look at the first question from this particular question you'll exactly get the idea of how we use nnest equation for the h cells all the types of questions which can be asked from Nest equation I'll be solving all those types of questions here only right that was a nice one yeah okay look at this particular question guys try to understand the question is we are given with an electrode look at the electrode carefully you have taken an iron Rod you have taken an iron
rod and this particular iron Rod is introduced in a solution containing feso4 right simple so basically you have taken an iron Rod let me just uh make it clear to you imagine this is a container and in this particular container what have we kept we have kept feo4 now this feo4 would have got dissociated as Fe di positive plus s so4 d negative now in the same container you have introduced an iron Rod this is the iron Rod which is introduced into a solution containing its own ions I'll be calling this whole set I'll be
calling this whole setup as electrode so this is that iron electrode right and my dear students what exactly I'm going to do look at the question the question is asking calculate the oxidation potential of this half cell when the concentration of feo4 is kept 0.1 M so question says when the concentration of electrolyte in the container is 0.1 mol at that point of time what will be the oxidation potential of this half cell yeah now have a look have a look people first of all initially as for the equation the concentration of feso4 kept 0.1
M so this has to be zero this has to be zero now imagine that this F4 it gets 100% dissociated into its ions and when this feo4 gets 100% dissociated into its ions and be left with nothing I be left with nothing now people 1 Mo gives 1 Mo 0.1 M gives 0.1 M this has to be also 0.1 M if I ask you what is the concentration of Fe diosi in the solution you'll say the concentration of Fe di positive in the solution is 0.1 M now you must be thinking why did I calculate
the concentration of Fe di positive my students I need that I need that concentration of V positive to write QC have a look see as for the question what do I need to calculate I need to calculate oxidation potential of this half cell when the concentration of feo4 is 0.1 Mo to calculate the oxidation potential I'll make sure that the electrode behaves like the anode because at anode only oxidation takes place right at anode only oxidation takes place now have a look what has to be the reaction at anode oxidation oxidation means loss of electrons
see this is iron with oxidation state zero this is iron with oxidation state plus2 so iron is changing its oxidation state from 0 to plus2 increase in the oxidation state oxidation loss of electrons so basically the reaction has to be Fe solid is getting converted into what Fe solid has to get converted into Fe dioso Aquis and with this you'll be writing two electrons so this is the reaction which will take place when this particular electrode will behave like theode now if I ask you how many moles of electrons are exchanged in the n in
this reaction I'll say two moles of electrons are exchanged okay two moles of electrons are exchanged tell me what about QC value what about QC value start with the product it's concentration of Fe D posit ra^ 1 divid by this is solid Unity now do you understand why did I calculate the concentration of Fe positive because I needed here I needed it here so the concentration of Fe positive in the solution is nothing that is 0.1 perfect so QC value is 0.1 right n value two now what do I need to calculate I need to
calculate the oxidation potential so use the nnest equation oxidation potential of the half cell is equal E X standard oxidation poal of the half cell minus 0.0591 divided by n that is going to be log of QC perfect so this is something which I need to calculate eox is equal e not Ox look at this particular value is this sop or SRP see this is plus 2 to0 plus 2 to 0 means decrease in the oxidation state decrease in the oxidation state means reduction so this is SRP but do I need SRP or sop we
need sop so this particular term is sop right this is SRP so just reverse its sign so it's plus 0.44 plus 0.044 minus 0.0591 divide by n value n value as for the equation is two then it's going to be log of QC QC value we have already calculated that 0.1 which means 10^ minus 1 now it's a matter of calculation only this is 0.0591 / by 2 and it's log of m n n log M so minus1 comes to the front log 10 is 1 so just solve this value and get the answer in
volts that's going to be the oxidation potential of this particular half cell right okay whatever you get from it that's going to give you the oxidation potential of the half cell which you were supposed to calculate imagine that you were supposed to calculate the reduction potential of the same half cell since you calculated the oxidation potential just reverse its sign you'll be getting the reduction potential of the same half cell as well when the concentration of f04 is taken as 0.1 m is this clip let me know once in the chats people quickly quickly PR
is saying it's 0.046 right okay it is 0.0 I think 47 right this is the oxidation potential of this particular half cell so its reduction potential will be minus times this okay this sort of equation I believe you can solve easily from now onwards look at the next question look at this particular question calculate the oxidation potential of this particular half cell your half cell is H2 gas this side you have got HCL so HCL in the container would have dissociated as H POS plus CL Nega so I'm making it in the simpler format like
this imag imagine that in this container you have kept HCL and this HCL would have got dissociated as H POS plus CL NE the concentration of HCL which you have taken that is 0.1 mol so initially this will be 0 this will be 0o now finally this is 0 this will be 0.1 M and this is 0.1 M as well because the stor is 1 is to 1 is to 1 right simple 1 mole gives 1 mole so 0.1 is going to give 0.1 yeah now people try to understand imagine this is a hydrogen rod
this is the H2 Rod here and you already know how this H2 Rod is made with the help of platinum right and the pressure at which H2 was introduced on Platinum that is 10 ATM that is 10 ATM right okay now guys try to understand as for the question what am I supposed to calculate oxidation potential of this hydrogen electrode this is the hydrogen electrode basically I need to calculate its oxidation potential to calculate the oxidation potential I'll make sure the electrode behaves like the anode I'll make sure the electrode behaves like the anode and
at anode oxidation takes place tell me when hydrogen electrode behaves like the anode oxidation takes place what has to be the reaction few minutes back I told you H2 gas it gets converted into what 2 * H positive Aquis and with that you'll be writing two electrons this is the reaction which will happen when the hydrogen electrod behaves like the anode right now my dear students if I ask you how many moles of electrons are exchanged here you'll say two moles of electrons are exchanged okay TR to calculate the QC value QC value has to
be it's concentration of H positive raised to the^ 2ide by what divide by partial pressure of divide by partial pressure of H2 rais power 1 now what is the concentration of H positive in the container the concentration of H positive in the container is nothing that's 0.1 molar 0.1 means 10^ minus 1 so it's 10- 1^ 2 divid by what is the par pressure of H2 that's 10 so this is 10^ minus 2 / 10 the value comes out be 10^ minus 3 so you got the QC value you got the QC value now what
are we supposed to calculate oxidation potential use the earnest equation oxidation potential of the half cell is equal standard oxidation potential sop of hydrogen that's zero sop of hydrogen is 0 - 0.05 91 divide by n value is 2 and it's log of QC QC is just 10us 3 again the matter of calculation only - 0.0591 / 2 log of M power is n log n so minus 3 comes to the front log 10 is 1 just solve it and get the answer in volts this will be the oxidation potential of this particular half cell
when the concentration of HCL in the container is 0.1 mol is this clear is this clear yes I'll teach everything conductance Co Ros everything I'll teach but that will be in the next session in this session I'll be completing complete EMF EMF part will be completed today today in the next session Paradise laws conductance c r law that will be done all the things clear perfect let's have a look on one more question let's have a look on one more question yeah okay look at this particular question similar type of question guys similar type of
question you are given with a half cell right you are supposed to calculate its oxidation potential we supposed to calculate it oxidation potential now oxidation potential you have to calculate in order to calculate the oxidation potential you'll make sure that this electrode behaves like the anode at anode oxidation takes place at anode oxidation takes place oxidation means increase in the oxidation state oxidation state here is zero and the oxidation state is minus1 so what should be the reaction should I write the reaction as cl2 G gives CL or CL NE gives cl2 you tell me
that what should I write should I write cl2 gives CL neg2 or CL Nega gives cl2 quickly what should I exactly write what do you think what do you think since I'm making sure that the electrode behaves like the anode right at anode oxidation takes place oxidation means increase in the oxidation state so through the reaction I have to show that oxidation state is increasing so for that purpose I have to write CL Nega CL Nega it gives what it gives cl2 gas right so now it becomes min-1 to 0- 1 to 0 means- 1
to 0 increase in the oxidation state means oxidation loss of electrons loss of electrons now this is cl2 so this has to be 2 * CL tell me one thing one cl2 when it loses one electron it gets converted into CL now as for the equation this is cl2 this is two so this has to be two as well right so you'll be wrting two electrons with it you'll be wrting two electrons with it I hope this is clear okay so this is the reaction now since n value you got n value is two now
write the QC QC is going to be it's going to be partial pressure of cl2 divide by concentration of Cl negative rais power 2 what is the partial pressure of cl2 10 ATM divide by what is the concentration of Cl Nega in the solution in the solution since in the solution you have taken HCL this HCL would have got dissociated as H POS plus CL stet is 1 is to 1 is to one right so one mole of HCL gives one Mo of cl2 0.1 M HCL will give 0.1 M CL Nega right so you
got the concentration of Cl Nega as well that is simply 10 ra^ minus 1 but this is ra^ 2 so it's going to be 10 / 10^ minus 2 the value will come out to be 10^ 3 so this is QC now what do I need to calculate I need to calculate the oxidation potential of the half cell so I'll write oxidation potential of the half cell is equal standard oxidation potential e xus 0.05 91 divided by this is going to be n and now it's going to be log of what log of QC I
believe all the terminologies you already have now so e Ox is equal e Ox what is e Ox standard oxidation potential look at this -1 to 0 -1 to 0 -1 to 0 increase increase means oxidation so this sop and that sop I need here so it's - 1.36 - 0.05 91 divide by n value is 2 log of 10 ra^ 3 log of 10 ra^ 3 means P log 10 log 10 is one just solve this part get the answer of the question get the oxidation po of the half C which you were supposed
to calculate right electrons will add why electrons will add at anode oxidation takes place oxidation means loss of electrons right that's why electrons are on right side okay and this is C negative CL negative right it's not CL or something I hope it's clear okay guys all right let's try to solve one more question let's try to solve one more question look at this particular equation see what the question says a solution containing 4 into 10^ minus 4 molar cr27 DGA and 2 into 10^ minus 2 m CR Tri positive ions shows a pH of
one calculate the reduction potential of this half cell so as for the question we are given with a half cell and half cell is like this hell is like this Platinum solid cr27 D Nega right CR Tri positive this is your half C correct now my dear students what you need to calculate you need to calculate the reduction potential of this half cell you need to calculate the reduction potential of this half cell in order to calculate its reduction potential I'll make sure this half cell behaves like the cathode I'll make sure this half cell
behaves like the cathode because at cathode only reduction takes place right now now try to understand what is the oxidation St of chromium here it is plus 6 oxidation state of chromium here it's + three so plus 6 to + 3 plus 6 to + 3 plus 6 to + 3 decrease in the oxidation state decrease means reduction and I have to write the reduction reaction only so basically cr207 D Nega Aquis it has to get converted it has to get converted into CR Tri positive Aquis then only I can represent reduction because reduction means
decrease in the oxidation state plus 6 to plus three decrease decrease means gain of electrons now when cr27 DGA gets converted into CR Tri positive how many electrons are gained plus 6 to+ three change is three three is the change for one atom but I have two atoms of chromium so 3 2 are six I'll say six electrons are being gained here right and then it gets converted into CR Tri positive make sure this is 2 CR so it has to be 2 * CR this side as well okay now people if I ask you
whether this particular reaction is balanced or not look at it carefully is it balanced look at it carefully is it balanced the reaction is not balanced because you got oxygen here no oxygen on this side there is one hint that is given as for the question pH of the solution is given to us as one if pH is one that means solution is acidic so you are going to balance this particular reaction acidic Medium you're going to balance this reaction in acidic medium and how do you balance the reaction in acidic medium seven oxygen on
this side no oxygen on right side so right side is oxygen deficient so add seven water molecules on this side now after this balance hydrogen 14 hydrogen on this side so it has to be 14 H positives on this side as well now the reaction is balanced reaction is balanced at the same time if I ask you how many moles of electrons are exchanged six moles of electrons are exchanged in the net cell reaction okay tell me the value of QC start with this particular thing it is concentration of chromium Tri positive rais stric that's
true this is in liquid form it's 's Unity nothing to do with that divided by it's going to be concentration of cr207 DGA raise paret one and this this is going to be concentration of H positive rais stric equ that's 14 this is how you write the Q expression correct now tell me one thing concentration of CR Tri positive is given this is given but concentration of H positive is not given but pH of the solution is given to me as 1 and you already know if pH is equal to 1 what does that mean
that means concentration of H positive is 10^ minus 1 M right if pH is X if pH is X so concentration of H positive becomes 10- X right so put the values here C Tri positive concentration is given to me I believe 2 into 10 -2 so 2 into 10^ minus 2 ra^ 2 divided by concentration of cr27 D negative given to me as for the equation 4 and 10us 4 concentration of H positive 10^ - 1 ra^ 14 right I'll say this particular term this particular term gets canceled out so the final value is
10^ 14 so this is q q c this is QC right what was I supposed to calculate as for the equation I was supposed to calculate the reduction potential now use an Earnest equation E red is equal e not red standard reduction potential this is the standard reduction potential which is given see nothing is mentioned nothing is mentioned whenever whenever nothing is mentioned you'll be assuming it as standard reduction potential and nothing is mentioned here right so assume it as SRP always so SRP is 1.5 - 0.05 91 divide by n values 6 log of
QC QC value is 10^ 14 right so M ra n is n log M so it is going to be 1.5 - 0.0591 / 6 log of m^ n is n log M so 14 log 10 log 10 is 1 so this particular term you have to solve the answer is going to be in volts so this is the reduction potential of the half cell which you were supposed to calculate I hope this is clear to everyone let me know once in the charts quickly is this clear is this clear let me know once in
the charts quickly people everyone everyone perfect okay let's try to solve one more question look at this particular question read the question carefully first I want you guys to read the question and let me know in the chats whether you can solve this question or not read it a zinc Rod is dipped in 0.1 mol solution of its salt at 25° Centigrade assume the electrolyte is 20% dissociated calculate the electrode potential they're asking you to calculate electrode potential they're not mentioning clearly whether to calculate oxidation potential or reduction potential whenever this kind of the scenario
arises you're supposed to calculate electrode potential that means they are asking you to calculate the reduction potential okay they're asking you to calculate the reduction potential so can't you do it what do you think what do you think about this question can it be done quickly in the charts can it be done see guys look at the questions clearly imagine this is the container and in this particular container what do we have we have got salt of a zinc let's assume that we have got zinc sulfate imagine we have got zinc sulfate now tell me
how zinc sulfate under goes dissociation zinc di positive plus s so4 d negative like this okay now you have introduced a zinc Rod here this is the zinc Rod which you have introduced so zinc Rod you have introduced into a solution containing its own ions so you'll be calling this whole setup as this whole setup is your zinc electrode right now as per the question you have to calculate the reduction potential of the zinc electrode when you have to calculate the reduction potential of the zinc electrode when 20% electrolyte is dissociated into its ions when
20% electrolyte is dissociated into its ions at that point of time what will be the reduction potential of this electrode now guys have a look see this is time T is equal 0 at time T is equal 0 what is the concentration of zinc sulfate that's given to me the concentration of zinc sulfate is 0.1 M so C value is 0.1 M right so this will be 0 this will be zero now if you remember we have discussed in equilibrium after some time this will be C minus C Alpha and this is going to be
C Alpha and this is going to be C Alpha right do you remember if this is C minus C this has to be ca this has to be ca toet is 1 is to 1 is to 1 okay now how much percentage of zinc alphate should get dissociated 20% if 20% is getting dissociated that means Alpha value is 0.2 Alpha value is 0.2 correct so tell me what will be the concentration of zinc di positives in the container when 20% electrolyte is dissociated when 20% electrolyte is dissociated what is the concentration of zinc diosi in
the container the concentration of zinc diosi is C Alpha C value is given to us 0.1 Alpha value will be how much 0.2 so 0.1 into 0.2 makes it 0.02 or you can write it as 2 into 10^ -2 mol so this is the concentration of zinc dioso in the container when 20% electrolyte has got dissociated into its I hope this is clear now tell me one thing what are we supposed to calculate we are supposed to calculate the reduction potential for the reduction potential to be calculated we'll make sure that electrode behaves like the
cathode because at cathode only reduction takes place right now what should be the reduction reaction reduction means gain of electrons so I would say the zinc di positives which are there in the container they'll collide with the rod right the zinc die positives which are there in the container they will be colliding with the rod will be taking two electrons from the rod and getting converted into zinc solid so this has to be the reaction now if I ask how many moles of electrons are exchanged how many moles of electrons are exchanged so basically I'm
asking you the N value n value is to right okay what about QC value QC is equal start with the product it's zinc solid it's AC masses Unity so 1 divided by it's going to be concentration of zinc di positive retric so it is 1 divid by what is the concentration of zinc dios what do we need to calculate we need to calculate the reduction potential of the half cell when 20 % electrolyte is dissociated now tell me tell me one thing what is the concentration of zinc D positive when 20% electrolyte is dissociated that
is 0.02 right that is 0.02 this value comes out to be 50 right this value comes out be 50 so QC value I got QC value I got right now if I got the QC value I'll be simply writing the nnest equation Nest equation says that Eed is equal reduction potential of the half cell is equal e not R standard reduction potential check this particular value is this SRP of or sop plus 2 to 0 plus 2 to 0 means decrease in the oxidation state reduction so this is SRP right and I need SRP only
so - 0 so it is- 0.76 so this is e not R minus 0.0591 divide by n value is 2 and it's going to be log of QC QC value is how much QC value is 50 50 can be written as 5 into 10 okay so now it's just a matter of calculation I'm just breaking it a bit then calculation part you'll do on your own this 0.0591 divid by 2 log of M into n log of M into n is log of M plus log of n log of 5 is 0.69 plus log of
10 is 1 right just do this calculation and get the value of reduction potential of this particular half cell in Vols right is this clear is this particular question clear to you will you be able to solve this sort of question as well someone is saying sir what you are doing is not live I know you're looking at the time this time is wrong right I know it is 7:17 at this point of time but over here it's mentioning as 647 then long it's like this perfect okay let's have a look on one more type
of the question see if the question is asked in the need from the electrode potential this can be the toughest question which can be asked in the neat examination right okay this can be your tough question which can be asked in the neat examination because in this question you'll be using the concepts of equilibrium as well and have a look this particular how exactly this particular equation is to be solved understand properly the question is calculate the reduction potential calculate the reduction potential at pH equal to 140 calculate the reduction potential at pH equal to
40 for this half cell for this half cell at 25° C ASP is given e not of something is also given so basically as for the question we are given with the copper electrode we are given with the copper electrode and we need to calculate it's reduction potential now in order to calculate the reduction potential should I make sure this copper electrode behaves like the cathode or anode I'll make sure it behaves like the cathode because at cathode only reduction takes place and I'm supposed to calculate reduction potential only right so let's assume that this
copper electrode is behaving like the cathode when it behaves like the cathode what's going to happen reduction reduction right reduction means gain of electrons so I'll say copper D positives in the solution they'll be gaining two electrons and they will be getting converted into copper solid this is the reaction that's going to happen this is the reaction that's going to happen this is your reduction reaction this will only happen when the copper electrode behaves like the cathode right so I'm making sure that copper electrode behaves like the cathode now tell me one thing what is
the n value n value you already got what about QC value QC value is going to be 1 / concentration of C diosi but my dear students as for the question do you see anywhere the concentration of Cu dios given to us do you see concentration of Cu dios anywhere given no right so we have to calculate concentration of C di positive with the parameters from the parameters which are given to us whatever parameters are given to us just use them and get the concentration of Co positive because we have to write QC okay now
what all things I'll be doing have a look as for the question as for the question pH is given to is 14 right now if pH is 14 you know at 25° C pH plus P this value is 14 at 25° Cen must be knowing this right if pH is 14 from here I'll be calculating p p will be zero right P will be zero if p is zero what will be the O Negative concentration it's going to be 10^ minus 0 that is going to be one O Negative concentration I could calculate from here
but why do I need O Negative concentration have a look my dear students KSP of Co whole twice is given tell me how the C whole twice will undergo dissociation the C hold twice how it will undergo dissociation it will undergo dissociation as CU di positive plus what plus 2 * o Nega this is how it will undergo dissociation right now if I ask you to write the KSP expression can't you write the KSP expression it's going to be concentration of C di positive ra power stric coefficient multiplied by concentration of O Nega rais power
stric coefficient this is how you write KSP expression okay now as for the equation is KSP value given absolutely it's given to me as 10us 19 is equal concentration of Co positive ra^ 1 what about O Negative concentration O Negative concentration is given to us as one m i mean this was not given I calculated it now you would understand why did I calculate right because I was supposed to get the concentration of C posi so finally I got the concentration of Cu POS as 10 -9 M perfect why did we calculate the cuos concentation
to get the QC it's 1 / 10^ - 19 so the value has to be 10^ 19 okay so this is your QC this is your QC perfect this is your QC clear since you calculate QC you have n value what are we supposed to calculate we are supposed to calculate the reduction potential use the nest equation E red is equal to e not red standard reduction potential look at this particular value + 2 to 0 plus 2 to 0 decrease decrease means reduction this SRP only right so 0.34 minus 0.0591 divide by n value
n value is 2 and it's log of QC and QC already you know stand ra^ 19 right yes it's done right so it is 0.34 - 0.0591 / 2 log of M ra power n log of M ra n is n log M right so it's going to be 19 log 10 log 10 is 1 just solve this particular part get the answer in Vols that's the reduction potential of the half cell which I was supposed to calculate is this clear let me know once in the charts quickly if this is super clear to everyone
yeah everyone in the chats people everyone everyone means everyone quickly do you need some lectures of organic chemistry as well do you need some Le because physical is almost done right we almost complete physical chemistry tell me in the chats with yes or no let me know in the charts with yes or no quickly organic or inorganic which one organic or inorganic which one I want everyone of you to say it just write it it's just going to take two seconds to write it just write it what what you want exactly organic or inorganic in
inorganic chemistry we have already complete bonding if you remember right okay so I'll take some of the chapters of inorganic Also let's start then in in organic with the first chapter periodic classification okay all right CH got it let's move on to one more question now let's move on to one more question and then we shall be discussing few more Concepts just a second can you solve this particular question look at it carefully look at this question look at this question right now look at this question look at this question you given with a reaction
in this particular reaction If You observe one elron is gained here gain of electrons is a reduction so this is a reduction right this is SRP that's given right calculate the reduction potential in neutral solution we are supposed to calculate the reduction potential now in this particular question in this particular question you're not supposed to write the reaction the reaction is already given to us right so directly do one thing write QC expression right QC expression start with the product it's partial pressure of NO2 raise partc that's one H2 is a liquid state nothing do
with that divide by concentration of NO3 negative rais stric coefficient and it's concentration of H positive rais it stric coefficient correct now now partial pressure of NO2 we already know how much is that one bar divided by concentration of NO3 Nega that's one mol right what is h positive concentration well H positive concentration is not given in the equation but you are given with the pH if pH is 7 if pH is 7 that simply tells you H positive concentration is 10 ra- 7 m right it is 10 ra- 7 m so this is 1
multiplied by 10^ - 7 ra^ 2 the overall value at the end comes out to be 10^ 14 so this is your QC this is your QC right if you got the QC value how many moles of electrons are exchanged one mole of electron n value is one now write the nnest equation so I'll be writing reduction potential of the half cell is equal e not red standard reduction potential which is 0.78 minus 0.05 91 divide by n value n value is 1 log of QC QC value is what QC value is 10^ 14 perfect
so it is 0.78 - 0 0.0591 right it's going to be log of M forn which is 14 log 10 log 10 is 1 just solve this equation and get the answer exactly in what get the answer exactly in volts I believe that you can solve each and every question in which you are supposed to calculate oxidation potential and reduction potential of what of the half cells right so let me know once in the chats let me know once in the chats quickly let me know once in the charts quickly is everything done so till
now we exactly saw how to write the nest equation for half cells and how do we calculate the oxidation potentials and reduction potentials of the half cells under nonstandard conditions right what are you saying sir rate rate reaction must be negative must be minus change or not chemical I'm not getting what you're trying to say just be clear with the words I'm unable to understand what you asking all right guys so let's move on now let's move on to one more topic that is nerest equation for complete reversible cells nnest equation for complete reversible cells
now guys try to understand so till now I was showing you how to write the nnest equation for half cells right why do we need n equation for half cells to calculate the oxidation or reduction potential of the half cell under non-standard condition condition now you'll be given with a full Cell full cell means complete galonic cell you mean right in which you have got two electrodes and both the electrodes are connected externally as well as internally perfect and imagine imagine like this let's say this is the Sol bridge on the left side of the
solid Bridge there's anode and the anode is like this zinc solid let's say it gives zinc dioso Aquas right on the right side there is cathode so let's say this copper dios Aquis it gives copper solid so first of all this is your Daniel cell right this is your Daniel cell now let me show you how do you this is your Daniel cell it's a Gonic cell basically it's one complete cell in which two electrodes are used it's a it's one complete cell in which two electrodes are used basically now the point is how do
we write the nest equation for complete cells okay my dear students why do we need the nest equation for complete cells to calculate EMF of a complete cell under non-standard condition to calculate EMF of a complete cell under non standard condition what do I need I would need the earnest equation how exactly you're going to write it first thing I'll be writing the reaction at anode what will be the reaction taking place at anode zinc getting converted into zinc di positive 0 to plus2 increase means oxidation you know it already you have discussed this so
reaction at anode is like the zinc solid it gives zinc D positive Aquis and with that you'll be writing two electrons this is the reaction at anode similarly if I ask you what about the reaction at cathode at cathode reduction takes place copper D positive gives copper solid so it is copper D positive Aquis right plus two electrons it gives copper solid so this is the reaction at cathode now in both the reactions electrons are balanced now you can directly add these two reactions you can directly add these two reactions when you add these two
reactions what do you get what do we get so two and two gets cancelled so my net reaction becomes zinc solid plus copper D positive Aquis it gives zinc d I positive Aquis plus what plus copper solid this is the net reaction this is the net reaction if I ask you how many moles of electrons are exchanged you'll directly say two moles of electrons are exchanged in the net cell reaction now after that I would want you guys to write the QC expression QC is going to be concentration of zinc di positive rais stric coefficient
that's one divided by concentration of Cu di positive ra for stric equ that's one right so this is how you get the QC after after you get the QC then calculate the E not cell as well how do you get the E not cell that is your e of cathode minus E of anode so SRP of cathode minus SRP of anode SRP of cathode minus SRP of anode right yes now my dear students after writing all these things after getting all these things here after getting all these things what do I need to write at
the end I'll be writing the nnest equation and nnest equation I'll be writing in this format e cell which is EMF of the cell under non standard condition is equal e not still - 0.0591 / n then it's going to be log of QC see e c you would have calculated right because e not cathode e anode it will be given to us as for the equation right so e cell we can calculate n value we can calculate QC we have calculated so put all the parameters here and get the value of e cell basically
right so this is how you use your nnest equation to calculate EMF of the cell right perfect is this clear happy birthday sir whose birthday is it is it my birthday I have no idea so guys let me know once in the chats all the things are clear to them are all the things clear did you get the idea of how nnest equation is used all right so let's exactly try to see some questions in which through which you'll get the idea how nnest is equation is used for the complete cells for example this is
the equation this is the equation on your screen let's see how this sort of equation is to be solved calculate the EMF of the C you are given with a cell in which you have used two electrodes right which are separated with the help of solid Bridge anode cathode right now you are supposed to calculate the EMF of this particular cell over you're supposed to calculate MF of the cell now my dear students it is one complete cell and you already know how do we calculate EMF of a complete cell we do it with the
help of Nest equation and you exactly know the steps to be followed you exactly know the steps to be followed so first of all I'll be writing the reaction taking place at anode tell me the what will be the reaction at anode tell me what will be the reaction at anode so at anode oxidation will take place pin is getting converted into SN die positive so I would say SN solid will be getting converted into SN D positive Aquis and with that you'll be writing two electrons this is the reaction at anode similarly reaction at
cathode reaction at cathode PB di postive is getting converted into PB solid so the reaction has to be so the reaction has to be PB di positive has has to get converted into PB solid right so the reaction has to be I would say PB di positive Aquis plus two electrons plus two electrons it has to get converted into PB solid so this has to be the reaction at cathode now you got reaction at anode you got reaction at cathode if I ask you what about the net reaction what about the net reaction so you'll
be directly adding them up when you add them up this two and two gets canceled out right so my net reaction has to be SN solid plus PB die positive Aquis it gives SN di positive Aquas plus PB solid so this is your net reaction okay now you are going to write the QC expression so start with the product it's going to be concentration of SN D positive ra stric coefficient this is solid active mass is one divided by it's going to be concentration of PB positive right retric coent now as for the question I
we given with the concentration of 10 di POS and PB pos2 concentration of SN D POS is given to us as one M divided by concentration of PB posi is 10us 3 m so the overall value comes out be 10^ 3 so this is QC so QC you calculate right after calculating QC you'll be calculating e not cell how do we get the E not cell e not cell is going to be e of cathode minus E not of anode right now which one is your cathode and which one is your anode cathode is lead
anode is 10 right so SRP of cathode SRP of lead this is the SRP of lead which is- 0.13 minus SRP of anode SRP of 10 SRP of 10 is- 0.14 the value exactly comes out to be how much 0.01 volt so this is your e not cell as well so you have calculated e not cell as well perfect now my de students all the parameters which are needed in the nnest equation you have already calculated them now you can directly use the nnest equation E cell is equal e not cell 0.01 minus 0.0591 /
by n value that's 2 N value is two because two moles of electrons are exchanged there right so n value is two so it's going to be / n log of QC QC value is already 10^ 3 so finally it has to be 0.01 - 0.05 91 / 2 log of M Power n which is n log M so 3 and log 10 is 1 right so solve this equation get the answer exactly in volts so EMF of the cell is going to be this particular term over here yeah is this clear to everyone quickly
let me know once's in the chat people is it clear perfect let's try to see few more questions so that you can Master it properly for example guys you have this sort of equation look at this particular equation carefully as for the question as for the question this is the solt bridge on the left side of it you have got anode on the right side of it you have got cathode so what are we supposed to calculate we are supposed to calculate EMF of the cell EMF of this complete cell EMF of this particular complete
cell so first of all what I'll be writing I'll be writing reaction at anode at anode at anode oxidation takes takes place your anode is your hydrogen electrode and you know when hydrogen electrode behaves like the anode this has to be the reaction H2 gas gives 2 * H POs to Aquis plus 2 electrons so this has to be the reaction right already discussed so H2 is getting converted into H positives so this has to be the reaction at anode similarly reaction at cathode at cathode you can see Iron Tri positive is getting convert into
iron di positive so I would say iron Tri positive Aquis will be gaining electron and will be getting converted into Fe diosi acis perfect so this is the reaction at cathode we got reaction at anode reaction at cathode now if I ask you whether the electrons are balanced in both the reactions or not the electrons are not balanced yet so this is two this is one so it's time to balance the electrons so multiply this particular reaction by number two so this has to be two times this has to be two times even this has
to be two times now you can add these two reactions directly and when you add these two reactions what do we we get we get something like this H2 gas plus 2 * F Tri positive Aquis what does it give it gives 2 * H positive Aquis and with that you get two * Fe di positive Aquis this is your net cell reaction this is your net cell reaction okay now if I ask you how many moles of electrons are exchanged you directly say 2 moles of electrons are exchanged in the net cell reaction so
write the expression for QC now write the expression for QC start with the product it's going to be concentration of H postive ra power 2 and it's going to be concentration of it is going to be concentration of Fe di positive concentration of Fe positive rais power two divided by partial pressure of H2 gas raise par to isometric one and with it concentration of f Tri positive raise to isometric now concentration of H2 sorry concentration of H positive H positive concentration of is how much 10- 3^ 2 comes out be 10^ - 6 right F
die positive concentration 0.01 which is 10us 2^ 2 comes out be 10us 4 divided by partial pressure of H2 gas 10^ minus 1 concentration of f positive 0.1 that means 10^ minus 1 ra^ 2 that means 10^ minus 2 in the numerator you have got 10^ - 10 in the denominator you have got 10^ minus 3 so the overall value is 10- 7 so this is QC you got the QC as well okay since you got the QC now it is the time to calculate e not cell e not cell has to be equal e
not cathode SRP of cathode SRP of this particular electrode is it given well it is Fe di positive gives F Tri positive so increase in the oxidation state so this is sop but I need SRP SRP will be 0.17 SRP of cathode minus SRP of anode anode is your hydrogen electrode it's Sr RP is z right e not cell is 0.17 volt now all the parameters which are needed to write the nest equation we have wrote that so what I'll be doing now I'll be using the nest equation I'll say e cell is equal e
cell - 0.0591 divid n value is 2 and it's going to be log of QC QC value is how much QC value is 10^ minus 7 right now it's just a matter of calculation 0.17 0.0591 / 2 log of m n is n log M so - 7 comes to the front log 10 is 1 you can solve it and get the answer exactly in volts right I hope this sort of equation you can easily solve again from now onwards is it clear let me know once in the chats quickly okay guys okay look at
this particular question look at this particular question now see guys the EMF of the cell is given EMF of the cell is given 0.81 volt calculate the charge on the metal calculate the charge on the metal okay see what the question says basically see what the question exactly says As for the question is concerned m is getting converted into MN positive right m is getting converted into MN positive so basically as per the question we are supposed to calculate the charge present on the metal ion right we have to get this particular n value the
charge present on the metal ion now again I'll be using the same procedure whatever I have been doing till now same procedure similar procedure I'll be doing and definitely you will be able to solve this question as well and see how exactly first of all what has to be the reaction at anode M gives MN positive so the reaction has to be M solid it gives MN positive Aquis plus how many electrons plus n electrons this is the reaction at anode similarly reaction at cathode H positive gives H2 so the reaction has to be 2
* H positive Aquis since hydrogen electrode is behaving like the cathode and when it behaves like the cathode the reaction has to be like this this your H2 perfect so you got the reaction at anode you got the reaction at cathode as well now my dear students if I ask you whether the electrons are balanced or not the electrons are not balanced in both the reactions here you have got n electrons here you got two electrons so I'll I'll be multiplying this reaction by number two I'll be multiplying this reaction by number n right so
after multiplying what do I get this is two times this is two times and this is 2 N times this will be 2 N times right this will be 2 n * perfect and this is also going to be 2 N electrons and this is going to be n * H2 okay now is the time to add these two reactions when I add these two reactions what do I get I get 2 * Metal Solid plus 2 n * H positive Aquis it gives what it gives 2 * MN positive Aquas 2 * MN positive
Aquas plus plus n * H2 gas so this is your net reaction now if I ask you how many moles of electrons are exchanged in the net cell reaction will you say n or 2 N I'll say 2 N moles of electrons are exchanged I'll say 2 N moles of electrons are exchanged in the net cell reaction now tell me one thing whether you can write the expression for QC or not absolutely it's going to be concentration of MN positive rais power St 2 this is going to be partial pressure of H2 gas rais toet
that's n divided by this is solid leave it this is H positive concentration rais that's 2 so this is how you write QC this is how you write QC now tell me one thing whether the concentration of MN positive is given or not 0.02 mol which means 2 into 10^ minus 2 ra^ 2 means 4 into 10^ minus 4 right partial pressure of H2 that's one concentration of H positive that's also one one rais per anything is one so QC value I directly got okay now as for the question let's try to calculate e cell
e not cell is e not of cathode cathode is your hydrogen electrode it's SRP Z minus SRP of anode SRP of this particular electrode look at this m gives MN positive M gives MN positive so 0 to plus 7 increase in the oxidation state means oxidation so this s o but what do I need I need SRP so this is- 0.76 so this value is equal 0.76 volt so this is your e cell as well right so I believe all the parameters are calculated now it is the time to use the earnest equation what you'll
be writing you'll be writing e cell is equal e cell right a cell is equal e cell minus 0.0591 divided by n but here you are not going to use n here you'll be writing 2 N because 2 N moles of electrons are exchanged then you'll be writing log of QC correct so e cell value as for the equation that's 0.81 minus sorry is equal e not cell is 0.76 so 0.81 minus 0.76 how much that comes out to be that comes out to be 0.05 right 0.05 now multiply that 0.05 with 2 and divide
that with 0.0591 will be equal to minus log of QC right I just simplified things nothing else I just simplified the things so it is going to be 0.1 divided 0.0591 is equal minus log of QC right minus log of QC uh I think one term is missing that's n right that's n QC as for the question is given to us QC as for the question is given to us how much QC I mean we have calculated QC and where the QC is QC is 4 into 10 ra^ how much 4 into 10^ minus 4
right now do one thing this this minus is here right this minus take it on the other side Take It on the other side now do a bit of calculation Now log of m n is log m + log n so log of 4 is 0.6 and this is going to be log of 104 is this is going to be -4 so this overall value comes out to be - 3.4 this overall value comes out to be - 3.4 - 3.4 so minus M gets canceled out so n value will be equal 3.4 * 0
0591 divided by 0.1 and approximately approximately this one is 1 approximate calculation this one is approximately two so approximately n value is coming out be 1.7 but what is n basically n is the charge on the metal ion right so n value is coming out be 1.7 can charge be 1.7 charge cannot be 1.7 right it has to be the integer I mean it has to be the natural number so which one is the closest natural number that's two right so I would say n value is coming out to be two that means the charge
on this metal ion over here the charge on this metal ion over here has to be exactly two so that's something which we were supposed to calculate is this clear now is this clear to everyone people quickly perfect I believe you can solve these sort of questions from now onwards okay I believe you can solve these sort of questions from now onwards let's look at one more topic that's again important what is it nerest equation at equilibrium what is meant by nnest equation at equilibrium what is first of all meant by equilibrium stage let's try
to understand it my dear students if you remember we have discussed Daniel cell long back right we have discussed Daniel cell I'll just use the Daniel cell to make you understand what this equilibrium what equilibrium conditions exactly are try to understand this is the container right this one container and and this is one more container over here which electrolyte you have you have got zinc sulfate in this particular container which electrolyte you have you have got copper sulfate in this particular container you are keeping zinc rod and here you are introducing copper Rod here you
are introducing copper Rod so you have got two electrodes zinc electrode copper electrode now my dear students you're connecting them with the help of what with the help of Emer right and this is your solid Bridge so this cell I'll be calling as the Daniel cell right this cell I'll be calling as Daniel cell and in Daniel cell we have discussed in Daniel cell your zinc electrode behaves like the anode and your copper electrode behaves like the cathode that's something which we know already we have discussed that right we have discussed that your copper electrode
it behaves like the cathode and your zinc electrode it behaves like the anode okay now guys tell me one thing for the the copper electrode to behave like a cathode right for the copper electrode to behave like the cathode can I say SRP of copper should be more than that of SRP of zinc because at cathode reduction takes place at cathode reduction takes place can I say copper electode can only behave like cathode if its SRP would have been more absolutely absolutely imagine that I'm giving you arbitary numbers imagine that SRP of cathode here is
5 volts and SRP of anode here was 1 volt imagine just to make you understand things first imagine the SRP of copper electrode was 5 volts the SRP of zinc electrode was 1 volt now which one has got more s SRP copper the one which has got more SRP will easily undergo reduction and reduction happens at cathode so your copper is behaving like cathode and your your zinc is behaving like anode right now okay now guys tell me one thing tell me one thing well these are the arbitrary SRP values which I gave you just
to make you understand right just to make you understand just to make you understand now tell me one simple when the cell will start working when the cell starts working what happens at anode oxidation takes place oxidation means loss of electron right the zinc Rod is made up of zinc atoms the zinc Rod it is made up of zinc atoms and those zinc atoms will undergo oxidation those zinc atoms will get con converted into zinc die positives and those zinc di positives will enter into this solution and when zinc die positive enter into this solution
at the same time from the salt bridge S4 dtive will come and neutralize that so just tell me one thing when the cell works when the reactions takes place at anode and cathode due to the reaction at anode what is happening to the concentration of zinc sulfate is the concentration of zinc sulfate in the left container coner increasing or decreasing what do you think is the concentration of zinc sulfate in the left container increasing or decreasing since in the left container what's happening oxidation zinc atoms are oxidized getting converted into zinc diosi that zinc dios
enters into this container and when the zinc dios enters into the Container at the same time from the salt bridge s so4 d will come will neutralize it so in the left container I must say in the left container I must say the concentration of zinc sulfate increases with time right the concentration of zinc sulfate increases with time in the left container whereas in the right container in the right container what happens reduction reduction of what since in this particular container you have got two types of ions copper D positive s so4 d negative those
copper di positives will collide with the rod will take two electrons from the rod will get converted into copper solid and that copper solid gets deposited on the rod due to which in this container in this container the concentration of copper d D positives the concentration of copper D positives in the right container is it increasing or decreasing is decreasing So eventually I'll say the concentration of electrolyte in the right contain is decreasing so when the cell starts working the concentration of electrolyte in the left container that is increasing concentration of electrolyte in the right
container that is decreasing correct and a few minutes back I told you few minutes back when I taught you nnest equation I told you one simple thing when the concentration of electrolyte increases reduction potential of electrode increases reduction potential of electrode increases with the increase in the concentration of electrolyte reduction potential of the electrode increases with the decrease in the concentration of electrolyte reduction potential decreases this is something which I discussed with you a few minutes back my de students initially the arbitary value which I gave you what was the reduction potential of zinc that
was one volt what was the reduction of copper that was for example 5 volts now when the cell starts working concentration of electrolyte hair increases if concentration of electrolyte increases in the left container so I would say the reduction potential of zinc will start increase the reduction potential of zinc will start to increase right so so the reduction potential of zinc will start to increase when the cell will start working the uction potential of zinc will start to increase when the cell starts working right so earlier it was 1 volt now it will increase with
time similarly the concentration of copper sulfate in the right container it decreases therefore the reduction potential of copper will decrease with time so reduction potential of zinc increase with time and reduction potential of copper that decrease with time can I say a stage will come at which the reduction potential of both the electrodes will be same a stage will Arise at which the reduction potential of both the electrodes will be the same same right or in general I'll say a stage will come at which the electrode potential of both the electrodes will same will be
the same at that point of time will there be any potential difference between the electrodes there will be no potential difference between the electrodes and that particular stage in the cell where the potential difference becomes zero because of same electrode potential of both electrodes that particular stage is something which you call as equilibrium stage I hope you got to know what I exactly said so what I'm going to conclude I'll say at equilibrium I'll say at equilibrium at equilibrium the reduction potential of both the electrodes have become equal if reduction potential of both electrodes have
become equal that means their oxidation potentials will be also equal so in general I'll say electrode potentials are same at equilibrium electrode potentials of both the electrodes should be same same therefore the potential difference will be equal to Z if potential difference is equal to Z if e cell is equal to0 Delta G for the cell Delta G for the cell is minus n of e cell right if e cell is zero so Delta G for the cell becomes zero and at this particular point of time will there be any current drawn from the cell
I'll say no current is drawn from the cell because potential difference of the cell has become zero right potential difference of the cell has become zero now my dear students if you look at your nnest equation your if you write your nnest equation at equilibrium if you write the nnest equation at equilibrium so first of all nnest equation is nothing that is a cell is equal a not cell- 2.303 P / NF and it used to be log of QC this was the general form of the nnest equation but if I write the nest equation
if I write the nest equation at equilibrium at equilibrium this term becomes zero and at equilibrium your QC is replaced by KC or KQ you can see right at equum QC and K are equal right now if I write the nest equation at equilibrium you'll be writing it like this a cell is equal 2.303 RT / NF and then it's going to be log of K EQ so this is the this is how you exactly write the nest equation Nest equation at equilibrium this is how you write the nest equation at equilibrium perfect so with
the help of this nnest equation at equilibrium we can calculate a not cell with the help of this particular expression already we we already know how to calculate cell that's e not cathode minus E not anode that's one way of calculating e not cell this is one more way of calculating you not cell just make sure the cells cell comes at equilibrium at that point of time you can write this equation through which you can calculate right now my dear students you can use few more things as well see if you use T is equal
25° Cen which is 298 kin right if you take the value of f as 965 kums if r value is taken as 8.314 jewles per Calin per mole at that point of time your nerest equation looks like this at that point of time your nnest equation how it looks like you'll be writing a Cel is equal a cell is equal 0.0591 divide by n 0.0591 divid by n then it's going to be log of KQ right this is your nest equation when the temperature is 25° Centigrade right that means 298 kin so this also this
expression also you can remember now my dear students there is one more thing there is one more thing do you remember I gave you one equation long back that was Delta G is equal Delta G Plus RT lwn of QC this was the equation which I gave you long back now if I write the same equation at equilibrium you know at equilibrium at equilibrium your Delta G is zero and QC is replaced by what QC is replaced by KQ right now if you write this particular equation at equilibrium so Delta G is z and QC
is replaced by KQ so so from this particular equation what do I get I get Delta G is equal minus RT lwn of KQ right this is one more expression to calculate Delta G one of the expression to calculate Delta G was minus NF e cell so this is one more expression of calculating Delta G well you can convert it in terms of what in terms of logarithmic form as well you can say Delta G is equal - 2.303 this is RT and it's log of k e q log of KQ perfect so these are
certain Expressions which are valid when the cell is at equilibrium right perfect guys is every single thing perfect tiller I believe every single thing is perfect till here right okay for example I'm giving you a question let me make a question like this calculate the equilibrium const for the following calculate the equilibrium constant for the following let's say the reaction is like this zinc solid plus cadmium di positive Aquis let's say it gives cadmium solid plus zinc di positive ATS let's say this is the reaction which is given to me and for example for example
e not cell is equal let's say e not cell is equal uh just a second let's say en not cell is equal 0.46 volts okay you have to calculate equilibrium constant you have to calculate equilibrium constant can you give it a try and solve this question see guys first of all you have to calculate equilibrium constant well in order to calculate the equilibrium constant this cell has to be at equilibrium this cell has to be at equilibrium right and when the cell attains equilibrium I'd say a cell is equal to0 if a cell is equal
to0 if a C is equal 0 at that point of time you can use the equation which we got few minutes back only e not cell is equal 0.0591 / n and then it's going to be log of KQ right a not cell is given what is n value n value will be two for this reaction n Valu is two from here you can easily calculate KQ so one equation one unknown nothing to do right these are very simple simple equations similarly this is one more simple question question similarly these are basic basic equations right
right these I believe you can do on your own calculate Delta G for the reaction when concentration of every ion is one M temperature is 298 kin so guys one thing you have to agree here if concentration is 1 M temperature is 298 kin believe the pressure is one bar that means the conditions which are given to us these are standard conditions so we have to calculate Delta g under standard conditions basically we have to calculate Delta g under standard conditions and Delta g under stand conditions is what that's basically Delta G and Delta G
is What minus NF e c minus NF e c so first of all what do I need to get I need to get e not cell basically I need to get e not cell so what is e not cell e not cell is something which is e not of code minus EOD of anode what is EOD of cathode that's SRP of cathode minus SRP of anode now what is cathode what is anode have a look this is SRP of tin this is SRP of copper which one has got more SRP this one has got more
SRP right uh just a second I think there is something wrong in this question the values are not up to the mark the values are not right either the reaction is given in the reverse format or either the reaction is not right or the values are not right just a second see copper is getting converted to Copper diosi so copper has to undergo oxidation and Tin is undergoing reduction if tin has to undergo reduction if this SN di positive has to undergo reduction so its SRP should be more right SRP of SN di positive this
is SRP this is also SRP SRP of SN di positive should be more but here it's mentioned as less so it's a wrong question basically the values are wrong okay the values are wrong perfect did you get what I'm trying to say see in the reaction copper is undergoing oxidation SN die positive is undergoing reduction for SN die posi undergo reduction for SN die positive to undergo reduction the SRP of SN diosi should be more but here SRP of SN dios is less so there are there is some discrepancy in the values right anyways leave
the values aside leave the values aside I'm just showing you the approach of the similar type of equation which can be of this type e cellule calculate as e cathode minus E anode this is SRP of cathode and this SRP of anode so from here you can calculate not cell in not cell f is 965 N means number of moles of electrons exchanged right here you'll get the answer in Jewels you'll be dividing with th then you'll be getting the answer in kilj I believe this is clear I believe this is clear to everyone okay
perfect so guys there is one more important topic which we are going to cover now that is concentration cells okay concentration cells this again one important topic from which question can be asked right so let's get to know what exactly these concentration cells are how many types of concentration cells we have how do we exactly write the nest equation for the concentration cells let's get to know about all these things one by one first of all in case of concentration cells what do we do my dear students in case of concentration cells let me tell
you a few points then I'll make you understand one by one in case of concentration cells let me tell you two same electrodes are used two same electrodes are used till now if you talk about the gvic cell in case of galvanic cell we used to take two different electrodes we used to connect them externally we used to connect them internally then we used to get one complete cell right so in the galvanic cells still now whatever galvanic cell we have studied we used to take two different electrodes and connect them but in case of
concentration cells we are not going to take two different electrodes we are going to take take two same electrodes right Point number one so two same electrodes will be taken in case of concentration cells number one number two number two EMF is EMF is or let me write it like this EMF of the cell EMF of the cell is due to is due to either is due to either concentration is due to concentration difference concentration difference of anode and cathode or the partial pressure difference or the partial pressure difference partial pressure difference between the anode
and Cath what exactly it means you'll get the idea in some time first of all I would want you guys to take a note of these statements in case of concentration cell we'll be using two same electrodes now EMF is generated in case of concentration cells due to due to either the concentration difference in anode and cathode or due to partial pressure difference in anode and cathode what this particular statement means you'll get the complete information of it in some time just wait for it you just tell me one thing my dear students if you
are taking two same electrodes if you are toing taking two same electrodes for example you taking zinc electrode on one side this is one more zinc electrode only so these are two zinc electrodes you connecting them externally you are connecting them internally so you are getting a cell if I ask you if you are using to same electrodes what will be the value of e cell e cell is basically e of cathode minus E of anode so SRP of cathode minus SRP of anode now cathode and anode they are same zinc and zinc so SRP
of cathode SRP of anode since both electrodes are same so their SRP values will be also same so first point which you need to know that e not cell in case of concentration cells do remember this particular value will be equal to zero this particular value will be equal to zero right so in case of concentration cells I'll be using two same electrodes right if you are using two same electrodes you are connecting connecting them externally as well as internally you are getting a complete cell now what about e not cell for this particular complete
cell what about e not cell for this cell over here I would say e cell will be equal e cathode minus E anode now cathode anode both are zinc electrodes their SRP value is same right so SRP of cathode SRP of anode will be the same because both are zinc electrodes so EOD cell will be equal to zero so do remember whenever from now onwards you will see two same electrodes connected externally and internally the first point which should come into your mind that is the E not cell value will be equal to zero okay
now guys let's try to analyze this a bit more in detail but before that before that let me tell you one more point before that let me tell you one more Point try to understand what exactly I'm going to say try to understand what exactly I'm going to say okay have a look people have a look and understand in order to understand these concentration cells in detail let me first of all tell you we divide these concentration cells into two categories one is going to be one is going to be electrolytic concentration cell which we
have to discuss in detail the second one is going to be electrode concentration cell electrode con concentration cell both these cells are to be discussed in detail but right now I'm just giving you the idea of few things which you are going to remember in case of electrolytic concentration cells what do we do the concentration of anode and cathode the concentration of anode and cathode are kept different the concentration of anode and cathode are kept different but in case of electrolytic concentration cells sorry in case of electrode concentration cells in case of electrode concentration cells
partial pressure partial pressure of what partial pressure of anode and cathode partial pressure of anode and cathode are kept different are kept different right in case of electrolytic concentration cells concentration of anode and Cath Cod will be kept different but do remember partial pressure of anode and cathode will be kept same a partial pressure of anode cathode will be kept different but concentration but concentration of anode cathode will be kept same this is the point which I want you guys to take a note of first of all now it is the time to discuss these
concentration cells in detail my dear students okay so our first topic right here is going to be electrolytic concentration cell so Mark The Heading as Mark The Heading as electrolytic electrolytic concentration cell electrolitic concentration cell this is the first important topic have a look people first of all in case of electrolytic concentration cells what do I exactly do I'll be taking two same electron I'll be taking two same electrodes let me represent it over here imagine that this is one container this is for example one more container in this particular container let's say we have
got MN positive ions in this particular container also we have got MN positive ions let's assume the concentration of MN positives in the left container is T1 mol let's assume the concentration of MN positive in the right container is for example C2 mol okay now my dear students here what exactly I'm going to do I'm keeping a metal rod here also what am I doing I'm keeping a metal rod so if I ask you whether the electrodes which I have taken whether these electrodes are same or different these are same electrodes this is M electrode
this is also M electrode now my dear students I am going to connect these externally with the help of Emer right and here you are introducing a solid Bridge so what did we get we got one complete cell which is made up of two half cells right whenever you see this kind of the cell whenever you see this kind of the complete cell in which two same electrodes are used you'll be first of all calling it as the concentration cell and in case of concentration cells in case of concentration cells your e not cell value
has to be right now zero right now which type of concentration cell this is my dear students no doubt you have taken two you have taken two same electrodes where is the difference the difference lies in the concentration of electrolytes the difference lies in the concentration of electrolytes whenever you see whenever you have such type of the concentration cell in which the concentration of anodic and cathodic container is different you will be calling this particular cell you will be calling this particular concentration cell as the electrolytic concentration cell okay now the first point how do
I represent this cell my dear students this is the solid bridge this is the solt bridge on the left side of the solt bridge I'll be writing the anode at anode your M solid is getting converted into M positive your M solid is getting converted into MN positive and the MN positive concentration in the left container how much is that that is C1 M right similarly on the right side of it I'll be writing the cathode at cathode your MN positives will get converted into M solid so MN positives which were there in the right
container whose concentration was C2 mol right they will be getting converted into M solid perfect so first of all this is how you represent the electrolytic con concentration C in which two same electrodes are used but only difference is their concentration is different right this is your electrolytic concentration cell now my dear students the point is what condition should be followed What condition this particular cell has to follow so that it can behave like a normal galic cell what kind of condition it has to follow let's try to make that condition let's try to make
that condition which electrolytic concentration cell has to follow so that it can behave like the Gonic cell right so that it can work basically so that it can produce current basically okay now try to understand first of all for this particular cell to be working what should happen at anode oxidation should happen right so first of all what has to be the reaction at anode at anode your M should get converted to MN positive so the reaction at anode has to be M solid should get converted MN positive with concentration C1 plus n electrons this
has to be the reaction at anode now similarly what should be the reaction at cathode what do you think what should be the reaction at cathode at cathode reduction should take place MN positive should get converted to M so the reaction has to be MN positive Aquis with concentration C2 M it should gain an electrons and should get converted into M solid right these are the two reactions which should happen first of all reaction at anode reaction at cathode now if I ask you whether electrons are balanced in both the reactions or not just check
it out if I ask you whether electrons are balanced in both the reactions or not absolutely in both the reactions electrons are balanced when you add them up n electrons n electrons will get canceled out this m solid M solid will get canceled out so what will be my net reaction my net reaction here is MN positive with the concentration C2 it should get converted into MN positive with the concentration C so this has to be my net cell reaction okay now my dear students if I ask you what about the QC value QC is
going to be concentration of M and positive that is C1 divid concentration of MN positive that is C2 C1 divide C2 simple if I ask you what about the E not cell value e not cell value will be equal to zero only because we are using two same electrodes right now comes a point now comes a point see guys if I write nerest equation for this electrolytic concentration cell how the nerest equation looks like you will write a cell is equal a cell a cell is zero right minus 0.0591 / n and it's going to
be log of QC and QC is just c1/ C2 right you can do one more thing you can do one more thing you can write the same equation like this as you can write eel is equal mult multiply this minus with this log it becomes 0.0591 / n now it becomes log of C2 / C right now it becomes log of C2 / C1 now my dear students if you remember how told you in the last session if you remember for any complete cell to be working for any complete cell to work like a normal
gmic cell it e cell value should be e cell value should be positive then only Delta G will come out to be Nega right we have discussed that in the last session so for this particular cell to be working I'll write for this particular cell to be working a cell value has to be a cell should be greater than zero right a cell should be greater than zero then only this cell will be working then only with this cell is going to produce current then only this cell is going to behave like a normal galonic
cell in which current will go from cathode to anode right perfect now guys the point is when this eel value will come out to be positive eel value will only be positive if if log of C2 / C1 will be positive now the point is when this log of C2 by C1 will be positive log of x is positive when X is greater than one so log of C2 by C1 will be positive when C2 / C1 is greater than 1 right which is only possible which is only possible if C2 is greater than C1
if C2 is greater than C1 so my dear students if you try to analyze things properly if you try to analyze things properly can I say I got to know something really interesting here I'll write for for an electrolytic concentration cell for an electrolytic concentration cell to be working for an electrolytic concentration cell just a second for an electrolytic concentration cell I'll write involving involving which electrodes metal metal metal metal involving metal involving metal metal and electrod for the electrolytic concentration cell involving metal Metaline electrodes to be working to be working like a normal galmic
cell like a normal galmic cell in which current is from cathode to anode in which current is from cathode dang to be working like a normal gmic cell to be working like a normal GIC cell C2 value should be greater than C1 what is C2 C2 was the concentration of electrolyte in cathode C1 is the concentration of electrolyte in anode so I'll say concentration of electrolyte concentration of electrolyte concentration of electrolyte in cathode should be greater than that of concentration of electrolyte in anode this is the condition which the cell has to follow which this
electrolytic concentration cell has to follow then only I can say then only I can say its e cell value is positive therefore the cell will be behaving like a normal gmic cell in which electrons will move from anode to cathode and current will shift from cathode to anode is this point clear to everyone let me know once in the chats is this clear to everyone so this is the first question which is asked what is the question the question is for the electrolytic concentration cell involving metal metal ion electrodes to be be working like a
normal galmic cell the concentration of electrolyte in cathode should be greater than that of concentration of electrolyte in anode which is obvious also if the concentration of electrolyte and cathode is more since you are using two different two same electrodes sorry right if you are using two same electrodes how come potential difference will generate here potential difference will only generate if concentration of electrolyte in cathode is more right if the concentration of electrolyte in this container is more why is that because more the concentration more the reduction potential so this concentration should be more this
concentration should be less if this concentration is more if this concentration is more that means reduction potential of this electrode is more and reduction potential of this electrode is less if reduction potential of this electrode is more this electrode is less that means potential difference again got created potential difference got created and whenever there is potential difference current will be there right then only I'll say the current will shift current will show its direction from cathode to anode right it's obvious in case of normal galonic cell the direction of current is from cathode to anode
now how come this particular cell will behave like a galvanic cell if the direction of current will be from cathode to anode when can be the direction of current from cathode to anode if this particular electrode behaves like the cathode when this electrode behaves like the cathode if it reduction potential is more when its reduction potential will be more when the concentration of electrolyte in this container is kept more as simple as that okay so this was our first point now comes the second point in electrolytic concentration cell there is something called as electrolytic concentration
cell involving involving hydrogen electrodes there is one special type of cell which is what you call as electroly IC concentration cell involving what involving hydrogen electrodes now what this cell exactly looks like let's have a look see guys imagine that you have taken two hydrogen electrodes you have taken two hydrogen electrodes okay you have taken two hydrogen electrodes it is just the difference in the electrolytic concentration is different but the partial pressure of hydrogen gas is same so let me quickly represent this particular cell directly so this is salt bridge on the left side it's
going to be anode so H2 gas H2 gas at the pressure of for example P bar gets converted into H POS gets converted into H postive Aquis right whose concentration for example is C mol and this is platinum solid here so this is your anode similarly your cathode in your cathode H positives H positive Aquis whose concentration is C2 mol it will be getting converted into H2 gas whose pressure is p bar bar again and here also I've used Platinum solid right so let me quickly show it to you also this is for example your
one hydrogen electrode this is one more hydrogen electrode here you have got H positives with concentration C1 here also you have got H positives with concentration C2 M right here you have kept hydrogen Rod H2 rod and you already know how this H2 Rod is made a is H2 Rod the partial pressure the pressure at which H2 was introduced the pressure at which H2 was introduced here that was P bar and here also the pressure is p bar only perfect now you are connecting them externally with the help of Amer and internally with the help
of Sol Bridge so I got a complete S I got a complete C in this particular complete cell tell me one thing whether the electrodes are same or different electrodes are same so it's a concentration cell right now are the concentration of electrolytes same or different concentration of electrolytes is different so it is electrolytic concentration cell involving which electrodes involving hydrogen electrodes right that's the topic which you wrote electrolytic concentration cell involving hydrogen now let's get to know when this particular cell will be working let's make a condition let's make a condition which this particular
cell has to follow such that this particular cell works like a normal galic so let's try to frame that condition first of all for this particular cell to be working like a normal galmic cell at anode what should happen ox ation so first of all the reaction at anode should be H2 gas whose pressure is p bar whose pressure is p bar right H2 gas it will get converted into 2 * H positive whose concentration is C1 plus 2 electrons this has to be the reaction at anode this has to be the reaction at anode
similarly what should be the reaction at cathode at cathode reduction should take place H positive should get converted into H2 so the reaction has to be 2 * H positive whose concentration is C2 M right it should gain two electrons and it should get converted into H2 gas whose pressure is how much whose pressure is again P bar only so this has to be the reaction at cathode this has to be the reaction at anode this has to be the reaction at cathode what should be the net reaction what should be the net reaction just
add these two reactions I students when you add these two reactions which all things will get canceled out two electrons two electrons canceled right okay H2 H2 canceled because their pressure is same so my net reaction has to be 2 * H positive with the concentration C2 mol it gives 2 * H positive with the concentration C1 mol this is my net reaction in this particular net reaction if I ask you how many moles of electrons are exchanged in the net cell reaction you'll say n value is equal 2 if I ask you what about
the QC value QC has to be concentration of H positive that is C1 ra^ 2 / C2 ra power 2 so it is C1 it is C1 / C2 whole rais bar two this is your QC since two same electrodes are used so it's a concentration cell in case of concentration cells e not cell value is taken to be zero perfect now my dear students after this what exactly am I going to do what exactly am I going to do I'll be using now the nnest equation and in case of nnest equation in case of
nnest equation I'll say e cell is equal to e cell e not cell is zero already - 0.059 91 divid n what is n value n value is 2 and here it is log of QC and QC is nothing that is c1/ by C2 raised to the^ 2 correct I believe everything is clear till here now dear students try to understand few more things try to understand few more things I'll write something like this esel is equal Asel is equal minus 0.0591 / 2 log of m ra^ n which is n logm so 2 comes
to the front two comes to the front it is log of C1 / C2 so 2 and two gets canceled out so you'll be writing Asel is equal minus 0.0591 it is log of M byn now which is log of M minus log of n right log of M minus log of n okay now if I ask you what is the C1 and what is the C2 see guys it is - 0.0591 log of C1 what is C1 exactly C1 is the concentration of H positive in anode right minus log of C2 C2 is the
concentration of H positive and cathode right C2 is the concentration of H positive in cathode now if I further solve it I'll be getting something like this e cell is equal see take this minus inside it's going to be 0.0591 right now it is minus log H positive minus log H positive means pH but this is pH of I similarly this is minus in the middle minus in the middle take this minus inside so minus log of positive which going to be pH so it's pH of what it is going to be pH of cathode
so this is the general expression my dear students which you are going to remember from now on this is something which you are going to remember from now on and this particular equation to calculate e cell it is valid for all the electrolytic concentration cells which involves the hydrogen electrodes okay now what exactly were we supposed to do we were supposed to form a condition which this cell should follow such that it behaves like a normal galmic cell and in case of normal GIC cell the current is from cathode to anode right so for this
particular cell to be working for this particular cell to be working like a normal galonic cell right in which the direction of current is from cathode to anode for this particular cell to be working I'll say e cell value should be greater than zero e cell value will be only greater than Z if if pH of anode will be greater than that of pH of cathode right so I will say for for the cell to be working for the cell to be working like what like a normal galonic cell like a normal galonic cell or
a cell to be working like a normal Gonic cell I would say pH of anode should be greater than that of what it should be greater than that of pH of cathode then only the difference between the two will come out to be positive then only e cell will be positive once e cell is positive I'll say Delta G is negative if Delta G is negative cell reactions will be spontaneous if Cell reactions are spontaneous that means that anode oxidation will happen at cathode reduction will happen electrons will move from anode to cathode and current
will show its direction from cathode to anode right so for this particular cell to be working do remember pH of anode should be greater than that of pH of cath is this clear to everyone let me know once in the chats quickly let me know once in the chats quickly tell me the answer of this particular question tell me the answer of this particular question what do you think what do you think about this one look at this particular cell on both the sides you have used zinc electrode right so electrodes are same so it's
a concentration cell number one now which type concentration of zinc diosi is C2 and the concentration of zinc diosi is C1 so concentration is kept different so it is electrolytic concentration cell it is electrolytic concentration cell involving which type of electrodes involving metal metal electrodes involving metal Metaline electrodes now this particular cell few minutes back only we have discussed this in detail now look at the options look at the options carefully choose the correct option the cell reaction will be spontaneous if C1 is greater than C2 now few minutes back only I told you for
the electrolytic concentration cell involving metal Metaline electrodes to be working like a normal galmic cell the concentration of electrolyte in cathode should be greater than that of anode what is the concentration of electrolyte in cathode it C1 right so C1 value should be greater than C2 C1 value should be greater than C2 absolutely C1 value should be greater than C2 then only this particular cell will behave like a normal galonic cell in which the direction of current will be from cathode to anode right so this particular statement is correct cell reactions will be spontaneous right
see it's simple it's evident if C1 is greater than C2 C1 here is concentration of electroly in cathode C2 is concentration of electrolyte in anode if concentration of electrolyte in cathode is greater than that of concentration of electrolyte in anode then only e cell will come out to be positive and once e cell is positive Delta G is negative if Delta G is negative cell reactions are spontaneous it will behave like a normal Gonic cell right so first option is correct second if C2 is greater than C1 if C2 is greater than C1 current will
flow from cathode to anode this not correct this not correct for the current to flow from cathode to anode that means for the cell to be working like a normal Gonic cell concentration of electrolyte in cathode should be greater than that of concentration of electrolyte in anode so this cannot be the answer look at the next one if C1 is greater than C2 if C1 is greater than C2 cathode concentration greater than anode concentration current will flow from cathode absolutely it will work like a normal Gonic cell correct if C1 is equal to C2 if
C1 is equal to C2 my dear students if C1 is equal to C2 that means C2 by C1 value will be equal to 1 if C2 by C1 is 1 so log of one is zero if log of 1 is zero that means e cell is zero if e cell is zero there'll be no current right the cell will not work the cell will not work this also correct done is this clear is this particular concept clear to you let me know quickly in the chats is this particular concept clear see few minutes back only
I told you I gave you this particular result right this was your electrolytic concentration cell involving metal metal and electrodes if C1 and C2 are equal if C1 C2 are equal so log of 1 is equal to zero therefore e cell is zero once a cell is zero that means the Delta G is zero so leave that part aside okay I believe every single thing is clear till here okay just a second look at this particular equation look at this particular equation what do you think about this one first of all try to identify which
cell this is are you using two same electrodes absolutely two same electrodes so concentration cell now which type electrolyte concentration electrolyte concentration different electrolyte concentration different so believe it it is an electrolytic concentration cell partial pressure is same partial pressure is same so it is absolutely electrolytic concentration cell involving which electrodes involving hydrogen electrodes do you remember few minutes back only we have discussed this electrolytic concentration cell involving hydrogen electrodes its e cell is equal 0.0591 multiplied by pH of anode minus pH of cathode few minutes back only we have discussed it right what we
have to calculate we have to calculate X so pH of anodic container we have to calculate okay what is eel value it is 02364 is equal 0.0591 pH of anode is X needs to be calculated pH of cathode in the cathode what is the H postive concentration that's one m so I'll say pH of cathode is equal P stands for minus log so minus log of H positive concentration of cathode so which is minus log of H positive concentration of cathode is one m so the value is zero so this particular value is zero so
x value from here will be equal to 02364 / by 0.0591 the value will be approximately equal to Cod right so what was this x x was the pH of the anodic container which we were supposed to calculate is this clear to everyone is this particular Point Clear to everyone let me know once in the chats quickly I'll be giving you one question here as the homework right I'll be giving you one question as the homework look at the question carefully but be careful with this one in this one concentrations are different and partial pressures
are same concentration different partial pressure same or let me solve this question wait let me solve this question have a look see what the question exactly is the question is as far as this particular question is concerned this is your solid bridge on the left side of it you are using anode on the right side you are using cathode right you using anode and cathod perfect now try to understand and analyze things we have to calculate eel and we have to comment on spontaneity how can we comment on spontaneity spontaneity we can comment only on
only when we have the eel value right so first of all it's time to get the eel and if eel is positive that means Delta G is negative cell reactions are spontaneous this cell will be working like a normal GIC cell if e cell is negative Delta G is positive cell reactions are nonspontaneous okay so first of all you have used same electrodes and both are hydrogen electrodes in which concentration are different partial pressure is same so this is again electrolytic concentration cell involving hydrogen electrodes right how do we calculate it Asel its as Cel
is equal 0.0591 multiplied by pH of anode minus pH of cathode right so first of all let's calculate pH of anode pH of anode is going to be minus log of H POS ion concentration in anode now what is h positive ion concentration in anode in anode you have used h204 in anode you have used h2so4 which would have got dissociated as 2 * h postive plus s so4 d Nega the concentration of h24 you have kept as 0.05 M 0 0.05 m initially so this would have been zero this would have been zero now
this h204 would have been completely dissociated so this is zero now one mole gives two moles 0.05 will give 0.1 M this is going to be 0.05 M so I would say the concentration of H positive in the anodic container that is 0.1 M so calculate the pH of the anodic container pH of anode is minus log of H positive of anode H positive concentration of anode is 10 minus 1 M value is one pH of anode is 1 if pH of anode is 1 now calculate the pH of cathode pH of cathode is going
to be minus log of H POS concentration of cathode which is minus log what is h positive of cathode it is HCL in cathode you have used HCL so HCL would have dissociated as H POS plus CL Nega 1 is to 1 is 1 to isometry so I can say 1 mole of HCL gives 1 Mo of H POS 10us 3 will give 10us 3 only so concentration of H posit in Cod is 10us 3 m only so the value comes out be three so you calculated the pH of anode you calculate the pH of
cathode right you calculated the pH of anode you calculated the pH of cathode so this value is one this value is three so 1 - 3 the value will be overall what the value will be overall negative so first of all you calculated eel after calculating eel then you got the sign of eel as well which is negative and you know when e Cel is negative you know when e Cel is is negative at that time Delta G will be positive if Delta G is positive I'll say cell reactions will they be spontaneous or non-spontaneous
I'll say cell reactions will be non spontaneous will this particular cell be working like a normal galmic cell this particular cell won't be working like a normal gmic cell right is this clear is this clear people quickly in the chats quickly let me know in the chats everyone everyone in the chats quickly all right now we have got one more concentration cell what is that that is electrode concentration cell do you know you know what is electrode concentration cell now in case of electrode concentration cell we'll be keeping the concentration same in anode and cathode
but we keep the partial pressures different for example for example I'm making the cell here I'm writing first of all Platinum solid right and here I'm writing let's say H2 gas whose partial pressure is P1 it gets converted into H positive whose concentration is for example c m this is the solid bridge on the right side I'll be writing cathode let's say in the cathod container there are HED positives whose concentration is C right that is getting converted into H2 gas and the partial pressure of H2 gas here I'm keeping as P2 bar and here
I'm using again the Platinum solid look at this particular cell carefully in this particular cell again you are using two same electrodes again you are using two same electrodes so it is the concentration cell it e cell has to be zero now is the concentration same or different concentration is same but partial pressure is different whenever you see this kind of the in which you will be using two same electrodes concentration of anode and cathode will be same but partial pressure will be different my dear students you will be calling this particular cell as electrode
concentration now what condition this cell has to follow so that it can behave like a normal Gonic cell so that the dire of current will be from cathode to anode so that the cell reactions will be spontaneous let's try to make those conditions as well so for this particular cell to be working like a normal Gonic cell for this particular cell to be working like a normal Gonic cell what should be the reaction at anode at anode oxidation happens H2 is getting converted into H positive so I'll say H2 whose pressure is P1 it has
to get converted into 2 * H postive whose concentration is C plus 2 electron this has to be the reaction at an has to be the reaction at anode right similar reaction at cathode H positive has to get converted into H2 so at cathode reduction has to happen so 2 * H positive with the concentration c m it has to gain two electrons then only it will be getting converted into H2 gas whose pressure there is P2 bar right so you wrote the reaction which has to happen at anode and cathode now are the electrons
balanced electrons are balanced in both bya students if electrons are balanced you can directly add these two reactions and when you add these two reactions what is the net reaction which all terms will be canceled out 2 H POS 2 H POS cancel because their concentration is same right so my net reaction is going to be H2 with partial pressure P1 it gives H2 with partial pressure P2 now tell me how many moles of electrons are exchanged I'll say two moles of electrons are exchanged in the net cell reaction okay n value is 2 now
if I ask you what about QC QC value is going to be start with the product it is partial pressure PR of H2 hair which is P2 divide by partial pressure of H2 hair that is P1 so QC we got since it's a concentration cell so it's e not cell value has to be zero it's a cell has to be zero now my dear students one more thing e not cell has to be zero we got to know that now write the earnest equation a cell is equal a not cell that's 0us 0.05 91 divide
by n value n value is 2 log of QC QC is nothing that is p2/ P1 p2/ P1 I believe it's clear till here I believe it is clear till here now tell me one simple thing me one simple thing if I multiply this minus with this log what I'll be getting exactly I'll be getting Asel is equal 0.0591 / 2 it's going to be log of P1 by P2 it's going to be log of P1 / P2 right so this is the result which I got till now this is the result which I got
for as now for this electrode concentration cell to be working like a normal galvanic cell its e cell should be also greater than zero right so e cell value should be greater than zero then only this cell will be working like a normal galonic cell then only the cell reaction will be spontaneous then only electrons will move from anode to cathode then only current will go from cathode to anode right so e cell has to be greater than zero because if e cell is zero then only Delta G for the reactions will be negative then
only I'll say the cell reactions are spontaneous then only I'll say the cell reactions are spontaneous now tell me one thing when this particular e Cel will be greater than Z I'll say e Cel will be only greater than Z if this particular term is greater than Z when will be log of P1 by P2 greater than 0 when will be log of P1 by P2 greater than 0 when P1 / P2 value will be greater than one that means when P1 will be greater than what then P1 will be greater than P2 what was
P1 and what is P2 look at it carefully P1 was the partial pressure of anode and P2 is the partial pressure of cathode so can I say something like this for the ab cell for the ab cell which is basically gas gas iron cell it is basically the cell name is for the above cell involving let me write it like this for the abos cell involving which electrodes involving gas gas ion electrodes right for the ab cell involving gas gas and electrodes to be working to be working like a normal galmic C to be working
like a normal gmic cell that means for its cell reactions to be spontaneous P1 should be greater than P2 P1 is the partial pressure of anode directly say partial pressure of anode has to be greater than that of partial pressure of cathode partial pressure of anode has to be greater than that of what it has to be greater than that of partial pressure of cathode right this is the condition which the cell has to follow then only the cell reactions involved in the cell will be spontan is this clear is this clear to everyone let
me know once in the chats quickly let me know once in the chats quickly quickly quickly quickly guys quickly look at this question will you be able to solve this question only thing is only thing here is see pressure is 2 ATM here it's 1 atm so pressure is different concentrations are also different so you won't be using any formula directly you'll be using the general procedure you'll write the reaction at anode reaction at cathode net reaction then you'll get the actual QC right actual QC not cell is zero then you'll be using the nnest
equation you'll be at the end getting e cell and if e cell comes out be positive cell reaction will be spontaneous if they come out to be negative they'll be non-spa is this clear is is this clear to you will you be able to solve this question or you want me to solve this question do you want me to solve this question or you'll be doing it on your own because it's a simple question nothing different or difficult to do here let me know quickly will you be solving this question I say yes or no
in the chats quickly yeah so I would want its answer in the comment section then if you guys are saying that you will solve it then right after the session I would want its answer in the comment section and I'll exactly see who all are the ones who will at least try this question right then answer can be right or wrong leave that part aside at least trying part I want you guys to do okay danir every book is best when it comes to pyqs because pyqs are same everywhere whatever book you want to consult
consult for pyqs right on every book pqs will be same only what kind of question this is which book is best for pyqs every book is best for pyqs right because pyqs they'll never change whatever publication you have got it I think you asked the same question yesterday in the J channel as well yeah guys one last Topic in the EMF part one last Topic in the AMF part what is that one last Topic in the AMF part that is thermodynamics of the cell that is thermodynamics of the C they can ask a question related
to thermodynamics of the cell as well what does that mean they can ask you they can ask you calculate Delta G for the cell which already you know right how to do they can ask you calculate entropy change calculate entropy change they can ask you calculate enthalpy change these are the three things which they can can ask you related to thermodynamics part of the cell they can ask you when the galvanic cell is working what will be the Delta G for the galvanic cell what will be the entropy change right what will be the calculate
the entropy change right calculate the enthalpy change these are the three things which they can ask now how to deal with these three things how do we get them my dear students if you remember in thermodynamics I've given you one equation if you remember the equation was like this DG is equal VDP minus sdt do you remember this equation and I had told you this particular equation is valid only for reversible processes it is valid only for the reversible process when I taught you gives free energy variation with pressure and temperature when I asked you
the Gibs free energy variation I mean when it taught you the Gibs free energy variation with pressure and temperature at that point of time I use this particular equation a is equal to VDP minus sdt my dear students if I write the same equation at constant pressure if I write the same equation at constant pressure at constant pressure can't I say DP value is zero change in pressure is zero if DP is Zer I'll say DG is equal to minus sdt minus SD but my dear students when it comes to reactions when it comes to
reactions in case of reactions in case of reactions we do not Define entropy we Define entropy change for the reaction right in case of reactions we do not Define gives free energy what do we Define we Define gives free energy change for the Rea in case of what in case of reactions in case of reactions you do not Define entropy you define entropy change of the reaction similarly you define what you define gives free energy change of the reaction okay now for a particular reaction instead of this particular equation I can write D of D
of G instead of G I'll be writing Delta G for the reaction is equal it's going to be minus instead of s it's going to be Delta s for the reaction multipli by what multiplied by DT perfect now why am I doing exactly all this why am I doing all this see guys see try to understand what exactly we'll get here can I write something like this uh just a second okay can I write D of D of Delta G for the reaction D of Delta G for the reaction divided by DT it comes out
to be equal minus delt s for the reaction okay now tell me one thing what about Delta G for the reaction that is minus NF cell right so minus comes out out n comes out F comes out it becomes D of e c / DT right and this particular term is valid at constant pressure you kept pressure constant in the beginning right and this particular term is equal to minus * delt s for the reaction so minus Min gets cancelled out so the first result which we got which you are going to remember from now
onward whenever they ask you calculate the entropy change in the Gonic cell right you'll be writing this expression n if be upon DT at constant my dear students this particular result you will be using to calculate what to calculate entropy change in case of the galonic cell n number of moles of electrons exchanged f farad is constant 965 this D by DT this is something which you call as temperature coefficient of the cell temperature coefficient of the cell which can be positive as well as negative which can be positive as well as negative if you
ask me the units of temperature coefficient d by DT this is change in EMF change in temperature EMF is volt temperature is Calin so its SI unit is going to be volt per Calin this temperature coefficient perfect so this is how you are going to calculate Delta s for the cell reactions similarly if they ask you to calculate Delta G for the reaction Delta G is simple minus NF C right so two parameters are done if they by chance ask you about Delta H for the reaction you know your Delta G as per thermodynamics Delta
G is equal Delta minus t Delta right now when it comes to the cell reactions you'll write RR everywhere perfect Now Delta G value you know Delta s value you know from here you can calculate Delta H let's get to know what will be the value of delta H let's try to understand see guys we know Delta G is equal Delta H minus t Delta s right so from here you can say Delta H will be equal Delta G + T this is Delta s right so I would say Delta H for the reaction is
equal Delta G is minus NF e cell right plus T Delta s is NF D upon DT NF D upon DT at constant pressure so let's make a finalized result so Delta for the reaction is equal if I take NF common if I take NF common it's going to be what it's going to be minus E cell plus it's going to be D upon DT which is what you call as temperature coefficient of the cell at constant pressure multiplied by T so this is the general result which you can remember also to calculate Delta H
for the galmic cell so they'll be asking few questions they'll be asking few questions first they can ask Delta G for the cell for the galvanic cell or they can ask Delta s for the galonic cell or they can ask Delta H for the Galvin C and these are the equations which you can use use to calculate Delta G Delta s or Delta H or sometimes they might ask you about the efficiency of the galonic sometimes they can ask you about the efficiency of the galonic cell as well and how do you calculate the efficiency
of the galonic cell it is simple mod of Delta G of the cell divided by mod of Delta for the Cell multipli 100 this is the general expression which you can use to calculate the efficiency of the to calculate the efficiency of the gvic perfect provided you know Delta G value you know Delta value just divide them multiply with 100 get the efficiency of the galonic cell as well okay perfect perect guys now now here there can be a question asked to I'm writing the topic as variation variation of EMF with temperature how EMF varies
with temperature this can be asked to how EMF varies with temperature right my dear students I'm going to divide this topic into two categories here I'll be dividing this topic into two categories it's simple it's very simple those galvanic cells those galvanic cells whose temperature coefficient is positive this is my first category of Gonic cells and those GIC cells whose temperature coefficient whose temperature coefficient is negative right so I've got two types of galvanic cells first those galvanic cells with temperature coefficient positive and those galvanic cells with temperature coefficient negative let's see how their EMF
varies with temperature does EMF increase on increasing the temperature or does it decrease let's get to know see so first of all I'm talking about that particular galonic cell whose temperature coefficient is positive whose temperature coefficient is positive right now my dear students for example in this particular cell whose temperature coefficient is positive if you increase the temperature if you increase the temperature of the cell upon increasing the temperature what will happen to DT value change in temperature change in temperature is final minus initial final temperature minus initial temperature this term will be positive simple
when you increase the temperature DT will be positive change in temperature is positive now this overall term is positive this overall term is positive and already we know since we increase the temperature upon increasing the temperature DT value is positive so DT value is positive denominator is positive if denominator is positive and the whole term is also positive for the whole term be to be positive I'll say de value should be positive as well I'll say de value it should be positive as well de positive means when de will be positive when change in EMF
will be positive if the final EMF will be greater than initial EMF so what exactly you concluded from here you saw upon increasing the temperature EMF is it increasing or decreasing E2 is greater than E1 E2 is greater than E1 I'll say upon increasing the temperature the EMF it increases right the EMF it increases so directly you can say e cell is directly proportional to temperature for those cells which have got temperature coefficient positive right which have got positive temperature coefficient I hope this is clear similarly those galvanic cells those galvanic cells which have got
temperature coefficient negative how their AMF varies with temperature see for example if you increase the temperature here upon increasing the temperature DT value will be positive absolutely DT is positive so denominator DT it is coming out to be positive denominator is positive but whole term overall term has to be negative since denominator is positive and overall term has to be negative for the overall term to be negative I say this de value has to be negative this de value it has to be negative then only the overall value will be negative now tell me one
thing when this de will be negative D is change in EMF it is only negative if E2 will be less than E1 right change in EMF will be only negative if final EMF will be less than initial EMF so upon increasing the temperature what is happening to EMF EMF is increasing I'll say e cell a will be inversely proportional tempure but it's valid for those cells which have got temperature coefficient as negative is this clear is this clear to everyone let me know quickly in the chats if this is super clear to everyone right quickly
so what you exactly these are the two things which you have to remember at the end nothing else e cell is directly proportional temperature for those cells which have got temperature coent positive e cell is inversely proportional temperature for those cells which have got temperature coefficient negative that's it okay perfect guys this every single thing clear here everything is dealt with detail everything is done with detail okay and here you have to remember how many results delt s result you have to remember Delta G result you have to remember and Delta H result you have
to remember if you do not want to remember the Delta H it's okay also because once you get Delta G value once you get Delta s value you can easily calculate DT from this particular equation one and the same thing okay perfect guys do you want to solve one question on this particular topic do you want to solve one question on this particular topic let me know once quickly do you want to do that okay let's have a look on the question which can be asked from here I'm writing it I'm writing the question okay
just a second one simple question I'm giving you the question is like this the EMF of the cell it is given to us as 1.08 volt EMF of the cell is given to us is 1.08 volt temperature coefficient of the cell D upon DT at constant pressure this temperature coefficient of the cell is given to me as -5 multiplied by 10^ - 5 Vol per kin okay what are we supposed to calculate calculate Delta G Delta H and Delta s and Delta s for the reaction and the reaction is like this cadmium D positive plus
two electrons it gives cadmium solid right and temperature this reaction is carried out at temperature 25° C this is one of the questions which can be asked from this top look at the question carefully the reaction is given so first of all how many moles of electrons exchanged two so n value is two first of all n value is two this is something which you got now Delta G Delta G Delta G is nothing that's minus n of e right so minus n value is two that is constant 965 0um right eel value is also
given 1.08 volts so every value is taken as per sa system that means Delta G value will be also coming as per sa system as per s system the value of delta G is Jews so the final answer will be in Jews number one number two delt s is simply going to be equal NF D upon DT NF D upon DT at constant pressure so what is n value n value is two constant 0um d by DT what is d by DT temperature coefficient that is - 5 into 10 ra^ minus 5 so by solving
this particular equation you'll be getting the value of delt s and Delta units you already know Delta s is basically Q BYT right entropy change Q BYT so it has to be jewles per Kelvin Now Delta H Delta H you have to calculate how you'll be calculating Delta H either you'll be using the direct result which I gave you or you can use this equation directly Delta G is equal Delta H minus t Delta s now my dear students Delta G already you calculated Delta s you calculated temperature is given converted in kelvin so from
here you can calculate Delta nothing else right Delta value will be also in Jews right these sort of questions can come these sort of questions can come and there has been I mean in your neat examination question has never been asked from thermodynamics of the cell so this time they might ask it from here so I believe we have completed the EMF part EMF part of electrochemistry is completed now it is time to enter into one more part of the chapter that is electrolytic part and that electrolytic part we shall be starting on Thursday day
after tomorrow so Thursday Thursday at 600 p.m. we'll have one more session in which we shall be completing the electrolytic part of this particular chapter as well okay perfect whe I think this much is enough for today hey guys good evening and welcome back again to your an academy NE English Channel I hope all of you doing great having a good time so my dear students let me know in the chats if all of you can hear me if I'm perfectly audible visible to every one of you let me know in the chats quickly with
the thumbs UPS good evening people good evening and welcome back good evening and welcome back so can you all hear me all perfect yes all perfect all perfect people all right so my dear students as you all must be knowing it is going to be the last session of the chapter electrochemistry today right and in this particular session what all things we are going to cover let me tell you that so in this particular session we are going to start with electrolysis number one we'll see the products of electrolysis we'll see the Faraday first law
of electrolysis Faraday's Second Law different types of questions which can be asked from these topics and we shall be talking about the conductance part as well right which involves those terminologies I'm sure you would have heard it already resistance resistivity conductance conductivity etc etc right and at the end we shall be discussing about the Koh Ross Law as well okay lecture duration will be 3 hour hours yeah in 3 hours I believe all these topics will be will be like cleared properly with all the problem patterns okay perfect so uh let me know once in
the chats are you guys done with the first three sessions of the chapter are you guys done with the first three sessions of the chapter let me know once in the chats quickly are you guys done with that quickly let me know in the chats are you all done with the first three sessions of the chapter all right perfect so basically in the first three sessions we have completed the AMF part right whatever we we were discussing till the last session that was AMF part of the chapter and I believe I have discussed all the
topics related to EMF all the problem patterns now you can consult any study material any book and I believe all the problem patterns you can easily solve if you have followed the first three session right and by following this particular session after completing this particular session I believe you can solve all the types of the problems which can be asked from the chapter electrochemistry provided you have made the notes and everything properly yeah perfect guys so let's get going then let's get started with something called as electrolysis okay let me first of all write the
definition of electrolysis then exactly I'll make you understand what this electrolysis is all about okay electrolysis electrolysis my dear students it is the process it is basically the process the process of the process of decomposition of an electrolyte the process of decomposition of an electrolyte into its into its elements into its elements by the supply of by the supply of direct current by the supply of direct current what it means you'll get the idea in some time first take a note of the statement number one number two the device the device in which the device
in which electrolysis is carried out is carried out that device is called as is called as electrolytic cell electrolytic cell that device is called as electrolytic cell Point number two take a note of this point number three point number three if you remember in case of galonic cells the cell reactions used to be spontaneous right they used to be spontaneous in case of your danial cell galonic cell Etc right in case of electrolysis I'll say reactions are made reactions are made spontaneous reactions are made spontaneous by the help of by the help of direct current
reactions are made spontaneous by the help of direct current which are otherwise nonspontaneous which are otherwise which are otherwise non-spontaneous okay one more point I'm mentioning here in the electrolytic cell we'll be using two electrodes one is your anode one is your cathode and do you remember in case of electrolytic cell the anode carries the positive charge and the cathode carries the negative charge whereas in case of your galvanic cell if you remember in case of galvanic cell the rod of anode used to carry the negative charge and the rod of cathode used to carry
the positive charge here it's reverse anodic Rod ha carries the positive charge cathodic Rod hair carries the the negative charge right and all the other things are same at anode oxidation takes place which involves loss of electrons and at cathode reduction takes place which involves the gain of electrons now let's try to understand all these things in detail I believe you have taken a note of all these things okay so let's try to understand all these things in detail now my dear students in order to make you understand the electrolysis I'll be taking one example
with the help of which you'll particularly understand what this electrolysis is all about okay what this electrolysis all about so I'm going to do the electrolysis of molten NAC and automatically you'll get the idea what this electroly exactly is okay so first of all I will be taking a container over here so imagine this is the container right my dear students in this particular container let's say I am keeping molten NAC in this particular container I'm keeping molten NaCl now first of all you must be thinking what this molten NAC is all about imagine that
you have taken NaCl solid in the container NAC solid in the container and you have heated up you have heated it up and you have heated NAC solid till the extent it gets converted into liquid right so imagine you have taken solid na in the container and you have started heating it up there will be a time when this NaCl solid will get converted into NAC liquid and that na liquid I'm something is something which I call as molten NAC so I have got molten NAC in this particular container okay now my dear students one
thing if there is molten na in the container can you let me know how many types of ions will be there in the container there'll be two types of ions one is going to be your na positive and one is going to be your CL negative correct now dear students what exactly we shall be doing I'll be inserting two rods here in the solution these are the two rods which I inserted in this particular solution and these two rods I'm connecting with the battery these two rods I'm connecting with the battery over here okay that
Rod which is connected with the negative termal of the battery will carry the negative charge and that Rod which is connected with the positive ter of the battery will carry the positive charge and as I told you the negatively charged Rod here is what I call as cathode and the positively charged Rod here I'll be calling as I'll be calling as anode right and you know at cathode what happens at cathode reduction takes place and reduction is nothing it is just the gain of electrons and similarly at anode I must say oxidation takes place right
and O oxidation involves loss of electrons okay now my dear students let's exactly get to know what when you connect these rods with the external battery let's get to know what happens at anode and cathode let's get to know what happens at anode and cathode so first of all let's try to analyze let's try to analyze the things which are going to happen at cathode cathode is your negatively charged Rod can I say this negatively charged Rod is going to attract the positively charged ions from the solution absolutely this negatively charged Rod is going to
attract the positively charged ions from the solution towards itself so This na positive is attracted by this particular rod and similarly CL Nega is attracted by this particular positively charged Rod perfect so can I say this na positive is going towards cathode This na positive is going towards cathode and at cathode what happens reduction happens so the na positive which is going towards cathode I must say This na positive at cathode will undergo reduction reduction involves gain of electrons so I must say at cathode your na positive at cathode your na positive it will undergo
reduction it will gain one electron and will get converted into na solid right and this particular na solid it will start getting deposited on this particular Rod it will start depositing on this particular Rod the first question which is asked the first question which is asked the products of electrolysis what is happening at cathode when you do the electrolysis of molten in AC you'll directly say at cathode it is the sodium metal which is being deposited it is the sodium metal which is being deposited in the similar way if I ask you what is happening
at anode at anode oxidation takes place right at anode CL negative will go and will undergo oxidation right CL negative at anode will undergo oxidation oxidation involves loss of electrons so the reaction has to be 2 * CL right it will get converted into cl2 gas and with this you'll be getting two electrons as well perfect so my dear students the question that asked here the question that's asked here what is happening at cathode what are the products of electrolysis you'll directly remember at cathode it is basically the sodium metal which is being deposited sodium
metal is being deposited and if I ask you what is happening at anode at anode it is the gas which is getting liberated so at anode cl2 gas is getting liberated right at anode it is the cl2 gas which is being which is being liberated okay this is the first point which you need to remember in case of this particular electrolysis right I hope you got the idea of what how this electrolysis exactly happens now my dear students tell me one thing the device the device in which electrolysis takes place the device in which electrolysis
takes place what do you call this device as you call this as electrolytic cell right you call this is electrolytic cell my dear students can I say can I say in this electrolytic cell an external voltage an external Supply is used an external source is used which is sending the current in the solution then only these reactions are happening absolutely right you have used an external battery here you have used the battery here right so what this battery exactly does it is sending the current in the solution due to which these reactions are happening over
here right so with the help of with the help of current the reactions are happening so the reactions which were supposed to be nonspontaneous we are making them spontaneous with the help of this external current with the help of this particular current okay so can I say the nonspontaneous reactions are are made spontaneous right we are making sure the reactions which were otherwise being non-spontaneous which were otherwi being nonspontaneous we are making those non-spontaneous reactions to happen with the help of with the help of the current okay with the help of the current perfect this
particular cell in which in which the cell reactions are happening with the help of the with the help of the current with the help of the Curr which goes into the solution this type of this overall process is something which you call as electrolysis I believe it's clear to everyone yeah is it clear to everyone let me know once in the chats quickly let me make you understand one more let me make you understand one more look at this particular scenario let's do the electrolysis of Molton pbbr2 let's do the electrolysis of Molton pbbr2 you'll
understand it in a better way see first of all this is the container which I have and in this particular container what have I kept I have kept pbbr2 pbbr2 which is molon molten pbbr so basically in the container what kind of ions do we get we'll be having PB di positive ions and B are negative ions in this particular container right so insert two rods here you have inserted two rods here now connect these rods with the battery right this is negative for example this is positive negative over here is called as cathode this
positive over here is called as anode right now can you let me what is going to happen can you let me know what's going to happen at cathode at cathode reduction takes place reduction means gain of electrons so this particular cathode this negatively charged Rod is going to attract PB die positive towards itself and this positively charged Rod is going to attract BR negative towards itself right so I would say at cathode PB die positive will undergo reduction at cathode reduction takes place and at cathode PB dulu is going so PB dios will go to
cathode and undergo reduction right so what has to be the reaction the reaction has to be PB die positive will undergo reduction will gain two electrons and will get convert into PB solid right similarly what is happening at anode what is happening at anode I'll say at anode BR negative will go and will undergo oxidation oxidation means loss of electrons so the reaction has to be two br ne2 will give br2 and with that you'll be writing two electrons so just tell me in the nutshell what is happening at cathode and anode so you'll directly
say at cathode at cathode lead is being deposited at cathode lead is being deposited and at anode it is br2 which is being liberated right simple simple guys right now I'm doing the electrolysis of molten electrolytes perfect I believe you can do the electroly of anything any molten electrolyte you can do now my dear students there comes a point since I've told you how to do the electro of what how to do the electrolysis of molten electrolytes now you can keep the electrolyte in aquous form as well what about Aquis NAC what about the electroly
of Aqua cuso4 right what about the electroly of Aqua pbbr2 what about them there are few things which you need to know before going into the details of the electrolysis of Aquis electrolytes right okay let's try to understand in simple way in simple way first of all I want you guys to remember two reactions one reaction is something which I'll be calling as oxidation reaction of water another reaction I'll be calling as reduction reaction of water these are two reactions which I would want you guys to remember first of all one is oxidation reaction of
water one is reduction reaction of water okay have a look have a look my dear students have a look understand if I ask you what is the oxidation state of this oxygen here it is min-2 what is the oxidation state of oxygen here it is zero so min-2 to 0 means increase in the oxidation state increase in the oxidation state is something which you call as oxidation oxidation involves loss of electrons so in this particular process when water will be getting converted into O2 what will happen loss of electrons will happen right loss of electrons
will happen and this particular reaction you are going to remember directly from now onward that to in the balanced format this reaction is called as oxidation reaction of water in the similar way in the reduction reaction of water what is going to happen electrons are not going to be lost electrons are going to be gained have a look if you look at this particular hydrogen its oxidation state is plus one this hyd its oxidation state is 0er + 1 to 0 decrease in the oxidation state plus 1 to 0 decrease in the oxidation state decrease
in the oxidation state is something which you call as reduction and reduction means gain of electrons have a look electrons are being gained here electrons are being gained here this particular reaction I'm calling as the reduction reaction of water so first take a note of these two reactions they are important why they are important you'll get the idea in some first one is called as oxidation reaction of water and which lost of electrons is happening second one is called as the reduction reaction of water in which gain of electrons is happening right in which gain
of electrons is happening perfect right guys now there is a term that's called as discharge potential how do you define the discharge potential how do you exactly Define the discharge potential it is defined as the potential the potential at which an ion is discharged at the electrode the potential at which an ion it is discharged it is discharged at the electrode let me make it simple for you let let me make it simple for you right if I make it simple for you just a second just a second let me make it simple for you
let me write it like this the work the work to be done by the electrode the work to be done by the electrode in order to in order to attract the oppositely charged ion in order to attract the oppositely charged ion towards itself towards itself see discharge potential actual definition is this actual definition is this the potential at which the ion is discharged at the electrode right but just to make you understand in simplified manner I mentioned it like this now what it means exactly let's get to know that see guys for example for example
I have to do the electrolysis let's say I have to do the electrolysis of Aqua NaCl imagine that I have to do the electroly of aquous NaCl right my dear students this is the container which we have and in this particular container what I'm keeping I'm keeping Aqua na if I'm keeping Aquas NAC if I'm giving Aquis NAC that means there is NAC as well as water right so how many types of ions will be there there'll be na positive there'll be CL Nega since there is water also in the container so there'll be H
positive ions there'll be O negative ions as simple as that right now my dear students imagine you have introduced two rods here these are the two rods which you have introduced correct now you are connecting them with the external battery connected them with the external battery this Rod gets negative this Rod gets positive negative 1 that's your cathode positive one that's your anode okay that's your anode perfect now guys try to understand this cathode this negatively charged Rod I mean this cathode it carries a negative charge right now this negatively charged Rod it will try
to attract the positively charged ion towards itself so it will try to attract this one as well as this one towards itself perfect similarly the positively charged Rod will try to attract this one as well as this one towards itself right this is clear now tell me one thing since this negatively charged Rod is attracting is attracting na positive as well as H positive towards itself so basically both na positive as well as H positive both na positive as well as H positive they would want to undergo reduction here at cathode they would want to
undergo reduction at cathode similarly CL Nega as well as O Negative both the ANS would want to go to anode and both the anions would want to get oxidized at anode right perfect see we have got two cats here perfect so both these Cates they would want to go to the cathode and they would want to undergo reduction similarly both these anions would want to go to anode and they would want to undergo oxidation but but among these two ions among these two ions which one will actually undergo reduction among these two ions which one
will undergo oxidation that depends on that depends on the discharge potential of these I that depends on the discharge potential of these eyes right let me tell you that get ion among the two that get among the two that get among the two which has got lower discharge potential which has got lower discharge potential will undergo reduction among the two that c that anion that anion among the two which has got lower discharge potential will undergo oxidation at an in simpler words if I if if you guys want to understand this in simpler words I
wrote a statement here the work to be done by the electrode in order to attract the oppositely charged ion towards itself have a look this electrode it is going to attract any positive as well as H positive right so when this electrode will be attracting This na positive towards itself that means the electrode is performing some work here it's attracting the na positive towards itself the electrode is attracting H positive towards itself it is performing some work right so wherever the amount of work to be done by the electrode is less wherever amount of work
to be done by the electrode is less I'll say that particular IE will have the Lesser discharge potential right so so among these two ions again I'm telling you among these two cats among these two Cates that K which has got lower discharge potential that ctin wherein electrode has to perform lesser amount of work to attract it that ctin will undergo reduction okay similarly look at these two these are ANS basically they would want to go to anode they would want to undergo oxidation but I'm telling you that anine whose discharge potential is less will
undergo oxidation will undergo oxidation so I would say among these two anions among these two anions that that anion that anion for which electrode has to do lesser amount of work to attract it towards itself that anion will undergo oxidation I hope this is clear now how come you can tell whether among these two which one has got lesser discharge potential among these two which one has got lesser discharge potential for that purpose I want you guys to remember one series over here this is something which I I want you guys to remember one is
the discharged potential series of cats and one is the discharged potential series of anions imagine imagine in the container imagine in the container there is H positive and there is na positive okay now which one has got lesser discharge potential H positive has got lesser discharge potential than that of any positive so if I ask you which ion among the two will undergo reduction at cathode you'll directly say it h positive which will undergo reduction at cathode right because it discharge poti this right actually how do we define discharge potential the potential at which the
ion is discharged at the electrode that potential at Which ion is discharged at the electrode perfect and this ion gets discharged at lower potential basically that's why it will undergo reduction at cathode similarly if you look at the ANS if you look at the ANS in the case which I gave you few minutes back Aquis na Right There Are CL negative ions and O Negative I there were CL negative ions and O negative ions now among these two ions which one has got lesser discharge potential it's CL Nega right among CL Nega and O Negative
which one has got lower discharge potential it is CL negative so CL negative will undergo oxidation at anode CL negative will undergo oxidation at anode as simple as that correct perfect C it discharge potential is less it gets discharged at lower potential basically okay or I can tell it in simpler ways in order to attract CL Nega towards itself anode has to go anode has to do lesser amount of work right I hope this is clear now try to understand all these things in detail now try to understand all these things in detail now see
guys for example for example the first topic which I'm mentioning here what is that just a second just a second for example I'm writing the heading as electrolysis of try to understand this electrolysis of aquous NAC electroly of Aquis Na and my dear students the electrodes which I'm using they are inert electrodes they are iner electrodes which I'm using okay now tell me one thing now tell me one thing let's say this is your electrolytic cell in this electrolytic cell you have got Aquis na that means there'll be na positive there'll be H positive there'll
be CL negative there'll be o negative in this container so two types of Cates two types of an this is a rod this is one more Rod connected with the external battery right this is negative this is positive negative is your cathode positive is your anode perfect now this negatively charged Rod it will try to attract na positive as well as H positive towards itself so both these ions would undergo I mean both these ions they would want to undergo reduction at cathode but that cat will undergo reduction whose dis charge potential is less now
compare na positive and H positive na positive and H positive which one has got less discharge potential na positive and H positive which one has got lesser discharge potential na positive and H positive which one has got lesser discharge potential it is H positive which has got lesser discharge potential my dear students if H positive has got lower discharge potential can I say it is going to be H positive which will undergo reduction here absolutely right absolutely now among clga and O Nega which one will undergo oxidation at anode CL negative it discharge potential is
less so it will undergo oxidation at anode right so if I ask you what will be the reaction at anode now at anode CL is undergoing oxidation it will lose electrons so the reaction has to be 2 * CL it gives cl2 plus 2 electrons right now if I ask you what will be the reaction at cathode what will be the reaction at cathode at cathode reduction will take place reduction of what reduction of of H positive this H positive has come from where this H positive has come from where H positive has basically come
from water so I would write the reduction reaction of water here I would write the reduction reaction of water here right my dear students and I hope you remember what about the reduction reaction of water f is back only I told you what about the reduction reaction of water look at this particular slide this is the reaction this is the reaction which is what you call as reduction reaction of water and this is going to happen this is going to happen at cathode this is going to happen at cathode right this is going to happen
at cathode let me know once in the chats if it is clear so this is going to happen at the cathode perfect so at cathode it's two * water plus 2 electrons it gives H2 + 4 * o Nega this is the reaction which is taking place at cathode so in short tell me one thing when we do the electrolysis of Aquis NAC when we do the electroly of Aquis NaCl at cathode what is happening at cathode it's H2 gas which is being liberated it is H2 gas which is being liberated and at anode what
is happening at anode something was being deposited I guess or gas was liberated gas was liberated which one at anode cl2 gas was liberated so at anode you got to know cl2 gas is liberated at cathode you got to know H2 gas is liberated right I hope this is clear to everyone clear guys let me know once in the chats let me know once in the chats quickly right so the reaction has to be 2 * water plus 2 electrons 2 * water plus 2 electrons gives H2 plus 2 * o this is the reduction
reaction which is happening at the cathode and it's your H2 gas which is being liberated at the cathode so these are the two things which I want you guys to remember for example for example for example you are doing the electroly of Aqua cuso4 let's imagine that you are doing the electrolysis of Aqua CS4 imagine this Rod carries the negative charge imagine this Rod carries the positive charge right and in the solution how many types of ions will be there in the solution how many types of ions will be there tell me that in the
solution how many types of ions are there you are doing the electroly of Aqua coso4 so there there are cu positive ions there are s so4 d negative ions there are H positive ions and there are O negative ions correct so this particular cathode this particular cathode and this is your anode right at cathode reduction will happen now this negatively charged Rod will attract CU positive as well as H positive towards itself but among these two among these two which one will undergo reduction here the one which has got lesser discharge potential the one which
has got lesser discharge potential now now tell me among CU positive and H positive which one has got lesser discharge potential it is cu positive so D CU positive is undergoing reduction at cathode so the reaction has to be CU positive plus two electrons gives what it gives copper solid so basically here it is copper solid which is being deposited it is copper solid which is being deposited similarly you have got two types of anions o Nega s so4 now both would want to go to the anode and both would want to undergo oxidation but
among these two which one will undergo oxidation the one which has lesser discharge potential that's o Nega o Nega has got lesser discharge potential so it will undergo oxidation at anode but if I ask you from where this O Negative has come you will say this O Negative has come from water this O Negative has come from water right so O Negative has to go to the anode and will undergo oxidation but this O Negative has come from the water so you'll write the oxidation reaction of water this was your oxidation reaction of water and
due to the oxid of water what is happening O2 gas is liberated O2 gas is liberated and at cathode copper solid is deposited so these are the products of electrolysis which you have to remember is it done is it done guys is it clear let me know once in the chats quickly let me know once in the chats quickly yeah say yes or no in the chats quickly I believe you got the concept right the electrodes which are used during the electrolysis those were you inert electrodes right those were you inert electrodes like Platinum the
actual topic my dear students which we have to discuss what is that the actual topic which we are going to discuss that is paraday laws from which questions frequently come in your examination right Faraday it has given two laws here in electrolysis so we are going to discuss these electrolysis these farad two laws in detail right so tell me once in the chats whether you can do the electroly of all the electrolytes with the help of discharge potential Series right whatever electrolyte I give you with the help of discharge potential series of Cates and ANS
can you do the electroly of every electrolyte right will you be able to guess what what is happening at cathode and what is happening at anode just tell me that just tell me that perfect then let me move on to farad's First Law of electrolysis have a look people so till now we got to know at the electrodes either metal is being deposited or gas is being liberated right on the inodes we saw either metal getting deposited or gas being liberated now what farad first law suggests that farad first law tells you that see on
passing current on passing current in an electrolytic Solution on passing current in an electrolytic solution the amount of substance the amount of substance deposited or liberated the amount of substance deposited or liberated at the electrode that is directly proportional to the charge which goes into the the solution that is directly proportional to the charge which goes into the solution basically you connected the electrodes with the battery what battery does it is sending charge into the solution it is sending charge into the solution more the charge that goes into the solution more will be the amount
of substance getting deposited or liberated at the electrode so mathematically I can say it like this the mass of substance the mass of substance in Gams the mass of substance deposited or liberated this w represents the mass of substance which is being deposited or liberated that is directly proportional to the that is directly proportional to the charge which goes into the solution that is directly proportional to the charge which goes into the solution now my dear students if you remove the proportionality sign here you'll get a constant which I represent with zed so it's Z
multiplied by Q what this Zed exactly is this Zed is called as electrochemical equivalent the Z is called as electrochemical equivalent right and it is equal e / 965 what is this e e is the E is the equivalent weight of same substance which is being deposited or liberated e is the equivalent weight of same substance which is being deposited or liberated so I'll say as per far's first law mass of the substance deposited or liberated is equal in stop Z you're going to write e divide by what divide by 965 multiplied what multipli by
q and Q is the charge and and charge is nothing that is current multiplied by time current multiplied by time so this is the first result which I would want you guys to remember from now on first result which I want you guys to remember with the help of this particular result with the help of this particular result we can easily calculate the mass of substance which is being deposited at at electrode or we can easily calculate the mass of the gas the mass of the gas which is being liberated at at the electrode perfect
which is being liberated at the electode this is the result number one now result number two result number two have a look can write it like this mass of the substance deposited or liberated in grams is equal what is e e is the equivalent mass of the same substance which is being deposited or liberated and equivalent mass of the substance which is being deposited or liberated is equal m mass of the same substance which is being deposit or liberated divided by n factor of same substance right and in the denominator what do we have 965
in the numerator we have got I multiplied by T if you take this m Mass here in the denominator if you take this molar mass here in the denominator so what do we get w/ M mass of substance deposited or liberated divided by m mass of the same substance given mass by m Mass that's called as moles so n represents the number of moles of the substance which are being deposited or liberated and represents number of moles of substance which are being deposited or liberated it's equal I MTI t ided n factor of same substance
which is being deposited or liberated multiplied by 965 this is something which I want you guys to remember again right this is something which I want you guys to remember again perfect after this my dear students one more thing for example this was your question number one this was your question number two for example if they ask you to calculate the the volume of the gas liberated at STP if they ask you how much volume of the gas is liberated at the electrode at STP so for that purpose to calculate the volume of the gas
liberated at STP first of all you'll be calculating the moles of the gas liberated and once you calculate the moles of the gas liberated at the after that you'll say volume of the gas liberated will be equal to moles of the gas liberated multiplied what multiplied by 22.4 L right if they ask you to calculate the volume of the gas liberated at STP this is how you do it first you will calculate the moles of the gas liberated multip 22.4 L if they do not mention STP if they do not mention STP then in the
question they'll give the pressure they'll give the temperature then you'll be using the ideal gas equation and calculate the volume of the gas liberated right I'll try to do certain questions with you so that you get familiar with the types of the questions which can be asked from these results okay but before that before that there's one more thing which I I would want to discuss with you one more simple thing which I would want to discuss with you my dear students in this particular reaction if I ask you how many moles of electrons are
exchanged in this particular reaction as you can see 12 moles of electrons are exchanged right if I ask you if this is a balanced chemical equation first of all if this is a balanced chemical equation and I'm asking you to calculate the N factor of a from the balanced chemical equation you can easily calculate n Factor how exactly let's say n factor of will be equal moles of electrons exchange that is 12 divid by stoet coent of a that's one n factor of a you directly say as 12 for example if I'm asking about n
factor of b n factor of B how many moles of electrons exchanged 12 divide by stri coent 2 so 12/ 2 is 6 12 divid 2 12/ 2 is 6 so 6 I'll be calling as n factor of B similarly what is the n factor of C 12/ 3 that's 4 12/ 4 3 12/ 2 sorry 12 6 that is 2 so these are the respect to n factors of these reactants and products so basically basically whenever you'll be given with a balanc chemical equation from the Balan chemical equation you can easily talk about the
N factors of different substances what you have to do moles of electrons exchanged divide by stric coefficient of reactant or product the one whose n factor is to be calculated remember this particular thing as well now why am I telling you this thing why am I telling you this thing see guys the types of the questions will make you understand why am I exactly telling you all these things look at the question carefully the simplest of all the questions the simplest of all the questions a solution of Aqua cuo4 a solution of Aqua coso4 is
electrolyzed for 10 minutes is electroly for 10 minutes with a current of 1.5 amp so basically a current of 1.5 amp is going into the solution for 10 minutes what will happen electrolysis will happen right and during electrolysis due to electrolysis a reaction would have happened at cathode a reaction would have happened at anode perfect reaction would have happened at cathode and reaction would have happened at anode as well now the question is asking what is the mass of copper deposited at cathode they're asking you what is the mass of copper which is deposited at
cathode so first of all my dear students you should know you should know when we do the electrolysis of Aqua cuso4 what is the reaction at cathode if you remember few minutes back only told told you at cathode when we do the electroly of Aqua coo4 at cathode this is the reaction through which copper is getting deposited right right people now you will directly use Faraday's first law we have to calculate the mass of copper deposited right tell me first of all what is the n factor of copper what is the n factor of copper
here in the reaction is equal moles of electrons Xchange two divide by stri coent one so n factor of copper is two right n Vector of copper is two now I'll be using the equation mass of copper deposited in G is equal equivalent mass of copper MTI I * T divide by what 965 I believe you are done equivalent mass of copper is mol mass of copper 63.5 G divided by n factor of copper n factor of copper is 2 in the denominator you have got 965 what is the value of current 1. 5 amps
of current is passed into the solution for 10 minutes but use the time in seconds so multiply it with 60 right one equation you're done right I think after solving this you'll be getting the value exactly around 0.3 G so what is the 0.3 G exactly you can say when we do the electrolysis of Aqua coo4 when a current of 1.5 amp is passed into the solution for 10 minutes you will say 0.3 G 0.3 G of copper would have got deposited at the cathode which is something I was supposed to calculate I believe it's
clear let me know once in the charts if it is clear so 0.29 G is the answer yeah quickly guys quickly let me know fast so that we can end the chapter fast cuz we have a lot of things to discuss so let me know quickly in the chats if it is clear yes or no everyone everyone in the chats sir when do you take coordination compounds when we when am I going to take coordination compounds you tell me guys Which chapter do you want the next after this Electro chemistry Which chapter do you want
the next majority wins majority wins tell me which chapter do you want the next quickly after Electro which one do you want to do here quickly quickly everyone just say it everyone nothing is wrong in it I can see majority is looking for atomic structure okay let me go with atomic structure next will be atomic structure right all right look at the next question guys look at the next question see or just wait let's not do this question first let me see just a second let's do this particular question first look at this question carefully
the question is what is the volume of oxygen what is the volume of o2 liberated at anode at STP during the electris of Aqua coo4 so we are doing the electro of aqua coo4 and you know when we do the electroly of Aqua CS4 which gas is liberated it is O2 gas which is liberated this is something which we know right O2 gas is liberated now they asking you to calculate volume of o2 gas liberated at STP so what do I calculate first first of all I'll calculate the moles of o2 liberated use the formula
I * T IDE by n factor of what n factor of O2 which is being liberated n factor of o2 multi 965 right now tell me number of moles of o2 liberated is equal current how much current of 2 Amp is passed for how many seconds for 8 minutes so 8 multipli 60 then only it comes in seconds n factor of o2 n factor of o2 is moles of electrons exchange 2 divide by stet coefficient that's 1 by two the value is 4 so n factor of o2 is 4 multip what multip 965 so from
this particular equation what do we get we'll be getting the moles of o2 produced but am I supposed to calculate moles or volume I'm supposed to calculate volume so volume of o2 produced will be equal moles of o2 produced multiplied by 22.4 L right this is something which you'll be calculating from here and you'll have to multiply it with 22.4 you'll be getting the volume of o2 produced is that clear is that clear to you people if you want to have a look on the exact values so moles are 0.2 and volume of o2 is
0.05 L yeah perfect I believe this is clear to everyone right let's try to solve a few more question look at this particular question guys look at this particular equation look at this particular equation as for this particular question question is concerned a current of 96.5 amp a current of 96.5 amp is passed for 10 seconds to 1 lit of a solution of 0.5 mol Aqua copper sulfate calculate the pH of the solution calculate the pH of the solution TR to understand what the question says as far as this particular question is concerned we are
again we are again doing the electris of Aqua coo4 right that's something which we are doing we are doing the electrolysis of aqua coo4 and my dear students we already know that when we do the electrolysis of Aqua CS4 we already know the reactions which happen at anode and cathode this is the reaction at cathode this is the reaction at anode right now at cathode what is happening at cathode it is just copper is being deposited now at anode what is happening at anode water is undergoing oxidation due to which first O2 gas is produced
at the same time H positives are also produced right H posit positives are also liberated at anode at anode water under goes oxidation due to which O2 gas is produced O2 gas is liberated and H positives are also generated and those HED positives will enter into the solution the H positives which we get at anode the H positives which we get at anode these H positives will definitely enter into the solution yeah these H positives will enter into the solution my dear students can I say this is the reaction which is going to take care
of the pH this is the reaction which is going to take care of the pH because through this particular reaction H positives are generated and those H positives are going into the solution right perfect what am I supposed to calculate I'm supposed to calculate pH try to understand one thing I'm supposed to calculate pH what is pH basically P stands for minus log so minus log of H positive concentration we have to calculate so can I say in order to calculate pH we should have we should know what is the total H positive concentration in
the solution after the electrolysis is done once the electrolysis is complete after that you'll find H positives in the solution now we will have to calculate the concentration of H positives in the solution after the electrolysis complete right how do I calculate the H positive concentration concentration is nothing it moles per unit volume so basically it is going to be number of moles of H pos2 divide by volume of solution in liters can I say in order to calculate the H positive concentration I should know the number of moles of H pos2 which are liberated
into the solution now how do I calculate number of moles of H postive liberated with the help of arod first law I'll say number of moles of H positive liberated is equal I * T divide by what divide by n factor of H positive multip what multip 965 correct so number of moles of H positive which are liberated in the solution is equal what is current current is 96.5 amp for how much time 10 seconds divided by what is the n factor of HED positive n factor of HED positive moles of electrons exchanged divide by
stric coent 2 by 2 is 1 this is one multipli what 965 the value comes out be 10 ra minus 2 so I'll say these many moles of H pos2 have gone into the solution in the solution in the solution you'll find these many moles of H positive once you get the moles of H positive tell me what will be the concentration of H positive in the solution moles of H POS in the solution divide by volume of solution in liters so it is 10^ minus 2 divided by what is the volume of Solution that's
1 liter so the value comes out be 10-2 only so 10^ minus 2 molar is the concentration of H positive in the solution if you got the concentration of H positive in the solution you can easily calculate pH of the solution P stands for minus log it is minus log of H positive so pH is going to be equal minus log of 10-2 the value exactly comes out be two so 2 is the pH of the final solution which we get after the electrolysis is done let me know once in the chats if it is
clear quickly guys quickly quickly in the chats quickly in the chats similar sort of question similar sort of question give it a try read the question first then let me know whether you can solve this or not read this question first read this question quickly the question is very simple one farad of electricity one farad of electricity one farad of electr is passed through 10 L of an Aqua Solution of Na so as for the question you are doing the electrolysis of Electro electroly of Aqua na AC and since you are doing the electroly of
Aquis na AC what are we supposed to calculate we are supposed to calculate pH of the solution my de students don't you remember when we do the electrolysis of Aquis na don't you remember the reactions at anode and cathode the reaction at anode the reaction at cathode now tell me one thing which reaction among the two should I consider to calculate the pH of the solution you tell me which reaction among the two should I consider to calculate the pH of the final solution after the electrolysis is done you tell me is it going to
be reaction at anode or cathode my dear students at anode nothing is happening happening only cl2 is being liberated but if you look at the cathode at cathode o negatives are also produced at cathode o negatives are also produced and those o negatives will enter into the solution yeah those o negatives will enter into the solution now see how am I going to tackle this question first of all if I ask you what about the N factor of O Nega what about the N factor of O Negative quickly it going to be equal moles of
electrons exchanged divide stet coefficient so 2ide 2 is 1 this is the n factor of O Negative okay now my dear students the first thing which I'll be doing I'll calculate the moles of h o neg moles of O Nega which are being produced which are being liberated into the solution moles of O Nega as for the formula I multip by T divided n factor of o2 * 965 but do you see current is not given time is not given current is not given time is not given okay but I multiplied by T is what
we call as Q instead of I into T instead of I multiplied by T I can Q so I'll say n factor of O NE my dear students will be equal what is q q is 1 farad 1 farad means 965 kums 1 farad is 965 kums divide by n factor of negative that's 1 divide by 965 the value is coming out to be one so you got to know one mole of O Nega is liberated into the solution right during this electrolysis once this electrolysis is done you'll find one mole of o neg in
the resulting solution if you got the moles of O Nega can't we get the O Negative concentration in the container it has to be equal moles of O Negative in the container divid by volume of solution in liters now tell me how many moles of O neg do we have one volume of solution in liters that's 10 lit the value is 10^ minus1 M 10^ minus 1 m so this is the concentration of O Negative in the final solution once the electrolysis is complete okay if you got the O Negative concentration if you got the
O Negative concentration you calculate P P stands for minus log it's minus log of O Negative concentration so it is minus log of 10^ minus 1 the value comes out be 1 so you got the P if you got the P you know at 25° C pH plus p is always equal to 14 at 25 so pH from here will be equal 14 - 1 which comes out be 13 is this clear is this clear people is this clear let me know in the chat quickly done look at this particular equation look at this particular
equation calculate how much current is required calculate how much current is requir to produce H2 gas at the rate of 1 ml/ second under STP read the question carefully calculate how much current is required to produce H2 gas at the rate of 1 ml/ second at STP so as for the equation we are doing some electrolysis right we are doing some electrolysis due to which what is happening what is happening H2 gas is getting produced now as for the question as for the question we have to do electrolysis in such a way in such a
way that 1 ml of H2 gas should get produced so how much ml of H2 gas should get produced 1 ml of H2 gas should be liberated in how much time in 1 second right we have to perform electrolysis in such a way that 1 ml of H2 gas should be liberate should get liberated in 1 second in order to liberate in order to liberate 1 ml of H2 gas in 1 second how much current should enter the solution that's something which you're supposed to calculate how much current should enter into the solution such that
1 M of H2 gas is liberated in 1 second at STP correct there are many ways of solving this question right the kind of approach which I've been using till now I'll use the same approach see guys this is the condition is STP right how much volume has to get liberated 1 ml if I ask you how many moles of H2 are supposed to be liberated how many moles of H2 are supposed to be liberated it's going to be 1/ 2240 so these many moles of H2 should be liberated in how much time in 1
second now you already know write like this number of moles of H2 liberated has to be equal it has to be equal I MTI T IDE n factor of H2 * 965 right now my dear students first of all you should understand how H2 gas is produced how H2 gas is produced how H2 gas is produced if you remember H2 gas is only produced when water at cathode under goes reduction when water at cathode under goes reduction due to the reduction reaction of water due to the reduction reaction of water H2 is produced right due
to the reduction reaction of water H2 is produced do you remember that do you remember that due to reduction reaction of water H2 is produced so you can write either the reduction reaction of water or in simpler ways you can write the reaction in this manner you can either write reduction reaction of water right or you can write the reaction in this manner perfect the choice is all yours answer will be same okay now what is the n factor of H2 here moles of electrons exchange divide by stric so two so number of moles of
H2 number of moles of H2 liberated we already know it that's 1 upon 2240 so number of moles of H2 liberated is 1 upon 2240 is AAL current we have to calculate Time 1 second n factor of H2 that is 2 multipli by 965 0 one equation one unknown don't you get the I value from here don't to get the I value from it and as far as I remember the value of I will come out to be 8.61 amp so what is meant by this 8.61 amp just tell me that just tell me that
what is meant by it it means that 8.61 amp should pass into the solution then only 1 ml of H2 gas will be produced in 1 second at C yeah is it clear is it clear guys quickly look at this question just tell me can you solve this question on your own can you solve this question on your own quickly let me know in the chats let me know in the chats quickly can you solve this question on your own then let me find one more question one more different type of question which can be
asked from the same this is something which you can solve now right it is all similar pattern similar pattern it is just volume is given in liters here so you do not have to divide with 2 to 40 you have to divide the 22.4 I'm giving you this as a homework you are going to try it and write the answer in the comment section at the end in the last session I gave one question only six people gave the correct answer only six people give the answer in the last last session right but I want
you guys to everyone to solve this question and write the answer in the comment section okay now let me give you one more question the question is like this I'm writing the question the question is for example the density the density of copper is equal 8.95 G per CM Cube or G per ml find the charge find the charge required find the charge required to Plate an area find the charge required to Plate an area of 100 cm square of 100 cm square to a thickness of to a thickness of 10^ min-2 CM 10^ min-2
CM using copper sulfate solution using cuso4 solution look at the question carefully and give it a try look at the question carefully and give it a try first read the question first read the question first read the question sorry okay guys have a look I'll solve this question you just have a look on and remember the approach the question is the density of copper is 8.95 G per CM Cub find the charge required to Plate an area of 100 cm squ to a thickness of 10 ra for -2 CM using copper sulfate solution my dear
students when we do the electrolysis of Aqua c04 when we do the electroly of Aqua C4 what happens at cathode at cathode CU die positive gains two electrons gets converted into copper solid and that copper solid is deposited on the cathode right let's assume that this is your let's assume that this is your cathode it's a plate it is your cathode for example this is your cathode and at cathode what is happening at cathode what is happening reduction is happening reduction of what CU positive so CU di positive from the solution is gaining two electrons
getting converted into copper solid this is your cathode this is your cathode on which this copper solid is being deposited so this is a plate on which copper solid is being deposit copper solid is being deposed on this plate right as for the question is concerned as for the question is concerned the area of cross-section of this particular plate the area of cross-section of this particular plate is 100 cm square dear students if the area of cross-section of this plate is 100 cm squ that means the copper which is being deposed here that has to
cover the area of that has to cover the area of 100 cm square right so the copper which is being deposited it has to cover the area of 100 cm square right since we are doing the electrolysis of copper sulfate what is happening at cathode reduction is happening copper solid is being deposited we have to continue the electrolysis we have to continue the electrolysis till till number one till the entire surface gets covered with the copper and after that the thickness of the copper the thickness of the copper on this plate should be the thickness
of the copper on this plate should be 10 ra^ minus 2 cm till then we have to continue the electrolysis till then we have to continue electrolysis I hope you're trying to you're getting this right we have to continue the electrolysis till the entire surface of the plate till the entire Sur surface of the electrode is covered and at the same time first entire surface should get covered after that the thickness of the copper on the plate should be 10-2 CM if I ask you if I ask you indirectly indirectly how much volume of copper
how much volume of copper are we depositing on this plate how much volume of copper are we depositing on the plate tell me that area of cross-section area of cross-section multip thickness area of cross-section is 100 cm squ thickness is 10us 2 so 100 * 10-2 is going to be 1 cubic cm so this much volume this much volume of copper basically we are depositing on this plate right this much volume of copper we are depositing on the plate correct now tell me is the density of copper given to you absolutely how much 8.95 G
per CM Cub can I say density of copper is nothing it is mass of copper deposited divid by volume of copper volume of copper in cm Cube this is basically 8.95 G per CM Cube correct agreed what is the volume of copper which is being deposited 1 cm Cub 1 cm Cub is being deposited so 1 multiplied with this so I'll say mass of copper deposited has to be 8.95 G so first of all you got to know the mass of copper which you are depositing so we have to deposit 8.95 G of copper right
we have to deposit 8.95 G of copper in order to deposit 8.95 G of copper what is the charge what is the charge which should go into the solution calculate the charge that should go into the solution okay calculate the charge that should go into the solution so that 8.95 g of copper is deposited so I'll directly say mass of copper deposited as per parad first law has to be equal equalent mass of copper MTI I * t/ 965 correct so what mass of copper we are going to deposit 8.95 G is equal equivalent mass
of copper is mol mass of copper divide by what divide by n factor of copper calculate the N factor of copper electrons exchanged divide by stric coefficient so this is two in the denominator it's 965 right I multip by 2 I can call directly as charge q i multiped by T is something which you call as Q perfect so from this particular result can't you get the value of Q yeah can't you get the value of Q can't you get the value of Q okay guys is it clear is it clear to everyone quickly in
the chats quickly everyone everyone Hi Wasim sir how are you I'm doing good bro how are you doing glad to know that RIT I think this Believers was my last year student you the same who used to follow me on YouTube on the last platform right there was one regular student whose name was Believers maybe you're the same anyways have a look so from this particular equation we can easily get the Q value right and Q from here you are going to get in colums right perfect so I believe you can easily solve these sort
of questions now right I believe you can easily solve these s of questions so let's move on to the Faraday second law of elect let's move on to the farad second law of electris first of all tell me in the chats whatever types of questions I showed you in the parad first law will you be able to solve all the types will you be able to solve all the types of questions say yes or no in the chats I believe every single thing is clear yeah perfect now let's have a look on farad Second Law
of electrolysis what this farad second law of electrolysis States let's have a look on the definition first then I'll make you understand okay if a current is passed in two or more electrodic solutions containing different electrolytes the mass of substance deposited or liberated in different cells will be in the ratio ratio of their equivalent masses or the number of equivalents deposited in all the cells will be the same now what is meant by this whole statement let's try to understand see guys as far as this particular question is concerned imagine that imagine that I have
got for example three electrolytic cells like this let's say I've got three electrolytic cells like this assume that in this particular container there is one electrolyte for example AB just to make you understand in this particular container there is one more electrolyte for example CD let's say in this particular container there is one more electrolyte for example EF so we have got three different electrolytes present in three different containers okay now what exactly I'm doing let's put the rods inside these are the two rods in the solution two rods in this particular solution two rods
in this particular solution right let let me connect these rods let me connect these rods and at the end let me connect them with the external battery as well I've connected them with the battery as well so first thing my dear students that Rod which is connected with the negative terminal will carry the negative charge and that Rod which is connected with the Positive terminal will carry the positive charge if this is positive this has to be negative this is negative this has to be positive this has to be negative and this has to be
positive right okay tell me one thing we have got we have got more than one electrolytic cells containing different electrolytes in series this is a series combination this is first of all the series combination right this is the series combination and you know in the series combination current Remains the Same you know it so can I say same amount of current is going into this solution same amount of current is going into this solution same amount of current is going into this solution so in all the three cells in all the three cells what will
happen electrolysis will happen due to electrolysis something would be happening at electrodes either metal will be deposited or gas will be liberated that's something right let's assume let's assume my dear students let's assume let's assume let's assume at this particular electrode let's assume a is being deposited and B is being liberated for example just to make you understand let's assume at this particular electrode at this particular electrode what is happening exactly at this particular electrode just a second guys just a second just a second uh just a second let's assume that on this particular electrod
let's assume that on this particular electrod a is being deposited and here for example B is getting liberated let's say here C is is getting deposited and here D is getting liberated for example here on this electrode let's say e is getting deposited and E is getting deposited and here f is getting liberated this is something which I'm assuming now as for farad second law is concerned whenever you'll be having two or more than two electrolytic cells connected in series containing different electrolytes the gram equivalence the gram equivalence deposited or liberated at every electrode will
be the same that means I can say the gram equivalence of a deposited at the first electrode will be same as that of will be same as that of gram equivalence of B liberated at the second electrode which will be same as that of gram equivalence of C deposited which will be same as that of gam equivalence of D liberated which will be same as that of gam equivalence of e similarly which will be same as that of gam equivalence of f perfect perfect this is something which we call as far Second Law gram equivalence
deposited or liberated at every electron will be the same only perfect now I hope you guys remember how do we calculate gram equivalence gram equivalence of the substance is equal do you remember mass of substance divide by equivalent mass of the substance or you can say number of moles of the substance multiply n factor of the substance this was the result to calculate gram equivalence studied in reduxx studied in reduxx right perfect okay guys now at the same time one more thing one more thing then I'll uh do certain questions so that you can understand
it properly see guys for example I'm taking one electrolytic cell I'm taking one electrolytic cell which contains an electrolyte which contains an electrolyte right it is connected with the battery due to which this gets negative this gets positive right negative is what you call as cathode here this your anode let's assume at cathode let's assume at cathode some substance a is getting deposited and let's assume at anode gas B is getting liberated just to make you understand now you can use farad's Second Law here as well you can say gam equivalence of a deposited ha
should be equal to gam equivalence of B liberated here you can write it like this gram equivalence of a deposited on the first electrode will be always equal to the gram equivalence of B liberated whether I mean at the same time you can do one more thing you can write the same statement in terms of Milli equivalence as well you can say Mill equivalence of a deposited will be equal m equal of B liberated right and my dear students if you got to know how to write these statements you done you can solve any question
of your need standard from farad Second Law let me show you one let me show you one look at the question let's see what the question is first of all the question is during electrolysis of Aqua coso4 we are doing the electrolysis of Aqua coso4 the mass of metal deposited is equal to 63.5 G calculate the mass of gas released and volume of the gas released at STP these are two three questions yeah try to understand try to understand see guys as far as the question we are doing the electrolysis of Aqua coo4 right and
when we do the electrolysis of Aqua coo4 at cathode copper gets deposited at cathode copper gets deposited and at anode O2 gas gets liberated this is something which we already know at cathode copper gets deposited at anode oo gas is released what farad loss stats what farad loss says what parad law says can't I say can't I say the number of gram equivalents of copper deposited should be equal the number of gram equivalents of what the number of gam equalent of o2 produced just equate their gram equivalence nothing else now tell me one thing gam
equivalence of copper can be written as mass of copper deposited divide equalent mass of copper similarly gal of o2 produced will be equal mass of o2 produced in G divid by equivalent mass of o2 yeah I think we are done aren't we done now my dear students try to understand what is the mass of copper which is being deposited calculate the mass of gas released okay the mass of copper which is being deposited that is 63.5 g so 63.5 G of copper is deposited divide by equivalent mass of copper equivalent mass of copper is molar
mass of copper divided by n factor of copper what is the n factor of copper 2/ 1 that's two is equal mass of o2 produced that's something which we have to calculate mass of o2 produced divided by equalent mass of o2 will be equal mol mass of o2 divided by n factor of o2 n factor of o2 will be equal electrons exchanged divide by stri coefficient so 2ide 1 by 2 that comes out to be 4 this one this one cancel the value comes out to be two right 4 1 48 so that 8 is
here and this is mass of o2 perfect the value comes out be 16 G so you got to know 16 G of o2 is liberated you got to know 16 G of o2 is liberated at anode 16 G of o2 is liberated at anode so we were supposed to calculate the mass of gas released we got to know 16 G of o2 is released and at the same time we have to calculate volume of gas released to right yeah right yes is saying sir please show effect of external cell on galvanic cell if possible I
think that is something which I've taught already check the first part check the first part that is something which I've discussed with you in the first part what happens when external batter is connected with the galvanic cell with reverse polarity I believe I have have discussed that guys have I discussed that in the first part I think yes okay let's solve this question first then I'll see whether I have taught that or not see so mass of o2 produces 16 G what else we have to calculate volume but before volume let's calculate moles of o2
produced moles of o2 produced will be equal mass of o2 produced divide by mol mass of o2 so these many moles of o2 are produced so volume of O produce has to be moles multiplied by 22.4 L the value has to be 11.2 so 11.2 L of O is produced yeah clear C yes yes I'll teach Co Rosa as well in this lecture only my dear students I believe you can solve these sort of questions okay I believe you can solve these sort of questions let me first of all check whether I have taught you
external this or not just a second happening just a second guys this is our playlist where is Electro chemistry part one because that is important topic I think I I would have discussed that electrochemistry part one this is part one right this is your Daniel cell this is salt bridge is your Daniel cell close this wait was the representation uh this was the first lecture where is the second one guys just give me a second okay this is the topic right here when I'm making you understand the EMF this is the external battery which I
have connected with the Daniel cell in reverse polarity here you can check this one you can check this one again yeah you can check this particular topic this is the part two here you can see this is your Daniel cell and external batteries connected with the Daniel cell with reverse polarity and through this particular diagram through this particular concept I have taught you how do we calculate EMF of the C cell basically okay right with the help of potential meter okay so this is taught perfect leave it aside now where were we we were discussing
this this question is clear this question is clear perfect let me move on let me move on to the next question look at this particular question read the question carefully and let me know whether you can solve this or not just read it carefully first just read it carefully pre electrolytic cells containing zinc sulfate ag3 and copper sulfate respectively are connected in series a current of 2 ampers was passed until 1.08 G of silver was deposited at cathode of celb how long did the current flow okay what mass of copper and zinc was deposited read
the question tell me can you solve this question on your own if you can do then I'll move on to the next topic tell me whether you can solve this question on your own or not read it quickly you solve it okay I'll solve it see guys as for the equation is concerned we have taken three electrolytic cells so imagine this is one electrolytic cell imagine this one more and imagine this is one more so we have got three electrolytic cells the first one is containing zinc sulfate the second is containing H3 and the next
one is containing what it's containing cuso4 all right now there are two rods here two rods are here as well and two rods are here as well now we will be connecting them with the external battery right this is the battery here negative this is positive perfect one is negative one is positive the negatively charged Rod I mean the negative ter of the battery is connected here so it carries a negative this carry is positive this is negative this is positive this is negative and this is positive perfect now guys ne1 is called as cathode
here on this particular cathode as for question what is getting deposited zinc here at this negative at this cathode what is getting deposited silver and here what is getting deposited over here copper is getting deposited perfect copper is getting deposited right and these are the reactions through which these these elements are getting deposited zinc dipositive from the solution is getting two electrons getting converted into zinc solid and that zinc solid is deposited here similarly AG positive in the solution will be getting one electron and will be getting converted into AG solid that a solid is
deposited here similarly copper D positive getting two electrons getting deposited as copper here perfect now in the question in the question a current of 2 ampers was passed okay current of 2 ampers is passing in every every electrolytic solution current of 2 Amp was passed until 1.08 G of silver was deposited so as for the question it is given that 1.08 G of silver is deposited right is deposited first is asking you how long did the current flow how long did the current flow right he's asking you t t value how long did the current
flow in what duration of time current flowed so look at the second cell my dear students in the second cell if I use the farad first law I can say in the second cell mass of silver deposited has to be equal equivalent mass of silver * I * t/ 965 now what is the mass of silver deposit as for the question it's 1.08 G is equal equivalent mass of silver is mol mass of silver divide by n factor of Silver N factor of silver electrons exchanged divide by stric coent that's one multip what multip 965
now I current current is 2 Amp multipli by time so one equation one unknown can't you calculate T from here e can be calculated and as far as I remember T value will be something 4825 seconds right so this is the time duration in which current has to flow right in which the current has to go into the solutions so the question was asking how long did the current flow current current has flown this long for 82.5 seconds number one number two what mass of copper and zinc was deposited what mass of copper and zinc
was deposited now use farad Second Law what farad second law states farad second law states that the gram equivalence deposited or liberated at every electrode will be the same right so first of all I'll write like this gram equivalence of zinc that is mass of zinc divide by equalent mass of zinc gam equivalance of zinc deposited has to be equal to gram equal of copper deposited sorry gram equal of silver deposited which is mass of silver divid by equalent mass of silver right G equal of zinc deposited should be equal to gram equence of silver
deposited correct gram equal of zinc deposited is mass of zinc deposited divid by equalent mass of zinc gram equal of silver deposited is equal to mass of silver deposited divid by equalent mass of silver now dear students have a look what is the mass of zinc which is being deposited that's something which has to be calculated divided by equalent mass of zinc is M mass of zinc divid by n factor of zinc n factor of zinc is two here is equal what is the mass of silver which is being deposited 1.08 G as for the
question divide by equivalent mass of silver that is mol mass of silver divide by n factor of silver so from this particular equation you can easily calculate the mass of zinc which is deposited and that to the answer will be in Gs correct similarly after this once you get the mass of zinc deposited you can you can equate gam equivalence of AG with g G equal of silver sorry gam equence of AG with gam equal of copper from there you can calculate the mass of copper deposited is this question clear now is this question clear
so T value is 4 82.5 seconds mass of zinc deposited is 0.325 mass of copper deposited you can calculate through this particular equation you have just equated the gam equivalence of AG deposited with the gram equivalence of copper deposited that's all is this clear to everyone is it clear to everyone people perfect so I believe you can solve the faradise first law and second law electrolysis questions easily now comes one more part of the chapter that is electrolytic conductance electrolytic conductance this is something important this is also important so have a look people what it
means what kind of questions can be asked just a second let let me see whether I have added questions here or not these questions are added okay so my dear students let me Mark The Heading as terminologies involved terminologies involved in electrolytic conductance terminologies involved in electrolytic conductance or let me just write terminologies involved in electrolytic conductor you can also call called it as you can also call it as conductivity cell you can also call it as conductivity cell now what this conductivity cell first of all is all about try to understand what exactly I'm
going to talk about yeah have a look people since we have discussed the electrolytic cell okay again I'm going to make the electrolytic cell here again I'm going to make the electrolytic cell try to understand guys this is important see this is first of all let me make the electrolytic cell here this is one container this is one container and my dear students in this particular container what do we have imagine that in this particular container we have got an electrolytic solution there is electrolytic solution in the container so in short I'll say there is
electrolyte in the container there is electrolyte in the container which electrolyte let's call this is AB electrolyte there is AB electrolyte in this particular container okay now what exactly I'm doing I'm going to place two electrodes here right I'm going to place two plates basically two metal plates I'm placing which are your electrodes so here this is your one metal plate which works like one electrode right similarly this is this is one more this is one more metal plate which works like one more electrode perfect I've taken two electrodes here now my dear students you
are connecting these electrodes with the help of battery with the help of battery this electrode this plate got the negative charge this plate got the positive charge correct now my dear students first of all imagine can I say this is the part of the electrode which is in contact with the solution yes this is the part of the electrode which is in contact with the solution so basically this is the part of the electrode which is dipped into the solution similarly this is the part of the electrode which is dipped into the solution as well
perfect my dear students let me make one diagram here have a look let me connect this with this let me connect this with this let me connect this with this and let me connect with this with this do you see do you see any diagram here do you feel something different here do you feel something different here try to understand just a second just a second what did I get what did I get what did I get it's a cube it's a cube right what this cube is containing basically what is present inside this Cube
what is present inside this Cube can I say in this Cube electrolytic solution is present basically a solution is present which contains electrolyte in this Cube let me take this cube out from the container imagine I'm taking this cube out of the container when I take the cube out of the container the cube will look like this let me show it to you I've taken the cube out of the container the cube is looking like this right what is this particular one it is the part of the plate it is the part of the electrode
it is the part of the electrode which is inside the solution this is the part of one more electrode which is inside the solution right this electrode carries negative this electrode carries positive and in between these electrodes what do I have I've got a solution which contains an electro which contains an electrolyte okay right in this Cube you have got an electrolytic solution perfect which electrolyte I had taken AB so basically in this Cube what you'll find in this Cube you'll find a positive ions and B negative ions right first of all my dear students
this particular cell which I made over here this particular cell is something which I call as conductivity cell this is something which you call as conductivity cell correct yeah this is something which I call as conductivity cell now my dear students Let Me Assume certain things let's assume the distance between these two electrodes let's assume the distance between these two electrodes is L CM let's say this area of cross-section of the electrode that area of cross-section of the electrode which is inside the solution right that means this one this area of cross-section of the electrode
which is inside the solution right let's call this as a CM squ a cm square is the area of cross-section of the electrode which is inside the solution which is in contact with the solution yes yeah so this particular cell is something which you call as conductivity cell now tell me one thing this particular plate carries negative charge this particular plate carries positive charge can I say this negatively charged plate is going to attract B negative towards itself absolutely and this positively charged plate is going to attract sorry negatively charged plate is going to attract
a positive towards itself correct and this positive charged plate is going to attract B negative towards itself now tell me one thing when a positive will be moving towards negatively charged plate when this a positive will be moving towards negatively charged plate during the moment of a positive can I say there will be some B negatives which will be attracting the same a posi towards itself yeah similarly when this B negative will be going towards positively charged plate when this B negative will be going towards positively charged plate during its moment M can you say
there will be some a positive which will be attracting this B negative towards itself absolutely do you see any obstruction in the moment of the ions absolutely there's obstruction in the moment of the ions and that obstruction in the moment of the I that is the reason why we say this particular electrolytic solution shows the resistance so what is resistance of this cell it is simply the obstruction during the M of the I right what is the resistance of this particular cell it is the obstruction during the moment of the I when the ions are
moving towards the respective electrodes perfect so my first terminology my dear students that is resistance that is resistance which is denoted by R which is denoted by r R I'm not going to write this definition because you already know it obstruction in the moment of the I right how do you calculate this R tell me that how do you calculate this R you know it already R is equal row L by a r is equal row L / a perfect r equal r l by a if I ask you first of all what are the
units of resistance resistance units are ohm resistance unit is O right how do you represent it this is how you represent it right this how you represent it clear to everyone now my dear students in this particular expression what did we get here we got this term row what is row row is called as resistivity row is called as resistivity resistivity is your row so tell me this row can I say row is equal r a / L resistivity row row is equal r a / L resistivity row is equal r a / L how
do you define this resistivity first of all how do you define this resistivity how do you define this resistivity understand my dear students imagine that imagine that the area of the cross-section a is 1 cm squ this particular area imagine it's 1 cm squ imagine that the distance between the electrodes is 1 cm right distance between the this is 1 cm if this cross-section area is 1 cm squ distance between them is 1 cm can you let me know what will be the volume of this Cube volume of this Cube will be 1 cm Cub
volume of this Cube will be volume of this Cube will be 1 cm Cub this is the volume of the cube right now tell me one thing when a is = 1 cm squ L is equal 1 cm can I say at that point of time your row is basically equal to R can I say your row is equal R perfect can I say resistivity is basically the resistance resistivity is basically the resistance it is basically the resistance shown by this particular electrolyte when when the distance between these plates is 1 cm and the dipped
area of cross-section the dipped area of cross-section that is 1 cm square or I can Define it like this as well since the volume of the cube is 1 cm Cube if the volume of the cube is 1 cm Cube if I ask you what is the volume of the electrolytic solution in this container I'll say volume of the electrolytic solution in the container right now is 1 cm Cube I'll say resistance shown by 1 cm cube of electrolytic solution resistance shown by 1 cm cube of electrolytic solution right that's something which I call as
resistivity perfect right so it is the resistance it is the resistance shown by what by 1 cm Cub of electrolytic solution can you let me know what will be the units of resistivity units of resistivity tell me that R is OHM a is cm square this is ctim so it has be ohm CM this is basically the unit of resistivity ohm CM clear so this was my second terminology that is resistivity talking about my third terminology talking about my third terminology just a second my third terminology is it is conductance it is conductance conductance is
basically represented by G how do you define this conductance conductance is defined simply as inverse of inverse of resistance shown by the conductivity cell right inverse of resistance now tell me one thing tell me one thing 1id by R is R is row L / a so this is a here perfect this is a here so my dear students tell me just one thing tell me this one thing look at this particular expression look at this particular expression okay let me tell you the units of this conductance it is 1 by R the unit of
conductance is going to be either you'll write ohm inverse or you'll write M Mo or you'll write it as this the unit of conductance right and Simon it's denoted by S that's all okay now similarly my four terminology that is conductivity conductivity conductivity is represented by Kaa right now the point is how do we Define this conductivity how did we Define conductance conductance was inverse of resistance so how do I Define conductivity conductivity or you call it as or you call it as specific conductance you call it as specific conductance tell me how do you
define it specific conductance is nothing it is it is inverse of I'm sorry guys I think I'm going to fall ill there is going to be old cff soon okay conductivity how do we define conductivity it is defined as inverse of inverse of resistivity inverse of resistivity now tell me one thing 1 divid by what is row exactly equal row is equal row is equal r a / L row is equal r a/ l so this is R this is a and this is L here this is L here right you can write it like
this 1/ r multip l by A Perfect this is one of the Expressions which I would want you guys to remember because I'll be using frequently this expression in the questions your conductivity Kaa that's equal to 1/ R multi L by a 1 by R multip L by a second thing second thing guys have a look I'm talking about conductivity specific conductance that is your Kaa that's your Kaa so Kaa is equal 1 by R is something which you call as conductance L by a is something which you call as cell constant this is conductance
this term is called as cell constant right so I'll write something like this conductivity conductivity is always equal conductance conductance multiplied by what multiplied by cell constant multiplied by cell constant right your cell constant is what your cell constant is L by a this is cell constant this is conductance and this is conductivity now if I ask you what are going to be the units of conductivity units of conductivity see conductivity is equal to conductance multip cell constant so units of conductivity is going to be this is conductance you can write it as Simon cell
constant this is centimeter this is ctim square so it's going to be centimeter inverse so you can write it as Simon cimet inverse it is basically the unit of what it is the unit of conductivity right perfect it is the unit of conductivity s CM inverse perfect now my dear students one more thing one more important thing before talking about the other terminologies one more important thing which you should remember from now on imagine this is your one electrolytic solution this is your one more electrolytic solution two electrolytic solution I have in two containers right
imagine the resistance of the first solution is RA the resistance of the first solution is RB right imagine the conductance of the first solu sorry conductivity of the first solution is K and conductivity of the second solution is KB these are two solutions solution number one solution number two now what you are doing you are mixing it you're adding them into a final container so in this final container we got 1 + 2 perfect you got 1 + 2 my dear students remember one important thing you're not going to say resistance of the final solution
is equal r1+ I mean R A Plus RB this is a wrong statement you are not going to do the question in this format but you will do the question in terms of conductivity you will say conductivity of the final solution that is equal Ka a plus KB you'll solve the question in terms of conductivity not in terms of resistance right you take a note of this point it's important from from this particular Point you'll get one question from this particular Point you'll get one question if you are adding two solutions uh having the resistance
R RB conductivity k kab b right the final solution which you get you are not going to say it's resistance is R plus RB you're going to solve the question in terms of conductivity you can write conductivity as conductivity of the final solution as conductivity of a plus conductivity of b mean conductivity of a plus conductivity of yeah right now guys before doing the other stuff before doing the other stuff let me talk about one more important terminology what is that that is molar conductivity molar conductivity molar conductivity is represented by Lambda M molar conductivity
is represented by Lambda M let me first of all write its definition then I'll make you understand molar conductivity Lambda it is defined as it is defined as the conductance the conductance shown by all the ions the conductance shown by all the ions when when one mole of electrolyte when one mole of electrolyte is present when one mole of electrolyte is present in a given volume of solution in a given volume of solution try to understand what it means exactly see guys few minutes back we made a conductivity cell here imagine this is the conductivity
cell okay imagine this is the conductivity cell can I erase all this part can I erase all this Parts say yes or no I just need some space here can I erase okay leave it as such I'm just I'm erasing this part I hope you got to know how this conduct cell was made I'm erasing this part all right see guys there is something important which I'm going to discuss with you now this is your conductivity cell and this conductivity cell it in what what is there in this cell electrolytic solution corre imagine imagine volume
of the electrolytic solution volume of the solution here in the container is B CM Cub it can be anything 10 cm Cube 1000 CM Cube 50 cm Cube whatever this is the volume of the solution V CM Cube I'm assuming that in a given volume of solution in a given volume of solution I'm assuming that in a given volume of solution 1 mole of electrolyte is present I'm assuming I'm assuming in a given volume of solution one mole of electrolyte is present so one mole of electrolyte is present in a given volume of solution my
dear students whenever one mole of electrolyte is present in a given volume of solution the conductance of all the ions shown at this particular point of time is something which you call as molar conductivity so molar conductivity it is the conductance shown by all the ions when one mole of electrolyte is present in a given volume of solution I hope it's clear I hope it is clear it is the conductance shown by all the ions when one mole of electrolyte is present in a given volume of solution correct now guys I'm directly going to give
you the result directly I'm not dering the result I'll be directly giving you the result molar conductivity of the solution is always equal Kaa multipli by th000 divided by marity of the solu this is the result which you guys will be using to calculate mol conductivity of the solution when you have to calculate molar conductivity in Simon cm squ per mole then you have to calculate the M conductivity in Simon cm squ per mole this is the equation which you'll be usinga multip th000 mity of the solution for example you have to calculate the molar
conductivity inide M Square per mole if you have to calculate the M conductivity in Simon M Square per mole this is not the relation which you'll be using you'll be using the relation like this Kaa Kaa conductivity divided by th000 multiplied by marity of the solution perfect my dear students I want you guys to remember these two expressions is this clear is this clear I hope you got to know what molar conductivity exactly is say yes or or no in the chats say yes or no in the chats okay all right there is one more
similar terminology that's called as equivalent conductivity which is represented by Lambda EQ which is represented by Lambda EQ now how do you define this equivalent conductivity let me write its definition it is similar it is similar equivalent conductivity is defined as it is defined as the conductance shown by the conductance shown by all the ions when 1 g equivalent when 1 g equivalent of electrolyte when 1 g equivalent of electrolyte is present is present in a given volume of solution the conductivity cell which I took few minutes back imagine in the given volume of solution
there is 1 g equivalent of electrolyte present when when 1 g equivalent of electrolyte is present in a given volume of solution whatever will be the conductance shown by ions at that point of time in the conductivity cell that is something which I'll be calling as equivalent conductivity as simple as that there is one difference only instead of one mole of electrolyte you are using one g equivalent of electrolyte nothing else rest everything is same okay only difference is instead of one mole of electrolyte you are using 1 g equivalent of electrolyte that's that's it
now the point is how do we calculate it Lambda EQ one of the result Kaa ,000 ided normality this is the equation which you will be using to calculate the equivalent conductivity to calculate the equivalent conductivity in Simon cm squ per equivalent if you'll have to calculate equivalent conductivity under these units this is the expression which you'll be using right but if you will have to calculate the equivalent conductivity in Simon M Square per equivalent what is the formula which you'll be using it's going to be Kaa divided th000 mtip by normal perfect correct now
my dear students if I ask you if I ask you if I ask you what is going to be the relation between molar conductivity and equivalent conductivity if I ask you what is the relation between molar conductivity and equivalent cond M conductivity and equivalent conductivity what is going to be the relation see since one of the relations I told you Lambda EQ is equal Kaa ,000 / normality this one of the results which I told you right which I gave you now tell me one thing how write it like this Kaa MTI th000 divid instead
of normality I'll write marity multip n Factor you know it normality is equal to marity multip by n Factor of solute solute is your electrolyte so normality is marity of the solution multip n factor of electrolyte correct this particular term my dear students this particular term what do we call this as we call this as Lambda M so it's going to be Lambda mide M factor from this particular result you got equivalent conductivity is always equal m conductivity of the solution divide by n factor of electrolyt and factor of electrolyte this particular result is also
important is this clear is this clear everyone let me know once in the chat quickly is it clear perfect right let's try to do a few questions on the same Concepts which we have discussed then we'll move on to few other Concepts too look at this particular question look at this particular question guys the specific conductance specific conductance means conductivity of 0.1 M solution of mgcl2 is given the cell electrodes of 1.5 CM squ area and placed 0.5 CM apart in the solution how much current should flow if potential difference between electrodes is 5 volts
what are we supposed to calculate we are supposed to calculate current right we are supposed to calculate current use the ohms law I equal v/ R right potential difference is 5 volts divided by resistance of the cell we do not know so we have to calculate resistance of the cell right we have to calculate the resistance of the cell few minutes back I give you one result important result I told you at time that is Kaa is equal to 1 / R * L by a right now tell me can't you calculate R from here
1/ Kaa MTI L by it's going to be 1 / this is specific conductance how much it it is 0.9 multiplied distance between electrodes distance between electrodes is 0.5 CM divided by area of cross-section of the electrodes that is 1.5 cm square so from here you can easily calculate your what resistance once you get the resistance once you get the resistance value put the value of resistance in this particular equation you'll get the current are you getting this are you getting this people right so resistance is coming as 0.075 ohm and current is coming out
to be 0.135 aamp I believe it's clear I believe it's clear to everyone okay let's solve one more question let's solve one more question look at this you can easily solve this as well now these are simple simple questions the resistance of 0.01 normal solution of an electrolyte is 210 oh resistance is given cell constant is given calculate conductivity and equivalent conductivity of the solution so we have to calculate conductivity as well as equivalent conductivity conductivity is Kaa which is equal 1 / R * L by a right so Kaa is going to be equal
to 1id r r is the resistance which is 210 ohms L by a is the cell constant cell constant is given to us as 0.88 CM inverse right when you solve this you get the answer in cim s CM inverse this your conductivity this was the first question second is calculate equivalent conductivity equivalent conductivity is equal you have calculate equivalent by the way we have to calculate equivalent conductivity in Simon cm squ per equivalent these has these are supposed to be the units in the equation so Lambda AQ is going to be Kaa * th000
IDE by normality right divide by normality perfect Kaa you have already calculated normality of the solution is also given put it here and get the equivalent conductivity of the solution in Simon and cm square per equivalent done these are just the formula based ones right just the formula based ones so Kaa you have calculated put it here and get the equalent conductivity that's it nothing else to do these are the basic basic questions now these are the basic basic questions now okay let's have a look on this question the equivalent conductivity the equivalent conductivity of
0.1 normal solution of ci2 is 100 and its units you already know it is Simon cm square per equivalent if the cell constant is this much how much current will flow when the potential difference between the electrodes is 5 volts right can you give it try can you give it a try see guys what are we supposed to calculate again we have to calculate current current is nothing it potential difference voltage divid by R right well potential difference is given to us as 5 Vols again now we have to calculate R how do we calculate
R exactly how do we calculate R exactly how do we calculate R exactly quickly how do we calculate R I told you Fus back only your Kappa is equal to 1/ R * L by a correct so R you can calculate as 1/ Kaa * L by A Perfect now as for the question this particular term cell constant this is given to me correct but this Kaa is not given to me so I have to calculate Kaa as well I to calculate Kaa as well from where I calculate the conductivity for that purpose equivalent conductivity
is given equivalent conductivity is given to me as 100 Simon cm squ per equivalent this is the equivalent conductivity which is given to me now you know equivalent conductivity is equal Kaa * th000 divide normality right this is equal 100 tell me one thing is the normality of the solution given to me yes it's given to me right if normality is given from here it is given to me as 0.1 from here you can calculate Kaa conductivity can be calculated if conductivity is calculated put it here cell constant already you know you'll get the resistance
if you got the resistance put it here you'll get the current that's something which you were supposed to calculate right is it clear people is it clear to everyone is it clear to everyone everyone let's talk about let's talk about first of all the effect effect of dilution effect of dilution on conductivity molar conductivity and equivalent conductivity what happens to conductivity molar conductivity and equivalent condu ity with dilution let's have a look people all right see first of all if I particularly talk about conductivity if I particularly talk about conductivity how do you exactly define
conductivity we have already defined it I hope you remember how did we Define conductivity conductivity was defined as it was defined as the conductance the conductance shown by one cm cube of an electrolytic solution right the conductance shown by 1 cm cube of electrolytic solution correct try to understand what exactly I'm going to say my dear students first of all I'm going to make the conductivity cell again I'm going to make the conductivity cell again so first of all this is the cube which I'm making over here from this particular Cube you'll get the idea
what happens to conductivity okay so imagine this is the conductivity cell which I'm again making and these are the plates these are these are the plates which are basically inside the Electro ltic solution these are the plates which are basically inside the electrolytic solution I hope you remember from where we got this conductivity cell right now my dear students inside this particular Cube we have got the electrolytic solution I hope you remember this is something which we have already discussed let's call this as the negatively charged plate let's call this as the posi charged plate
perfect if you remember from where we got this Cube if you remember from where we got this Cube we got this Cube from the electrolytic cell basically we took this cube out and this shaded this shaded part what is this shaded part it is the part of the electrode that is inside the solution it is the part of the electrode that is inside the solution actually this is the actual size of the electrode this is basically the actual size of the electrode right this is the actual size of the electrode and this part of the
electrode is inside the solution I hope this is clear okay and my dear students if you remember the distance between these two electrodes is represented by what it's represented by L cm and that area of cross-section which is inside the solution that is represented by a CM squ perfect perfect my dear students how do we Define the conductivity it is the conductance shown by 1 cm cube of the solution it is a conductance shown by 1 cm cube of the solution imagine imagine the volume of this particular solution the volume of this electrolytic solution for
example for example is 100 cm Cube it is 100 cm Cub let's assume that in this 100 cm Cube electrolytic solution let's assume that there are there are for example some 500 ions present there are are some 500 ions okay so if in 100 cm Cube there are 500 ions present can you let me know in 1 cm Cube how many ions will be present 500 by 100 that comes out be five ions so if in 100 cm Cub solution there are 500 ions that means in 1 cm cube of solution there are five ions
present perfect so that means that means whatever will be the value of conductivity whatever will be the value of conductivity that is due to that is due to these Five Pines correct now try to understand imagine that you are adding some extra solvent here imagine that you are adding some extra solvent here so you are doing the dilution that means you are doing the dilution you're adding some extra solvent Here My Dear students on adding the extra solvent what will happen to the volume of solution volume of of solution will increase initially the volume of
solution was 100 cm Cub right now due to dilution let's assume that the volume of the solution reached 1000 CM Cube imagine due to dilution volume of the solution reached 1000 CM Cube what did we do we just added solvent did I do anything to the solute did I did I did I add some more electrolyte here no the amount of el in the container is same it is just I'm adding more solvent okay after the addition of more solvent the volume of the solution increased volume of the solution increased to 1,000 CM Cube now
in 1,000 CM Cub how many total ions will be there total ions will be still 500 only because I did not add some electrolyte I did not add more electrolyte electrolyte is same now tell me this was the case of before dilution and this is the case of after dilution tell me one thing if in 1,000 CM Cube there are 500 ions tell me in 1 cm cube of solution how many ions will be there how many ions will be there let me not call it as 5,000 let me call it as just a second
otherwise calculation will be wrong let's say after dilution the volume of the solution is 500 cm Cub after dilution so 500 cm cube of solution contains 500 ions so 1 cm cube of solution will contain 1 ion 1 cm cube of solution will contain 1 ion my dear students can I say due to dilution due to the addition of extra solvent the number of ions per unit cube of the solution the number of ions per unit cube of the solution decreased can I say something like that initially before dilution in 1 cm cube of solution
there were five pins right so conductivity was due to 5ins now due to dilution due to dilution volume of the solution increased now for example volume of the solution now is 500 cm now in 500 cm Cube I must say there are still 500 ions only right so in 1 cm cube of solution there is one ion so due to dilution what is happening number of ions per unit volume number of ions per unit volume were initially five and now due to dilution number of ions per unit volume is one so I would say due
to dilution due to dilution number of ions per unit volume volum of the solution decreases right I'll say due to the due to dilution due to dilution I would say number of ions number of ions in 1 cm cube of solution 1 cm cube of electrolytic solution that decreases and my dear students it's understood if number of ions in one CM cube of solution decreases what will happen to conductivity conductivity automatically will decrease so do remember conductivity conductivity conductivity of the electrolytic solution it decreases with dilution is this point clear to everyone this was our
first point this was our first point okay let me know once of the charts if this point is clear number of ions in 1 cm cube of solution are decreasing if number of ions are decreasing that automatically tells you that conductivity of the solution is decreasing right because what is conductivity it is the conductance shown by all the ions in 1 cm cube of solution if in 1 cm cube of solution number of ions are decreasing due to dilution that automatically tells you the conductivity is decreasing with dilution perfect so this was your conductivity part
now let's have a look on let's have a look on molar conductivity now what happens to the molar conductivity what happens to the m conductivity with dilution what happens to the m conductivity with dilution first of all how do you define the M conductivity do you remember how do we Define the M conductivity it is the conductance shown by all the ions when one mole of electrolyte is present in a given volume of solution I hope you remember let's say this electrolytic solution which is there in the container let's say volume of this electrolytic solution
is for example CM Cube volume of this particular electrolytic solution is 1 sorry V CM cub and in this given volume of solution let's assume that there is one mole of electrolyte presence there is one mole of electrolyte pres so I'm assuming one mole of electrolyte is present in this given volume of solution so whatever conductance this particular solution shows right now that is something which I call as molar conductivity that is something which I call as molar conductivity now what happens to mol conductivity with the dilution that is the discussion imagine that in this
particular solution the electrolyte which we have used for example that the strong electrolyte for example that is a strong electrolyte I'm assuming that I'm assuming that I'm assuming that in this particular given volume of solution I'm assuming that in this particular given volume of solution there is one mole of strong electroly present there is one mole of strong electroly present strong electroly is the one which gets completely dissociated into SS so if one mole of strong electrolyte is in this particular solution that one mole of electrolyte would have got completely converted into its ions now
my dear students what all types of ions will be there in the container there'll be cats and anions in the container there'll be catons as well as anions in this particular container perfect now imagine you are doing dilution if you are doing dilution that means you are adding some extra solvent if you are doing if you are putting some extra solvent in the solution what is happening to the volume of solution volume of the solution is increasing if volume of the solution is increasing if volume of solution is increasing the distance between the oppositely charged
ions that will increase and if the distance between oppositely charged ions increases the force of attra rtion between them decreases and if the force of attraction between these ions is decreasing due to dilution I'll say due to dilution since force of attraction between these ions is decreasing I'll say due to dilution the ions will start moving freely comparatively and more the moment of ions more the moment of ions more has to be the M conductivity done understood right clear guys clear so I had one mole of strong electrolyte in this solution and whatever conductance this
solution was showing that was called as molar conductivity of this solution now you are adding some extra solvent due to the addition of extra solvent what is happening to the distance between the ions it is increasing force of attraction between the ions it's decreasing now the ions can comparatively move freely right ions can comparatively move freely if the ions can comparatively move freely I would say molar conductivity of the solution automatically increased right so do remember in case of strong electrolytes in case of strong electrolytes with dilution with dilution molar conductivity of the strong electrolyte
that increases what is the reason reason is the distance between the ions that increases right I'll say the reason is distance between the oppositely charged I that increases hence force of attraction between the ions decreases and eventually the ions start moving comparatively freely I start moving comparatively freely I hope this point is clear to everyone is this point clear to everyone quickly in the chats yeah now my dear students as I told you with dilution molar conductivity completely increases it keeps on increasing with dilution Moler conductivity keeps on increasing so if I keep on doing
the dilution if if I keep on adding the solvent molar conductivity of this solution will keep on increasing M conductivity will keep on increasing and my dear students there will be a time there will be a time when this particular solution shows maximum M conductivity and when this particular shows maximum M conductivity when the distance between the ions will be when the distance between the ions will be infinite at that time there'll be no force of attraction between the ions at that time ions can completely move freely okay so with dilution with dilution motor conductivity
of this solution of this electrolytic solution it is increasing right if you keep on continuously doing the dilution mot conductivity will keep on increasing the ions will go far the ions will go far the ions will go far right okay there will be a time when these ions will be at infinite distance apart at that point of time force of attraction between the ions will be zero so at that point of time there is maximum free moment of these ions so if there is maximum free moment of ions I'll say m conductivity value value of
this particular solution will be maximum and at that point of time when the mol conductivity of the electrolytic solution attains its maximum value we say that that dilution at which the molar conductivity of the solution attains its maximum value that dilution is something which we call as infinite dilution let me know once in the chats if you got the idea of what infinite dilution is so I'll write it over here if I keep on doing the dilution continuously then there will be a point which is what we call as infinite dilution there'll be a time
which is what you call as infinite dilution so at infinite dilution distance between the ions is infinite force of attraction zero ions will move freely I'll say Lambda M molar conductivity attains its maximum value attains its maximum value and the maximum value of M conductivity the maximum value of mol conductivity is something which you call as molar conductivity at infinite dilution or it is represented like this Lambda M infinite perfect is this clear is this clear same thing happens to equivalent conductivity as well same thing happens to equivalent conductivity as well equivalent conductivity also increases
with the dilution and at infinite dilution equivalent condu tivity attains its maximum value which is what you call as equivalent conductivity at infinite dilution right perfect I'll say equivalent conductivity also increases and hence attains its maximum value which is what you call as equivalent conductivity at infinite dilution or you can represent it like this perfect perfect guys so in case of strong lorid if the electrolytic solution contains a strong electroly with dilution ITS m conductivity as well as equivalent conductivity increases because ions they go far they go far so when you keep on doing the
dilution continuously at infinite dilution there'll be maximum distance infinite distance between the oppositely charged ions in the container due to which no force of attraction between the ions due to which ions start moving freely so the maximum M conductivity value or the maximum equivalent conductivity value will be there at infinite ution perfect my dear students if you want to understand it with the help of graph if I plot a graph for a strong electrolyte and the graph is between Lambda m versus Under root of C where C is the concentration of electrolyte in the container
where C is the concentration of electrolyte in the container so first of all what is C concentration concentration means moles per unit volume perfect tell me one thing this graph I'm plotting for the strong first of all I'm plotting for the strong electrolyte okay at this particular point of time what is root C value root C value here is zero at this point root C value is z if root C is z that means C is z that means C is Z when will be c0 c will be zero at infinite dilution when volume of
the solution volume of the electrolytic solution had become infinite right so this point represents your infinite dilution and infinite dilution molar conductivity of the strong electrolyte right molar conductivity of the electroic solution that attains its maximum value and that maximum value is Lambda not M so at this point this point represents your infinite dilution this point represents infinite dilution at infinite dilution M conductivity attains this maximum value which is Lambda n let's assume that this point is Lambda M this point is Lambda not M now tell me one thing this is infinite dilution at which
volume of solution had become infinite now if you go in this direction in this positive x-axis Direction what is happening to root C value if you go in this direction root C value is increasing if root C is increasing that means C is increasing if root C is increasing that means C is increasing P increasing means volume of the solution is decreasing at this point of time volume of the solution was infinite now you are going in this direction T is increasing right that means V is decreasing volume of the solution is decreasing initially the
volume of the solution was infinite Now volume of the solution is decreasing if volume of the solution is decreasing that means the oppositely charged ions in the container they are coming closer if they are coming closer distance between them increase distance between them decreases if distance between the oppositely charged ions decreases force of attraction between them increases now the ions cannot move the way they were right the the way they were moving there what will happen to M conductivity M conductivity will decrease so M conductivity value it will keep on decreasing with time perfect but
this particular graph it is valid for what it's valid for strong El right okay now tell me one thing since this particular graph can you tell me what will be its equation its equation see there is an intercept there is a negative slope so it will follow this type of the equation Y is equal - mx + C this particular graph will follow this particular equation y- mx + C now along y AIS what are you plotting along Y axis you are plotting Lambda m is equal minus m is the slope slope of the curve
let's say slope of the curve is represented by a what is X along xais you are plotting root C and what is this particular c c is The Intercept intercept is the distance from here to here which is Lambda which is Lambda this is the equation which correctly represents the behavior of strong electrolyte when the behavior of strong electrolyte the behavior of molar conductivity of strong electrolyte when it under goes dilution perfect this particular equation is valid for a strong electroly when it under goes dilution I hope this is clear I hope this is clear
right I hope this is clear now my dear students when you talk about the weak electrolytes when you talk about the weak electrolytes imagine that in this container you have got a weak electrolyte imagine that you have got one mole of V electrolyte you have got one mole of a v electroly in a given volume of solution okay right imagine that you have got one mole of V electroly in a given volume of solution so whatever conductance this particular solution will be showing what you call as M conductivity right now if there is weak electrolyte
in the container what will happen to the dilution let's check that to let's check that too for example there's a weak electrolyte in the container for example there is weak electrolyte in the container if there is weak electrolyte in the container tell me about weak electrolyte weak electrolyte gets partially ionized that does not get 100% dissociated into its ions now my dear students with dilution with dilution if you remember with dilution degree of dissociation degree of disso of a weak electrolyte what happens to it with dilution the degree of dissociation of a weak electrolyte that
increases correct that increases with dilution degree of dissociation of a weak electrolyte that increases as for o Falls dilution law right now if the degree of disor of a weak electrolyte increases can I say number of ions number of ions in the solution they're increasing if number of ions in the solution are increasing I directly say molar conductivity of the solution increases as well as equivalent conductivity of the solution increases so mol conductivity as well as equivalent conductivity of the weak liic solution that also increase with dilution but the reason is different reason here is
as the dilution is done degree of disor of the weak electroly increases more ions will enter into the solution more the ions more will be the M conductivity as well as equalent conductivity right and Molar conductivity at infinite dilution molar conductivity at infinite dilution that's what you call us Lambda n similarly equivalent conductivity at infinite dilution that is Lambda EQ now my dear students if you try to draw a graph if you try to draw the graph between Lambda m versus root C for a weak electrolyte for a weak electrolyte the graph comes like this
the graph comes like this it's a hyperbola right the graph comes like this right the graph comes like this now tell me one one thing at this particular Point root C value is z that means C is Z when is c0 I told you who back when is c0 C is zero when volume of the solution is infinite that means this particular point it represents infinite dilution it represents infinite dilution since I told you with dilution since I told you with dilution the degree of dissolution keeps on increasing right if you continous keep on doing
the dilution can I say there will be a time when the weak electrolyte would have got 100% ionized and that point when the weak electrolyte would have got 100% ionized that dilution I call as infinite dilution right if this point represents infinite dilution so I would say at this particular point of time the electrolyte would have got 100% ionized the electrolyte would have got 100% converted into its ions so at infinite dilution there will be maximum ions in the container so molar conductivity as well as equivalent conductivity would have attained its maximum value which is
what you call us molar conductivity or equivalent conductivity at infinite dilution right perfect that's why the graph started somewhere here now my dear students as you go in the right direction root C is root C is increasing if root C is increasing that means C is increasing if C is increasing if C is increasing if C is increasing that means volume is decreasing if volume is decreasing if volume is decreasing if volume is decreasing volume of the solution is decreasing what does that mean what does that mean that means the degree of ionization of the
V electrolyte what is happening to that or let me make it simple I told you when you do the dilution of the V electrolyte number of ions in the container increases right due to dilution volume of the solution increases so in short I'll say when the volume of the solution keeps on increasing with dilution number of ions in the solution keeps on increasing right now the volume of the solution is decreasing in this direction volume of the solution is decreasing if volume of the solution is decreasing what does that mean that means number of ions
in the solution will be decreasing if number of ions in the solution are decreasing I'll say m conductivity as well as equivalent conductivity they'll be decreasing also perfect but from this graph are you able to find the M conductivity at infinite dilution see from this particular graph you were able to find the Moler conductivity at infinite dilution with the help of this intercept but in this particular curve do you have any intercept there is no intercept there is no intercept so from this particular curve which is valid for a v electrolyte from this particular curve
you cannot calculate the molar conductivity at infinite dilution for a weak electrolyte but for strong electrolytes we can do with the help of the graph with the help of its intercept now the point is how to calculate the molar conductivity and equivalent conductivity at at infinite dilution for a weak electrolyte if it cannot happen through this particular graph if you are unable to get the molar conductivity at infinite dilution for a v electroly from the graph if you're unable to do that then how do we get the Moler conductivity at infinite dilution for a v
electroly my de students in order to get the molar conductivity at infinite dilution as well as equalent conductivity at infinite dilution for a weak electrolyte in order to get those values I'm going to introduce one more topic that is what you call as cohol Ros law that's what you call as cohol Ros law Co Ross Law so what this C Ross Law exactly States see you can read its definition this law states that the equivalent conductivity of any electrolyte at infinite dilution is the sum of equivalent ionic conductivities of cation and anion given by the
electrolytes at infinite dilution what it means exactly what it means exactly what it means let's try to understand first of all Whenever there is infinite dilution at infinite dilution 100% electrolyte is ionized at infinite dilution the distance between the oppositely charged ions in the solution that is infinite force of attraction between the oppositely charged ion that is zero so that that means I can say at infinite dilution ions they move freely right they move freely so at infinite d the ions move freely this is something which understood now my dear students try to understand try
to understand what exactly I'm going to say try to understand what exactly I'm going to say see for example in the conductivity cell let's say you have got an electrolyte like n you have got an electrolyte like n perfect like n first of all how this NAC would have got dissociated it would have got dissociated as na positive plus CL soent one it is one there it is one perfect now Co R La says that molar conductivity of this NAC for example at infinite dilution is the sum of is the sum of conductivities of its
ions basically so basically since this electrolyte at infinite dilution it's 100% ionized right number one it's 100% ionized there is no force of attraction between the ions ions are moving freely ions are moving freely so it is the ions basically which contribute it is the ions basically which contribute towards the molar conductivity of the electrolyte at infinite dilution so molar conductivity of the electrolyte at infinite dilution as for Co Ross Law is the sum of those contributions is the sum of those contributions which are made by the ions what it means have a look M
conductivity of n at infinite dilution see how do you write it as for your gold the stri is one write one first then write mol conductivity at infinite dilution or na positive plus this is one also right molar conductivity at infinite dilution of Cl Nega correct for example for example you have got electrolyte like this you have got for example B2 at infite dilution B2 first of all it would have got completely dissociated into its ions right and at the same time the force of attraction between the oppositely charged ions at infinite diretion will be
zero right so these ions will be moving freely now as per Co R law Lambda not M of B2 Lambda M or B2 will be equal the stri is 1 so 1 * Lambda M of b d positive Plus+ 2 * by Lambda M of Cl Nega correct perfect for example you have got electrolyte like this al2 so4 whole th how it would have got dissociated at infinite ution it is 2 * Al Tri positive plus 3 * s so4 d negative now if you want to write the molar conductivity at infinite ution for al24
whole th you can write it like this start with two that right molar conductivity at infinite dilution for Al Tri positive plus 3 * the mol conductivity at infinite dilution for s so4 i perfect can you write these statements for every electrolyte you should be able to write these particular statements correct you should be able to write these particular statements for the electrolytes now my dear students let me tell you one more thing since you are getting the Moler conductivities of electrolytes at infinite dilution like this if I ask you how do you get the
equivalent conductivity of the electrolyte at infinite dilution already if you remember if you remember I've already discussed one relation with you equivalent conductivity is equal equivalent conductivity is equal m conductivity ID n factor of the electrolyte we have discussed this relation perfect so if you get molar conductivity at infinite dilution of the electrolytes once you get the M conductivities you can easily calc their equivalent conductivities at infinite dilution equalent conductivity at infinite dilution will be equal m conductivity of the electrolyte at infinite ution divide by n factor of this particular electrolyte this particular electrolyte this
particular electrolyte I hope this is clear I hope this particular statement is clear right I hope this particular statement is clear perfect perfect guys one more thing one more thing for example for example I'm writing a statement like this let's say I already know equivalent conductivity at infinite dilution of Na positive let's say this is something I know I already know equivalent conductivity of Cl negative at infinite dilution in equivalent conductivities right I already know equivalent conductivity at infinite dilution of B positive B positive these are the terms which I already know correct now if
you want to write directly the equivalent conductivity of Na equivalent conductivity n when you directly want to write equivalent conductivity of NAC in terms of equivalent conductivity at infinite dilution for their ions in terms of their ions listen to me carefully imagine equivalent conductivity at infinite d solution for na positive you know this one you know this one you know right now I want you directly I want you directly to write the equivalent conductivity of at infinite dilution for n right that to in terms of the equivalent conductivities of their ions that to in terms
of their equivalent conductivities of their ions at infinite ution you do not have to you do not have to talk about these stric coefficients at that particular point of time you can directly write Lambda EQ of Na positive plus Lambda EQ of Cl Nega perfect for example here I was writing the M conductivity of B2 I was taking into consideration these tetric coefficients these tetric qus but if you are directly writing the equivalent conductivity at infinite dilution in terms of the equivalent conductivities of their ions then you do not have to talk in terms of
you do not have to write the stric coefficients here for example example let me make it clear let's say I'm writing equivalent conductivity at infinite dilution of B A2 of B cl2 if you want to write it in terms of equivalent conductivities of their ions you do not have to you have got nothing to do with the stric coefficient you'll directly write equivalent conductivity of b a d positive plus equalent conductivity of of Cl Nega nothing to do with their stric coefficients but if you are writing motor conductivity at infinite dilution for B2 right you
will be writing Lambda M of b a d positive plus 2 * Lambda M of Cl Nega I hope these two statements are clear to you I hope these two statements are clear to you yeah I hope these two statements are clear to you perfect guys and at the same time if you directly know the mol conductivity at infinite dilution of an electrolyte divide it with the N factor of electrolyte you will get directly the equivalent conductivity of the same electrolyte at infinite dilution one more important thing all these things are very very important guys
all these things are very very important these are very small small things wherein majority of the students they do the mistake okay perfect now let's exactly get to know how this Cal Ross SL is I mean how this is used basically where all do we use this C Ross Law number one number one Co Ross Law exactly is used to calculate the limiting molar conductivity of a v electroly limiting M conductivity of a weak electrolyte means the Ross Law is used to calculate the molar conductivity of a weak electrolyte at infinite dilution because from the
graph we could not get the M conductivity at infinite dilution of a Vite so it is the cold R law which is going to give us which is going to give us the mol conductivity at infinite dilution of a v electroly this is the first application to calculate to calculate you can write it as like this to calculate the limiting M conductivity to calculate the limit M conductivity of a v electrolyte this is the first application this is the first application of the c r law to calculate the limiting M conductivity of a veh electrolyte
to calculate the M conductivity at infinite dilution of the veh electrolyte okay now how exactly how exactly it's simple guys have a look understand properly for example for example mol conductivity at infinite dilution for HCL is given to us okay NaCl is also given to us us ch3 Co na is also given to us these are few parameters which are given to me and by looking at these parameters we have to calculate the molar conductivity at infinite dilution for ch3 Co correct how exactly I'll get this particular term from these terms Now understand simple thing
it is understand it properly so first of all Lambda not M of HCL I can write it as Lambda M of h postive plus Lambda not M of Cl correct because toet is 1 is to one and this value is already given to me as 426 7 cm squ per mole Lambda M of NAC I can write it as Lambda M of Na positive plus Lambda M of Cl NE and this particular term is given to me as 126 Lambda M of ch3 Co it is going to be equal Lambda M of ch3 Co plus
Lambda M of Na positive and this value is equal 91 so how many equations do I have how many equations do I have I have got three equations from these three equations I have to get this particular value Lambda M of ch3 Co first of all how do you break this ch3 Co I say basically I have to get this value Lambda M of I have to get basically the value of Lambda M of ch3 Co plus Lambda not M of H posi basically this is the value which I need to get this is the
value which I need to get perfect now my dear students how do you get this value how do you get this particular value out of these three equations this is equation one all this is equation two all this is equation three right how do you get it how do you exactly get it what do you need exactly you need Lambda not ch3 Co negative and Lambda not H positive so you need this term and you need this particular term perfect rest all the terms should get canceled rest all the terms should get cancel perfect so
I'll be adding equation 1 with equation 3 and I'll be subtracting equation two I'll be adding one with three and subtracting two at the end perfect if I subtract two this becomes Min this becomes Min this becomes Min perfect what do I get at the end I'll say this particular term this particular gets cancel this particular term this particular term gets canceled what do I get at the end Lambda not ph3 NE plus Lambda after doing this particular operation after doing this particular operation what am I left with I'm left with Lambda of H positive
plus Lambda of ch3 Co Nega perfect Lambda H posit plus Lambda ch3 Co negative what do I call this as I'll be calling this particular term as Lambda M of ch3 Co perfect how did I get this by doing this particular operation same operation you are going to do with these molar conductivities so 1 + 3 - 2 so 1 + 3 - 2 so it has to be 426 1 + 3- 2 whatever value you get from here that is going to give you the M conductivity at infinite dilution for this particular vehic electroly
did you get it did you get it properly let me know once in the chats quickly guys quickly everyone did you get this particular term let me give you one more question let me give you one more question let me see if you can solve this or not the question is like this I'm giving you Lambda M of NaCl this value I'm giving you as for example 126.5 s cm squ per mole this is the value which I give you second value I'm giving you as Lambda not M of HCL this is something which I'm
giving you again this value is something 425.31 Simon cm squ per mole and I'm asking you to calculate Lambda M for hi this is something which you are supposed to calculate can you give it a try can you give it a try my dear students same same thing you have to do same thing you have to do first of all Lambda not M of NCL can be written like this Lambda M of Na posit plus Lambda M of Cl Nega this particular term I'll be writing as Lambda M of H POS plus Lambda M of
Cl this particular term I'll be writing as Lambda M of Na POS plus Lambda M of I this particular term I'll be writing as Lambda M of H POS plus Lambda M of I this is something which to be calculated let's call this equation one let's call this as two let's call this as three from these equations from these equations from these equations in total there are 1 2 3 4 5 six six terms out of six terms out of six terms from these six terms I should get only Lambda not H posi plus I
Nega so there should be only H positive and there should be I negative rest all the things rest all the things should be cancelled out so what I should be doing I'll be just doing one thing I'll do equation 2 plus equation 3 minus equation 1 let's see what I'll be getting I'm subtracting equation number one I'm subtracting equation number one so Lambda not na positive Lambda na postive is cancel Lambda clga Lambda C Nega is cancelled so what I'm left with I'm left with Lambda H POS plus Lambda I so I got Lambda H
positive plus Lambda M of I ne this is something which I call as Lambda M of Lambda M of of hi now how did I get this basically how did I get this by adding second with third and subtracting first adding second with third and subtracting first do the same operation with their mol conductivity values so 2 + 3 so this plus this minus this so the answer has to be 425.31 - 126.5 Solve It Whatever value you get answer is going to be Simon cm squ per mole this is the molar conductivity at infinite
dilution for this particular electrolyte hi I hope this is again clear to everyone so this was the first application of the K Ros law right to get the limiting M conductivity to get the mol conductivity at infinite dilution for a weak electrolyte correct now what about its second what about its second application it second application is calculation of calculation of degree of dissociation for a weak electrolyte calculation of degree of dissociation for a weak electrolyte this is one more application part of C Ros law this one more application part of C Ros La my dear
students understand degree of dissociation degree of I'm directly going to give you the result I'm not going to derive it right I'll give you the result and we'll try to apply it in the questions okay directly I'm giving you the result degree of dissociation of a weak electrolyte as per Co Ross Law is a is a molar conductivity of a weak electrolyte at a given concentration divided by molar conductivity of the same weak electrolyte at infinite dilution at infinite dation what is the given concentration what is infinite divion St you know it already I'm directly
giving you the result I'm not deriving it again and again right see guys for example you have got weak electrolyte like this ch3 Co let's assume I got 0.1 m ch3 co perfect when the concentration of ch3 Co is 0.1 mol let's assume its molar conductivity at that point of time at that point of time for example to make you understand let's say it's 1 7 cm squ per Mo perfect now my dear students let's assume let's assume you already know the M conductivity of ch3 Co at infinite dilution let's say it's 100 cm squ
per mole perfect this particular term I'll be calling as molar conductivity at a given concentration this is something which you call as molar conductivity at infinite dition so M conductivity at given concentration divide by infinite dilution so this divide by this will give you the value of alpha for ch3 Co it will give you the value of alpha for ch3c remember it directly remember it directly I hope you got to know what is molar conductivity of a weak electrolyte at a given concentration I hope you got to know that now where do we use it
that is the point my dear students this Alpha this Alpha you can talk about it's one more application say for example you have got a weak electrolyte let's say you have got a weak electrolyte AB let's say you have got a weak electrolyte AB perfect if it is a weak electrolyte it will definitely be in equilibrium with its SS its signs are a positive Aquis plus b negative Aquis correct guys let's assume its degree of disso is Alpha and initially at time T is equal to Z its concentration is C its concentration is C I'm
assuming that the ionization constant of this V electrolyte is Ki KI represents ionization constant of the V electrolyte my dear students we can calculate we can calculate ionization constant of a weak electroly in terms of in terms of Lambda M and Lambda M we can do that how exactly they this is C this has to be zero initially this has to be zero now at equilibrium this has to be cus C Alpha this has to be C Alpha this has to be C Alpha if I want to write the expression for ionization constant of this
weak electroly expression will be simply C Alpha squid 1 - Alpha right it is C Alpha squ divid 1- Alpha now my dear students already Alpha you know Alpha you know Alpha is equal to what Alpha is conductivity of the weak electrolyte at a given concentration divided by m conductivity of the same weak electrolyte at infinite dilution so this is C Alpha is M conductivity of the same Vite at a given concentration divide by at infinite dilution what ra Power Ra power 2 this is C Alpha Square C Alpha Square divided by 1 - Alpha
1 - Alpha now when you solve it you get the ionization constant of the vle electrolyte like this see you are going to write as such this is molar conductivity at a given concentration of square right the square divided by divided by divided by it's going to be Lambda M multiplied by Lambda M minus Lambda m c correct this is how you get the ionization constant this is how you get the ionization constant of the weak electrolyte in terms of Lambda M and Lambda MC perfect or in short if you will be given with a
we electroly just calculate its Alpha first of all in terms of these two values if you got its Alpha value if you got Alpha put it here in this expression get the K value no need to get this one as well no need to write it no need to remember it perfect right guys is it clear is it clear is it clear let me know once in the chats let's try to do few questions on the same concept then only you'll try to understand where it is used exactly my dear students I'm going to write
a question I'm going to write a question the question is like this just a second the question is molar conductivity at infinite dilution of ch3 Co it is given to me as 39.7 Simon cm squ per mole this is the value which is given to me at the same time when the concentration of ch3 Co is kept 0 0.1 mol when the concentration sorry 0.01 mol this is when the concentration of ch3 Co is kept 0.01 Mol at that point of time at that point of time it's molar conductivity so it is mol conductivity at
a given concentration it is M conductivity at a given concentration for ch3 Co this particular value is given to me as it is given to me as 3.97 Simon cm squ per mole what am I supposed to calculate I'm supposed to calculate degree of dorion of ch3 Co and at the same time I'm supposed to calculate of solution these are the two things which I'm supposed to calculate can we give it a try can we give it a try my dear students the weakly that's given to me as ch3 Co its molar conductivity at infinite
dilution is given its mol conductivity at 0.01 mol concentration is also given we have to calculate Alpha as well as pH now what is Alpha basically Alpha is nothing it is M conductivity of ch3 Co at a given concentration divided by m conductivity of ch3 Co at infinite dilution this is equal this particular term it is given to me as 3.97 and this particular term it is 39.7 the value comes out be 10^ minus 2 so you got the alpha value you already got the alpha value perfect now second second thing we have to calculate
pH as well now tell me this ch3 Co how it would have undergone dissociation it would have undergone dision like this ch3 Co plus plus h positive correct ch3 Co plus h positive right people now initially I'm keeping its concentration is C which is equal to 0.01 mol this is 0 this is 0 now tell me at equilibrium this is going to be cus C Alpha this will be C Alpha this will be C Alpha so if I ask you what about the concentration of H positive here concentration of H postive in the solution will
be C Alpha C value already you know that is 0.01 Alpha value you got that is 0.01 so the value comes out be 10 rais power this is 10^ min-2 C Alpha is 10-2 so this 10- 4 so you got the H positive concentration if you got the H positive concentration can't you calculate pH pH is nothing that's minus log of H positive concentration so it isus log of 10us 4 so the pH of the final solution comes out to be four isn't this clear isn't this clear to everyone people yeah perfect perfect now one
last liation of the co Ross Law and with that we'll end the session okay guys is it clear to here let me know in the chats quickly the last application part of the co R law what is that that is calculation of solubility and solubility product for a sparingly soluble salt I hope you have studied ionic equilibrium which I did a few days back only on this channel right first of all sparingly soluble salt is the one one whose solubility is very very very less you know that right fly soluble salt is the one whose
solubility is very very very less so basically whenever I talk about the sparingly soluble salt it means the salt with very less solubility the salt with very less solubility so imagine that you have got a solution in which you have got sparingly soluble salt that sparingly soluble salt is very less soluble what does that mean that means the concentration of the salt in the solution will be very very very less right if you have got a sparingly soluble salt in the container for example in the solution for example that means the solubility of the sparingly
soluble salt is very less correct what does that mean that means the concentration of the salt in the solution will be very less concentration of salt in the solution is very less means concentration of the salt is approaching towards zero you have got a sparingly soluble salt my dear students sparingly soluble salt its solubility is very less right so its concentration in the solution will be very very very less so I would say concentration of the sparingly soluble salt in the solution will be approaching towards zero perfect right now molar conductivity of the sparingly soluable
salt I'm representing with Lambda M I'm representing with Lambda m do you remember when the concentration is zero when the concentration when the concentration exactly is zero at infinite dilution concentration is zero at infinite dilution if you remember at infinite dilution root C is z that means C is z so concentration is equal to Z represents infinite dilution concentration concentration is equal to zero represents what infinite dilution so can I say molar conductivity of a sparingly soluble salt will be taken equal to the m conductivity of it at infinite isolution both the things will be
taken same did you get this particular Point did you get this particular point so I believe this particular point is clear to everyone let me know in the chats if this particular point is clear this is important guys sparingly soluble sort it solubility is very very very less it's con conentration in the solution will be approaching towards zero so I'll say m conductivity of a sparingly soluble salt is same as that of its mol conductivity at infinite dilution right because at infinite dilution C value is zero now let me tell you there's one very important
result I'm not deriving it I'm just giving you the result we'll show its application molar conductivity at infinite dilution of a sparingly soluble salt it is always equal conductivity ulti th000 divided by solubility of the sply soluble salt this is one result this is one result which you have to remember this is one result which you have to remember and its units are simply going to be Simon cm square per mole whenever you will have a sparingly soluble salt sparingly soluble salt it's m conductivity at infinite dilution is always equal Kaa multi th000 solubility of
the sparingly Sol I'll show you its application we'll get to know how it exactly Works where do we use all this but before that my dear students do you remember your KSP solubility product how do you get the solubility product imagine that imagine that you have got the electrolyte in this format ax B Y imagine how it will undergo dissociation first of all this Y is going to be y positive here this x is going to be x negative here now we have got how many A's here X so this is X a + y
+ y b - x imagine it solubility is s so its concentration of a + y will be x * s this will be y * s something which we have discussed in ionic equilibrium if you remember if I want to write the expression for KSP I'll start with the product it's going to be concentration of a + YC coefficient concentration of B minus X rais for tetric coefficient so KSP expression has to be equal concentration of a + y That's x * s ra^ x this is y * s ra^ y you can write
the expression like this KSP is equal this is x^ x * y power y this is s^ x + s ra^ x + y so this is how you get the KSP expression correct this is how you get the KSP expression perfect now my dear students try to understand what one simple thing first of all if you will be given with a sparingly soluble salt let's say you got you are given with a sparingly soluble salt you are given with its molar conductivity at infinite dilution you are given with its conductivity as well so easily
you can calculate solubility right once you get the solubility you can easily write the expression for KSP right so whenever you are supposed to calculate KSP so first of all you have to calculate the solubility solubility you are going to calculate from the expression which I gave you right from that expression you'll get solubility put it here in this expression get the KSP of the sparingly soluble salt for example I'll be giving you certain questions let me see if you can solve those questions or not one two questions we'll do and we are done with
the chapter and then we are done with the chapter okay the first question which I'm writing on the screen and I would want you guys to solve it I would want you guys to solve it the question is like this the question is the specific conductance the specific conductance of a saturated solution of a saturated solution of agcl is equal it is given to me as 2.28 into 10 ra^ - 6 - 6 right Min - 6 perfect the equation is find find the KP find the KP of agcl find the KP of agcl if
Lambda M if Lambda M of agcl is equal 138.3mm squ per Mo look at the question carefully and let me know if you can solve this question or not what do you think what do you think guys quickly what do you think KP we have to calculate KP we have to calculate I'll just show you the procedure of solving this question well my dear students conductivity Kaa value it is given to us conductivity of agcl is given to us how much is that I'm just writing the value 2.2 we into 10^ - 6 right at
the same time molar conductivity at infinite dilution for this saturated solution of agcl this is 138.3mm squ per mole what am I supposed to calculate KSP I told you in order to get the KP we have to get the S value how do we get the S 2 minutes back only I give you a relation M conductivity at infinite dilution of this agcl can I write it as Kaa * 1,000 divide solubility perfect this value is given to me this value is given to me so from here you can easily calculate solubility you can easily
calculate solubility of agcl now my dear students tell me just one thing tell me just one thing this agcl how it would have undergone dissociation AG positive plus CL Nega if it solubility is s this will be S this will be S correct can you write the KSP expression here KSP is going to be concentration of AG posi power 1 concentration of c^ 1 so that's going to be S square at the end s value you will calculate from here and put the value of s here and S Square will give you the value of
KSP that understood this is the approach of solving the equation then calculation part I'm sure you can do on your own right I'm sure you can do that clear guys one more question I'm writing let me see if you can solve this or not the question is like this conductivity of agbr conductivity is of agbr it's given to me as 8.5 into 10^ minus 7 right it is given to me as 8.5 into 10- 7 can you tell me the unit of conductivity first of all Kaa 1 by R into L by a 1 by
R is conductance so Simon this is 7 cm inverse right perfect so conductivity of agbr is given at the same time at the same time some other parameters are also given molar conductivity at infinite dilution of AG positive that's also given to me how much that is 62 similarly molar conductivity at infinite dilution of of BR negative that's given to me a 78 and units already you know this is Simon cm square per mole and this will be S cm squ per mole 2 what are we supposed to calculate we are supposed to calculate solubility
and solubility product calculate solubility and solubility product of what of agbr of agbr can you give it a try can give it a try can you give it a try quickly everyone my dear students it's a very simple question first of all you have to calculate solubility right you have to calculate solubility so solubility to calculate solubility you can write molar conductivity at infinite dilution of agbr right how much that is as per the formula it's going to be Kaa MTI 1000 ided by solubility correct this is the expression which you have to use somehow
but as for the question Lambda not M of agbr is not given to me m conductivity at infinite dilution for agbr this is not given to me but I can use Co Ross SL I can say m conductivity at infinite ution of agbr it is basically molar conductivity of at infinite dilution of AG positive plus molar conductivity at infinite dilution of BR negative perfect I would say m conductivity at infinite dilution for agbr it has to be equal this particular value is given 62 this is 78 so how much the value comes out to be
62 + 78 that comes out to be 140 right Simon cm squ per mole so you got the mol conductivity at infinite dilution of what of agbr so this particular term you got this particular term you got conductivity of agbr is given so from this particular expression you can easily calculate solubility you can calculate solubility right perfect once you get the solubility of agbr you're done with one part of the question the second part says calculate KSP how do you write the KSP expression for agbr so first of all if I talk about agbr agbr
it would have got converted into AG positive plus BR negative if it is solubility is s this is also going to be S this is also going to be S so KSP expression is simply going to be what it's going to be s squ s you would have calculated from here put the value of s here and get the KSP right clear perfect so you can solve this question in this particular approach perfect so my dear students with this our one more chapter is done and dusted that is electrochemistry right right and the next chapter
do let me know in the comment section I want you guys to let me know in the comment section Which chapter do you want the next right that you are going to tell me in the comment section I hope your complete electrochemistry chapter is completely clear now please and please do watch this particular session again right because all the terminologies which we discussed at the end you might get confused afterwards right so do this particular session again once again at once .5x speed you can watch it so that all the all the important things all
the concepts would be in your mind all the time okay and now you can solve whatever module you have whatever study material whatever book you are solving right I'm guaranteeing you if you have would have followed these four sessions of electrochemistry I believe you can solve any coaching material whatever you have because I've not skipped a single point here okay perfect so with this I'll be take a leave I'll see you guys in the next session do let me know in the comment section Which chapter you want the next unacademy let's crack it
