[clears throat] Hey guys, good evening. Good evening and welcome back again to your anac academy English channel. I hope all of you doing great, having a good time. So people quickly 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 quickly in the chats. Everybody let me know quickly in the chats if all of you can hear me. If I am perfectly audible, visible to every one of you. Yes. Good evening people. Good evening and welcome back. Good evening and
welcome back. So what's up? How are you all doing? All well I'm seeing you after long I think 10 12 days. Yeah. All right. So as you all must be knowing today we are going to do one of the most amazing chapters of your chemistry right and the name of the chapter is nothing but the electrochemistry from which every year almost like three to four questions are asked okay and as you know I start every single session every single chapter from the basics from the scratch. So today also we shall be starting this particular electrochemistry
chapter from the basics from the scratch. If you have not studied the chapter before does not matter. If you have studied the chapter before does not matter. Right? This particular session is going to be for everyone. All right? Whosoever is preparing for NE 2026 or whosoever is preparing for even J 2026. You guys are most welcome in the today's session. Yeah. Uh people have started asking about the duration. Well, the duration of this particular session is going to be 11 12 hours because it's a big chapter. You know it already. It's a big it's one
of the biggest chapters of your chemistry. So the duration is going to be definitely some 11 12 hours for sure. But in between I'll be keeping on giving you certain breaks as well so that you can refresh yourself. Yeah. [clears throat] All right. All right. All right. So can you let me know in the chats from which part of the country you guys are watching me from right now? I just want to know from which state you guys are from. I just want to know that first. I just want to know that first. Quick. Everybody
in the chats. Everybody in the chats. Quickly people. Quickly. Everybody in the chats. I just want to know from which part of the country you guys are watching me from. Tamil Nadu, Telangana, Maharashtra, Delhi, AP, Hana, Kashmir, somebody is my Hamsai, Karnataka. Okay. All right. All right. Perfect. It's perfect. So, are you ready to be with me till the end? Yeah. Because as I keep on saying like after one two hours people like their jo fades down right. So let me know in the chats first of all are you guys ready to be with me
till the end. I just want to know that first. I just want to know that first. Quick, quick, quick, quick. And guys, tomorrow is Sunday. Yeah. So, tomorrow you can sleep properly because tonight it's going to be a sleepless night. Okay. Tomorrow since it's Sunday, so you guys can sleep till 11:12. Yeah. All right, then. So, let's get going. Let's get started with one of the most amazing chapters of your chemistry. My dear students, you know the name of this particular chapter is your electrochemistry. Okay, electrochemistry. As I promised already, I'll be starting from the
basics from the scratch. So definitely I'll be starting exactly from the basics. Exactly from the scratch. Right. And at the same time, the handwritten PDF of this particular session, the handwritten PDF of this particular session, I'll be sharing on my telegram. If you are not the part of my telegram yet, be the part of the telegram as soon as possible. The name of the telegram channel is W was chemistry official. W A S S I M P H A T Chemistry official is the name of the telegram channel on which I shall on which I
shall uh share the handwritten PDF my dear students. Okay. All right. Should we start? Yes. Yes. Masu you can also watch it. All right guys. So before starting this particular electrochemistry chapter there are few things there are few things few basic things which I would want to share with you. First of all and what are those basic things exactly? The first basic thing my dear students which all of you must be familiar with what is that? That is called as oxidation. That's called as oxidation. One of the basic stuffs from which I'm starting the chapter.
And how do we exactly define the term oxidation? My dear students, oxidation is nothing but it is defined as the loss of electrons. Oxidation is defined as the loss of electrons. Whenever you see an element losing electrons, whenever you see an element losing electrons, understand that element is undergoing oxidation at that point of time. So whenever an element would be losing the electrons since loss of electrons is called as oxidation that means that particular element which is losing the electrons that particular element is undergoing oxidation. As simple as that. So loss of electrons is what
we call as oxidation. Similarly my dear students if you ask me what is reduction exactly? If you ask me what is reduction exactly? The simplest answer that I'm going to give you reduction involves reduction involves the gain of electrons. So whenever you see any species whenever you see any specy gaining the electrons whenever you see any species gaining the electrons understand that particular species would be undergoing reduction. So in short in simple words loss of electrons is what we call as oxidation. Gain of electrons is what we call as the reduction. Okay. For example, my
dear students, imagine that you are taking let's say an element X. Let's say you are taking an element X. Okay. Let's assume that this X loses one electron and is getting converted into X positive. Let's say this X loses one electron and gets converted into X positive. So I would say in short this X is losing the electron and that particular specy which loses the electrons we say that particular specy under goes oxidation or you can see just one more thing if I ask you what is the charge on this X here you'll say zero
if I ask you what is the charge on this X here your answer will be + one so if I ask you is the oxidation state of X increasing or decreasing I would say the oxidation state of X is increasing inreasing from 0 to + one. Remember my dear students increase in the oxidation increase in the oxidation state increase in the oxidation state of an element is also termed as oxidation. So whenever you see an element whose oxidation state would be increasing remember that particular element would be undergoing oxidation. Okay. Whenever you see an element
whose oxidation state would be increasing remember that particular element would be undergoing oxidation and that would be definitely losing the electrons. Okay. In the similar way my dear students for example I'm taking an element Y let's say it's gaining one electron and after gaining Y and one electron it gets converted into Y negative. Okay. So first of all gain of electrons is what you call as reduction. So this Y is undergoing reduction here. Or in other words, if you want to see if I ask you what is the charge on this Y here, your answer
will be zero. What is the charge on this Y here? The answer will be minus one. So is the oxidation state of Y increasing or decreasing? I would say the oxidation state of Y is decreasing. Right? And let me tell you my dear students, decrease in the oxidation state. Decrease in the oxidation state of an element. Let me write it completely. Decrease in the oxidation state of an element is termed as what you call it as the reduction. You call it as the reduction. So two simple terminologies which I would want every one of you
to remember from now onwards. Point number one whenever you see an element undergoing increase in the oxidation state remember that element is undergoing oxidation that means that element would be losing the electrons. And whenever you see an element whose oxidation state would be decreasing remember decrease in the oxidation state of an element is called as reduction. Okay. And that element which under goes reduction that will be definitely gaining the electrons. Yeah. So in short increase in the oxidation state is termed as oxidation. Decrease in the oxidation state is what we call as reduction. Okay. Now
my dear students two more basic terminologies which you must be familiar with. What are those? There's a term called as oxidizing agent. There's a term called as oxidizing agent or you can call it as oxidant as well. There is a term called as oxidizing agent or you can call it as the oxidant as well. First of all, how do we define this term and how do we identify it? That is the point here. Have a look people. Let me tell you oxidizing agent is the specy. Oxidizing agent is the specy that oxidizes others that oxidizes
others but itself under goes reduction but itself under goes a reduction. That particular species which oxidizes others but itself under goes reduction. What do I call that as oxidizing agent? What is the meaning of this? You'll get the idea in some time. Before understanding you, before making you understand what oxidizing agent exactly is, before making you understand that, I'm going to write the definition of reducing agent first. I'm going to write the definition of reducing agent first. Reducing agent or you can call it as the reductant as well. Now, what is the reducing agent? What
is the reductant? I would say that particular species. I would say the specy that reduces others that reduces others but itself under goes oxidation but itself under goes oxidation. that particular species which reduces others but itself under goes oxidation we call that as the reducing agent. Now let's try to understand this oxidizing and reducing agent a bit more in detail. Try to understand people what I'm going to say. For example, for example, I'm taking a general common reaction. The reaction is something like this. Zinc solid plus copper dipositive aqueous. Let's say it gives zinc dipos
to aus and for example it gives copper solid. This is the reaction for example. Okay. This is the reaction. Imagine that I need to check which one is oxidizing agent here and which one is reducing agent. Try to understand carefully what I'm going to say. First of all if I ask you what is the charge on the zinc here? Your answer will be zero. Charge on copper is plus2. Charge on zinc is +2. Charge on copper here is zero. Now if I ask you is the oxidation state of zinc increasing or decreasing 0 to +2
means increase in the oxidation state. Increase in the oxidation state is what you call as oxidation. So I would say the zinc is undergoing oxidation here. The zinc is undergoing oxidation here. So the zinc would be losing the electrons. If this zinc under goes oxidation here. So this zinc is basically losing the electrons. And as I told you a few minutes back that species which under go oxidation that is what you call as the reducing agent. That species which under goes oxidation that's what we call as the reducing agent. As simple as that. In the
similar way people is the oxidation state of copper increasing or decreasing plus 2 to zero decreasing. And you know decrease in the oxidation state is what we call as decrease in the oxidation state is what we call as reduction. Reduction involves what? Gain of electrons. So I would say this copper dip positive would be gaining the electrons. And my dear students that particular species which under goes reduction is what you call as the oxidizing agent. So my dear students this is the way by means of which we can check whether the species is the oxidizing
agent or the reducing agent. As simple as that. Okay. Now people let me take one more example which will make things a bit more clear to you. For example, I'm taking a reaction like this H2 gas plus copper dipositive aquis. Let's say it gives 2 * H positive aquis and for example here I'm writing copper solid. I'm taking one general reaction here. Okay. Now my dear students I have to check which one is oxidizing agent which one is reducing agent. Again I'm going to use the same process. If I ask you what is the oxidation
state of hydrogen here it is zero. Oxidation state of copper here is +2. Oxidation state of hydrogen here + one. Oxidation state of copper here that is zero. Okay. So is the oxidation state of hydrogen increasing or decreasing? Increasing. Increase in the oxidation state is what you call as oxidation. So I would say this H2 is undergoing oxidation. Therefore it must be the reducing agent plus 2 to 0. Decrease in the oxidation state. decrease in the oxidation state is what you call as reduction and that particular species which under goes reduction is nothing but that
is the oxidizing agent. So this is how you identify the oxidizing agent and the reducing agent in a particular reaction. If I'm clear, let me know once once in the chats quickly. If you understood the concept, let me know quickly in the chats. Yeah. So in short, increase in the oxidation state is called as oxidation. Decrease in the oxidation state is called as reduction. That species which underos oxidation is called as reducing agent. That species which under goes reduction is called as the oxidizing agent. As simple as that. Okay. Perfect guys. Let's move on and
let's talk about few more things. Okay. Let me take a reaction here like this. For example, I'm writing a reaction MO4 negative plus F A D positive. Let's say it gives M and D positive plus FI positive. This is for example a reaction which I've written and I want you guys to check I want you guys to check which one is the oxidizing agent and which one is the reducing agent. My dear students, what is the first thing you'll be doing? Understand? You'll be calculating the oxidation state of magnes here. So let the oxidation state
of magnes be x. Oxygen shows minus2 oxidation state. There are in total four oxygen atoms. The net charge here is minus1. So x value when you solve it will come out be + 7. So this magnes is basically in + 7 oxidation state. This iron here is in plus2 oxidation state. This magnes in plus2 oxidation state and this iron here is in plus three oxidation state. Now if you look exactly what is happening magnes + 7 to +2 decrease in the oxidation state. Decrease in the oxidation state is what you call as reduction. So I
would say this M4 negative here it is undergoing reduction and that particular species which underos reduction that is nothing but your oxidizing agent. Now similarly plus2 to plus 3 increase in the oxidation state. Increase in the oxidation state is what you call as oxidation. So this Fe dipulos is undergoing oxidation. So it must be your reducing agent. So my dear students, this is how you guys are going to identify your oxidizing agent and the reducing agent in a chemical reaction. I believe I'm clear. Okay, I believe I'm clear. Now let's move on. Let's move on
basically to the first topic of our chapter. What is that? That is your electrochemical cell. That is your electrochemical cell. How do we exactly define the electrochemical cell? My dear students, electrochemical cell is nothing but it is a device. It is basically a device that converts. It is the device that converts. I would say either chemical energy the device which converts either chemical energy into the electrical energy or vice versa. So that particular device, that particular device, that particular cell which converts either chemical energy into the electrical energy or electrical energy into the chemical energy.
You'll be calling that particular device as the electrochemical cell. And my dear students, this electrochemical cell is actually of two types. One is called as one is called as galvvinic or you call it as the voltage cell. This is the first type of your electrochemical cell, galvvinic or voltage cell. And the second one that we have to discuss, that's going to be your electrolytic cell. That's going to be your electrolytic cell. Now, what is the difference between the two cells? First of all, both the cells are basically both these cells are basically the types of
electrochemical cell. Electrochemical cell is nothing but electrochemical cell is nothing but it is the device that converts either chemical into electrical or electrical into chemical. Right? Now this electrochemical cell I'm classifying into two types. One is going to be your galvanic or you call it as a voltic cell and the second one is going to be your electrolytic cell. Okay. Now what is the difference between these two? My dear students small small differences which you are going to note right now. Small differences which you are going to note I mean note right now. And with
time with time I'm going to explain those things one by one when I'll be teaching these two cells in very much depth. Okay. So first few points I would want you guys to write. Okay. And when I'll be teaching these two cells in detail there you'll understand the exact meaning of the points. Okay. See guys galvvinic cell what does it do? It converts chemical energy into electrical energy. Galvinic cell it converts chemical energy into electrical energy. How exactly? We'll we'll study that in some time. Okay. Point number one. Point number two in case of galvvinic
cells spontaneous cell reactions takes place. In case of galvvinic cells when I'll be teaching you the galvvinic cells. What you will see? You will see the cell reactions will be happening on its own. Cell reactions will be happening on its own. Right? And that particular process which happens on its own. Right? What do you call that particular process? Spontaneous process. So basically in case of galvvinic cells when I'll be teaching you in some time right g in case of galvvinic cells you'll observe one thing. You'll observe the cell reactions will be taking place on its
own. Right? So basically the cell reactions are spontaneous. And that particular process which is spontaneous you must be knowing it's delta G for the system at constant pressure temperature that is negative. So in case of your galvvinic cells right wherein spontaneous cell reactions takes place their delta G is going to be negative. And my dear students at the same time remember two more things in case of galvvinic cells we are going to use two electrodes. One is called as anode and one is called as cathode. One is called as anode. One is called as cathode.
The rod of the anode will be carrying negative charge and the rod of the cathode will be carrying positive charge. How this negative comes? How this positive comes? In some time you are going to understand. Remember right now in case of your galvvinic cells you are going to use two electrodes anode and cathode. The rod of the anode which will be carrying the negative charge and the rod of the cathode will be carrying the positive charge. And one more thing at anode remember always oxidation takes place and at cathode always a reduction takes place. At
anode always oxidation takes place and at cathode always reduction takes place. Rest you already know oxidation involves loss of electrons reduction involves gain of electrons. Now what happens in electrolytic cell again I'll be writing few points for electrolytic cell as well and slowly slowly I'll be explaining them one by one here. But dear students, electrolytic cell, let me tell you, it converts electrical energy into the chemical energy. It converts electrical energy into the chemical energy. Let me tell you, cell reactions are carried out. Cell reactions are carried out with the help of the help of
with the help of a DC source with the help of DC source. Cell reactions are carried out with the help of a DC source which are otherwise non-spontaneous which are otherwise non-spontaneous. So basically in case of your electrolytic cells cell reactions are not going to happen on their own. We are going to make the cell reactions happen with the help of the external DC source. Right? So which so basically in case of electrolytic cells the cell reactions are non-spontaneous which are made spontaneous with the help of what which are made spontaneous with the help of
external DC source again how exactly we'll get to know when I'll be teaching you this electrolytic cell in detail again my dear students again if I ask you since the cell reactions are first of all non-spontaneous so I would say delta G will be positive and in the cell also I'll be using two electrodes one is anode and one is cathode but what What is the difference here? The rod of the anode here carries positive charge. The rod of the cathode here carries a negative charge. The rest all the things are same. At anode oxidation
is going to take place and at cathode reduction is going to take place. At anode oxidation is going to take place and at cathode reduction is going to take place. Okay. DC means direct current. DC means direct current like your battery. Battery is a DC source. Yeah, I believe things are clear till here. Okay. So, these are the two types of electrochemical cells. One is called as your galvvinic cell or voltage cell and the second one is called as electrolytic cell. Now, my dear students, the first topic that we are going to discuss here what
is that? That is our galvvinic cell. That is our galvvinic cell. Okay, that is our galic cell. That is our galic cell. And as I told you, what does the galic cell do? Galvinic cell converts chemical energy into the electrical energy. It converts chemical energy into what? Into the electrical energy. Well my dear students the most common example the typical example of the galvonic cell is what you call as the Daniel cell. Daniel cell let me tell you it is the most typical example of what Daniel cell is the most typical example of the galic
cell. So basically we have to study this Daniel cell in detail. We have to study this Daniel cell in detail. And my dear students, what about Daniel cell? Let me tell you in case of your Daniel cell, we are going to use two electrodes. One is going to be your zinc electrode. One is going to be your zinc electrode and one is going to be your copper electrode. In case of the Daniel cell, we are going to use two electrodes. One is your zinc and one is your copper. Now my dear students if you compare
zinc with copper if you compare zinc with copper let me tell you zinc is considered to be more electropositive than copper. Zinc is considered to be more electropositive than copper. So I would say this is considered to be less electropositive than zinc. Okay. The one which is more electropositive, the one which is more electropositive that would be having the tendency to lose electrons that would be having the tendency to lose electrons and the one that loses electrons that is basically undergoing oxidation and you know oxidation it takes place at anode. I would say in case
of your Daniel cell which is the most common example of the galvanic cell. In case of your Daniel cell, your zinc electrode is going to behave like the anode at which oxidation is going to take place. Okay. And similarly your copper electrode add copper electrode remember gain of electrons is going to take place and you know gain of electrons is what you call as a reduction and reduction and reduction takes place at cathode. So there are two things my dear students which you have to remember here. In case of your Daniel cell, you guys are
going to use two electrodes. Zinc electrode and copper electrode. Zinc electrode is going to behave like the anode and copper electrode is going to behave like the cathode. Rest you know at anode oxidation takes place, loss of electrons takes place. At cathode reduction takes place that means gain of electrons takes place. Now before making you understand how this Daniel cell works, you should first of all know how the zinc electrode and copper electrode is made. You should first of all know how the zinc electrode and copper electrode is made. So let me tell you that
first. Let me tell you that first. Let me tell you that first my dear students. First of all, let me just do one thing. Let me take a container. This is a container. And in this particular container, I'm going to take a salt of zinc. In this particular container, I'm going to take a salt of zinc. For example, in this container, imagine that I'm taking zinc sulfate. I'm taking zinc sulfate. It is the salt of zinc. And this zinc sulfate in this particular container, it would have dissociated as zinc dipositive and SO4 diggative. So I
would say in this particular container, in this particular solution, you'll find zinc dipositives and SO4 dgative. Okay. Now if I do one thing, if I introduce a zinc rod, if I introduce a zinc rod here in the same solution, I've introduced a zinc rod in the solution that contains zinc ions. I've introduced a rod I've introduced a rod in the solution. I've introduced a zinc rod in the solution that contains what? That contains zinc ions. Let me tell you this complete setup here. This complete setup. This complete setup zinc rod immersed in a solution containing
zinc ions. This complete setup I'll be calling as zinc electrode. I'll be calling it as the zinc electrode. Similarly, my dear students, if you would want to make a copper electrode, you can do that too. So, what you have to do is you'll be taking a container. In this particular container, you are going to take a salt of copper which is for example copper sulfate. You are taking a salt of copper which is basically your copper sulfate. So this copper sulfate in the solution it would have dissociated as copper dip positive plus SO4 D negative.
Okay. And my dear students in this particular solution you are introducing a copper rod. You're introducing a copper rod. So you have introduced a copper rod in a solution containing the copper ions. You have introduced a copper rod in a solution containing its own ions containing copper ions. So this setup also I'll be calling as the electrode. And which electrode this is? This is the copper electrode and this is the zinc electrode and this copper electrode and zinc electrode it is used in which cell in case of Daniel cell. So basically now we are going
to start studying the Daniel cell. Exactly. So you can mark the heading here as the Daniel cell. You can mark the heading here as the Daniel cell. Just mark the heading as the Daniel cell. What is the Daniel cell? First of all, it is the most typical example. It is the common example of what? It is the common example of galvvinic cell wherein chemical energy gets converted into electrical energy. Now, we are going to see how this chemical energy is going to get converted into electrical energy. My dear students, understand what I'll be doing. I
have taken I have taken zinc electrode and copper electrode. Let me connect these electrodes externally with the help of a voltmeter. Let me connect them externally and at the same time I am going to connect them internally as well with the help of an inverted U-ype tube. This is the inverted U-ype tube here. This is the inverted U-ype tube. This is the inverted U-Type tube. And my dear students, in this inverted U-Type tube, what exactly I'm going to make? I'm going to keep in this inverted U-Type tube, I'm going to take I'm going to keep
an inert electrolyte. What is the example of the inert electrolyte? You can take K2S04. You can take NH4, NOO3, right? Etc. In this inverted U-Type tube, you are going to take the inert electrolyte and you are going to mix it. You are going to mix it with gelatin or auragar. You are going to mix this inert electrolyte with what? With gelatin or auragar. And when you mix the inert electrolyte with gelatin or auragar, what do you get? You'll be getting a jelly- like paste. And that particular jelly like paste is introduced in this inverted utype
tube. So in this inverted utype tube, you have got a jelly- like paste here. You have got a jelly- like paste here. Okay. Now if I first of all ask you what all ions would be present in this inverted utype tube. Imagine that you have kept K2SO4 here. If you have kept K2SO4 here, that means in this inverted U type tube, there will be S SO4 D negative ions as well as K positive ions. And my dear students, the ends of this inverted U-Type tube, you are going to seal it with the cotton balls. The
ends of the inverted U-Type tube, you are going to seal it with cotton balls. Now this inverted utype tube that contains an inert electrolyte mixed with gelatin and we got a jelly like paste which is introduced here in this inverted utype tube. So let me tell you this inverted utype tube is nothing called as it is called nothing but a salt bridge. So this inverted utype tube here I will be calling it as the salt bridge. Okay I'll be calling it as the salt bridge. Now my dear students understand one thing. Understand one thing. As
I told you already, the zinc electrode, it behaves like the anode. The copper electrode, it behaves like the cathode. Rest of the things you already know. At anode, what happens? Oxidation. Oxidation involves loss of electrons. At cathode, what happens? A reduction. And reduction involves gain of electrons. So basically, so basically what happens exactly at anode oxidation is going to take place. At anode oxidation is going to take place. And what happens exactly? What happens exactly at anode? See this rod over here? It is made up of zinc atoms. And those zinc atoms of which a
rod is made up of those zinc atoms of which a rod is made up of. Those zinc atoms will start losing electrons. They'll start getting converted into zinc diposites and you'll be getting the two electrons here. Since if I talk about the zinc rod, the zinc rod is made up of zinc atoms. Those zinc atoms will undergo oxidation. They will get converted into zinc dipositives and two electrons. Let me tell you these two electrons they'll be accumulated on the rod itself and this zinc dipositive it goes into the solution. It goes into the solution due
to which the solution is going to get the positive charge. Earlier this solution was neutral. The solution was neutral because there were equal number of zinc dip positives and SO4 d negatives present. Now from the rod a zinc atom is getting oxidized getting converted into zinc dip positive and two electrons. Those two electrons are accumulated on the rod and that extra zinc dipositive goes into the solution due to which the solution gets positive and since electrons are accumulated on the rod I would say the rod is going to get the negative charge. The rod is
going to get the negative charge. So my dear students, as I had already told you, the rod of the anode, it carries what? It carries a negative charge. Now what happens at cathode? What happens at cathode? At cathode, you know what happens? Reduction. Reduction means what? Gain of electrons. So what happens at cathode? Basically these copper diposit ions, they are basically present in the solution. These copper diposit ions which are in the solution, they are going to collide with the rod. The copper dipos2 ions in the solution they are going to collide with the rod.
Okay, copper dipos2 ions in the solution they are going to collide with the rod and they will take two electrons from the rod. They will take two electrons from the rod and will get converted into copper solid and that copper solid will get deposited here on the rod. That copper solid will get deposited here on the rod. Okay, am I clear till here people? Quick everybody. So at anode oxidation takes place at cathode reduction takes place. Now my dear students the copper diposites which were there in the solution they collided with the rod they took
two electrons from the rod got converted into copper solid. Now understand one thing since the copper diposit in the solution they are colliding with the rod taking two electrons from the rod. If electrons are taken out from the rod that means this rod is going to get with charge. This rod is going to get the positive charge. This rod is going to get the positive charge. Right? And initially there were equal number of kines and annions in the solution. Initially there were equal number of kines and annions in the solution. Now imagine that one of
the kine from the solution. It collided with a rod took two electrons from the rod got converted into copper solid and that copper solid got deposited on the rod due to which the number of copper diposit ions in the solution will decrease. Due to which the number of copper dipos ions in the solution will decrease. Earlier there were equal number of kines and an ions. Now one of the kine collided with the rod took two electrons from the rod got converted to copper solid and that copper solid got deposited on the rod. So in this
solution copper dip positives they are decreasing with time. If they are decreasing that means the solution will be getting which charge negative charge effectively effectively it's going to get the negative charge. Earlier kines and ends were equal now kines are decreasing. So solution is effectively going to get the negative charge. Now my dear students since electrons are accumulated on this rod because of the oxidation of the zinc atoms. Since electrons are accumulated on this rod because of the oxidation of zinc atoms here, right? And the path is already made. The path is already made. Can
I say there will be a moment of electrons from anode to cathode? Absolutely. There will be the moment of electrons from anode to cathode. But if I ask you is it a normal moment of electrons or this is the directional moment of electrons. I would say this is the moment of electrons in a particular direction. And you know when charge moves in a particular direction current is produced. So if electrons are moving from anode to cathode if there is directional movement of electrons from anode to cathode so definitely this cell is going to produce current.
That is the use of this particular cell. That is the use of this particular cell. It converts your chemical energy into what? Into electrical energy. So the first thing which came to our mind right now. What is that? That is the direction of electrons in case of the Daniel cell is from anode to cathode. That means zinc to copper. Right? And if the direction of electrons is from zinc to copper that means anode to cathode. So the direction of current is always reverse. Direction of current is always reverse to the direction of electrons. I would
say the direction of current is going to be from copper to zinc. This is the first conclusion my dear students which you have to remember. This is asked what is the direction of electrons in case of Daniel cell. What is the direction of current in case of Daniel cell? So electrons they move move from anode to cathode. Current flows from cathode to anode. Okay. Now my dear students if you look carefully if you look carefully what happened on this rod from this rod zinc atom got converted into zinc dipositive and that zinc dipos to entered
into the solution then again one more zinc atom will under go oxidation one more zinc atom will undergo oxidation from the rod right so one one zinc atom it will be keeping on coming from the rod towards the solution due to which what will happen to the thickness of zinc rod conclusion number two the thickness of zinc rod The thickness of zinc rod, the thickness of zinc rod is it going to decrease or increase with time? The thickness of zinc rod is going to decrease with time or you can say this zinc rod it slowly
slowly dissolved in the solution. It dissolves in the solution. Right? Similarly, what is happening to the thickness of this particular rod? Remember, if I talk about the copper rod thickness, copper rod thickness is going to increase with time. It is going to increase with time. Okay, it's going to increase with time. This is the conclusion number two. My dear students, conclusion number three. This was the reaction taking place at anode. This was the reaction taking place at cathode. If I would want to write a net reaction, if I would want to write a net reaction,
let me tell you in order to write the net reaction, the first thing which should be in your mind, what is that? In order to write the net reaction, the electrons should be balanced in both the reactions. In order to write the net reaction, the electrons they must be balanced in both the reactions. If I ask you are the electrons balanced here? Yes, two here, two here also. Since the electrons are balanced, so you can directly add these two reaction. If the electrons were not balanced, then you were supposed to balance the electrons first and
then you could have added the reactions. But here in this particular case, the electrons are balanced in both the reactions. So you can directly add these two reactions and when you add these two reactions you'll be getting zinc solid plus copper dipul to aquis. What does it give? It gives zinc dipul to aquis and with this you are getting copper solid. So this is our net reaction which is taking place in this Daniel cell. Okay. And how many electrons got cancelled? Two electrons got cancelled. So I'll say n value is equal to two. What does
this n signify? N signifies the number of moles of electrons. The number of moles of electrons exchanged between anode and cathode. The number of moles of electrons exchanged that is represented by n which is nothing but that is two. Okay. Which is nothing but two. Perfect. Am I clear with this? Am I clear with this people? everyone in the chats. So some conclusions which are very important. Number one, electrons move from anode to cathode. Current from cathode to anode. Zinc rod dissolves with time. The thickness of copper rod is going to increase with time. Right?
And n value in this particular reaction is nothing but two. And this particular cell what does it do? It converts your chemical energy into the electrical energy. Okay. Now my dear students, what is the main function of this salt bridge? The salt bridge which we have used here. What is the function of it? What is the function of it? So if you exactly talk about the functions of the solid bridge if you exactly talk about the functions of the salt bridge. Just give me a second. Let me use a different slide only. Okay. So if
we talk about the functions of the solid bridge the first function of the solid bridge which all of you must be knowing now what is that? It completes it completes the internal circuit. As you saw here, the salt bridge is doing nothing but it is completing the internal circuit right here in this particular case. Now guys, point number two that is important. What is that? It maintains It maintains the electrical neutrality. It maintains the electrical neutrality [clears throat] of both the solutions. It maintains the electrical neutrality of both the solutions. How exactly? Have a look.
See as I told you when the extra zinc atom when the extra zinc ion comes into the solution the solution gets a positive charge. When copper dipositive from the solution is reduced the solution gets a negative charge. But remember as soon as the positive charge enters in the solution as soon as the extra positive charge gets developed in the solution at the same time from the salt SO4 dgative will come and will neutralize it. Similarly, as soon as the negative charge is developed in the solution, at the same time, K positive is going to come
from the salt bridge and will neutralize it. So, what is the main function of the salt bridge? The main function of the salt bridge is that it maintains the electrical neutrality in both the solutions. It maintains the electrical neutrality in both the solutions. Okay, this one of the important one and the most important use of the salt bridge. What is that? Let me tell you, it avoids It avoids the liquid junction potential. It avoids the liquid junction potential. Now what is the liquid junction potential and how the salt ridge avoids the liquid junction potential? Let's
try to understand that. That's again a very important point. So my dear students, what exactly I would be doing here? Let's say I'm taking a bigger sized container like this. I'm taking a bigger sized container like this and I'm dividing this particular container into two chambers. I'm dividing this particular container into two chambers with the help of a semi-permeable membrane. I'm dividing this container into two chambers with the help of a semi-permeable membrane. But dear students, in this left chamber, in this left chamber, imagine that I'm keeping zinc sulfate. In this left container, I'm keeping
zinc sulfate, which I believe got dissociated already into zinc dipositives and SO4 diggatives. And the right chamber that you see here in this right chamber, I'm taking the salt of copper. I'm taking copper sulfate in the right chamber which already got dissociated as copper dipositive plus SO4 dionate. Friends, if you talk about the left chamber, this left chamber contains equal number of kines and anines due to which this left solution is electrically neutral. Similarly, this particular right solution, this solution which on the right side, which in the right chamber, it again contains equal number of
kines and an ions. So, I would say this solution which is in the right chamber, it is also electrically neutral. So basically both the solutions are right now electrically neutral. Now what exactly am I going to do? For example, I'm introducing I'm introducing a zinc rod here. And here I'm introducing a copper rod. I'm introducing a zinc rod. Here I'm introducing a copper rod. This is something I'll be calling as the electrode. This is also something I'll be calling as the electrode. So I have got zinc electrode and copper electrode. Now I'm connecting them externally
with the help of a voltmeter. So if I ask you is the circuit complete from outside? Yes, from the outside the circuit is complete. Is the circuit complete from inside? Inside also as you can see you are getting a closed loop here. You're getting a closed loop here. So the circuit is complete. Basically the circuit is closed. The circuit is closed. So here in order to close the circuit you don't have to use a salt bridge because the circuit is already closed. You're getting a loop here. You're getting a closed loop. You're not supposed to
use the salt bridge here to complete the internal circuit as the internal circuit is already completed because these solutions are in direct contact with the help of a semi-p permeable membrane. Okay. Now my dear students what's going to happen since this is your anode this is your cathode and you know at anode oxidation takes place due to which what happens one of the zinc atom will undergo oxidation will get converted into zinc dipositive and that zinc dipositive enters in the solution due to which this solution this left chambered solution it is effectively going to get
the positive charge. Similarly, there are copper diposites here. They are going to collide with the copper rod, take two electrons from the rod, get converted into copper solid and that copper solid gets deposited on the rod. Due to which the number of copper diposites in the solution will decrease due to which the solution gets which charge? The solution gets a negative charge. This solution got positive. This solution got negative. They are going to attract each other and they are going to accumulate here at the junction. They are going to attract each other and they'll accumulate
here at the junction. Similarly due to again oxidation and reduction these positives and negatives will keep on coming. These positives and negatives will keep on coming and they'll keep on getting accumulated at the junction. Now my dear students on the left side of the junction can I say the left part of the junction is at positive potential and the right part of the junction is at negative potential. The left part of the junction is at positive potential. The right part of the junction is at negative potential. So can I say at this junction a potential
difference is getting created and this potential difference which is getting created here at the junction what do you call this potential as you call this as liquid junction potential you call this potential difference here which got created as liquid junction potential. So if I do not use a salt bridge, this liquid junction potential gets created. But imagine if you had used a solder bridge. If you had used a salt bridge, what would have happened? As soon as positive charge got was getting developed here in the solution. As soon as the solution was getting the positive
charge at the same time from the solid bridge SO4 diggative would have come neutralized it. As soon as this negative charge got developed in the solution at the same time K positive would have come and neutralized the solution. due to which there would have been no positive no negative getting developed in the solutions. If there are no positives, no negatives getting developed in the solutions, I would say this potential difference at the junction would not have got created. Okay. So I would say the main and the important function of the solbridge is it avoids the
liquid junction potential. I believe this is also clear to everyone. Now my dear students let's talk about let's talk few things about the inert electrolyte that we use let's talk about few more points about the inert electrolyte that we use in the solbridgeidge let me tell you the inert electrolyte in the solid bridge the ions of the inert electrolyte the ions of the inert elect must not must not participate In the net cell reaction in the net cell reaction my dear students as you see this was our net cell reaction in this net cell reaction
do you see any K positive or SO4 D negative there is no K positive or SO4 D negative in the N cell reaction so remember that electrolyte is used right that electrolyte is used as the inert electrolyte whose ions won't participate in the net reaction point number Point number two. Point number two. The ionic mobilities ionic mobilities of kion and nion. The ionic mobilities of kion and nion must be the same. Must be the same. What does that mean? That means the speed with which an annion from the salt bridge enters the solution with the
same speed kion should also enter into the same solution. So their ionic mobilities their speed must be the same. One more thing the transportation number what is transportation number? You'll get the idea in some time. The transportation number of kion and nion must be the same. transportation number of the kion and nine must be the same. In short, remember one thing. What is that? The speed with which kion and nine from the salt bridge are going to enter in the solution, they must be the same. They must be the same. The speed with which kine
enters into the solution and enter nine enters in the solution, they should they should enter into the solution with the same speed at the same time. Okay. Are all these things clear? Let me know once in the chat quick. Let me know once in the chat quick people. Everybody. Yeah. Now there is one more very very very important point which I would definitely want you guys to remember. What is that? This is the point. This is the point point guys. Just read the statement once. Just read the statement once. The statement is pretty simple guys.
What is the statement? If Ag positive ions, Hg dipositive ions, PB dipositive ions, PL thalium positive ions, if they are present in any of the solutions, if they are present in any of the solutions, if they are present in any of the solutions, at that point of time you are not going to use which inert electrolyte? You are not going to use KCL as the inert electrolyte. If any of these ions are present in the solutions in the anodic solution or cathodic solution if any of these ions is present at that point of time you
guys are not going to use KCL as the inert electroly what? Because what KCL does? KCL when reacts with these ions what's going to happen a precipitate gets formed. Precipate gets formed at the mouth of the salt bridge. plate like AgCl, Hg2 CL2, PBCL2, P ICL they get it gets formed at the mouth of the salt bridge right due to which it'll be really difficult for the ions to migrate into the solution okay so whenever my dear students remember it directly as a point whenever you have got a Daniel cell in which the ions like
Ag positive HG d2 PB diit are present in that particular galvanic You guys are not going to use the inert electrolyte as KCL. Why? Because this KCL it forms precipitate with these ions that gets accumulated at the mouth of the salt bridge due to which the migration of ions towards the solutions. Right? That is going to be difficult. So remember this particular point as the memory based point. Okay. It seems that I'm preparing for J. I'm but I'm doing chemistry from here and I think this is more than enough for me. Yes, this is definitely
going to be more than enough for me for you. Okay, lecture plus the DPP which I keep on sharing after the session. Just solve that. That's more more than sufficient. Don't worry. All right. Now, my dear students, there is one more important thing. What is that? That is the effect of external battery on the galvvinic cell. Effect of external battery on the galvanic cell. What happens exactly? Let's try to understand. For example, I'm taking a Daniel cell. Let's say I'm taking a Daniel cell. So this is your one of the containers that contains zinc sulfate.
Okay. This is one more container. What does it contain? It contains copper sulfate. In this particular container you are introducing a zinc rod. In this particular container you are introducing a copper rod. So how many electrodes do I have? I've got two electrodes. One is my zinc electrode and one is my copper electrode. Let me connect these electrodes externally with the help of a voltmeter. Let me connect these electrodes internally with the help of a salt bridge. So I got a complete galvvinic cell. I got a complete galvvinic cell. Now my dear students, what exactly
am I doing? What this galvvinic cell does? This galvvinic cell produces current. This galvic cell, it produces current. Okay, think over it logically. If this gammic cell is producing current, that means there must be the potential difference between these electrodes due to which current is produced. You know, as per your Ohm's law, current flows due to what? Potential difference. And in the galaxy cell, current is flowing from cathode to anode. You know that. So there must be potential difference between these two electrodes. By means of which current is produced by this particular cell. So definitely
between these two electrodes there must be the potential difference. Okay. Now guys, what exactly am I going to do? Here, this rod carries a negative charge. This rod carries a positive charge because this is anode, this cathode. Okay. Now, here I'm going to take an external battery. External battery. One battery I'm taking. This is a positive end of the battery. This is the negative end. Positive end of the battery. Negative end. Okay. I took a battery here. I took a battery here people. And here for example I'm keeping a switch. I'm keeping a switch as
well. Okay. I took a battery. Now what exactly am I doing? I'm connecting this particular wire. This particular wire. I mean I'm connecting the negatively charged rod with the positively charged plate here. And this I'm for example joining with the negatively charged plate. Try to understand what am I doing? I took the external battery. The positive end of the external battery is connected with the negative one. The negative end of the external battery is connected with the positive one. Now what happens exactly? There are some conclusions which I would want you guys to remember. Okay?
I'm not going to go into the details of all this, right? Because we have not discussed emf yet. Okay? I'm not going to go into the details of all this but some important three conclusions I would want you guys to remember for now. What are those conclusions? Conclusion number one. My dear students, when the voltage of the external battery was kept less than 1.1 volt, when the voltage of the external battery was kept less than 1.1 volt. At that time it was observed that At that time it was observed that the electrons were going from
zinc electrode to copper electrode and current was going from copper electrode to zinc electrode. This was the first conclusion. This was the first conclusion. What this conclusion is when the external battery voltage was kept less than 1.1 volt electrons I mean current was flowing from copper to zinc. Second point my dear students when the voltage of the external battery was kept exactly 1.1 volt at that time there was no current in the cell there was no current in the cell no current was generated from the cell that means the cell was not working at this
particular point of time the third point when the voltage of the external battery was kept greater than 1.1 volt everything reversed What reversed exactly? Electrons were now flowing from copper to zinc and current was now going from zinc to copper. That means the cell was now behaving as the electrolytic cell. It was now behaving like the electrolytic cell. These are the three important conclusions. They can be asked directly in the form of the memory based thing. Right? What when external battery is connected with the galvvinic cell with reverse polarity, what happens exactly? Okay, what happens
exactly? So remember these three points. If I'm clear, let me know once in the chats so that I can move on. Let me know quickly in the chats so that I can move on. All clear. Everybody quick quick quick in the chats. All right people, now comes the point. What is that? That is how do we represent a galleric cell? How do we represent how do we represent a galic cell? This is again one important thing and just one thing my dear students I would want you guys to remember. What is that? Try to understand
in order to represent a galvanic cell in the middle always I'll be drawing two lines. In order to represent a galvvinic cell in the middle I'll be always using two lines. And my dear students on the left side these two lines these two lines in the middle what does it represent? It is going to represent these two lines are going to represent a salt bridge. These two lines are going to represent a salt bridge. On the left side of the salt bridge, what do we have? We write anode. On the right side of the salt
bridge, we write cathode. On the left side of the salt bridge, we write anode. On the right side of the salt bridge, we are going to write the cathode. Yeah, sir. Letters cannot be seen. I think it's all fine. I can see. Okay. So first thing guys, these two lines are going to represent the solid bridge. On the left side of it, you are going to write the anode. On the right side of it, you're going to write the how exactly how exactly try to understand. For example, I have got a galvvinic cell whose net
reaction is like this. Zinc solid plus copper dip to aquis. It gives zinc diples to aquis. And for example, you have copper solid. Let's say this is a net reaction which is taking place in some galvvinic cell. This is a net reaction which is taking place in some galvvinic cell. And after looking at this net reaction, I need to represent a cell. How exactly I'm going to represent it? Try to understand all the things what I'll be telling you. First of all, oxidation state of zinc zero. Oxidation state of copper plus2. This zinc is in
plus2. This copper is in zero oxidation state. Is the oxidation state of zinc increasing? Yes. Increase in the oxidation state is what you call as oxidation. So zinc is undergoing oxidation. Similarly plus 2 to zero decrease in the oxidation state. Decrease in the oxidation state is what you call as reduction. And you know oxidation takes place at anode and your reduction it takes place at cathode. Oxidation takes place at anode and your reduction it takes place at cathode. So how do I represent it exactly? Now these two lines are going to represent the salt bridge.
On the left side of the salt bridge I'll be writing anode. At anode do you see oxidation is taking place? Zinc solid is getting converted into zinc dipos to aquis. I would say zinc solid is getting converted into zinc dipos to aquis. Now if I ask you is the phase same here or different? This is in solid phase. This is in aquisphase. So phase is different. When phase is different you use the term slash in between them. When the phase is different, you use this term slash. If the phase was same, if the phase was
same, then you could have used comma instead of slash. Okay. Similarly, on the right side, what do we write? On the right side, we write cathode. And at cathode, reduction takes place, right? Reduction of what? Copper dipositive into what? Copper solid. So, at cathode, I would say copper dipositive aquis is getting converted into what? Copper solid. So, tell me, is the phase same or different? Again, the phase is different. So, use the slash in the middle. So you have represented this particular cell properly. I believe it's clear. Now for example, let's say you have got
a cell whose cell reaction is this one. You have got a cell whose cell reaction is this one. Okay, you got a cell whose cell reaction is this one. Now after looking at the cell reaction, I would want to represent the cell. How exactly am I going to do it? Again, oxidations of iron zero. Copper here is in plus2. Air iron is in plus2. Air copper is zero. Now 0 to +2 means increase in the oxidation state. Increase in the oxidation state is what you call as oxidation. And you know oxidation takes place where oxidation
takes place at anode. Similarly plus 2 to 0 decrease in the oxidation state. Decrease in the oxidation state is what you call as reduction. What you call as reduction and reduction takes place where? Exactly at cathode. Okay. Now my dear students these two lines in the middle they are going to represent what? They are going to represent a solid bridge. Now let me tell you on the left side of the solid bridge what do we write? We write anode. At anode FA solid. At anode as you can see FA solid is getting converted into what?
It is getting converted into FA DOS to aquas. Is the phase same or different? Different. So use a slash. Right? Now similarly on the right side on the right side what do we have? On the right side what do we have? On the right side we write the cathode and you know add cathode what happens? reduction copper dip positive is getting converted into copper solid. Copper dipositive aquis is getting converted into copper solid. So is the phase same or different? Again the phase is different so use the slash in the middle. And if the phase
is same then use comma. If the phase is different use the slash. I believe I'm clear. Okay. Now for example, let's say you got the cell whose cell representation is given to me like this. Let's say you got the cell whose cell representation is given to me like this. I mean the cell the net cell reaction is given to me like this. By looking at the net cell reaction I have to represent the cell. I have to represent the cell. So again my dear students the similar thing I'll be doing. The similar thing I'll be
doing. Oxidation state of zinc here is zero. Right? Oxidation state of hydrogen here is + one. This is +2. This is zero. Now 0 to +2 increase. Increase means oxidation. And you know oxidation takes place where? At anode. This is undergoing reduction. Okay. So this is going to be my cathode. So you identified your anode and cathode. So in the middle two lines soluble on the left anode oxidation and oxidation means your zinc solid is getting converted into what? It is getting converted into zinc dipos to aquis. The phase is different again slash. Now here
h positive aquis h positive aquis is getting converted into what? Is getting converted into H2 gas phase is different. Use the term slash. Till now my dear students whatever electrodes we made till till now whatever electrodes you saw everywhere it was solid aquis it was solid aquis right for example for example here in this particular case here in this particular case it was solid aquis it was solid aquis right let's say take the second case take the second case it was solid aquis it was sorry not this one where is that where is that where
is Okay, here in this particular case it was solid aquis. It was solid aquis. So whenever your electrodes are basically solid aquis whenever your electrode is like this solid aquis. So it's a normal electrode. Basically it's a normal electrode which is made under normal conditions. This is again a normal electrode. But whenever you have got an electrode wherein the phase is where is the wherein the phase is not solid aquis it can be gas aquas it can be aquis aquis wherever you see an electrode where the phase is not solid aquis there you will have
to use platinum solid as well there you will have to use platinum solid as well okay remember this particular point directly Remember this particular point directly. Whenever you see an electrode wherein the phase is not solid aquas, it is something apart from solid aquis. It can be aquas aquas, it can be gas aquas, anything right? At that point of time you'll have to use platinum solid as well. Now let me take one more example with that things will get clear. See guys this we have got a galvvinic cell whose net cell reaction is for example
this one. Now first of all when you calculate the oxidation state of chromium here it will come out to be plus 6. Oxidation state of iron here +2. This chromium here is in plus three. And this iron here is in plus three. Now few students must be thinking how come the oxidation state of chromium is six. Let the oxidation state of chromium is x. There are two chromium atoms. Oxidation state of oxygen is minus2. There are seven oxygen atoms. The net charge is minus2. When you solve this the value of x will come out to
be + 6. Okay. Now + 6 to + 3 decrease. Decrease means reduction. So this CR27 D negative is undergoing reduction. If it under goes reduction that means it's going to behave like the cathode. Similarly plus 2 to + 3 increase. Increase means oxidation. Increase means oxidation. And you know oxidation takes place where? Exactly at anode. Now since you identified your anode and cathode. So in the middle salt bridge on the left side of the salt bridge. Anode at anode oxidation is taking place and as you can see Fe dipositive aquis is getting converted into
what? It is getting converted into fee tripositive aquis. Now if I ask you is the phase same or different? The phase is same. So use comma in the middle not the slash. Use comma. Now people will tell me one more thing. Tell me one more thing. Is the phase solid aquas? No. The phase is aquas. Whenever you see the phase anything apart from solid atoms you'll be using a term platinum solid here as well. So I'm using a term platinum solid. Now on the right cathode add cathode reduction CR27 D negative gets converted into CR
try positive CR27 D negative gets converted into what? Crositive aque. Now again the phase is different. Sorry the phase is same. So use comma in the middle. Now it is not solid aquis. It is aus aquis. So use platinum solid. So whenever you see the electrode apart from solid aquis you have to use the platinum solid as well at that point of time. If I'm clear let me know once in the chat till now. Did you understand how exactly the cells are represented? Did you understand how exactly the cells are represented? Believe yes. perfect guys.
Now comes one more important thing. What is that? Now the point is something very very very important guys which you have to understand in detail because the entire chapter is based on the things which I'm going to tell you now. So have a proper eye have a proper eye here on the screen. How to write the net cell reaction for a given cell? How to write a net cell reaction for a given cell? See guys, try to understand carefully what I'll be saying. First of all, let's say I'm taking a cell here. The cell is
like this. Zinc solid zinc dipostor solid bridge copper dipostor ais copper solid. Now first of all these two lines they represent what? Salt bridge. You know on the left side of the salt bridge what do we have? We have anode. On the right side of the salt bridge what do we have? We have cathode. Now if I ask you what happens at anode you know at anode oxidation takes place. Yeah. At anode what happens? Oxidation takes place. At anode oxidation takes place. Oxidation means loss of electrons. So basically at anode your zinc is getting converted
into zinc diit. Your zinc is getting converted into zinc diit. So 0 to +2 0 to +2 0 to +2 increase in the oxidation state. Increase in the oxidation state is called as oxidation which means loss of electrons. I would say at anode zinc solid would be getting converted into zinc dip positive and with this you'll be getting two electrons as well. Now similarly what happens at cathode at cathode reduction takes place see plus 2 to 0 plus 2 to 0 decrease in the oxidation state which means reduction gain of electrons. So I would say
this copper diit this copper dip positive it would be gaining how many electrons? It would be gaining two electrons and then getting converted into copper solid. Now in both the reactions are the electrons balanced? The electrons are balanced in both the reactions. If electrons are balanced in both the reactions just add them up. When you add them up two electrons are cancelled. So the net reaction is copper solid plus sorry zinc solid plus copper dip positive aquis it gives me zinc dipos to aquis and with this you'll be getting copper solid as well so this
is my net reaction now if I ask you how many electrons got cancelled two that means the value of n here is two that means the value of n here is two now there is one more thing that is a reaction quotient reaction quotient represented by qc reaction ction quotient reaction quotient which is represented by QC. Now how do we exactly write the expression of reaction quotient? Well my dear students few things I would want you guys to remember here. What are those things? Whenever you see any reactant or product in Whenever you see any
reactant or product in aquisphase you are going to use the term concentration for that. Whenever you see any reactant or product in aquas phase you are going to use a term concentration for it. Whenever you see a reactant or product in gaseous phase you'll be using the term pressure for it. And whenever you see any reactant or product in pure solid or pure liquid form, it is not going to be considered. It's not going to be considered. These are the three things which I would want you guys to remember. Now my dear students, in order
to write the expression for QC, remember you'll be always starting from the product. You'll be always starting from the product. This product is an aqua state. the one that's in aqua state what do we use we use the term concentration so that's the reason I wrote concentration of zinc di positive raised to the power the stoometry coefficient that's one here multiplied by this copper it's in solid state nothing do with this divided by come to the reactant side this is solid nothing do with this this is aquis so use its concentration concentration of copper I
positive raised by its stoometric coicient so this is how you write reaction at anode reaction at cathode net reaction, N value and QC expression. I believe I'm clear with this and people if I'm clear with this let me take few more examples with that things will get more clear to you. For example, I'm taking a cell like this platinum solid H2 gas H positive aquis solbridge copper dipos to aquis copper solid. This is your cell first of all. This is your cell first of all. Yeah. Now what all things we can conclude from the cell?
These two lines represent the solid bridge. On the left, what do we have? We have anode. On the right, what do we have? We have cathode. At anode, what happens? At anode, what happens? Oxidation. At anode, what happens? Oxidation. Oxidation means loss of electrons. So basically as you can see at anode H2 is getting converted to H positive. So 0 to + 1 0 to +1 means increase in the oxidation state. Increase in the oxidation state means oxidation. Oxidation means loss of electrons. So when H2 gets converted into H positive when H2 gets converted into
what? When H2 gets converted into H positive, it would be definitely losing electrons. H2 would be losing electrons. Now the point is how many? First of all these are two hydrogen atoms here. I'll write two hydrogen atoms. Now 0 to + 1 0 to + 1 what is the change in oxidation state 0 to + 1 the change in oxidation state is one is the change for one atom I have two H2 will be losing two electrons and then it'll be getting converted into H positives here this is the reaction at anode now cathodic reaction
all of you already you know what happens at cathode at cathode reduction takes place plus 2 to0 decrease in the oxidation state means reduction gain of electrons So copper dipul to aquis would be gaining two electrons and getting converted into copper solid. Electrons are balanced in both the reactions. So just add these reactions. When you add these two reactions, your net reaction becomes H2 gas plus copper dipositive aqueous. It gives twice H positive aquis plus copper solid. Since two electrons got cancelled, so I would say n value here is two. Now how do I write
the reaction quotient expression? I will be starting from the product. This is an aqua state. The one that's in aqua state you are going to use its concentration. Concentration of H positive raised par stoometric equation two solid nothing to do with this divided by H2 gaseous phase. So pressure of H2 gas raometric equation one aquis use the term concentration concentration of copper diometric equation one. So this is how you guys are going to write the expression for QC for this particular reaction as well. So tell me once in the chats. So people tell me once
in the chats. So people tell me once in the chats if I'm clear to everyone quickly. >> [snorts] >> Got it. Okay. Let me take one more example and with that I believe things will get more clear. Let me take one more example. With that things will get more clear. Understand guys? For example, I'm taking a spell like this platinum solid. Feosto aquis. Feosto aquis bridge Mn4 negative aquis MN diositive aquis and here I'm using platinum solid. This is the cell for example. Now what do we have to do? We have to do all the
stuff whatever I've taught you till now. So first of all, first of all people on the left side of the salt bridge what do we have? We have anode. On the right side of the salt bridge what do we have? We have cathode. Now what happens at anode? First of all I would say at anode oxidation is going to take place. At anode oxidation is going to take place. Oxidation means loss of electrons. See at anode F dipositive gets converted into F positive. Fe dipositive aquis gets converted into what? Gets converted into Feos to aquis.
So plus2 to plus three. The change in the oxidation state is one. So I would say Fe dipositive would be losing one electron getting converted into F tripositive. Similarly what will be happening at cathode? As you can see at cathode reduction takes place. Mn4 negative gets converted into M and di positive. M4 negative it gets converted into M and DI positive. Now if I ask you what would be the oxidation state of magnes here it'll be + 7 + 7 to +2 decrease in the oxidation state decrease in the oxidation state is what you call
as reduction reduction means gain of electrons so M4 negative would be gaining five electrons final minus initial the value comes out be five it'll be gaining five electrons and then would be getting converted to M& dipos2 right now my dear students if I ask you are the electrons balanced in both the reactions No, the electrons are not balanced. What am I supposed to do? I'll be multiplying this reaction by number five. So that this becomes five times. This becomes five times. This becomes five times. Now the electrons are balanced. Now you are going to add
these two reactions. When you add these two reactions, my net reaction becomes 5 * F positive aquis plus MO4 negative aquis. It gives 5 * F trip positive aquis plus MN diositive aquis. Now if I ask you what is the N value here? Your direct answer should be five. N is five. N is five. Just give me a second guys. There is some electricity cut. Just a second. Just be like this. Okay. Let me get the generator on. Do you like the dark mode better? Do you like the dark mode better? Then I can continue
like this only. If you like the dark mode better. No, I don't like it. No. No, I don't. Is it actually good? Okay. All right. Let's move on. Now, since we added the reactions and since five electrons got cancelled, so I said n value is nothing but five. Now, you tell me one thing. Is this reaction balanced? The reaction is not balanced. See four oxygen atoms here. No oxygen atom on this side. the reaction is not balanced. Now the next thing after this is going to be you will be balancing the oxygen atoms with the
help of water molecules. You'll be balancing the oxygen atoms with the help of water molecules. Okay, that is something we have to discuss in redox reaction chapter but right now I'm just teaching you. So remember it for now. So in order to balance oxygen we have to add water molecule on that side which is oxygen deficient. How many oxygen atoms on this side? Four. How many oxygen atoms on this side? None. So, right side is oxygen deficient. How many? Four oxygen deficiencies. So, add four times water molecules on this side which is oxygen deficient. Now
try after after balancing oxygen you have to balance oxygen. Sorry. After balancing oxygen, you have to balance hydrogen. Do you see any hydrogen here? No. Do you see any hydrogen here? Yes. How many? Eight. So this side is hydrogen deficient. This side is hydrogen deficient. And I'll be adding some 8 H positives on that particular side which is hydrogen deficient. Now the reaction is balanced. Now the reaction is balanced. Now we can easily write the expression of QC. You will be starting with the product. This is an aqua phase. The concentration of Fe positive raised
power stoometric equation that is five M and D positive in aquis phase the concentration of M and D positive raise power stoometric question one this water is in liquid phase nothing to do with that divided by come to the reactant side this H positive since it is an ion ions are supposed to be in aquas phase only right that's why I wrote aquis here okay so this is going to be concentration of H positive raised power stoometric equation that's Eight Fei positive it's an ion concentration of FI positive raised power stoometric equation 5 M4 negative
again in aquas phase if it is an aquas phase so it's going to be concentration of M4 negative raised power stoometric equation that's one so this is the expression for QC which I was supposed to calculate this is the expression for QC which I was supposed to calculate am I clear with this everybody Is it clear people? Is everything You have to balance oxygen with the help of water molecules. We did not start nervous equation yet. Just wait. We did not start the nerous equation yet. All right people. So I believe things are clear here.
Now now, now we are going to enter into the new part of the chapter. What is that? That is electrode potential. But electrode potential will be starting after a 15minut small break. So that you can have a tea or coffee. Okay. I'll be also back on time. Okay. All right. A small break. Go have a tea and come back on time guys because now we are going to enter into the real phase of the chapter. See you in some time. Is everyone back? All right. Yes. Most of you can watch it. You can watch it.
Don't worry. So tell me quickly in the chats if all the things which we have discussed till now are they absolutely clear to you. Say it once in the chats. Say it once in the chats. If all the things discussed till now are super clear to you or not. Everyone everyone All right people. So let's move on now to one more amazing topic. What is that? Something very important from which questions are frequently asked. Okay. And this is nothing but electrode potential. So first of all, how do we define the term electrode potential? Let me
write its small definition and then I'll make you understand what this electrode potential is all about. My dear students, electrode potential is defined as the potential difference. It is defined as the potential difference that gets created. The potential difference that gets created between the rod and the solution. The potential difference that gets created between the rod and the solution. And let me tell you it is represented by capital E. It is represented by capital E. Now let's try to understand what exactly it means. See guys for example imagine that there's a container and in this
particular container for example I am taking let's say zinc sulfate so the zinc sulfate in the container assume that it has dissociated completely its ions ions are zinc dip positives and SO4 d negatives the zinc sulfate that has been taken in a container that has got dissociated completely into its ions. Ions are zinc dip positives and SO4 dgatives. So right now in the solution there are equal number of kines and annions present. So I would say this solution right now is electrically neutral. Now people for example I'm introducing a zinc rod here in this container.
I'm introducing a zinc rod here in this container. So can I say the zinc rod is placed in a solution containing its own ions. The zinc rod is placed in a solution containing what? Containing its own ions. So this complete setup I'll be calling as the electrode. This complete setup is called as the electrode. For example, when this electrode behaves like the anode, when this electrode behaves like the anode, what happens? When this electrode behaves like the anode, oxidation takes place. Oxidation takes place. So basically the rod is made up of zinc atoms. Those zinc
atoms of which rod is made up of those zinc atoms will undergo oxidation and those zinc atoms will get converted into zinc dip positives and with that you'll be getting two electrons and you know these two electrons will be accumulated on the rod and this zinc dipositive that goes into the solution and when the zinc dipositive goes into the solution the solution gets the positive charge the solution gets a positive charge and similarly there are electrons accumulated on the rod. I would say the rod gets the negative charge. My dear students as you can see
the rod got negative charge solution got positive charge. So can I say between the rod and the solution between the rod and the solution potential difference got created. Absolutely between the rod and the solution between the rod and the solution potential difference got created. And this potential difference which gets created between the rod and the solution is what you call as the electrode potential. Read the definition. The potential difference that gets created between the rod and the solution is what you call as the electrode potential. Yeah. Now my dear students again I'm taking the same
zinc electrode. Again I'm taking the same zinc electrode. So for example this is again the zinc sulfate which got converted into zinc dip positives and SO4 dgatives. So there are equal number of kines and anines present right now in the container. So the solution is right now electrically neutral. So here you are introducing a zinc rod. Here you are introducing a zinc rod. Now my dear students assume that this zinc electrode now behaves like cathode. If the zinc electrode behaves like cathode, what's going to happen? Reduction is going to happen. Reduction. Reduction means gain of
electrons. Reduction means gain of electrons. So in this solution, you have got zinc dipositive ions. These zinc dipositive ions 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 the zinc diposit ions in the solution, they are going to collide with the rod. They are going to take two electrons from the rod. and will get converted into zinc solid. So the zinc dipositive ions they collided with the rod took two electrons from the rod. So if electrons are taken from the
rod the rod will be getting the positive charge. The rod will be getting the positive charge right? When the zinc diposites collided with the rod took two electrons from the rod got converted into zinc solid and that zinc solid gets deposited on the rod. So tell me what happens to the number of zinc diposites in the solution. Earlier there were equal number of kines and an ions. Now one of the kine took two electrons from the rod got converted into zinc solid and that zinc solid got deposed on the rod. So in this solution the
kines are decreasing with time. So the solution is effectively going to get with charge. The solution is effectively going to get the negative charge. Do you see on the rod there is positive charge on the solution there is negative charge. So again I must say a potential difference got created. A potential difference got created between the rod and the solution. And this potential difference which gets created between the rod and the solution. The potential difference that gets created between the rod and the solution is what you call as the electrode potential which is represented by
capital E. Which is represented by capital E. So electrode potential which is represented by what? Which is represented by capital E. Now my dear students this electrode potential is basically of two types. One is called as oxidation potential. One is called as the reduction potential. See here the electrode behaved like anode. Here the electrode same electrode it behaved like what? It behaved like cathode. When the electrode behaved like anode rod got negative solution got positive. When the same electrode behaved like the cathode rod got positive solution got negative. So first of all in general the
potential difference between the rod and the solution. In general the potential difference between the rod and the solution you call that as the electrode potential. But this electrode potential wherein rod gets negative solution gets positive. This electrode potential is called as the oxidation potential. And this electrode potential this electrode potential is called as the reduction potential. So when the rod gets negative solution gets positive. This electrode potential is called as oxidation potential. And similarly when the rod gets positive solution gets negative. This electrode potential is called as the reduction potential. So that's why I said
electrode potential I'm dividing into two categories. Oxidation potential and reduction potential. Okay. Now people this oxidation potential there is one more way of representing it. You can write it like this. You can write it like this. EO X oxidation potential. Reduction potential you can write like this. E R E D or or oxidation potential you can represent like this. E of M gives M N positive. Reduction potential you can represent like this. E of M N positive gives M. Now check carefully. Oxidation state of metal here is zero. 0 to + N. 0 to +
N. Increase means oxidation. Oxidation means oxidation potential plus n to0 decrease in the oxidation state. Decrease in the oxidation state is something we call as reduction. So this term represents your reduction potential. So your oxidation your electrode potential I classified into two parts. One is oxidation potential, one is reduction potential. Right? Now my dear students when the same oxidation potential or reduction potential or in general if I say when this electrode potential or this electrode potential is measured under standard conditions is measured under standard conditions. Now you must be thinking what are standard conditions? Standard
conditions are nothing. Standard conditions are nothing. when pressure is kept one bar, temperature is kept constant generally 25° centiggrade and concentration of the solution is kept one molar. These conditions are what you call as the standard conditions. So basically basically if I do one thing if I keep the pressure here as one bar if I take the concentration of the solution one molar if I keep the temperature constant generally 25 that means I have got the electrode under standard conditions and remember when electrode potential is measured under standard conditions when electrode potential is measured under
standard conditions what do you call that as you call that as the standard electrode potential which is represented by E not which is represented by E not. So what is E not? E not is basically the standard electrode potential. The potential difference between the rod and the solution which is measured under standard conditions which is measured when pressure is kept one bar temperature constant and when concentration is kept one molar. Right? So you call the electrode potential as the standard electro potential. So this also I can classify into two types. One is going to be
your standard oxidation potential. One is going to be your standard reduction potential. Or you can represent it like this. E not of E not of M gives M N positive. This represents your SOP. This represents your SOP standard oxidation potential. Similarly, the standard reduction potential you can represent it like this. E not of M N positive gives M right or you can call it directly as the standard reduction potential. So standard electrode potential I classified into two parts. One is standard oxidation potential one is your standard reduction potential. Okay. Now my dear students remember for
an element for an element remember SOP is always equal to minus * its SRP SOP is equal to minus time its SRP for example what does that mean that means if I write E not of zinc gives zinc die positive if I say this value is equal to 0.76 volt if I say this value is equal 0.76 volt first of all look here 0 to + 2 increase means oxidation so this is your SOP this is the SOP now you can write the similar thing like this as well E not of zinc dipositive gives zinc
zinc diitive gives zinc so +2 to zero decrease reduction so this srp so the srp of the same is going to be -0.76 volt okay So remember this particular point as well. Remember this particular point as well. Okay. SOP is always equal to minus time SRP. Now comes one more point. Now comes one more point. My dear students SOP or SRP of an element or let me write directly of an electrode. SOP or SRP of an electrode is measured is measured with the help of with the help of with the help of a reference electrode
with the help of a reference electrode and that reference electrode used here is called as the standard hydrogen electrode. The point is how do we calculate how do we calculate the SOP values and SRP values of different elements? We calculate the SOPs and SRPs of different elements with the help of a reference electrode which is what you call as the standard hydrogen electrode. Okay, which is what you call as the standard hydrogen electrode. Now, now what the standard hydrogen electrode is, how it is made and what all things we have to remember about the standard
hydrogen electrode. Try to understand carefully guys what I'll be saying. Mark the heading as standard hydrogen electrode first. Standard hydrogen electrode. Standard hydrogen electrode. S H E it is a reference electrode which is used to calculate the SOP and SRP values of different elements. Okay. How the standard hydrogen electrode first of all is made. Have a look here. For example, I'm doing one thing. I'm taking a container. Let's say this is the container. In this container my dear students for example I'm keeping H positive ions I'm keeping H positive ions in the container okay let
the concentration of H positives here in the container is 1 molar 1 mole per liter 1 molar okay now now for example I'm taking a U type tube like this an inverted Uype tube like And this in this inverted U-Type tube, I'm taking a platinum wire. This is a platinum wire which I'm taking here. This is a platinum wire. Let me tell you this is your platinum wire which is for example coated with coated with platinum black. I have taken a platinum wire here. Now my dear students from this opening I'm introducing H2 gas. I'm
introducing H2 gas at a pressure of one bar or 180. I'm introducing H2 gas at a pressure of one bar or almost 1. Now people when H2 gas is introduced here when H2 gas is introduced this this H2 gas which is being introduced it gets absorbed on the surface of this platinum wire. It gets absorbed on the surface of this platinum wire. It gets stickked on the surface of the platinum wire. Now when the entire surface of the platinum wire gets covered with hydrogen molecules. Now from outside does it look a platinum wire to me
or hydrogen wire? From outside if I see it now it looks like a hydrogen wire to me. So can I say this hydrogen wire is introduced in a solution containing its own ions. This hydrogen wire is introduced in a solution containing its own ions. So this electrode which we have made here this is what we call as hydrogen electrode. This is what we call as hydrogen electrode. Since concentration is 1 molar, pressure is 180 1 bar or 180m. For example, I'm keeping the temperature 25° centigrade. That means the conditions which I've used here they are
standard conditions. So I'll be calling this electrode as the standard hydrogen electrode. I'll be calling this as the standard hydrogen electrode. I hope you got to know how the standard hydrogen electrode is made. Now my dear students, this hydrogen electrode again it can behave like anode. Again it can behave like cathode. Now think over it. When this hydrogen electrode behaves like anode, what is going to happen? Oxidation is going to happen. The H2 gas molecules which are there in the rod, they are going to undergo oxidation. H2 gas which is there on the rod under
goes oxidation gets converted into H positives and with that you'll be getting two electrons. This is the reaction which you'll have to remember when hydrogen electrode behaves like the anode. This is the reaction that's going to happen. If the same hydrogen electrode behaves like the cathode, if the same hydrogen electrode behaves like the cathode, then what's going to happen? H positives in the solution, they will be colliding with the rod. they'll be taking the electrons from the rod. So just reverse the same reaction. It'll be 2 * H positive plus two electrons gets what gets
H2 gas. So the hydrogen electrode it can behave like anode. It can behave like cathode. Okay. Remember irrespective of the fact hydrogen electrode behaves like anode or behaves like cathode. Let me tell you the SOP as well as SRP. The SOP as well as SRP of hydrogen electrode is taken to be 0 volts. Is taken to be zero. Okay. It is taken to be zero. It is taken to be zero. The standard oxidation potential and the standard reduction potential of the hydrogen electrode it is taken to be zero. It is taken to be zero. Or
you can write the same thing like this as well. You can write E not of H2 gives H positive 0 to + one increase oxidation. So this is your SOP. Similarly E not of H positive gives H2. This is your SRP. Both the values are taken to be what? Both the values are taken to be zero. When it comes to the hydrogen electrode SOP SRP of hydrogen electrode is taken to be zero. Yeah. Perfect. I believe things are clear till here. Well, you must be thinking, are there some other reference electrodes? This hydrogen electrode, it's
a reference electrode. You must be thinking, are there some other hydrogen electro? Are there some other reference electrode? Absolutely, there are some other reference electrodes as well. What are the other reference electrodes that we have? What are the other reference electrodes that we have? One is called as caluml electrode. This is one of the reference electrodes like your standard hydrogen electrode. Right? Another is silver silver chloride electrode. These are another two reference electrodes which are used. Basically the standard hydrogen electrode it is comparatively difficult to make. The standard hydrogen electrode is comparatively difficult to make.
Right? So the other reference electrodes which are comparatively easy to make that is your calaman electrode. Okay. Your silver silver chloride electrode. These are the other reference electrodes which we have. Now my dear students since I'm discussing electrode potential with you now you must be thinking what kind of knowledge what kind of information this electrode potential gives us. So here comes a point that is significance of electro potential that is significance of electrode potential. What is its significance? Where do we use this? Where do we use this? Try to understand properly what I'll be saying.
Pretty much simple guys. Have a look. Let me tell you this electrode potential. Electro potential is of two types. Oxidation potential, reduction potential. Remember oxidation potential gives the tendency of an element to undergo oxidation. Oxidation potential is G it gives the tendency of an element to undergo what to undergo oxidation. Whatever tendency element has to undergo oxidation that is checked by its oxidation potential value. Similarly point number two a reduction potential gives the tendency of an element to undergo reduction. Reduction potential it gives me the tendency of an element to undergo reduction. Oxidation potential gives
me the tendency of an element to undergo oxidation. A reduction potential gives me the tendency of an element to undergo reduction. Now remember two points which I'm going to tell you now very very very important points. Remember more the SRP of an element more the SRP of an element more will be the tendency of an element to undergo reduction. more the SRP of an element more will be the tendency of an element to undergo reduction that element which under goes reduction what do we call that as oxidizing agent so better will be the oxidizing agent
and remember oxidizing agent it's oxidizing power better the oxidizing agent more will be its oxidizing power so this is the first statement which you guys need to remember more the SRP of an element More will be the tendency of an element to undergo reduction. Better oxidizing agent more oxidizing power. Remember the statement like this only. Okay. Second point my dear students. More the SOP of an element, more is the tendency of an element to undergo oxidation. More the SOP of an element, more is its tendency to undergo oxidation. Better will be the reducing agent. Better
will be the reducing agent. and more will be the reducing power. These two statements are very very very important. So you need to remember them on priority my dear students. So oxidizing power and SRP they are directly proportional. Oxidizing power and SRP. Okay. Perfect guys. So more the SRP, more the tendency of an element to undergo reduction, better the oxidizing agent, more the oxidizing power. Clear with us everyone. Clear to everyone. Quickly in the chats people quickly everybody quickly quickly quickly. Everyone so can I conclude two statements here? What are the statements? Since SRP and
oxidizing power they are directly proportional. So I'll say oxidizing power is directly proportional to SRP. If it is directly proportional to SRP, it will be inversely proportional to SOP. Correct? Similarly, your reducing power. Reducing power is directly proportional to what? Your reducing power is directly proportional to SOP and inversely proportional to SRP. remember these two statements and see now see now how how I'm going to utilize these concepts and how I'll be framing the questions simple guys try to understand very properly what I'm going to say for example I'm taking four elements A B C
and D these are the four elements which we have A B C D okay let's say here I'm writing E not of M gives Mn positive E not of M gives MN positive. What does it represent? SOP or SRP. First tell me that M gives MN positive 0 to + N increase oxidation. So this represents SOP. Let's say this value is 1 volt. This value is 2 volt. This value is 3 volt. This value is 4 volt. So tell me which one has got more SOP. This more the SOP more the tendency to undergo oxidation
better the reducing agent more the reducing power. So this D will be having maximum reducing power here. Got it? For example, I'm writing like this. E not of M gives MN positive. What is it again? Again it's SOP only. But now I'm keeping the values. For example, -1 volt, -2 volt, -3 volt, -4 volt. Now tell me which one has got more SOP. Now this has got more SOP. more the SOP more the tendency to undergo oxidation better the reducing agent more the reducing power it's reducing power here would be more okay similarly for example
I'm writing E not of MN positive gives M is it SOP or SRP MN positive gives N plus N to zero decrease reduction so this is SRP let's say this value is 1 volt this value is 2 volt This value is 3 volt. This value is 4 volt. Now, which one has got more SRP? As you can see, this one has got more SRP. More the SRP, more the tendency to undergo reduction. More the SRP, more the tendency to undergo reduction. Better the oxidizing agent, more the oxidizing power. So, it's oxidizing power more. For example,
I'm writing something like this. Again I'm writing E not of M N positive gives M. Okay, let's say I'm taking the values as -1 volt, -2 volt, -3 volt, -4 volt. Tell me which one has got more SRP. Now I would say this one has got more SRP. More the SRP, more the tendency to undergo reduction, better the oxidizing agent, more the oxidizing power. It's oxidizing power will be more. Am I clear with all this? Everybody in the chats, am I clear with all this? Everybody in the chats quickly, quickly people. Everybody, did you understand
the concept in a better way? Everyone in the jazz people What the hell? Perfectly done. Perfect. Let's try to see few questions on this electrode potential so that you'll get used to all the things. Look at the first question. You are given four reactions. You have to check which of the following is the strongest reducing agent. Strongest reducing agent. Okay guys, have a look understand. Reducing agent. Reducing agent is the one. Which of the following is the strongest reducing agent? Reducing agent is the one which under goes oxidation. Step by step I'm telling you. Reducing
agent is the one which under goes oxidation. So that means the one that under goes oxidation easily will leave the better reducing agent. Now which one under goes oxidation easily? The one which has got more SOP. So you have to answer the question in terms of what? You have to answer the question in terms of SOP. What are these values? Are these SOP values or SRP values? If you look all the reactions in every reaction reduction is happening. Gain of electron is happening. Mg positive is gaining electron. So mg dipositive is undergoing reduction. So this
is the srp of this mg dip positive. Zinc dip positive gaining electrons. So this zinc dip positive is undergoing reduction. Right? SRP of zinc dip positive. Nickel dipositive is undergoing reduction. So this SRP this is SRP. So these are basically the SRP values. These are basically the SRP values of what? Of all these ions which we have. But we have to answer the question in terms of SOP. So what I'll be doing I'll be reversing all these reactions. When I reverse the reactions the first reaction becomes mg solid gives mg di positive plus2 electrons. Okay.
Now its e not value will it be minus 2.37 or plus 2.37? It'll be plus 2.37 volt. Similarly zinc solid gives zinc diit plus two electrons. Its E not value will be plus 0.76 volt. Similarly, nickel solid gives nickel die positive plus 2 electrons. It's E not value will be plus 0.73 volt. I'm just reversing the signs. That's it. Fe solid gives Fe tri positive plus 3 electrons. Its E not value will be plus 0.04 volt. Now tell me are these SOP values now or SRP values? So what do you think is it SOP now
or SRP? See magnesium is losing electrons. So magnesium now is undergoing oxidation. Magnesium is undergoing oxidation. Zinc is undergoing oxidation. Nickel is undergoing oxidation. Iron is undergoing oxidation. So these are basically the SOP values. These are basically the SOP values. SOP values of what? SOP values of all these metals here. Now which one has got maximum SOP? This is the maximum SOP. So I would say magnesium has got maximum SOP. If magnesium has got maximum SOP, more the SOP, more the tendency to undergo oxidation, better the reducing agent. So which one is the better reducing
agent? It's going to be magnesium. Yes, I believe it is clear to everyone. One more question of the similar pattern so that things will be like very very very much clear to you. We have to give the oxidizing power. We have to give the oxidizing power. Understand? Oxidizing power. Which has got more oxidizing power? The one that is the better oxidizing agent? The one that is the better oxidizing agent has got more oxidizing power. Now oxidizing agent itself under goes what? Reduction. So the one that under goes reduction easily. Better oxidizing agent. Now which one
under goes reduction easily? The one that has got more SRP. The one that has got more SRP. So you are going to solve the question in terms of SRP. Now tell me are we given with SOP values or SRP values? Everywhere gain of electrons is happening. Gain of electrons is what we call as reduction. So everywhere SRP is given. Now cobalt tripositive gaining electron. So this cobalt tripositive is undergoing reduction. Serium plus4 is undergoing reduction. Pb +4 undergoing reduction. Bi plus3 undergoing reduction. Now which one has got maximum reduction potential? Take the values. Which one
has got maximum reduction potential? This is the maximum reduction potential value right. Reduction potential means reduction which is undergoing reduction cobalt trip positive. So cobalt tri positive it has got maximum reduction potential right. So if cobalt triositive has got maximum SRP it will under go reduction easily better oxidizing more oxidizing power. So cobalt tri positive has got the highest. After cobalt tri positive which one is the next value 1.67 that is the SRP of what? PB +4. So PB+ 4 is going to be the next. So it's option B that is going to be the
correct answer of the question. Did you understand everybody? If yes then tell me the answer of this question. The reducing power you have to give me the order of reducing power. Now more reducing power means better reducing agent. Better reducing agent means it should undergo oxidation easily. Which under goes oxidation easily? which has got more SOP. So the question has to be solved in terms of SOP. What are these values? These are SRP values. As you can see everywhere reduction is taking place. So this is SRP of zinc dip positive. SRP of calcium dipositive, SRP
of magnesium diposit, SRP of this. But I have to solve the question in terms of SOP. So reverse the reactions. So first reaction becomes zinc solid gives zinc di plus two electrons. Now this is not the reduction reaction. Now this is the oxidation reaction. And which one is undergoing oxidation? Zinc is losing electrons. So, zinc is undergoing oxidation. So, its E not value will be plus 0.76 volt. Similarly, this was the E not of zinc. Similarly, what would be the E not of calcium? SOP of calcium SOP of calcium will be plus 2.87 volt. SOP
of magnesium SOP of magnesium will be + 2.36 volt. Similarly, SOP of nickel would be plus 0.25 volt. Now tell me which has got maximum SOP. These are all SOP values. These were the SRP values of these ions which were undergoing reduction. Now these are the SOPs of these metals which are undergoing oxidation. So which one has got maximum SOP? Now this one calcium. So if calcium has got maximum SOP maximum SOP means maximum reducing power. Now after calcium the next is going to be magnesium. Next is going to be magnesium. So it's going to
be option B. Did you understand? How this sort of a question is to be solved. Everyone people everyone. Did you understand? Yeah. perfect guys. Perfect. Perfect. Perfect. Now let's move on to one more topic. What is that? That is your EMF. EMF. How do we exactly define the term EMF? How do we exactly define the term emf? Let me tell you in simple words emf is defined as the potential difference between the electrodes between the electrodes when the cell is not in use. That means when no current is withdrawn from the cell when no current
is drawn from the cell. See guys first of all you know your Daniel cell. What does the Daniel cell do? Daniel cell produces current. Daniel cell produces what? Daniel cell produces current. Now current flows only due to potential difference. In case of your Daniel cell, you have got zinc electrode and you have got copper electrode. You know that this is your zinc electrode. This is copper electrode and connected externally as well as internally. This is your Daniel cell. What does it do? It produces current. Current flows due to what? Potential difference. So that means my
dear students, there will be potential difference between these two electrodes. There will be potential difference between these two electrodes due to which current will be produced. There will be potential difference between these two electrodes due to which current will be produced. Now the maximum potential difference between the electrodes the maximum potential difference between the electrodes when the cell is not in use when no current is drawn from the cell you call that maximum potential difference as the emf. You call that maximum potential difference as the emf. Okay, I'm not going into much details of EMF.
Perfect. I'll just teach you all the things which can which can be asked in your exam. Nothing more, nothing less. Okay? Otherwise, let me tell you this emf is measured by an instrument called as potentiometer. This emf is measured by an instrument called as potentiometer. But let's not talk about that. Let's not exactly see how potentiometer is used to calculate the EMF and all right that's of no use to us for our examination purpose. Okay just remember one thing emf is nothing but it is the maximum potential difference between the electrodes when the cell is
not in use. Yeah you can remember it like this as well. You can remember it like this as well. For example, I got a battery. I got a cell from the market. Let's say I got a cell from the market. This is your battery which I got from the market. The battery is right now in the wrapper only. It is wrapped. Okay, it is wrapped. So the battery is right now not in use. It is not in use. On the wrapper they have mentioned 5 volt. On the wrapper they have mentioned the voltage of the
battery is 5 volt. So right now the cell is not in use. It is not in use. Now when I put the same battery in this remote for example the battery starts working. The battery starts working. So with time with time when the battery starts working with time the potential difference between the electrodes here in this battery will be 4 volt then 3 volt then 2 volt then there will be a time and it will be 0 volt. Then when it is 0 volt that means the battery will be dead. You can remember it like
this. Okay. That means at 0 volt the battery is dead. At this particular voltage the battery was working. At this particular voltage, the battery was working. At this particular voltage, the battery was working. Here the battery was not in use. The cell was not in use. Now, which one is the maximum potential difference among all these? 5 volts. This is the maximum potential difference when the cell is not in use. And the maximum potential difference between the electrodes when the cell is not in use, you call that as the emf. So this 5 volt is
what I call as the emf of the cell. Rest all the other voltage apart from 5 volts this 4 volt 3 volt 2 volt they are called as cell voltage. Maximum potential difference is called as emf all the other voltage are called as normal cell voltage. Okay you can remember it like this. You can remember it like this. In reality if we have to understand emf then we have to study potentiometer first. Okay. Then after studying potentiometer then I'll I can tell you what emf exactly is. For that you have to take a galvvinic cell.
you have to connect that with the external battery in reverse polarity right then only we can talk about those things but that's of no use to us right now so just remember the potential difference between the electrodes when the cell is not in use right when no current is drawn from the cell is called as emf now my dear students one more thing when emf when emf is measured when emf is measured Under standard conditions the maximum potential difference between the electrodes when the cell is not in use that's called as emf. When emf is
measured under standard conditions you call that as the standard emf which is represented by e not cell. You call that as the standard emf which is represented by e not cell. First of all emf is represented by what? EMF is represented by E cell. Your standard EMF is represented by E not cell. So if you have your gonic cell in your mind, if you have your Daniel cell in your mind, the maximum potential difference between the electrodes when the cell is not in use, that's called as EMF. When that maximum potential difference is measured under
standard conditions, when that maximum potential difference is measured under standard conditions, you call that as a standard emf which is represented by E not cell. And remember my dear students, this E not cell is nothing but it is equal E not of cathode minus E not of anode. This E not cell is nothing but it is E not of cathode minus E not of anode. What is this E not of anode? E not of cathode here. E not of cathode represents the SRP of cathode. E not of anode again represents SRP of anode. So SRP
of cathode minus SRP of anode. That gives me the E not cell. That gives me the standard EMF of the cell. Well, my dear students, you can do one more thing as well. You can convert this result in one more form. See, I use the term SRP of cathode here. SRP of cathode. Instead of SRP of cathode, you can write minus * SOP of cathode. You can write minus* SOP of cathode. In the middle the sign is minus. This is SRP of anode. You can write minus * SOP of anode. So - plus the result
becomes SOP of anode minus SOP of cathode. You can remember the result like this as well. You can remember the result like this as well. But no need to remember two results, right? Just remember this particular result. You can make one more result. See this result. In this particular result, both are SRPs. In this particular one, both are SRPs. You can keep one as SOP, one as SRP. Right? You can keep one as SR SOP, one as SRP. So you can form other results as well. But our priorities this particular result that is E not
cell is equal to E not cathode minus E not anode. E not cathode is SRP of cathode. E not anode is SRP of anode. Simple. This particular result is important. Now guys try to understand where I use this particular result. Have a proper eye on the screen now because something which I'm going to tell you now that is something very very very important. Okay. So try to understand what all things I'm going to tell you now. For example, I'm writing something like this. E not of a positive gives a is equal to 1 volt. Here
I'm writing E not of B positive gives B is equal to 3 volt. If I ask you how many electrodes do I have right now? Can I say I have two electrodes? I have two electrodes. For example, for example, I am connecting these electrodes externally as well as internally with the help of salt bridge. I'm connecting these electrodes externally as well as internally with the help of salt bridge. Can I say after connecting these electrodes, I will be getting a complete galvvinic cell. A complete galvvinic cell consists of two electrodes. I took two electrodes. I
connected them externally as well as internally. So I will be getting a galic cell. Now the point is how do you calculate the standard emf of that galvvinic cell. And you know standard emf of the galvvinic cell is nothing but that is e not cathode minus e not anode. Srp of cathode minus srp of anode. Now you guys are going to tell me one thing. You guys are going to tell me one thing. See is this SOP or SRP? + 1 to 0 decrease reduction. So this SRP plus one to zero decrease reduction. So this
SRP which one has got more SRP this one more the SRP more the tendency to undergo reduction and reduction takes place at cathode. So this will be your cathode and this will be your anode. Since you identified your cathode and anode since you identified your cathode and anode. So what was this E not C? This was SRP of cathode. And this was SRP of anode. This was your SRP of anode. SRP of cathode is how much? SRP of cathode is 3 volt minus SRP of anode. SRP of anode is 1 volt. The value comes out
be 2 volts. So this 2 volts is going to be the standard emf of the cell. Simple. Did you understand? Let me take one more example so that things will be more clear to you. For example, here I'm writing E not of M gives MN positive. Let's say this is - 5 volt. Here I'm writing E not of N negative gives N. It is equal to 5 volt. Okay. So how many electrodes do I have right now? How many electrodes do I have right now? I again got two electrodes. connect them externally as well as
internally. I will be getting a complete galvvinic cell. I will be getting a complete galvvinic cell. Now we have to calculate the standard emf of that complete galvvinic cell. We have to calculate E not cell. First of all tell me one thing. What is this particular value? Is this SOP or SRP? See 0 to + N. 0 to plus N increase oxidation. So this is SOP. -1 to 0 increase -1 to 0 - 1 to 0 increase oxidation so this SOP do I have any result I mean do I use the result in terms of
SOP or SRP I use the result in terms of SRP so convert this SOP into SRP so it's SRP will be + 5 volt its SRP will be minus 5 volt right so I got the SRP values now tell me which one has got more SRP first or second has got more SRP more the SRP more the tendency to undergo reduction and reduction takes place at cathode. So this is going to be your cathode. This is going to be your anode. Since you identified your cathode and anode, Since you identified your cathode and anode, now
what would be E not cell? Eode cell would be E not cathode. SRP of cathode minus SRP of anode. SRP of anode is - 5. So 5 - 5 comes out be 10 volt. So 10 volt is the standard AM of the cell. Understood? Understood people one more example with that things will be more clear. Let's say I'm writing E not of A gives A D positive. This is equal to minus 3 volt. Okay. E of B D positive gives B is equal to 4 volt. I've got again two electrodes. Again I've got two electrodes.
Again I'm connecting them externally as well as internally. So I am getting a complete galvvinic cell. Now 0 to +2 increase oxidation. So this is your SOP plus 2 to 0 decrease. Decrease means reduction. So this is your SRP. Now this is already your SRP. But here it's SOP. So convert this SOP into SRP. So it'll be plus 3 volt. This is SRP. Now compare the SRP values. Which one has got more SRP? This has got more SRP. More the SRP, more the tendency to undergo reduction. Reduction takes place at cathode. So this is your
cathode. This is your anode. Since you identified your cathode and anode, now you can easily check E not cell. And E not cell is going to be nothing but E not cathode minus E not anode. E cathode means SRP of cathode. SRP of cathode is 4 volt minus E not anode. SRP of anode. SRP of anode is how much? SRP of anode is 3 volt. So 4 - 3 comes out be nothing but 1 volt. So this is how if you'll be having two electrodes if you connect them externally as well as internally. If you
connect them externally as well as internally, you'll be getting a complete galvanic cell. And from a complete galvanic cell, you can easily calculate what? You can easily calculate the standard emf of the cell. If I'm clear, tell me once in the chats. Right people all clear. Okay. So, one question for you. Tell me the answer of this one. Sir N to N negative. Where was N to N negative? Just a second. Are you talking about this one? This is this N negative to N minus 1 to0 increase. Increase increase means oxidation. So this SOP SRP
will be minus 10. Is this this correct only? N to N negative will be 0 to - 1. 0 to - 1 means SRP decrease. Okay. All right. Tell me the answer of this question. As far as the question is concerned, the standard oxidation potential values are given. Standard oxidation potential values are given. But do I need standard oxidation potential values? No, I don't need SOP values. I need SRP values. So what would be the SRP of this one? SRP of the first one will be -0.76. SRP of the second one will be -0.41. I
got their SRPs. I got their SRPs. Now which one is more? This is more more the SRP more the tendency to undergo reduction reduction takes place at cathode. So this is cathode. This is anode. Basically your iron is going to behave like cathode and your zinc is going to behave like anode. Now what am I supposed to calculate? I'm supposed to calculate E not cell. E not cell is nothing but E not cathode minus E not anode. So this is SRP of cathode. SRP of cathode is -0.41 minus SRP of anode that is -0.76. The
value comes out be plus0.35 volt. Okay. So this is the E not cell. All clear with this. Everybody in the chats. All clear with this. Perfect. Okay. Now comes one something important. What is this important thing? See guys, if I ask you one thing, you tell me in the chats. Okay, if I ask you in order to make a galvvinic cell, how many electrodes are required? Just tell me the answer of this one. In order to make a galvvinic cell, how many electrodes are required? Your answer will be two electrodes are required. Your answer will
be two electrodes are required. If I give you any two electrodes, if I give you any two electrodes, okay? And if you connect them externally and internally you'll be getting a galvvinic cell, complete galvvinic cell. Now the point is will that galvvinic cell will work or not will that galvvinic cell works or not? That is the question I am telling you in order to make a galvvinic cell you need two electrodes. Connect them externally internally. No doubt you made a galvvinic cell but the point is will that work or not? Will that produce current or
not? Okay, that will be told by this particular topic that will be given by this particular topic. This particular topic is going to tell us whether the galvvinic cell will work or not. So first of all I am going to give you one result which I'm not going to derive because the derivation is of no use to us right now. Okay, I'm just going to give you the result and we'll see how it's applied. The result is delta G is equal to minus NF cell. This is the relation between delta G and emf of the
cell. If you write the same equation under standard conditions, If you write the same equation under standard conditions, then then instead of delta G, you'll write delta G KN is equal minus NF cell minus NF cell. The same equation which you have written under which conditions? Under standard conditions. Now guys have the proper eye on the screen and understand what exactly I'm going to tell you. If you take two electrodes, if you take two electrodes, if you connect them externally and internally, you'll be getting a complete galvvinic cell. Now will that galvvinic cell work or
not? That decides on just one thing whether the cell reactions in the cell whether they are happening on its own or not. That means in other words I can say whether the cell will work or not it depends on just one thing whether the cell reactions are spontaneous or not. Whether the reactions are happening at anode and cathode whether the reactions are happening on its own or not. If the reactions are happening on its own if the cell reactions are spontaneous the cell will work. If the cell reactions are non-spontaneous if they are not happening
at anode cathode so the re those so the cell will not work. As simple as that. Okay. So I would say something like this for a cell for a cell whose delta G for the reactions will be negative. Delta G for the reaction negative means delta G negative means reactions are spontaneous or any spontaneous process delta G for the system is negative. So for example, I've got a galic cell whose delta G is negative. Delta G negative means minus NF cell. This term is negative. When can this particular term be negative? This entire term can
be negative only if E cell is positive. If E cell is positive. So if you have got a galvvinic cell whose delta G is negative or you can say whose E cell is positive I would say in that galvvinic cell the cell reactions will be spontaneous. The cell reactions will be spontaneous. That means at anode automatically oxidation will take place at cathode automatically reduction will take place or in short I would say electrons will move from anode to cathode. Current will go from cathode to anode. So the cell works like a normal galvvinic cell. The
cell works like a normal galvvinic cell wherein wherein electrons they move from anode to cathode and current goes from cathode to anode. Okay. Is this particular point clear? The second point which is similar to this. What is that? If you have got a complete galvvinic cell whose delta G is positive whose delta G is positive. Delta G positive means minus NF E cell is positive. Minus NF E cell can be only positive if E cell is negative. If E cell is negative. So if you have got a galvanic cell whose delta G is positive or
E cell is negative or E cell is negative in that galvvinic cell reactions if delta G is positive that means cell reactions are non-spontaneous that means the reactions won't take place on its own if the cell reactions are non-spontaneous that means the cell does not work the cell does not work like like a normal gammic cell. Understood? Similarly, the third point, if delta G for the cell is zero, what does that mean? That means minus NF cell value is zero. It is possible only if E cell value is equal to Z. E cell 0 means
there is no potential difference between the electrodes. If there is no potential difference between the electrodes, the cell does not produce any current. And this state is called as the equilibrium state in the cell. The cell is present at its equilibrium state. This we have to discuss in detail in some time. Just remember it for now. Just remember it for now. Are all these things clear? If I quickly summarize this, if I quickly summarize this, let me tell you for any galvvinic cell to be working the cell reactions in the galvvinic cell must be spontaneous.
And for the cell reactions to be spontaneous, delta G for the cell should be negative or emf should be positive. That is a criteria for the cell to work. Am I clear? Am I clear my dear students? Everybody. Okay. Is this clear? This particular point we have to discuss in detail after some time. Now if this is clear guys, then there is there is something more which I would want to share with you. What is that? Whenever in a question you are supposed to calculate delta G, let me tell you the SI unit of delta
G. The SI unit of gives free energy change is nothing but it is jewel. It is jewel delta G. Sometimes in the question they'll ask you to calculate the nonPV work. Sometimes in the question they'll ask you to calculate the nonPV work or they might ask you to calculate the useful work. They might ask you to calculate the useful work. Whenever you are supposed to calculate useful work or non-p work that means you have to calculate the value of delta G. Sometimes they can play with the language of the question as well. Sometimes in the
question they might ask you to calculate useful work to calculate nonp work. That means indirectly they're asking you to calculate delta G. And delta G is nothing but minus NF cell. The third point delta G is an extensive property. Delta G is an extensive property and extensive properties can be added or subtracted directly. Delta G is an extensive property and let me tell you extensive properties can be added or subtracted directly. Whereas whereas if I talk about your E electrode potential or emf it is an intensive property. It is an intensive property. So these values
cannot be added or subtracted directly. Delta G can be added or subtracted directly because delta G is an extensive property. Okay. In see it is basically the point from thermodynamics chapter. Thermodyamics says that extensive properties can be added or subtracted. Intensive properties cannot be added or subtracted directly. Okay. So let me tell you delta G it is the example of your extensive property. So it can be added or subtracted whereas E cannot be added or subtracted directly. Okay. I believe I'm clear. Okay. One more thing. One more thing you need to remember it as well.
For example, you have got a reaction like this. A + A + B gives A positive plus B negative. Let's say this is a reaction general reaction. For example, its E not value is equal to let's say some X volt some X volt. Okay, its E not value will be some XV or let me make things little clear to you, little little easy to you. For example, I have got m n positive plus n electrons gives m. Okay, let's say it's e not value is equal for example 2 volt. Okay, remember when you multiply this
reaction by some number, when you multiply this reaction by some number, nothing happens to its E not value. Remove this point directly. When you multiply this reaction by some number, nothing happens to its E not value. When you divide this reaction by some number, nothing happens to it. C not valid. But when you reverse the reaction, you have to change the sign. Sign becomes minus2. Then remember these three points as well. Okay. So let me summarize this point. Whatever I've told you, the point is when a reaction is multiplied by some number happens to its
E not value. Okay, you can you can you can modify it. You can write it as when a reaction is multiplied or divided by some number. Okay, multiplied or divided by some number. Nothing happens to C not value. Perfect. But when a reaction is reversed, the sign of E not is also reversed. When a reaction is reversed, sine of E not is also reversed. Perfect. Now these points are very very very important when it comes to the questions. Okay. And what kind of questions can be asked from this topic? First one is the simplest among
all. First one is the simplest among all. The question is calculate delta G for the reaction when the concentration is equal to 1 molar temperature is 298 Kelvin. So that means we have to calculate delta G under standard conditions. Delta G under standard conditions is delta G not basically. So delta G not is minus NF not cell minus NF E not cell minus NF not cell. So first of all the cell that is given to me the cell that's given to me I need to calculate it E not cell. What is E not cell? E
not cell is nothing but that is E not cathode minus E not anode. E not cathode means SRP of cathode. Few minutes back I told you E not an anode means SRP of anode. SRP of anode. Now in this reaction copper is undergoing oxidation. Copper is undergoing oxidation. Oxidation takes place at anode. So this copper electrode is your anode. That means this tin is your cathode. So you identified your anode and cathode. Uh just a second guys. I think the values are wrong in the question. Because in the question they have shown copper undergoing oxidation
right if copper is undergoing oxidation so copper has to be anode that means the still has to be cathode if the still has to be cathode that means it srp should be more but here they have shown srp of this is less srp of this is more no so that means you the question will be like this it won't be minus it will be plus and this will be minus this will be the equation this will be the question. Okay, this would be the question. Did you understand? I'll give the break at 9 guys, not
now. Did you understand? Now I identified my cathode and anode. Now what is E not of cathode? SRP of cathode. SRP of cathode is given as 0.16 minus SRP of anode is given to me as -0.34. The value comes out to be 0.50 volt. If E not cell is 050 volt. Now tell me delta G KN is equal to minus. What is the N value for the reaction? If you look at the reaction carefully, I have taught you how to calculate N value. N value for the reaction is two. This f represents Faraday's constant and
the value of Faraday's constant you should be knowing 96500 kum. So f is a constant farad's constant whose value is 965 0 kum e not value is equal to 0.5. It is just a matter of calculation.5 into 2 is 1. So delta g will come out to be - 96500 and answer is going to be in jewles. Right? Since you have taken all the parameters as per si system. So delta G not will also come as per SI system and the unit of delta G as per SI system is jewel. Okay. Now comes one more
question guys. This is one of the type of the question which can be asked from just the formula. The second one you are given with three reactions. You are given with three reaction. Out of three reactions E value of two reactions is given to us. E value of two reactions is given to us. Okay. We have to calculate E value for the third reaction. We have to calculate E value for the third reaction. Let me tell you my dear students, the reaction for which E is to be calculated, the reaction for which E is to
be calculated, we are supposed to make that reaction out of those reactions whose E value is given. The reaction whose E value is to be calculated. We are supposed to make that reaction out of those reactions whose E value is given. Now understand if I call this is equation number one. If I call this is equation number two, I'm calling this is equation number three. I have to make third equation from first and second. Now think over it. How can you make third reaction from first and second? It's pretty much simple. If you add these
two reactions, Cu positive, Cu positive cancels out. It becomes Cu Di positive. 1 and one two electrons gives Cu solid. Cu dipos plus two electrons gives Cu solid. So basically third reaction we are getting when we are adding first and second. Now if you are thinking that instead of third you'll write E3 instead of one you are writing E1 instead of two you can write E2. If you are thinking like this that means you are gone you're doomed. Why? As I told you a few minutes back, E is not an extensive property. It's an intensive
property. And intensive properties cannot be added or subtracted directly. Intensive properties cannot be added or subtracted directly. Extensive properties can be added or subtracted directly. So that means my dear students, you are not going to solve this question in terms of E. You will be solving this particular equation in terms of delta G because delta G is extensive and you can add or subtract extensive properties directly. So how we are going to solve this? Instead of third you write delta G3 instead of one you'll write delta G1 instead of two you'll write delta G2. Now
delta G3 is nothing but minus N3 F3 is equal delta G1 minus N1 F1 delta G2 minus N_sub_2 F2. So first of all f FF everywhere canceled multiply through to minus sign. So E3 value will be equal N_sub_1 E1 plus N_sub_2 E2ide by N3. Now as per the question E1 is given E2 is given E1 is given E2 is given. But now what is this N1 N2 N3? This is the value of your N1. So N1 value is one. One electron here also one electron. So N2 value is also one. Here you can see two
electron. So N3 value is two. So n1 we know n2 we know n3 we know a1 we know a2 we know. So put the values get the e3 value. Did you understand? Did you understand people? Everybody did you understand? Let me give one more question of the similar format so that you'll understand it. You have got a reaction Fe2 plus 3 electrons gives Fe solid. Fe2 positive plus2 electrons gives Fe solid. This is gives here. Okay. This is gives arrow arrow and this is also arrow. So I'm basically given with three reactions. Correct? I'm given
with three reactions out of which E value for two reactions is given and E value for third reaction is to be calculated equation E value for two reactions is given E value for third reaction is to be calculated what am I supposed to do I'm supposed to make this equation out of these two calling this is equation number one calling this is equation number two now third equation I have to make from first and second now how can we make third equation In the third equation you have C of one electron here. So what you
have to do? You have to subtract these two. You have to subtract second from first. So if this is positive, this becomes negative. If this is plus two electrons, this becomes -2 electrons. This becomes minus Fe. So when you subtract these two, what do you get? You get Fe trip positive plus 3 electrons minus Fe dositive minus 2 electrons. This time this cancels nothing on this side. So three electrons - 2 electrons is one electron. So it becomes Fei positive plus one electron. Take this minus Fei positive on that side. It becomes plus FI positive.
Was I supposed to make this reaction only? Take it out. I was supposed to make this reaction only. Yeah, I was supposed to make this reaction only. How did I make this reaction? How did I made the third equation? Third equation I made when I subtracted second from first. When I subtracted second from first. Now, am I going to do the same operation with their E values? No. I'm going to do the same operation with their delta G values. So this is going to be delta G3 is equal delta G1 minus delta G2. Delta G3
is nothing but minus N3 F3 is equal. Delta G1 is - N_sub_1 F1 minus - plus N_sub_2 F2. So F FF everywhere cancels if you multiply with minus sign throughout. So what do you get at the end? you'll be getting E3 is equal N_sub_1 E1 minus N_sub_2 E2 divided by N3 value. Now Nsub_1, N2, N3 are given. This is our N1 value. This is our N2 value. This is our N3 value. So N1, N2, N3 are given. E1 E2 values are also given. Even E2 values are also given. Just put the values and get the
value of E3 which you were supposed to calculate. As simple as that. Yes, I believe I'm clear to everyone. I believe I'm clear to everyone. Let me solve one more question for you. Look at this particular question. Let me write these reactions. First of all, the first reaction is M4 negative gives what? It gives Mn4 D negative. Okay. Its E1 value is for example given to me as 1 volt. Second reaction is M4 D negative gives M O2 its E2 value is equal to 2 volt. Third one MN O2 gives M and di positive. Its
E3 value is for example equal to 3 volt. Mn4 negative gives M& die positive. It's E4 value is to be calculated. Now the reaction whose E value is to be calculated. The reaction whose E value is to be calculated. You are supposed to make that reaction. But before that in the first reaction what about the electrons? If you look carefully the oxidation state of magnes will be + 7. The oxidation state of magnes will be plus 6 + 7 to + 6 decrease reduction how many change in the oxidation state final minus initial it will
be one electron so basically m4 negative would be gaining one electron then getting converted into m4 d negative similarly oxidation state of magnes here will be + 6 and it will be +4 plus 6 to +4 again decrease reduction gain of electrons how many two because plus 6 to + 4 similarly this is + 4 here + 4 to +2 decrease reduction again two this is + 7 here + 7 to +2 so this has to be five electrons yeah guys I am basically supposed to make this particular reaction now think over it how will
you make this particular reaction how will you make this particular reaction the reaction which we have to make we need five electrons here how do you get five electrons 2 + 2 + 1 2 + 2 + 1 is 5. So basically I can do one thing. I can say fourth equation is made when you add first second and third and whatever operation you are doing with the reactions. Same operation you have to do with the delta G values. So this will be delta G4 is equal to delta G1 plus delta G2 plus delta G3.
Now delta G4 will be equal to how much? It'll be - n4 f4 is equal - n1 f1 minus n_sub_2 f2 - n3 f3. So f ff everywhere cancels out. Multiply with minus sign and at the end you'll get e4 e4 will be n_sub_1 e1 plus n_sub_2 e2 plus n3 e3 divide by what? N4. Now nsub_1 is 1, n2 is 2, n3 is 2, n4 is 5. So n1, n2, n3, n4 we have all the values. E1, e2, e3, e1, e2, e3 is given. Put the values and get the e4 value. That's all. Is this
particular concept clear? Did you understand how this sort of a question is to be solved? I would want answers from everybody. Will you be able to solve this sort of a question if it is asked? Everyone in the chats guys everyone in the chats quickly quickly quickly quickly after this topic I'll give you the break. Electrochemical series done In short, one thing you have to remember the reaction whose E is to be calculated, you are supposed to make that reaction out of those reactions whose E value is given. Perfect guys. Now let's have a look
on this electrochemical series. What is an electrochemical series? First of all my dear students simple when these elements when the metals these are metals first of all when the metals are arranged on the basis of their increasing SRP values. When the metals are arranged on the basis of their increasing SRP values. Okay. The series of elements which you get here is called as electrochemical series. You would have seen it already. You would have seen it already. When different metals are arranged on the basis of their increasing SRP values, the series of elements which you get
here is called as your electrochemical series. So this is your lithium, potassium, calcium, sodium, magnesium, aluminium, magnes. Right? This is your electrochemical series wherein I have arranged the metals as per their increasing SRP values. Now guys see what you have to do. This hydrogen is our reference. This hydrogen is our reference. And you know SRP of hydrogen is zero. SRP of hydrogen is zero. Right? That means all the elements which are before hydrogen all the elements before which are before hydrogen then SRP value will be negative. Simple I arranged these elements on the bas of
their increasing SRP values. So if SRP of hydrogen is zero that means the SRP of all these metals which are coming before hydrogen will be negative or in short I would say the metals which come before hydrogen their SRP is less. The metals which come before hydrogen their SRP is less. The elements which come after hydrogen their SRP is more. Okay, this is the first point which you have to understand. Before hydrogen you have got lesser SRP elements. After hydrogen you have got more SRP elements. Yeah. Now what kind of information this electrochemical series gives
us? First of all, this electrochemical series gives me the idea about the oxidizing power of an element which we have already discussed. The one that has more oxidizing power is the better oxidizing agent. The one which is the better oxidizing agent that will undergo reduction easily. The one that underos reduction easily that has got more SRP. Right? So your oxidizing power it is directly promotional to what? SRP and inversely promotion to SOP. This is something which we have already discussed. Something that we have already discussed, right? Reducing power. More the reducing power, better the reducing
agent. Better the reducing agent. More is its tendency to undergo oxidation. That means more would be SOP. Right? So reducing power is directly proportional to what? SOP. Inversely proportional to what? SRP. Yeah. Perfect. This is also something which we have already discussed. This is something which we have not discussed. Predicting displacement reactions. For example, if you look at this particular reaction, you have got a solution of copper. You have got a cooked solution of copper. And in this solution of copper, you are introducing zinc. In this solution of copper, you are introducing what? Zinc. Now,
what the zinc is doing? Zinc is displacing copper from its solution. Zinc is displacing copper from its solution. Right? Zinc is going to displace copper from its solution. So what do you get? You get here ZnO4 plus plus copper. You took an aqua solution of copper copper and in this solution of copper you are introducing zinc. What the zinc is doing? Zinc is displacing copper here and getting converted into zinc is displacing copper. So copper when being displaced, it gets converted into zinc sulfate. When this displacement happens, that's the point. When this displacement happens, try
to understand what is the oxidation of zinc zero. Oxidation of zinc here plus two. So zinc is undergoing oxidation. Zinc is undergoing oxidation. Zinc is undergoing oxidation. Okay. If this is undergoing oxidation, if this undergoing oxidation that means this is undergoing, this copper is undergoing copper dipos is undergoing reduction. Copper dipos is undergoing reduction. The one that under goes oxidation, the one that under goes oxidation easily, its SOP will be more. Its SOP will be more. If it SOP is more that means its SRP will be less, its SRP will be less. Can I say
that element whose SRP will be less? that element whose SRP will be less will undergo oxidation. That means I can say in other words that element whose SRP will be less that will displace the element from its solution. That element whose SRP will be less that will be displacing one one more element from its solution like this. Did you understand? Now which elements are there whose SRP is less? All those elements which come before hydrogen. All those elements All those metals which come before hydrogen in electrochemical series they can displace they can displace other metals
from their aqua solution. They can displace other metals from their aqua solution. They can displace other metals from their aqua solution. So which metal can displace which metal can displace other metal from its aqua solution? that metal whose SRP is less can displace the other metal from its aqua solution. Did you understand this particular point? Everyone in the charts guys quick quick then only and move ahead. Did you understand this particular point? So if I would want to summarize it, I can summarize it like this. the metal that comes before hydrogen that means I'm talking
about that metal whose SRP is less can displace other metal that comes after hydrogen. from its aquous solution. Okay, remember this particular point. Now I'm asking you one simple question. Let's say I'm writing a reaction. Copper solid plus zinc sulfate gives copper sulfate plus gives copper sulfate plus zinc. Will this reaction happen? Is this reaction going to happen? Can copper displace zinc? See, can copper displace zinc? Quickly, this is zinc here. Copper is here. Copper is after hydrogen. Zinc is before hydrogen. Zinc can displace copper from its solution. Copper cannot displace zinc from its solution.
Right? So, this reaction cannot happen. Copper cannot displace zinc from its solution. Did you understand? I say it in the chats Okay, one more important point. Those metals which come before hydrogen in electrochemical series can liberate hydrogen from the solution of an acid like HCl. See, you would have seen one reaction in your notebooks, textbooks, wherever. Mg solid plus HCl MG solid plus HCl it gives what? It gives MGcl2 with that hydrogen gas is produced. Have you seen this sort of a reaction? This sort of a reaction you would have seen many a times or
aluminium plus HCl. Aluminium plus HCl gives Alcl3 plus hydrogen gas. So this metal is introduced in an acidic solution. This metal is introduced in an acidic solution. And what is being liberated? Hydrogen gas is being liberated. So if you if you see carefully what is happening to metal it oxidation state is zero and it oxidation state is plus2 increase in oxidation state means oxidation. So this metal here is undergoing oxidation. I would say when metal under goes oxidation then only hydrogen gas will be produced. When metal under go oxidation then only hydrogen gas will be
produced. Now which metal under go oxidation easily? The one that has got more SOP. More SOP means lesser SRP. So that metal which has lesser SRP. That metal which has lesser SRP can displacement can can produce hydrogen gas right in an acidic solution that metal whose SRV will be less it can produce hydrogen gas when it's introduced in an acidic solution like HCl or in other words which metals have got lesser SRP those metals which come before hydrogen those metals which come before hydrogen in the electrochemical series can easily liberate hydrogen gas from a solution
of an asset like HCl. Got this particular point as well. But remember this one guys, remember this one. Remember this one. When a metal When a metal which will be having lesser SRP which comes before hydrogen. When a metal which will be having lesser SRP which will be coming before hydrogen. When that metal is introduced in an acidic solution of NH3 HNO3. When a metal which comes before hydrogen which has got lesser SRP when it is introduced when it is introduced in an acidic solution of HNO3 at that point of time hydrogen gas is not
liberated hydrogen gas is not produced at that point of time oxides of nitrogen are produced at that point of time oxides of nitrogen are produced remember this particular point okay because NO3 negative that is a better oxidizing agent than that of H positive. So remember it directly guys. See what all statements you have to remember. Number one, number one that metal which comes before hydrogen, it can displace other metal from its aqua solution which comes after hydrogen. Point number one. Point number two, that metal can displace I mean that metal can liberate hydrogen gas in
an acidic solution like HCl which comes before hydrogen. Okay. Point number three, the metal which comes before hydrogen if it is introduced in an acidic solution of HNO3 at that point of time hydrogen gas is not produced. At that point of time nitrogen oxides are produced without the liberation of H2 gas. Okay, I hope this clear. And one more point, the stability of your metal oxides, stability of your metal oxides, it depends on the SOP, directly promotional SOP or you can say inversely proportioned SRP. So that metal whose SRP will be more that metal whose
SRP will be more the stability of its oxide will be less. The stability of its oxides will be less. So basically those metals which come before hydrogen those metals which come before hydrogen they have got lesser SRP they have got lesser SRP that means their oxides are comparatively more stable and those metals which come after hydrogen their oxides would be those metals which come after hydrogen their SRP is more their SRP is more that means the stability of their oxide will be less that is the reason that is the reason why the oxides of all
these metals which come after hydrogen Their oxides are highly unstable and they easily decompose on heating. They easily decompose on heating. Okay, perfect. Now you need to just tell me one thing guys. What is the answer of this question? I need a stability of these oxides. Stability of oxide is directly proportional is inversely proportional to SRP. Which one has got least SRP among copper, magnesium, zinc and lithium? Lithium has got the least SRP. So its oxide will be maximum stable. So I would say the stability of B is maximum. Right? Now among among copper, magnesium
and zinc among copper, magnesium and zinc among copper, magnesium and zinc, magnesium has got less SRP. Right? Among copper, magnesium and zinc magnesium comes here. Zinc comes here, copper comes here. So magnesium oxide would be more stable than zinc oxide than copper oxide. Magnesium oxide, zinc oxide, copper oxide, right? Magnesium oxide, zinc oxide, copper oxide. This is the stability of these. If I am clear with all of this, tell me once in the charge quake. Yes. All clear. All clear people. All right. Now this particular topic that is the most important topic of the chapter.
Let's do this particular topic after a dinner break. So I'm giving you a dinner break of exactly 40 minutes. Okay. So I'm keeping the break. It is basically your dinner break till what is the time? It is 8:32. So but I would want every one of you to be back on time because something which I'm going to start now that is the actual stuff which is asked in the exam okay and I would want all of you to be back on time otherwise there is no fun in teaching if you would sleep. Yeah. Okay. So,
be back on time, guys. Take care. I'll see you. So, are you all back? Yes. Can you all hear me? All right. All right. So, are you done with your dinner, everyone? Are you done with your dinner? Are you all done? To study equilibrium, you have to go for mole concept first. Okay? So, study mole concept first, then you can go for equilibrium. Do not jump directly into equilibrium. And there is no need to spam now. I think you got your answer. All right. So tell me whatever things we discussed till now. Is every single
thing clear? Tell me people whatever things we discussed till now. Is every single thing clear? Every one of you let me know quickly in the chats. Yes. Yes. I'm done with my dinner. All right, guys. So, let's get going then. Let's get started with the most important topic of the chapter and you know the topic that is your nerist equation. Okay Narnist equation and guys since uh it's already been like 4 hours and 15 minutes. So let's do one thing. I'll be break this particular chapter into two parts because as far as I can see
the way things are going in detailed manner it won't be complete in 12 hours. Okay. All right. So, in this particular session, let's complete the AMF part. In this chapter, in this particular session, let's complete the AMF part. And tomorrow morning at 9:00 a.m. or 10:00 a.m., I'll post the uh remaining 3 to four hours, okay, of this particular chapter wherein I'll be doing the electrolysis part. I think that's going to be more convenient. Yeah, that is going to be more convenient. Sir, can I refer this video for boards? Well, honestly speaking, this video is
not for this session is not for boards. Okay, this is not for boards. So people, is it convenient? Then in the today's session, let's complete the AMF part. That's also a huge part. RDM apart. Okay. And the remaining 3 to 4 hours, I'll be posting tomorrow morning at 10:00 a.m. on this channel. >> [clears throat] >> Today's session will take still like 2 to 3 hours. Yeah, not less than 3 hours. No, the part two I'll be posting in the recorded format. Part two, the uh electrolysis part that I'll be posting in the recorded format.
Okay, because tomorrow you have got one more live session on the channel taken by Ambika ma'am. So I don't want to clash with that. Okay, let's move on then equation. My dear students, let's first of all know what this nerous equation is all about and from where do we get this? Try to understand what I'll be saying. I'm going to write an equation from thermodynamics and the equation is delta G is equal delta G KN plus RT lawn of QC. This is one of the equations which I have written from the chapter thermodynamics. Now from
this equation itself I am going to derive your actual nerous equation which you will be using in the questions. My dear students first of all you know your delta G. Delta G is nothing but minus NF cell. Delta G KN is minus NF cell plus RT law of QC. Okay. If I divide throughout with minus NF, If I divide throughout with minus NF, my dear students, this will become E cell is equal to E not cell minus RT divided by NF. And this is lawn of QC. Now let me do one thing. Let me write
the logarithmic form of the same equation. So I'll say E cell is equal A cell minus 2.303 RT divided by NF and it's going to be log of QC. Well, this particular equation which I made over here, you call this particular equation as the nest equation. You call this particular equation as the Nest equation. But my dear students, I'm not going to use this equation in the questions. I'm further going to simplify it. Let me further simplify it a bit. What simplification exactly? Let me do one thing. If I take the value of temperature as
25° centiggrade, that means 298 Kelvin. If I take the value of Faraday's constant as 96500 kum. If I take the value of R as 8.314 JW per kelvin per mole. If I put all these parameters into this equation, what do I get? I will be getting a cell is equal to a not cell minus 0.0591 divided by N and it's going to be log of UC. Okay, this equation is what you call as your Nest equation. This is your Nest equation. Now my dear students in this learn equation, in this n equation what are these
parameters which we have mentioned over here? Since QC you already know that is the reaction quotient. I've already told you how to calculate this. N number of moles of electrons exchanged. I've already told you how to calculate this. E not cell standard EMF of the cell that is E not cathode minus E not anode. Already told how to calculate this. E cell will be the emf of the cell under non-standard conditions. Okay. E cell would be emf of the cell under non-standard conditions. Now my dear students where all do we use this nerous equation? Where
all do we use this nus equation? Remember first of all this nerous equation is used to calculate this nerous equation is used to calculate the oxidation potential of the half cell. Half cell means electrode. This nerous equation is used to calculate the oxidation potential of the half cell of the electrode under non-standard conditions. under non-standard conditions and how exactly you guys are going to write nerous equation at that point of time when you will be supposed to calculate the oxidation potential you'll write ner equation like this instead of E cell you'll write EO X instead
of E not cell you'll write E O X minus 0.0591 0591 divided by N. Then it's going to be log of QC. This equation I'll be using when I will be supposed to calculate the oxidation potential of the half cell under non-standard conditions. Let me tell you this EO X it is nothing but it is the oxidation potential of what? Of the half cell under which conditions? Under non-standard conditions. Okay. What is this E not O X? E not X represents standard oxidation potential of the half cell. This N already you know it represents number
of moles of electrons exchanged number of moles of electrons exchanged and QC all of you are already familiar with what is that that is the reaction quotient. So when we are supposed to calculate the oxidation potential of the hawk cell oxidation potential of the electrode under non-standard conditions I will write the nest equation in this format. Okay application number one. Application number two. If for example I'm supposed to calculate If I'm supposed to calculate the reduction potential the reduction potential of the half cell half cell means electrode under which conditions? Again under non-standard conditions. If
I'm supposed to calculate the reduction potential of the half cell again under which conditions? Under non-standard conditions. So again I'll be using the nonus equation. How exactly I'll be using the nus equation? Instead of e cell I'll write e instead of e not cell I'll be writing e not re minus 0.0591 divided by n and it's going to be log of qc. Okay. This equation I'll be using where where do I use this particular equation? This equation is used to calculate the reduction potential of the half cell under non-standard condition. So this is what we
call as this ER is what we call as reduction potential of the half cell under which conditions under non-standard conditions. This E not is what you call as standard reduction potential. This N you already know number of moles of electrons exchanged right and QC is nothing but reaction quotient. So application number one to calculate oxidation potential of the half cell. Application number two to calculate the reduction potential of the half cell. Now application number three. Application number three is going to be to calculate the emf of the cell. To calculate the emf of a complete
cell of a complete cell under which conditions? under non-standard conditions and when you are supposed to calculate emf of a complete cell under non-standard conditions at that point of time the nerest equation is written like this E cell is equal A cell minus0.0591 0591 divided by N. And it's log of QC. Log of QC. Okay. And this E cell is what you call as EMF of a cell under which conditions? Under non-standard conditions? This E not cell is what you call as standard EMF of the cell. This is the standard EMF of the cell which
is nothing but E not cathode minus E not anode. N you already know number of moles of electrons exchanged number of moles of electrons exchanged in the net cell reaction and QC is something which already you know what is that that is the reaction quotient that is the reaction quotient now let's try to exactly get into the details of the nus equation till now I just gave you the nus equation number one number two I told you where all I mean under what conditions we use the renos equation to calculate oxidation portal of the half
cell to calculate reduction portal of the half cell and to calculate the emf of a complete cell under non-standard conditions these are its main three applications okay now we'll get to know how to write this nest equation one by one for half cells for complete cells for hydrogen electrodes etc etc try to understand what exactly I'm going to do so mark the heading here Mark the heading as narnest equation for the half cells. Nice equation for the half cells. Mark this heading guys. Okay. First of all, first of all, for example, you have got a
container. This is the container, guys. In this particular container, for example, for example, we have got MN positive ions. we have got mn positive ions and with that there are some nions as well with that there are some nines as well so basically I have taken a salt here and that salt got dissociated into kine and nines perfect so in this solution let me assume that there are equal number of kines and nn present if there are equal number of kines and nions present that means the solution right now is electrically neutral right now my
dear students Imagine that I'm introducing a rod made up of metal M. I've introduced a rod made up of metal M into a solution containing its own ions. So I've introduced a rod in a solution containing its own ions. So I'll be calling this particular setup as the electrode. Right? This is the electrode. First of all, this is the electrode. Now my dear students, you know this electrode can behave like anode. It can behave like cathode as well. This electrode, it can behave like anode as well as cathode. So when this electrode behaves like the
anode, when this electrode behaves like the anode, what's going to happen? I would say at that point of time, oxidation is going to happen. When the electrode behaves like the anode, I would say oxidation is going to happen. What happens during oxidation? I would say metal solid will start getting converted into MN positives. The reaction is going to be m solid getting converted into m n positive aquis and with that you'll be writing n electrons. This is the reaction taking place when this particular electrode behaves like the anode. Okay. Now my dear students if I
do one thing if I write the expression for reaction quotient you will have you'll have to start with the product. You'll write concentration of MN positive raised power stoometric coicient divided by this is solid nothing to do with this simple okay how many moles of electrons are exchanged in the overall reaction I would say n moles of electrons are exchanged perfect now if I write the nas equation I'll write nus equation like this since I made electrode to behave like anode at anode oxidation takes place so technically I want to calculate the oxidation potential of
this half cell under non-standard conditions the oxid eration potential of the half cell under non-standard conditions will be equal to E X that means the standard oxidation potential of the same cell minus 0.0591 0591 divided by N log of QC and QC is nothing but QC is concentration of MN positive. QC is nothing but QC is concentration of MN positive. Perfect. So this is with the help of this particular equation I can calculate the oxidation potential of this electrode under which conditions? Under non-standard conditions you just have to put the value of concentration of MN
posit in the solution. N value standard oxidation potential solve this get the oxidation potential right now guys understand one more thing understand one more thing imagine that the electrolyte which you have taken in this container imagine that the concentration of electrolyte in the container in the beginning is more more the concentration of electrolyte in the container more the concentration of electrolyte in the container That means more would be the number of MN positives in the container. The the electrolyte which you have kept in the container in the beginning. If its concentration is more if its
concentration is more that means more will be the number of MN positives in the container. If the concentration if the number of MN positives in the container are more that means this term its value would be more. If the value of this term is more that means the difference between these two will be less which means that oxidation potential of the half cell would be less. So I would say oxidation potential of the half cell oxidation potential of the electrode is inversely proportional to concentration of electrolyte present in the container. This is one important conclusion
which I would want every one of you to remember. Oxidation potential of the electrode. Oxidation potential of the electrode is inversely proportional to the concentration of electrolyte in the container. What does that mean? That simply means something like this. For example, this is zinc sulfate in the container. For example, here also you have got zinc sulfate in the container. Here you have introduced zinc rod. Here you have also introduced a zinc rod. Let's say concentration of zinc sulfate here is 10 molar. Let's say concentration of zinc sulfate here is 100 molar. Okay. So do we
have same electrodes? Absolutely we have same electrodes. Where the difference lies? The difference lies in the concentration of electrolyte. So wherever the concentration of electrolyte would be more oxidation potential will be less. So I would say over here the concentration of electrolyte is more in this container. So the oxidation potential of this half cell oxidation potential of this electrode will be comparatively less than that of this one. Okay. I believe this is properly clear to everyone. Now my dear students, one more thing. One more thing. There could have been one more possibility that this electrode
could have behaved like the cathode. This electrode, it could have behaved like what? It could have behaved like the cathode. Now you tell me when the same electrode behaves like the cathode, what's going to happen? Oxidation or reduction? When the same electrode behaves like the cathode, I would say reduction is going to happen. Reduction means gain of electrons. So mn positives in the solution they are going to collide with the rod. So I would say mn positive ions in the solution they will collide with the rod. They will take n electrons from the rod and
will get converted into m solid. Okay. Now if I write the qc expression qc is going to be start with the product. It's in solid form. So one divided by concentration of mn positive raised power stom that's one. Now my dear students since since you made this electrode to behave like cathode and add cathode reduction takes place. So technically what do I want? I want to calculate the reduction potential of the half cell. I want to calculate the reduction potential of this electrode which will be equal as per nist equation E minus 0.0591 / N
and it's log of QC. QC is nothing but one upon concentration of MN positive. one upon concentration of MN positive. Okay. One upon concentration of MN positive. Guys, if you understand carefully, if I say what if the concentration of electrolyte in the container is more. If the concentration of electrolyte in the container is kept more that means the concentration of MN positives in the container will be more. If the concentration of MN positives in the container is more that means 1 divid by concentration of MN positive. If the concentration of MN positive is more. So
1 divid by this this value would be less. If this value is less if this value is less. If this value is less that means this entire value will be less. If this entire value will be less that means the difference between these two would be more. So which technically means your reduction potential would be more. So reduction potential and the concentration of electrolyte do you see directly or inversely? They're directly proportional. So I would say the reduction potential of the electrode reduction potential of the half cell that is directly proportional to the concentration of
electrolyte present in the container. So more the concentration of electrolyte present in the container more is going to be the reduction potential of the electrode. This is one important conclusion guys. Yeah. Am I clear with this everyone? People am I clear with this? Am I clear with this people? Everyone right now, now now let's get to know how the same ner equation is written for the hydrogen electro. So I'm marking the heading as nerous equation for hydrogen electro. Nur equation for hydrogen electrode nerous equation for hydrogen electro. Let's try to understand what it means. See
guys, for example, I have taken a hydrogen electrode here and I believe all of you remember how hydrogen electro is made. But I'm not going into the details of that again. So let me just roughly make the hydrogen electrode. Let's say this is the container and this particular container contains H positive ions here. Okay. And here you have got hydrogen rod. Rest you know how this hydrogen rod is made. Let's say the pressure of the hydrogen here is for example for example 1 atm. I believe all of you remember how this hydrogen electrode is made
with the help of platinum. So first of all first of all the hydrogen electro which I've taken here even this hydrogen electrode it can behave like anode or it can behave like cathode. So first of all let's do one thing. Let's just make this let's make this hydrogen electrode to behave like anode. When this hydrogen electrode behaves like anode what's going to happen you know at anode oxidation is going to take place. So if this hydrogen electro behaves like anode, I would say it will under oxidation. And when the hydrogen electron under goes oxidation, what
will be the reaction? I would say H2 gas would get converted into twice H positive. And with that you'll be writing two electrons. Okay. This is the reaction. When this hydrogen electrode is going to behave like the anode okay now my dear students, how many moles of electrons do you see here? Two. That means N value that is equal to two. Okay. If I want to write the QC expression, QC expression is going to be concentration of H positive raised parametric equation 2 divided by partial pressure of H2 gas and the pressure of H2 gas
I have kept as 1 atm. So QC is nothing but that's concentration of H positive^2. Now people since I made this electrode to behave like anode and at anode oxidation takes place so technically I wanted to calculate oxidation potential of this electrode which is nothing but EX - 0.0 0591 divided by N log of QC. I hope you remember the standard oxidation potential as well as standard reduction potential of hydrogen electro is taken to be zero. We have discussed that. So this term is already equal to what? This term is already equal to zero. If
this term is zero, it is going to be minus0.0591 divide by N value is 2 log of QC. QC is concentration of H positive power stoometric equation that is 2. I would say oxidation potential of the hydrogen electrode is equal -0.0591 divid by 2 log of m raised^ n is n log m. So 2 comes to the front 22 gets canceled out. When 22 gets canceled out what am I left with? I'm left with only log of h positive concentration. Okay if I do one more thing I can write e o x is equal 0.0591.
If I take minus in the bracket it becomes minus log of h positive. Right? And you know my dear students it will be 0.0591 minus log of H positive is nothing called as pH minus log of H positive is nothing but it is called as the pH. So oxidation potential of the hydrogen electrode oxidation potential of the oxidation potential of the hydrogen electrode when the pressure of hydrogen gas is kept 1 atm is equal 0.0591 0591 multiplied by the pH of this particular solution which is here in the container. Am I clear with this? I
believe this is clear. Similarly, my dear students, if you want to derive an expression for the reduction potential, if you want to derive an expression for the reduction potential of the same hydrogen electrode, it will be simply just change the sign. It would be -0.05. 0591 multiplied with pH of the solution. Okay, this is the reduction potential of the hydrogen electrode. Perfect. When the pressure of the gas is kept as 1 atm. Am I clear with this? And if I'm clear with this, let's try to use all these concepts in the questions directly. Let's try
to use all these question, all these concepts in the questions so that you'll understand where all do we use all the stuff. Look at this question guys. Look at this question. Calculate the oxidation potential of the following half cell or the following half cell. If you look carefully, if you look carefully, in the container you have FSO4. In the container you have FSO4, right? And in this FS0O4 solution you have introduced an iron rod. Okay. So I would say you have basically taken a container and in this particular container you have kept FSO4. This FSO4
in the container would have dissociated as Fe D positive and SO4 D negative. Now you have introduced iron rod here. So I would say I've introduced iron rod in a solution containing its own ions. Okay. So this is an electrode and iron electrode. I will have to calculate its oxidation potential. If I want to calculate the oxidation potential of this iron electrode, you know oxidation takes place at anode. I'll make sure this electrode behaves like the anode. And when this electrode behaves like the anode, what's going to happen? When this electrode is going to behave
like an anode, oxidation is going to take place. Oxidation is going to take place. See it is zero. This is plus2 0 to +2 increase. Increase means oxidation loss of electrons. So reaction is going to be Fe solid will be getting converted into Fe dipos to aquis. With that you'll be writing two electrons. This is going to be the reaction at anode. This is going to be the reaction at anode. Perfect. Now what is the N value? First of all N value here is two. N value here is two. What is the QC value? QC
is going to be concentration of FI positive raised power stoometric question that's one yeah now what is the concentration of FI positive in the solution look carefully initially at time t is equal to0 what was the concentration of FSO4 that was 0.1 molar this would be zero this would be zero now imagine this has got completely ionized it has got completely dissociated into its ions so I'll be left with nothing 1 mole gives 1 0.1 would have given 0.1 this would be also 0.1 molar. So I can say concentration of FA dipos in the solution
is 0.1 molar. 0.1 means 10^ minus 1. So this is the value of QC. I got the value of QC. I got the value of N. Now what do we have to calculate? I have to calculate oxidation potential. So I just have to use an earnest equation. I can say oxidation potential of the half cell is equal to X standard oxidation potential minus 0.0591 0591 divided by N and it's log of QC. Now my dear students E O X is equal what is the SOP value? Is the SOP value given? Take it out. This is
plus 2 to 0. Plus 2 to 0 decrease. Decrease means reduction. So this is your SRP. But what do I need? I need SOP here. I need SOP here. So just change the sign. The sign is going to be plus 0.44 minus 0.0591 0591 divided by N value. What is the N value? What is the N value? N value here is 2 and it's going to be log of QC. QC is 10^ minus 1. So you can finally write it as 0.44 - 0.0591 / 2 log of M raised^ N is N log M. So
minus 1 comes to the front log 10 is 1. Just solve this and get the answer in volts. So this is going to give you the oxidation potential of this particular electrode, right? Which I was supposed to calculate. If I'm clear, let me know in the chats. Yeah, let me know quickly in the chats if this particular question is clear. My dear students, now whatever value will be coming after solving this equation, after solving this equation, whatever value will be coming, you just have to change the sign of that value. You just have to change
the sign. What would you get? You would be getting the reduction potential of the same electrode. Yeah, people are saying the answer is coming out to be 0.47. So if it is 0.47 volt, so this is the oxidation potential of the electrode. So I would say if you were supposed to calculate the reduction potential of the same electrode that would have been [clears throat] -0.47 47. As simple as that. Right. Perfect. Let's move on to one more question. Look at this question, guys. Calculate the oxidation potential of the following half cell. Again you are given
with the half cell. Half cell means electro. Again you are given with the electrode. And if you look carefully this is hydrogen electrode. And you are supposed to calculate its oxidation potential. You are supposed to calculate its oxidation potential. If the pressure was 18 atm. If the pressure was 180m then you would have used the direct result which I gave a few minutes back. E is equal 0.0591 into pH. But right now you cannot use that result because that result is valid only when the pressure is when the pressure of hydrogen gas is 180. Right?
But here the pressure of hydrogen gas is 1080. What do we do in this case? We have to calculate the oxidation potential. If we need to calculate the oxidation potential that means we'll make sure that the electrode behaves like the anode. And when it behaves like the anode, what would be the reaction? The reaction would be H2 gas gives 2 * H positive aqueous and with that you'll be writing two electrons. Okay. So what is the N value? First of all N value is two. What is the QC value? QC is going to be concentration
of H positive raised power stoometric coicient divide by pressure of H2 gas. Now if I ask you what is the concentration of H positive? Concentration of H positive is 10^ minus1 molar raised^2 divide by pressure of H2 is 10. So the value comes out be 10^ minus 3. So this is QC. This is QC. Okay. You got the QC value, you got the N value. What we have to calculate? Oxidation potential. So I would say oxidation potential of the half cell is equal E X - 0.0591 divided by N and it's log of QC. Now
people E O X is equal E O X for hydrogen electro SOP value is 0. So this is 0 minus 0.0591. N value we have calculated as 2 log of QC QC value is how much QC value is 10^ minus 3 right so it becomes -0.0591 0591 / 2 log of m raised^ n which is n log m. So minus 3 comes to the front log 10 is 1. So this answer will come in old correct. I hope this is clear till here. Right people look at this particular equation look at this particular equation. A
solution containing 4 into 10^ -4 M CR27 D negative and 2 into 10^ -2 M CR try positive ions shows a pH of 1. Calculate the reduction potential of the half cell. We have to calculate the reduction potential of the half cell. The half cell that's given to us is basically like this. We are directly given the reaction of the half cell. CR27 and D negative is getting converted into CR positive. So first of all if you want to show the half cell half cell you can show like this platinum solid CR 207 D negative
aque crositive aquas. This is your half cell. First of all you are supposed to calculate its what? You are supposed to calculate its reduction potential. You are supposed to calculate its reduction potential. If you are supposed to calculate its reduction potential that means you are supposed to make sure that this electrode behaves like the cathode because at cathode only reduction takes place. Now when this electrode behaves like the cathode reduction is going to take place. Now have a look oxidation state of chromium here is plus 6 and chromium is in plus three. So plus 6
to + 3 decrease means reduction gain of electrons. So the reaction is going to be the reduction reaction is going to be CR27 D negative will be definitely gaining some electrons and getting converted into CRI positive. Now first of all on this side we have got two chromium. So I'll make it as two chromium. So how many electrons are being gained here see + 6 to + 3 + 6 to + 3 change in the oxidation state is three. Three is the change for one atom I have two. So 3 2 are six. So the
R27 D negative plus 6 electrons here's 2 * Crosit. If I ask you is this the complete reaction? This is still not the complete reaction. Why is that? Because no doubt you're you have balanced chromium but have you balanced oxygen? You have not balanced oxygen. So on left side you have got seven oxygen. So on right side you'll be adding seven water molecules. Now on right side you have got 72 14 hydrogen's. So I'll be adding 14 H positives on this side. Already I've told you how do we balance oxygen and hydrogen. We have studied
that right now people there are six moles of electrons here. So n value is nothing but six. We got the value of n correct? We got the value of n. Now since I have to calculate the reduction potential. So n we got what is left? Reaction quotient is left. To calculate the reaction quotient value I'll be starting with the product. So it's concentration of CR positive raised^2. This water is a liquid state. So nothing to do with that divided by come to the reactant side. So first of all it's going to be concentration of H
positive raised^ 14 right. Concentration of CR2 O7 D negative raised power stoometry question that's one. Perfect. So this is the QC value. Now as far as the question is concerned what is the concentration of CR try positive? concentration of CR try positive is given to me as 2 into 10^ minus 2 raised^ometric question that's 2 divided by what is the H positive concentration well H positive concentration is not given to me but if you look carefully we are given with the pH if pH is equal to 1 that means H positive concentration would be equal
10^ - 1 molar simple H positive concentration is nothing but 10^ -1 and raised power 14 What is the concentration of CR27 D negative? It is given to me in the equation as 4 into 10^ -4. So this term this term gets cancelled. So the value comes out to be 10^ 14. So this is the QC value. I got the QC value. I got the N value and look at this value. E not is given. Now is this SOP or SRP? See plus 6 to + 3 decrease means reduction. So this is SRP. This is
basically SRP that is given to me. What am I supposed to calculate? As far as the question is concerned, I'm supposed to calculate the reduction potential. I'm supposed to calculate the reduction potential. Now, now people, if you want to calculate the reduction potential, then you have to write the non equation. E is equal it's going to be E minus 0.0591 divided by N. And it's going to be log of what? Log of QC. So E not R is the standard reduction potential SRP that's given to me as 1.5 minus 0.0591 divided by what is the
N value? N value is nothing but 6 and log of QC. QC is 10^ 14. Now it is just a matter of calculation and you'll be getting the answer in volts. That's all. Is it clear people? Tell me once in the chats if this is clear. Everyone tell me once in the chats if all of this is clear. Everyone, perfect guys. So, moving on to one more question. Look at this question carefully. Look at this question. We are given with the reaction. We have to calculate the reduction potential in neutral solution. Neutral solution means pH
will be 7. Yeah. So what do we do? How do we solve this question? Then still now whatever questions we solved in none of the question the reaction was given to us. Every time I was supposed to make the reaction but this is the question wherein the reaction is already given. So you are not supposed to make the reaction. The reaction is already given. Okay. How many electrons do you see? One. So n value is one. So if you write the reaction quotient write the reaction quotient start with the product it's pressure of NO2 because
it's in gaseous phase raise question one water is in liquid state nothing to do with that NO3 negative in aquas state so concentration of NO3 negative raised power stoometry coition one this is concentration of H positive raise partic now guys we almost done we are almost done how exactly pH is equal to 7 if pH is equal to 7 that means Means the concentration of H positive is going to be 10^ - 7 volt 10^ - 7 M. Now what is the pressure of NO2 that's given to me 1 bar divided by NO3 negative concentration
is 1 m H positive concentration is 10^ - 7 10^ - 7 square comes out be 10^ -4 so the overall value will be 10^ 40 I got QC I got N. What do I have to calculate? What do I have to calculate? I have to calculate the reduction potential. In order to calculate the reduction potential, you'll write the learners equation. E is equal to E R minus 0.0591 divided by N. And it's going to be log of what? Log of QC. Now people, what is this E not R? This is the standard reduction potential.
And standard reduction potential of the half cell is given to me as 0.78 minus 0.0591. N value is given to us in the equation that is 1. Log of QC. QC is 10^ what? 10^ 14. Right? Now it is just again the matter of calculation. So this is 0.0591. Log of M^ N is N log M. So 14 log 10 is 1. Just solve this particular term and get the answer in volts. Simple. So this will give me the reduction potential of this half step. Right? I believe it's clear. Similarly, let's try to solve this
particular equation. Let's try to solve this particular equation. Look at this question guys. The question is calculate the reduction potential at pH equal to 14. At pH equal to 14 for the electrode that is copper copper electrode at 25° centiggrade. Ksp of CO whole twice is given. Okay. E not value is also given. So how do we exactly approach this sort of question? The electrode that's given to us is copper electrode, right? And what are we supposed to calculate? We are supposed to calculate the reduction potential. So in order to calculate the reduction potential, in
order to calculate the reduction potential, I'll have to make sure the electrode behaves like the cathode because at cathode only reduction takes place because at cathode only reduction takes place. So when the electrode behaves like the cathode, a reduction is going to take place. Reduction means copper die positive would be gaining two electrons and would be getting converted into copper solid. Okay. What is the N value here? N value here is two. What is the QC value here? QC value will be 1 divid by concentration of copper dipositive. This is solid. Its ecto mass is
1 divided by concentration of copper dipositive here. Now people look carefully in the question. Am I given with copper dipositive concentration? No, not anywhere. I'm not given with the copper dipos concentration. I have to calculate the copper dipos concentration. How exactly? For that purpose, if you look carefully, Ksp of CO hold twice is given. So, first of all, Cu twice. Cule twice. How will it dissociate? Copper D positive plus 2 * O negative. If you would want to write the Ksp expression, Ksp expression is simply going to be concentration of copper diposit raised^ 1 concentration
of O negative raised^2. Right? Now what do I actually need? Actually I need copper dipositive concentration. So to calculate copper dipositive concentration I need Ksp value. I need O negative concentration. Ksp value is given but O negative concentration is not given. So how do I calculate O negative concentration now? How do I calculate O negative concentration? Now O negative concentration can be calculated with the help of this particular term. PH pH in the question is given to me as 14. If pH is equal to 14 since you know at 25° centiggrade PH plus P is
equal to 14. So if PH is equal to 14. So P will be 14 - 140. If P is 0 that means O negative concentration will be 10^ - 0 that means 10^0 which comes out to be 1 m. I got the O negative concentration. If I got the O negative concentration, Ksp value in the equation is given 10^ -19 is equal. This is concentration of copper D positive. O negative concentration is 1. So I got the concentration of copper D positive as I got the concentration of copper positive as 10^ - 19 molar. So
QC value came out to be 10^ + 19. If you got the QC, if you got the QC, what do we have to calculate? we have to calculate the reduction potential. So in order to calculate the reduction potential use the earnest equation e is equal e minus 0.0591 divided by n and it's a log of what log of qc. So my dear students e standard reduction potential that is given to me in the question that is 0.34 minus 0.0591 0591 divided by what is the value of N? N value is 2 log of QC QC
value is 10^ what 10^ 19 now just the matter of calculation 0.34 - 0.0591 0591 divided by 2 log of m raised^ n is n log m so 19 comes to the front log 10 is 1 again just a matter of calculation and you'll be calculating you'll be getting what you'll be getting the reduction potential of the electrode which I was supposed to calculate if I am clear with this tell me once in the charts quickly This done they can ask oxidation potl as well. If they ask oxidation protocol then reverse the reaction. That's it.
then reverse the reaction. Oh, is it cyclone in P in Tamil Nadu? Is there a cyclone? Okay, just stay safe. All right guys, look at this particular equation. Look at this one. Look at this one guys. The pressure of hydrogen required to make the pressure of hydrogen required to make potential. Potential means a reduction potential required to make a reduction potential of hydrogen electro zero. So basically in the question we have to make the reduction potential of hydrogen electro zero in pure water. We have to make the reduction potential of hydrogen electrode as zero in
pure water. And for that purpose in order to make the reduction potential of hydrogen electro zero what should be the pressure of hydrogen gas? That is the question. That is the question. That is the question. Now how do we solve this particular equation? First of all, if you look at the hydrogen electrode. Do you remember how hydrogen electro is made? These are the H positives kept and this used to be the inverted tube. This used to be the platinum wire. Over here H2 gas used to be introduced. Okay. So basically we have to calculate the
pressure of H2 gas. The pressure of H2 gas required required to make the reduction portion of hydrogen electrode is zero. Now in order to make the reduction potential of hydrogen electrode as zero let's do one thing let's make sure that hydrogen electro behaves like the cathode first when it behaves like the cathode the reaction is going to be 2 * H positive aquis plus 2 electrons gives what gives H2 gas this is the reaction which going to take place first okay this is the reaction going to take place now N value is how much n
value is two if N is two write the QC expression QC is going to be pressure of H2* divided by concentration of H positive^2 pressure of H2 I have to calculate divided by what is the concentration of H positive they mentioned pure water pure water at 25° centigrade what is the pH of pure water what is the pH of pure water at 27° centigrade sorry at 25° centiggrade that's 7 because water is neutral so concentration of H positive will be equal 10^ - 7 m so technically concentration of H positive is given 10^ - 7
so it square comes out be 10^ - Copy. This is QC. You got the QC value. You got the N value. Now as per the question, use the nerous equation. I can say E reduction potential of the hydrogen electrode is equal E not standard SRP minus 0.0591 divided by what divided by N log of what? Log of QC. And QC value is pressure of H2 divided by 10^ - 10^ -40. Now this is SRP of hydrogen electro that is zero already because you know SOP as well as SRP of hydrogen electro is zero. Now as
per the question we have to make the reduction potential of hydrogen electro zero as well. So this is zero this is zero. This means that log of partial pressure of H2 / 10^ -14. This value has to be zero. When is the log of X0? Log of X is 0 only when X is equal to 1. log of x log of x is 0 when x is equal to 1. So x here is partial pressure of h2 / 10^ -4. This value has to be 1. It means that partial pressure of h2 is equal 10^
-14. Let's say the units are in atm. So I would say the pressure of h2 gas should be 10^ -14 atm. The pressure of h2 gas should be 10^ -4 atm. Then only the reduction potential of this electrode. then only the reduction potential of this half cell would be zero. Am I clear with this? Did you understand how to deal with such kind of the question? Did you understand guys? Look at this one. Look at this particular question guys. Who was this polar? Who was this polar? Yeah. Okay. The question is The question is zinc
electrode is dipped in a 0.1 molar solution of a salt at 25° centigrade. Assume the electrolyte is 20% dissociated. Calculate the electro potential. Electro potential means a reduction potential. If in the question they ask you to calculate the electro potential that means you are supposed to calculate what? Reduction potential. See guys what the question says. Try to understand it's a simple basic question only not difficult. Imagine you have got a container. Imagine you got a container. In this container you are keeping a salt of zinc. You are keeping a salt of zinc. For example you
are keeping zinc sulfate. This salt of zinc. Now this zinc sulfate how it should under go dissociation zinc diitive plus SO4 D negative. Now you are keeping zinc rod here in the solution. So basically you got an electrode. Which electro this is? This is zinc electrode. So you have got a zinc electrode. Now we have to calculate the reduction potential of the zinc electrode. When when 20% is it mentioned 20%? Yes. When 20% zinc sulfate dissociates into it ions. So when 20% of zinc sulfate would have dissociated into its ions at that moment at that
point of time what would be the reduction potential of the electrode that is the question now first of all initially what is the concentration of zinc sulfate I have taken concentration of zinc sulfate in the container initially was 0.1 molar so this would be zero this will be zero now in the question they have mentioned that 20% zinc sulfate is dissociated if 20% zinc sulfate is dissociated ated that means alpha value degree of dissocion will be 20 by 100 that means 0.2 2. So at I would say when 20% zinc sulfate is dissociated how much
will be left? This would be C minus C alpha. This would be C alpha. This would be C alpha. So concentration of zinc diitive when 20% electrolyte is dissociated. Concentration of zinc diit electrolyte is dissociated is equal to C into alpha. C value is 0.1 and your alpha value is 0.2. The value comes out be 0.02. This is the concentration of zinc dip positive in the solution. When 20% electrolyte would have dissociated when 20% electrolyte would have dissociated. This would have been the concentration of zinc dipositive at that point of time. Now what do we
have to calculate? We have to calculate the reduction potential of the electrode when 20% electrolyte is dissociated. Now in order to calculate the reduction potential I have to make sure that the electrode behaves like the cathode because that cathode only reduction takes place and when the zinc electrode behaves like the cathode what's going to happen when this zinc electrode behaves like the cathode what is going to happen a reduction so I would say zinc diposites in the solution they are going to collide with the rod they will take two electrons from the rod and will
get converted into zinc solid this would be the reaction when this electrode behaves like the cathode because at cathode reduction takes place right now people what is the value of n here the value of n here is two that means two moles of electrons are exchanged if I ask you the qca value qc is going to be start with the product it's in solid state so 1 divided by it's going to be concentration of zinc diositive now what is the concentration of zinc diitive when 20% electrolyte is dissociated that is 0.02 02. So it comes
out to be 2 by 100 that means 50. So QC value came out to be 50. Yeah, QC value is equal to 50. If QC is equal 50, what was I supposed to calculate? I was supposed to get the reduction potential of the electrode. And reduction potential of the electrode is nothing but it is E not minus 0.0591 divided by N. And it's [clears throat] log of what? Log of QC. So what is E not R? This what you call as SRP. See plus 2 to 0 plus 2 to 0 means decrease reduction. So this
is SRP already which is given to me - 0.76 minus 0.0591 N value here is 2 log of QC. QC is 50. 50 can be written as 510. Now again a matter of calculation. So this will be -0.76 - 0.0591id 2 log of M into N is log m + log n. So log 5 plus log 10 log 5 is 0.69 and log 10 is 1. So this entire value comes out to be 1.69. This entire value comes out to be 1.69. Now again just the matter of calculation you just have to solve this and
get the reduction potential of the half cell. So I believe you got to know how do we calculate the oxidation potential as well as the reduction potential of the half cell of the electrode under non-standard conditions. If this is clear tell me once in the chats then I'll move on to the nurest equation for complete cells. Quickly guys, tell me quickly in the chats. Done. All right. Now, let's move on to one more thing. What is that? That is the nerness equation for complete cells. Nest equation for complete cells. Okay. So first of all when
we are supposed to calculate emf of a complete cell first of all a complete cell is made up of two half cells a complete cell is made up of two electrodes right whenever you are supposed to calculate emf of a complete cell under non-standard conditions under non-standard conditions at that point of time you can use the nuros equation and how nerous equation is used at that time you can use this particular equation directly E cell is equal E not - 0.0591 divided by N log of QC. This is how you write the Nest equation at
that point of time when you are supposed to calculate what? When you are supposed to calculate the EMF of a complete cell when you are supposed to calculate EMF of a complete cell under non-standard conditions. Now let's get to know how do we use this exactly for example let me make one general question a simple question I'm making here the question is like this I'm making a complete cell zinc solid zinc dipos aquis whose concentration is 0.1 molar in the middle I'm keeping salt bridge on the right side I'm keeping copper dipositive concentration copper dipositive
aquis whose concentration again I'm keeping 0.1 molar and this is for example copper solid okay now imagine E not of zinc gives zinc die positive is equal to 0.76 volt imagine E not of copper die positive gives copper is equal to 0.34 volt I'm asking you to calculate the emf of the cell I'm asking you to calculate the emf of the cell. Yeah, I'm asking you to calculate the emf of the cell. How we are going to solve this sort of a question in which you are supposed to calculate emf of a complete cell. My
dear students, the process is really simple. First of all, these two lines they represent the salvage. On the left side of the salvage, what do we have? We have anode. On the left side of the salvage, what do we have? We have anode. And at anode, what happens? Oxidation. At anode as you can see zinc gets converted into zinc dip positive. So 0 to plus2 increase means oxidation loss of electrons. So at anode I must say your zinc solid is getting converted into zinc dipositive to aqueous plus how many electrons? Plus two electrons right? This
is the reaction taking place at anode. Yeah. Now my dear students n value sorry this is the reaction takes place at anode. Now what is the reaction taking place at cathode? At cathode if you see a reduction is taking place. Copper diposit is getting converted into copper. I would say copper dipos to aquis would be gaining two electrons and then would be getting converted into copper solid. Now if I ask you are the electrons balanced in both the reactions? Yes the electrons are balanced. If the electrons are balanced you can add the two reactions. When
you add the two reactions the net reaction becomes zinc solid plus copper dipos to aquis. It gives zinc dipos to aquis plus copper solid. So n value here is two. n value here is two. Got it? N value here is two. Yeah. N value here is two. Now if N is two if I get the QC value QC is going to be concentration of zinc D positive raised per stoometric coation one divide by concentration of copper D positive raised per stoometric coation one what is the concentration of zinc D positive 0.1 what is the concentration
of copper D positive 0.1 the value comes out be 1 so QC value is equal to 1 okay QC value is equal to 1 now I got the QC I got the N I can calculate E not cell as well and you know E not cell is nothing but E not cathode minus C not anode. So it is SRP of cathode minus SRP of anode. Now this is your cathode SRP of cathode. SRP of zinc. Well, this is SOP of zinc. What do I need? I need SRP of zinc. So - 0.76 minus SRP of
anode. Uh sorry, I made a mistake here. This is basically SRP of cathode. SRP of cathode means SRP of copper. This is anode. This is cathode. So SRP of copper. SRP of cathode. SRP of copper. SRP of copper is given to me as 0.34 minus SRP of zinc is -0.76. The value comes out to be 1.1 volt. So you got E not cell, you got the value of QC, you got the value of N. Now you can directly use the NUS equation. You can directly use the NUS equation. You can say E cell is equal
to E not cell minus 0.0591 divided by N log of QC. Now in the question we got to know QC value is equal to 1. If QC value is equal to 1 that means log of 1 is equal to 0. This entire term becomes zero. So I would say E cell is nothing but E not cell and E not cell we have already calculated how much that is 1.1 volt. This is something I was supposed to calculate. Yeah. Am I clear? Did you understand how ner equation is used for the complete cells? Did you understand
how nervous equation is used for the complete salespeople? Everybody let's solve few more questions so that things are going to be like very smooth, very clear to you. Again the similar type of question which I believe you can easily solve. Again the similar type of question which I believe you can easily solve. We again have to calculate emf of the cell. Now in order to calculate emf of the cell what do I need to do? The same process which I have been following till now. What is the process? On the left side of the salt
we write anode. At anode what happens? Oxidation in gets converted into sni positive. So I would say the reaction is going to be S and solid gives S and DI positive aquis plus two electrons. This is the reaction at anode. Similarly, what would be the reaction at cathode? PB dip positive is getting converted into PB. So the reaction has to be PB dipositive. Aquis would be gaining two electrons and would be getting converted into PB solid. This is the reaction at cathode. You got the reaction at anode. You got the reaction at cathode. Now add
the two reactions because electrons are balanced. Right? So I would say two electrons would get cancelled. So N is equal to two. My net reaction becomes Sn solid plus PB dip positive aquis. It gives Snipos positive aqueous plus PB solid. This is my net reaction. If this is my net reaction now what we can do? We can get the QC expression. QC is going to be concentration of SN diitar stoometer question one divided by concentration of PB diitometer question one because this is in solid this is in solid nothing to do with them now what
is the concentration of SNI positive concentration of SN positive is 1 m divided by concentration of PB DI positive that is 10^ minus 3 the value comes out be 10^ 3 only so QC you got you got the N value now comes your E not cell so how do I calculate E not cell E not cell is again the name that is E not cathode minus E not anode E not cathode means SRP of cathode cathode is your lead and SRP of lead is given this is SRP PB di to PB plus 2 to 0
decrease means reduction so this SRP so -0.13 minus SRP of anode SRP of tin SRP of tin is also given that is also -0.14 the value comes out be 0.01 01 okay this is E not cell now E not you know QC you know N value you know put all the values in the Nest equation your Nest equation is simple that is E cell is equal E not cell minus 0.0591 / N and it's going to be log of what log of QC now my dear students E not cell value we have calculated 0.01 minus
0.0591 0591 n value we already got that is 2 log of qc qc is how much qc is 10^ 3 so again just the matter of calculation so this is 0.0591 0591 divided by 2 log of M raised^ n is n log m so 3 comes to the front log 10 is 1 again the matter of calculation get the answer in volt right am I clear people am I clear till here everyone All right, one more question and after that I'll be giving a break. Okay, one more question and with that I'll be giving a
break. Look at this question guys. Look at this question. The reaction is given to us. The question says if E cell is equal to E not cell at 25° centigrade concentration of FI positive is this much partial pressure of O2 is this much PH is X calculate X that is the question that is the question now first of all since the reaction is given can I write the QC expression because the reaction is given otherwise I was till now I was making the reactions but here in this question the reaction is already given so QC
is going to be start with the product so it's going will be concentration of Fe positive raised parametric equation 2 right this water is in liquid state so nothing to do with that divided by this is in solid state nothing to do with that I'll write partial pressure of O2 raised^ 1 and concentration of H positive raised^ 4 right now as far as the question is concerned PH is equal to X if PH is equal X that means concentration of H positive would be 10^ - 6 M concentration of H positive would be 10^ minus
X M. Now in the question, what is the FA tripositive concentration? How much is F positive concentration? Sorry, this is F tripositive. This is not positive. This trip positive. Okay, my bad. F trip positive concentration is given to me as 10^ -3 square comes out be 10^ - 6 divided by pressure of O2 gas is given to me as 10^ -2 okay concentration of H positive is 10^ - X raised^2 isometric question 4 now this 10^ -2 here 10^ -2 gets canceled out it becomes 10^ - 4id 10^ - 4x okay now since you know
the equation Since you don't know the ner equation, n equation says that e cell is equal e cellus 0.0591 divided by n and it's log of qc. This is your nas equation. Now as far as the question is concerned question is saying that if e cell is equal to e not cell when can e cell become equal to e not cell? E cell can become equal to e not cell if this term becomes zero. E cell will be equal to E not cell if this entire term becomes zero. When can this entire term become zero?
This entire term can become zero only if QC value becomes one. Then only log of 1 is zero. So I would say E cell would be equal to E not cell if QC value is equal to 1. If QC value is equal to 1. Right? If QC value is equal to 1. QC we already got how much is that that is 10 rais^ -4 / 10^ - 4x this value has to be equal to 1 it means that 10^ -4 is equal to 10^ - 4x so when you compare x value is nothing but x
value is nothing but one so this is something I was supposed to calculate right Perfect. All right. So this was how to write the nerous equation for the complete cells. Now before discussing the ner equation at equilibrium a small 15 minutes break I'm taking. So you guys also take it because a lot of students have been asking about the break. So break I'll be giving till 10. 45. But everyone be back because this is again one of the very important topics that has to be covered. Okay. All right. See you after the break guys. See
you after the break. Guys, can you hear me? Can you all hear me? Am I audible, visible to everyone again? What's happening in the chats? Do you want me to block all of you? All right, I've done engineering. I'm B tech civil. Okay, let's move on now. So now we are going to talk about the nerist equation at equilibrium. This is again one important topic and a simple to perfect but have a proper attention on the screen now so that you understand what is meant by the narnest equation at equilibrium. So first of all people
let's say I am again taking a Daniel cell here I hope you remember your Daniel cell and in case of Daniel cell how many electrodes do we use? We use two electrodes. What all electrodes are used in Danielson cell? The first electrode is going to be your zinc electrode. Let's say this is your zinc rod which is introduced in zinc sulfate solution. And similarly here I'm introducing a copper rod in what? In copper sulfate solution. So I have taken two electrodes. One is your zinc electrode and one is your copper electrode. Now my dear students,
again I'm connecting them externally as well as internally with the help of a salt bridge. Again I'm connecting them externally as well as internally with the help of what? With the help of salt bridge. Now my dear students you know at anode what happens oxidation. At cathode what happens? Reduction. This is your anode. first of all at which oxidation is going to take place and this is your cathode at which reduction is going to take place. If you remember one thing I have already discussed with all of you. What is that? I have told you
a reduction potential of an electrode is directly proportional to the concentration of electrolyte present in the container. Reduction potential of the half cell. Reduction potential of the electrode is directly proportional to what? It is directly proportional to the concentration of electrolyte present in the container. More the concentration of electrolyte in the container, more is going to be the reduction pot of the electrode. As simple as that. More the concentration of electrolyte in the container, more is going to be the reduction potential of the electrode. Now try to understand what exactly I'm going to say. My
dear students when this cell will start working at anode oxidation is going to take place. The zinc atoms from the rod they are going to get converted into zinc dip positives and those zinc dipositives will enter into the solution and as soon as zinc dipositive enters into the solution from the saltbridge SO4 d negative will come and will neutralize it. So zinc dip positive coming from the rod entering into the solution. SO4 d negative coming from here. So what does that mean? That means in this container the concentration of ions the number of ions are
increasing. That means in this container in this anodic container the number of ions the concentration of ions is increasing. Or in other words I can say the concentration of zinc sulfate in this left container it increases with time. It increases with time. Right? Similarly, Pu dip positives they collide with the rod. PU dipositives which are in this solution they collide with the rod they get converted into copper solid. That copper solid gets deposited on the rod. That copper solid gets deposited on the rod. I would say the number of ions in the solution they decrease
with time or I can say concentration of copper sulfate it decrease with time. Now since if the concentration of zinc sulfate is increasing that means if the concentration of zinc sulfate if the concentration of electrolyte in this container is increasing that means the reduction potential of this electrode is increasing. So with time the reduction potential of this electrode will keep on increasing. Similarly if the concentration of electrolyte in the right container decreases that means the reduction potential of this electrode this copper electrode will decrease with time. So on one side on one side I would
say reduction potential of zinc electro is increasing on other side reduction potential of copper electrode is decreasing. Can I say there will be a time when the reduction potential of both the electrodes will be equal. There will be a time when the reduction potential of both the electrodes will be equal. At that point of time there will be no potential difference between these electrodes. There will be zero potential difference. And when there will be zero potential difference between the electrodes, the current will stop. The current will not be produced from the cell. And that stage
in the cell is called as the equilibrium stage. That stage in the cell is called as the equilibrium stage. So in short, as this cell keeps on working, the concentration of zinc sulfate in this container will keep on increasing. That means the reduction potential of this electrode, the zinc electrode will keep on increasing. The concentration of copper sulfate decrease with time. So reduction potential of copper electrode decrease with time. On one side reduction potential of zinc electro increases. On other side reduction potential of copper electrode decreases. There will be a time when the reduction potential
of both the electrodes will be equal. At that point of time there will be zero potential difference between these electrodes and at that point of time the cell will stop working and that stage in the cell is called as equilibrium stage. So I must say in short at equilibrium at equilibrium the potential difference between the electrodes that is equal to zero. The potential difference between the electrodes is equal to zero. Right? Now if E cell is equal to Z since you know delta G is equal minus NF E cell delta G is equal to minus
NF cell. If E cell is equal to Z that means delta G also will be equal to Z. So these are the first two important things which you have to remember when equilibrium is established in the cell at that point of time E cell as well as delta G for the cell it becomes zero right people I believe this is clear till here now if you write the nerous equation the actual nerous equation that was E cell is equal E not cell - 2303 3 RT divided by NF [clears throat] it used to be log
of QC this used to be my actual nerous equation. Now if I write this nerous equation at equilibrium, if I write the same nerous equation at equilibrium, then instead of E cell, E cell is equal to Z. This E not cell comes on this side. So minus E not cell minus minus gets canceled out. So the equation becomes E cell is equal 2303 RT divided by NF log on. At equilibrium, we do not write QC. At equilibrium, QC is replaced by AEQ. At equilibrium, reaction quotient is called as the equilibrium constant. At equilibrium, reaction quotient
is called as the equilibrium constant. So, first of all, this particular result is valid when equilibrium will be established in the cell. Okay. Now if you put the value of R, if you put the value of R as 8.314 JW per kelvin per mole. If you keep the temperature as 25° centiggrade that means 298 Kelvin. If you take the Faraday constant exactly as 9650 kum if you put all these parameters into this equation the equation further reduces like this. ELL is equal 0.0591 divided by N and then it becomes log of K EQ. Okay. This
is the simplified form of the nerist equation at equilibrium. You believe this is clear? And my dear students, if this is clear, there is one more result which I would want to show you over here. What is that result? Let's have a look. Since the equation from where I have derived the nus equation that was delta G is equal delta G KN plus RT lawn of QC. This used to be the main equation. This used to be the main equation. Now people, if I write the same equation at equilibrium, if I write the same equation
at equilibrium, at equilibrium, your delta G is zero. And as I told you, QC is replaced by KQ. So put these in this particular equation. What you'll be getting? You'll be getting delta G KN is equal minus RT lawn of lawn of K EQ. Or you can write delta G KN is equal -2.303 RT log of KQ log of K EQ. So these two results are again important from which questions can be framed. I believe it's clear to everyone. I believe it's clear to everyone. Okay. So when equilibrium is established in the cell, when the
cell attains equilibrium, what happens exactly? You know it now. You know it now guys. Quick. when equilibrium is established in the cell at that point of time E cell becomes zero your delta G also becomes zero right and I believe you know how it happens now let's see what kind of questions can be framed from this nerous equation at equilibrium the First and the simplest question which you have on your screen. You are given with the cell reaction. E not is given. We have to calculate equilibrium constant. We have to calculate equilibrium constant. Now people
pretty much simple since we have to calculate the equilibrium constant. So I believe the cell is at equilibrium and at equilibrium E not cell is equal 0.0591 divided by N and it's log of KQ. Correct? What is the N value for the reaction? As you can see zinc is losing two electrons and cadmium dipositive is gaining those two electrons. So n value is given to me as two e not is given as 0.46. So this becomes 0.0591 divided by n value is how much? 2 log of k eq. So when you cross multiple you get
a log of k eq as 15.6 6 or I would say K EQ is equal NT log of 15.6 and how do you calculate NT log of 15.6 a simple trick is there 15.6 you can break as 15 + 0.6 okay you can break it as 15 + 0.6 this 15 you can directly write as 10^ 15 multiplied by this 0.6 0.6 6 is basically log of four. So it becomes 4 into 10^ 15. So this is the value of equilibrium constant which I was supposed to calculate. Yeah, I believe this is clear to everyone.
Now for example, look at this question. Look at this question guys. Look at this question. We are given the reaction as you can see in this particular reaction N value is equal to two. Okay. Equum constant is given we just need to calculate E not cell. So just simple E not cell is equal 0.0591 divided by N value is 2 log of K EQ. KQ is given to me as 10^ 6. Just one step question. This is log of M raised^ N is N log M. So 6 comes to the front. Log 10 is 1.
So it is 0.0591 multiplied by 3. This is the E not cell which I was supposed to calculate in volts. Right. Perfect. Similar sort of question guys. For a cell involving one electron, cell involving one electron means n is equal to 1. Okay. E not cell is also given how much 0.59 volt. What do we have to calculate? Equum constant K EQ. Now in order to calculate equilibrium constant I believe the cell is at equilibrium and when the cell is at equilibrium the equation which I'll be using that is E not cell is equal 0.0591
divided by N and it's a log of KQ now E not E not cell is how much 0.59 right is equal 0.0591 0591 divided by n value is 1 and this is log of k eq this is 1 * this is 10 * I would say log of k eq is equal to 10 right or I would say k eq is equal to n log of 10 and n log of 10 is nothing but it will be 10^ 10 directly it is basically 10.0 zero right so it will be directly 10^ 10 this is the
equum constant which I was supposed to calculate I hope this is like very much clear to all of you perfect now my dear students there comes one more important topic what is that is concentration cells this is one more important topic okay concentration cells So first of all what are these concentration cells? My dear students, if I ask you till now, whatever galvvinic cells you have seen, till now whatever galvvinic cells you have seen, did we use two same electrodes or two different electrodes? Whatever galvvinic cells we have seen till now, did we use two
same electrodes or two different electrodes? We used two different electrodes. But in case of concentrations and what we are going to do, we are going to use two same electrodes. Two same electrodes are used. Okay. Two same electrodes are used. Number one. Number two, E not cell in case of the concentration cell will be zero. Let's try to understand this. For example, here I'm taking zinc sulfate. And this is your zinc rod. Zinc rod introduced in zinc sulfate. Here also I'm taking zinc sulfate and you are introducing a zinc rod. So do we have two
same electrodes are two different? I've got two same electrodes. If I connect them externally as well as internally what do I get when I connect them externally and internally? I'll be getting a complete galvvinic cell. Now, now if I want to calculate E not cell, E not cell is going to be E not cathode minus E not anode. What is E not cathode? Out of these two, one will be cathode, one will be anode. So E not cathode means SRP of cathode. SRP of cathode means SRP of zinc minus E not anode is again SRP
of anode. SRP of anode is SRP of zinc. Now SRP of zinc is SRP of zinc minus SRP of zinc. This will come out to be zero. So whenever you have got a complete cell in which two same electrons are used the first point that should come into your mind that is E not cell then E not cell will be zero. So in case of the concentration cells wherein two same electrons are used remember E not cell value is equal to zero. E not cell value is equal to Z. Now guys how this works and
all for that I have to classify this concentration cell into two types. The concentration cell is basically of two types which we have to discuss. What are the two types of the concentration cells? The first concentration cell would be called as electrolytic concentration cell. Electrolytic concentration cell and the second one would be called as electrode concentration cell. electrode concentration cell. What happens in electrolytic concentration cell? Let me first of all tell you I'll explain this in some time. First of all, take a note of these points. In case of electrolytic concentration cell, concentration of the
electrolytes is kept different but the partial pressure of the electrodes is kept same. Concentration is kept different, partial pressure same. In case of electrode conceration cells I must say concentration is kept same concentration of electrolytes is kept same but partial pressure is kept different. Now let's try to understand them one by one. Let's try to understand them one by one. Okay. So first of all we'll be talking about the electrolytic concentration cell. First of all, we'll be talking about the electrolytic concentration cell electrolytic concentration cell. Let me first of all try to make this electrolytic
conserations. As I told you already in case of electrolytic concentration cell, what happens? Just a second. Just a second, guys. It's just give me a second. Okay, just give me a second. Okay. In case of electrolytic conservation cells, the first point which I'm going to write here, emf is generated. EMF is generated due to due to the difference in the concentrations of the electrolytes. Try to understand what it means. Try to understand what it means. See guys, for example, this is one of the containers which I have taken and this is one more container. Okay,
two containers I have. Let's say here in this particular container we have got MN positive ions. In this container also we have got MN positive ions. Okay. Let's say the concentration of MN positive in this container is equal to C1 M. The concentration of MN positive in this container is for example C2 M. Now for example you are introducing a metal rod M and in this container and in this particular container also you are introducing a metal rod M. So how many electrodes do you see? Two electrons you can see two electrodes. Now connect them
externally and internally. Connected them externally. Connected them internally as well. Okay. So in this particular cell do you see two same electrons used? Yes. So it is the concentration cell in which two same electrons are used. Now are the concentrations kept same or different? Concentrations are kept different. So it is the electrolytic concentration set. How do I represent it? First of all see guys in the middle I'll be using two lines which represent your salt bridge. On the left side I'll be writing anode and at anode your M solid will be getting converted into MN
positive. at anode m solid gets converted into MN positive aquis whose concentration is kept as C1 molar. On the right side we have got cathode and add cathode reduction takes place. Mn posit2 gets converted into M. At cathode Mn positive aquis which was there in the right container whose concentration was C2 molar. It got converted into M solid. So first of all you have represented this particular cell. you have represented this particular cell. So first of all the point is when will this cell work? What condition this cell has to follow so that it will
work the condition is pretty much simple. For any galvvinic cell to be working for any galvvinic cell to be working its cell reactions must be spontaneous. That means at anode that means at anode oxidation should take place and at cathode reduction should take place. Okay. At anode oxidation should take place. M should get converted into MN positive. I would say M solid should get converted into M N positive aqueous. With that you'll be writing N electrons. Okay. And its concentration is nothing but C1 M. What should be the reaction taking place at cathode for the
cell to be working at cathode? a reduction should happen. Mn posit2 should get converted into M. So I would say MN positive aquis whose concentration is C2 molar. I would say it should gain some n electrons and get converted into M solid. Okay. So in both the reactions the electrons are already balanced. If the electrons are balanced, let me directly add these two reactions. And when you add these two reactions, what will happen? This N and this N will get cancelled. This M solid, this M solid will get cancelled. So my net reaction is going
to be MN positive aquis gives MN positive aquis but their concentration is different. The concentration of this is C2 molar and the concentration of this one is C1 M. Now if I ask you what should be the reaction quotient? Reaction quotient should be C1 divided by C2. C1 divided by C2. Okay. If I ask you what should be the in cell value since two same electrodes have been used. So A not cell will be zero. in our salary zero. Now my dear students just try to understand one thing. Just try to understand one thing. If
I write the Nest equation, Nest equation says that E cell is equal in order cell that is 0 minus 0.0591 divided by N and it's log of QC. QC is how much? QC is C1 divided by C2. Or I can write it like this. E cell is equal to 0.0591 0591 / n multiply this log with minus it becomes log of C2 by C1 log of C2 divid by C1 correct this your E cell expression if this is your E cell expression now we can easily form the condition we can easily form the condition for
this cell to be working the condition would be I would say for the cell for the electrolytic concentration cell for the electrolytic concentration cell to be working to be working. Okay, by the way, which electrolytic conservation cell is this? For the electrolytic concerration cell, I would say involving involving metal metal ion electrodes involving metal metal ion electrodes for the electrolytic concentration cell involving metal metal ion electrodes to be working like a normal galvvinic cell. like a normal galvvinic cell I would say delta G should be negative or I can say E cell should be positive
or I can say E cell should be positive. You know that already for any gic cell to be working right its cell reactions must be spontaneous. That reaction must be spontaneous means delta G should be negative and if delta G is negative that means its E cell should be positive. We have discussed this. Now the point is when can this E cell be positive? This E cell can be positive only this if if this entire term is positive. When can be this entire term positive? Entire term is positive only if log of C2 by C1
will be positive. Now when is log of X positive? Log of X is positive only. Log of X is positive only when X will be greater than one. I would say when C2 divided by C1 would be greater than 1. Which tells you that C2 has to be greater than C1. Now what is C2 and what was C1? P2 is the concentration of electrolyte in cathode. P1 is the concentration of electrolyte in anode. So I would say concentration of electrolyte in cathode should be greater than that of concentration of electrolyte in anode. Then only then
only emf of the cell will come out to be positive and once emf of the cell comes out be positive then only delta G will come out be negative then only cell reactions will be spontaneous. Then then only at anode oxidation will happen at cathode reduction will happen. Electrons will move from anode to cathode and current will move from cathode to anode. So this is the condition my dear students which this cell has to follow then only its cell reactions will be spontaneous otherwise not right. Am I clear with this point? Is this particular point
clear? Perfect guys. Now there is one more thing here in this electrolytic concentration cell. Understand that also I'm marking the heading as electrolytic concentration cell involving involving hydrogen electrodes electrolytic concentration cell involving hydrogen electrodes. See how the cell looks like. First of all, uh duration of this particular session, I'll just take one more hour. Okay, I'll just take one more hour. Sir, did you work in Chennai with Shimon? Sir, me and Shimon, we started our teaching career together from Chennai. Yes. Yes. Yes. Yes. Right. Uh back in 2018, we started both of us and he's
a very good friend of mine. All right. Electrolytic conceration cell involving hydrogen electrons. Let's have a look on this now as well. First of all, how this particular cell looks like. See guys, in the middle I'm keeping two lines which represent your salt bridge. On the left side of the salt bridge, what do we have? We have got anode right? Let's say your anode is like this. H2 gas H positive aquis. Let's say the pressure of H2 gas is P bar and concentration of H positive aquas is for example C1 M and here I'm using
platinum solid. So the anode is my hydrogen electrode. Anode is my hydrogen electrode. Now if you want to have a look on the cathode, the cathode is again the hydrogen electrode only. H positive aqueous this is H2 gas. Let the concentration of H positive in the cathodic container is C2 molar. Let the pressure of this is P bar only. And this is platinum solid. This particular cell look at this particular cell. Look at this particular cell carefully guys. two same electrodes are used. The pressure is kept same but the concentration is different. So this is
electrolytic concentration cell involving which electrons? Involving hydrogen electrons. Now what is the condition? What is the condition? For this particular cell to be working. We have to frame that condition. We have to make that condition. To make that condition I must say for this cell to be working its cell reactions must be spontaneous. And when can be it cell reaction spontaneous? What is meant by that term? it cell reaction must be spontaneous. That means at anode oxidation should happen. At cathode reduction should happen. So at anode I must say your H2 gas should get converted
into H positives. And with that you'll be writing two electrons. What was the pressure of this H2 gas? It was P bar. What was the concentration of H positive? It was C1 M. Similarly reaction at cathode. Add cathode reduction should happen. H positive should get converted into H2. So the reaction has to be 2 * H positive aquis whose concentration is C2 M. It should be gaining two electrons and getting converted into H2 gas whose pressure is again P bar whose pressure is again P bar. Now tell me one thing are the electrons balanced in
both in both the reactions the electrons are balanced and since the electrons are balanced so you can cancel them up and value will come out to be two right but all things will get cancelled this H2 this H2 gets cancelled because the pressure is same so my net reaction becomes 2 * H positive aquis whose concentration is C2 molar it gives 2 * H positive aquis whose concentration is C1 molar now if I would want to write the expression for QC It becomes C1 squared divid by C2 square. So I will write C1 by C2
whole square or let me write it directly. What was C1? C1 is the concentration of H positive in anodic container. So this is the concentration of H positive in anode divided by what is C2? C2 is the concentration of H positive in cathode raised per what^2. Again if I ask you what about the E not cell value since it's the concentration cell wherein two same electrons are used. So E not cell has to be zero. E not cell has to be zero. Now my dear students as per your equation E cell will be equal E
not cell is 0 minus 0.0591 divided by N value is how much? 2 log of QC. How much is QC? QC is concentration of H positive anode divided by concentration of H positive in cathode whole raised power 2 E cell is equal -0.0591 0591 divided by 2 log of M ra^ n is n log m. So 2 comes to the front. So this becomes log of m by n. Log of m by n is nothing but log of m minus or let me just write it like this. Yeah. Minus log of n. Okay. So this
two this two gets cancelled. E cell will be equal 0.0591. Okay. 0.0591. Now take this minus inside it becomes minus log of H positive in anode minus * minus log of H positive in cathode minus log of H positive cathode. Perfect. So my net equation becomes E cell is equal 0.0591 minus log of H positive anode means PH of anode minus minus log of H positive means PH of cathode. So this is my final equation to calculate emf of the electrode sorry EMF of the electrolytic conceration cell involving hydrogen electrodes. Now for this cell to
be working for the cell to be working I must say for this cell to be working like a normal gic cell what should be the condition? The condition is pretty much simple for this cell to be working again delta G should be negative then only cell reaction will be spontaneous. Delta G negative that means E cell should be positive. How E cell will be positive? E will be positive only if PH of anode would be greater than that of PH of cathode. If PH of anode is greater than PH of cathode, then only this entire
term will come out to be positive. And once E cell is positive, automatically delta G becomes negative. And once delta G is negative, cell reactions will be spontaneous. And cell reaction spontaneous means electrons will move from anode to cathode. Current will flow from cathode to anode. That's all. Yeah. Right people is this particular point clear. So for the electrolytic concerration cell involving hydrogen electrodes to be working like a normal galvinic cell pH of anode should be greater than that of pH of cath. Let me first of all see whether you will be able to solve
some questions or not. Take this particular question. What do you think about this one? What do you think about this particular question? Do you see two same electrodes? Yes. So it's a concentration cell. Is the concentration same or different? Concentration is different. So it is the electrolytic concentration cell. Right? It's the electrolytic concentration cell. Now in the for the electrolytic concentration cell involving metal metaline electrodes to be working concentration of cathode should be greater than concentration of anode. Concentration of electroly and cathode. What is the concentration of electroloid and cathode here? C1. Right? So this
cell will work only if C1 is greater than C2. This is the condition for the cell reactions to be spontaneous. This is the condition for the cell reactions to be spontaneous. Correct? We have discussed it few minutes back only. Now the question is the cell reactions to be spontaneous if C1 is greater than C2. If C1 is greater than C2, cell reactions are spontaneous. Yes, correct. If C2 is greater than C1, if C2 is greater than C1, anodic concentration is greater than cathodic concentration. Current will flow from cathode to anode. There will be no current.
If C1 is greater than C2, if concentration of electrolyte in cathode is greater than that of concentration of electrolo in anode, cell reaction will be spontaneous. That means electrons will move from anode to cathode. Current will move from cathode to anode. Current will flow from cathode to anode. This is correct. Okay. If C1 is equal to C2, if the concentration of both the both the electrons is same, if the concentration of both the electrons is same, if C1 and C2 are same, this becomes one. Log of one is zero. That means E cell comes out
be zero and E cell comes out be zero means the cell does not work. E cell 0 means the cell does not work. No current is produced. So this is also correct. So A, C and D is the correct answer of the question. Clear? And if this is clear, can you check this particular question out? Just a second. Not this one. Yeah, this one also you can check. This one also you can check. If you look carefully, do we have two same electrodes? This is hydrogen. This is hydrogen. Yes, two same electrodes. Pressure is same,
concentration is different. So it is the electrolytic concentration cell involving hydrogen electrode. Yeah. This is the electro litic concentration cell involving hydrogen electrode. How do you calculate it E cell? Few minutes back only I gave you the formula to calculate EMF of the electrolytic concentration cell involving hydrogen electrodes. E cell is equal 0.0591. 0591 multiplied by PH of anode minus PH of cathode. So I have to calculate PH of anode and I have to calculate PH of cathode. So pH of anode will be what? See in the anodic container we have got H2SO4. If I
talk about your anode in the anodic container we have got H2SO4. How this H2SO4 would have dissociated twice H positive aquis plus SO4 diative aquis. What is the concentration of H2SO4 which we have taken? 0.05 molar. This would be zero. This would be zero. Now H2S04 it's a strong acid completely dissociated. So this will be zero. 1 mole gives two. So 0.05 gives 2 into 0.05 which comes out to be 0.1 m. So I would say in the anodic container the concentration of H positive is 0.1 m. Concentration of H positive in the anodic container
is 0.1 m which is 10^ minus 1. So I would say pH of anode will be simply equal to minus log of H positive concentration of anode. So it's going to be minus log of n^ minus1 the value comes out be 1. Okay. So this is the pH of anode. This is the pH of anode. Similarly look at the cathodic container. In cathodic container you have taken HCl. In your cathode we have taken HCL. So HCl would have dissociated as H positive plus CL negate. Initially the concentration of HCl is kept as 10^ minus 3
m. So this would be zero. This will be zero. Now this will be zero. This will be 10^ minus 3. So concentration of H positive in the cathodic container. Concentration of H positive in the cathodic container is equal 10^ - 3 M. So I would say PH of cathode will be simply minus log of H positive concentration in cathode. So it is minus log of 10^ minus 3. The value comes out to be three. You got PH of anode as well as PH of cathode. Now your E cell is equal E cell is equal to
0.0591. 05 91 multiplied by what is pH of anode 1 minus what is pH of cathode that is three the value will come out be positive or negative the value will come out to be negative and when E cell comes out be negative that means delta G will be positive and if delta G is positive the cell reactions will be non-spontaneous when delta G is positive the cell reactions will be non-spontton so the cell will not work that's something I was supposed to check right that is something I was supposed to check I believe this
is also clear to every one of you? Yeah. Okay. One more concentration cell is there my dear students. Let me make you clear with that as well. Or before doing that look at this question. Two line salt bridge on the left side. hydrogen electrode on the right side hydrogen electrode pressure is same pressure is same concentration is different so it is again electrolytic concentration cell involving hydrogen electrodes it is again electrolytic concentration cell involving hydrogen electrodes so it's E cell is 0.0591 this is pH of anode minus pH of cathode simple E cell is given
to be as 02364 is equal 0.0591 0591. What is the pH of anode? On the left side, you have got anode. So, PH of anode is X minus. What will be the PH of cathode? In the cathode, it contains H positive concentration is one. This is the H positive concentration in cathode. So, minus log of 1 is zero. So, when you solve X value will approximately come out to be four. What was this X? X was the PH of the anodic container. So, PH of anodic container is coming out to be four. That's all. Yeah.
Perfectly done. Okay. There is one more concentration cell which you have to be familiar with. What is that? What you call as electrode concentration cell. Electrode concentration cell. Now what happens in electrode concentration cell? In electrode concentration cell let me tell you emf is generated emf is generated by keeping by keeping partial pressure different emf is generated by keeping partial pressure different. Let me show it to you exactly. Let me show it to you exactly. For example, this is your salt bridge. On the left side of the salt bridge, you have got anode. Let's say
your anode is this H2 gas whose pressure is for example P1 bar. This is H positive to aquis whose concentration is C M. Okay. And here I'm writing platinum solid as well. On the right side I'm using one more electrode which again the hydrogen electrode. So H positive aquis whose concentration is C M this will be H2 gas whose pressure is P2 bar. Okay platinum solid. Now do you see two same electrodes used hydrogen hydrogen? Two same electrodes are used right people. I would say in this particular cell if you see carefully two same electrodes
are used. Now concentration is kept same partial pressure is kept different. That particular concentration cell in which in which pressure was same concentration different that was electrolytic air concentration is same but partial pressure is different. So this is electrode conceration cell. Okay this is electrode conceration cell. Well you can represent the cell in a different way as well. How can you represent it in a different way? Take do you see H positive whose conceration C? H positive whose conseration C. Instead of this entire term instead of this entire term write only H positive only one
H positive whose concentration is C. Now on the left side of this H positive there was H2 gas whose pressure was P1 bar right and this was platinum solid. on the right side of this H2 H positive there was again H2 gas whose pressure was P2 bar and this is platinum solid well you can represent the cell like this as well there is no need to represent a sol two lines here you can represent it like this as well okay instead of this entire part I just wrote one H positive and the left side of
it and the right side of it that's it now we have to see the condition condition which The cell has to follow so that its cell reactions will be spontaneous. So that the cell will work. Now for the cell to be working you know for the cell to be working cell reaction should be spontaneous. Cell reaction spontaneous means at anode H2 should get converted into at anode H2 gas whose pressure is P1 bar it should get converted into 2 * H positive aquis whose concentration is C m plus 2 electrons. Correct? Now add cathode. At
cathode H pos2 gets converted into H2. So I would say 2 * H posit2 aus whose concentration is C m it should gain two electrons and get converted into H2 gas. Okay now people electrons are balanced. Okay twice H positive twice H positive will get canceled out because their concentration is same. This H2 and this H2 won't get canceled out because their partial pressure is different. One is P1 and one is P2. So my net reaction becomes H2 gas whose partial pressure is P1 bar. It gives H2 gas whose partial pressure is P2 bar. Okay.
Now now now if I ask you what is the E not cell value? E not cell value is zero. You know that already. What is the N value? N value is two. If you want to write the QC expression, QC expression is going to be partial pressure of this H2 which is P2 divided by partial pressure of this H2 which is P1. correct which is P1. Now for this electrode concentration cell to be working I would say for the electrode concentration cell which we have right now to be working like a normal galvvinic cell wherein
electrons move from anode to cathode and current flows from cathode to anode. For this cell to be working its cell reactions must be spontaneous and cell reactions must be sp will be spontaneous only if E cell value will be greater than zero. If E cell value will be greater than 0. Now first of all what is the expression of E cell? Expression of E cell is nothing but E cell is equal E not cell which is 0 minus 0.0591 divid by N value is 2 log of QC. QC is P2 by P1. you see is
P2 by P1 or this minus when you multiply with log it becomes P1 by P2. This becomes P1 by P2. So for this cell to be working E cell should be greater than zero. E cell can be only greater than zero if this term is greater than 0. If log of P1 divided by P2 will be greater than 0. Now when can be log of X greater than 0? Log of X can be greater than 0 only when X is greater than 1. So I would say P1 divid by P2 should be greater than 1
which is simply telling you that P1 should be greater than P2. What was P1? What is P2? P1 was the partial pressure of anode. P2 is the partial pressure of cathode. So I can say partial pressure of anode should be greater than that of partial pressure of cathode. Partial pressure of anode should be greater than the partial pressure of cathode. Then only emf of the cell will be positive. And once emf is positive, delta G will automatically be negative. And once delta G is negative, cell reactions will be spontaneous. That means at anode oxidation will
happen. Add cathode reduction will happen. Electrons will move from anode to cathode and current will flow from cathode to anode. So this is the condition which the cell has to follow. Then only its cell reactions will be spontaneous. Okay. Am I clear with this people? Quake. Yes, E not is zero because of two same electrodes. There is one more topic left in the EMF part guys. Okay, it is not going to take a lot of time. It is not going to take a lot of time, right? It's a simple topic only. It is thermodynamics of
the cell. Thermodynamics of the cell. See sometimes in the question they'll ask you to calculate delta G for the cell. They'll ask you to calculate delta S for the cell. They'll ask you to calculate delta H for the cell. Gives free energy change. Right? Entropy change during the cell reaction. Enthalpy change during the cell reaction. These can be asked in your examination as well. These three thermodynamic functions. Calculate the gives free energy change. Entropy change during the cell reaction. enthalpy change during the cell reaction. Now, how do you calculate delta G for the cell reaction?
Uh, similarly, delta S for the cell reaction, delta H for the cell reaction. Let's have a look on that as well. See, one simple thing I'm going to do, I'm going to use one of the equation from the thermodynamics which is DG is equal to W DP minus SD. This is one of the equations of the thermodynamics which is used for reversible processes basically. reversible processes. Now my dear students, if I write the same equation at constant pressure, if I write the same equation at constant pressure, at constant pressure, I must say DP value will
become zero. If DP is zero, I would say DG is equal to minus S DT. So at constant pressure, DG is equal to minus S DT. Now for the reactions for the reactions you do not define gives free energy you define change in gives free energy for a reaction for a reaction you do not define entropy you define change in entropy for the reaction for reactions entropy change gives free energy change is defined so instead of g you guys are going to write delta G instead of S you guys are going to write delta S
so it becomes D of delta G is equal minus * delta S for the reaction multiplied by DT. So I would say delta S for the reaction will be equal to minus * D of delta G for the reaction divided by divided by DT. Correct? So it becomes minus * D upon DT of delta G. Delta G is nothing but minus N of E cell minus N of E cell or or I can further simplify it a bit. I'll say delta S for the reaction delta S for the cells delta S for the cell reaction
is equal to mines comes out NF comes out so it becomes NF ded by DT since we had kept pressure constant so it is at constant pressure. So this is one of the equations by means of which we can calculate entropy change during the cell reactions. When the galvvinic cell works when the galvvinic cell works right entropy change during the reaction is get by this particular formula and you already know f you already know this d by dt is called as temperature coefficient of the cell. This is called as temperature coefficient of the cell. And
if I ask you what are going to be its SI unit, this is emf, this is temperature. So it's volt per kelvin is going to be its SI unit. Okay. The entropy change units, it's SI entropy change unit. Entropy SI unit is going to be jewels per kelvin. So this is the formula with the help of which we can calculate delta S for the cell. Similarly, delta G for the cell you already know that is minus NF cell. Nothing to do with this particular point. This is something which we already know. Now similarly how do
we calculate delta H for the cell? Since delta G you know delta S you know you already have the result delta G for the reaction would be equal delta H minus T delta S for the reaction. What do I have to calculate? I have to calculate delta H for the reaction which becomes delta G for the reaction plus T * delta S for the reaction. Now delta G for the reaction is equal to minus NF cell plus E stands for temperature. Delta S for the reaction we have calculated already that is NF d upon DT
at constant pressure. Perfect. Now the final result will be delta H for the reaction will be equal if I take NF common. If I take NF common, it becomes minus * E cell plus D upon DT at constant pressure multiply by temperature. So this is the result with the help of which you can calculate enthalpy change during the cell reaction. So these are the three thermodynamic functions which are asked to us. Calculate delta S for during the reaction. During the cell reaction, calculate delta G during the cell reaction. calculate delta H for the cell reaction.
Right? All the three results are in front of you. Okay? At the same time, sometimes they might you they might ask you about the efficiency of the cell. Sometimes they might ask you about the efficiency of the cell, which is nothing but which is magnitude of delta G divid by magnitude of delta H multiplied by 100. They might ask you about this as well. Calculate the efficiency of the cell. Delta G divided by delta H multip 100 mod magnitude. Okay. Right. Now guys if you look at one more thing if you want to have a
look on variation of emf. Variation of emf with temperature. How this emf varies with temperature. How the emf of the cell varies with temperature. Since we have got two types of cells basically we have got two types of cells. The first cell whose temperature coefficient d by dt the first type of the cell whose temperature coefficient is positive and the second type of the cell whose temperature coefficient would be negative. First type of the cell whose temperature is positive. second type of the cell whose temperature equation is negative. So that particular cell whose temperature coefficient
is positive. If you increase the temperature of this cell whose temperature equation is positive, if you increase the temperature, I would say dt change in temperature will come out be positive. If dt is positive that means denominator is positive. The overall term has to be positive. The overall term has to be positive. Denominator is already positive. That means numerator also should be positive. Numerator positive means de value should be positive. de positive what does that mean? D positive means e2 should be greater than e1. What does that mean? That means on increasing the temperature emf
increases. On increasing the temperature emf increases. So that means emf of the cell is directly proportional to temperature. But this is not valid for all the cells. This is valid for only those cells whose temperature question is positive. Right? This is valid for only those cells whose temperature question is positive. Correct? Similarly, what about those cells whose temperature equation is negative? When you increase the temperature of such cells whose temperature is negative, when you increase the temperature, that means dt value will be positive. DT is positive. Overall term has to be negative. That means numerator
has to be negative. Numerator negative means DE value is equal to negative. When is D negative? D negative means E2 would be less than E1. That means EMF of the cell decreases. That means EMF of the cell decreases. So EMF and temperature are they now directly proportional or inversely? They're inversely proportional to temperature. But this is this particular thing is valid for only that particular cell whose temperature coefficient would be negative. Okay. whose temperature coicient would be negative. So emf and temperature they are related. They are directly proportional as well as inversely proostal depending on
whether the temperature coefficient of the cell is positive or the temperature coefficient of the cell is negative. Correct? One question with that will end the session. Okay. Look at this particular question. We are given with the reaction. Look at the reaction. N value is equal to two. N value is equal to what all things we have to calculate delta G delta H delta S see delta G is nothing but that is minus NF E cell simple N value is 2 F is 9650 and E cell is given to me in the question which is 1.018
018. When you solve this, the answer will be in jewles. Similarly, delta S delta S will be minus * NF D / DT at constant pressure. Was it minus times or plus? I think it was NF D by DT. It was not minus. It was NF D by DT. So, N value you know already that's 2. F value is 96500. D by DT is called as temperature equation. that's given to me that is -5 into 10^ -5 since all the parameters are used as per SI system. So delta S will come as per SI system
is going to be jewels per kelvin. Similarly your delta H delta H formula few minutes back we made NF this is minus E cell plus D upon DT at constant pressure multiplied by temperature. So n value is 2 f is 96500 minus e cell 1.018 018 plus d by dt is how much - 5 into 10^ - 5 and temperature in kelvin will be uh I think something is okay temperature in kelvin will be 298 kelvin again the matter of calculation you'll be getting the delta h that to in jones okay and my dear students
with this the emf part of this particular chapter is over there is one more part left in this chapter that is electrolysis part as I've told you I will break this chapter into two parts. This was part one and part two I'll be posting tomorrow morning. That will be 3 to 4 hour session. Okay, that will be the recorded part but that be the recorded session. All right, that would be the recorded session. So are all the things clear till here? Everyone once did you understand it properly? whatever things I have discussed with all of
you. So with this your AMF part of electro game stage chapter that means 70% of the chapter is done. The remaining 30% I'll be providing you in the morning exactly at 10:00 a.m. Right? That will be a 3 to 4 hour session a recorded session. Okay? That will be also detailed. So tomorrow by 1 or two you'll be done with this particular chapter. All right, perfect. And do let me know in the comment section of this particular video which chapter do you want the next. Okay, do let me know in the comment section of the
video. I would want you guys to write in the comment section which chapter do you want me to tell you next. And at the same time one more thing on Saturday I'll be taking the session on Saturday but you are going to let me know in the comment section. Okay, ch with this I'll be n leave I wanted to take the 4hour session live the part two of this chapter but tomorrow it's unbecom session okay tomorrow it's unbecome session so I cannot I mean the whole schedule will get altered It's one of the same thing.
It's one of the same thing. Don't worry. Okay. Anyways, the PDF of this session wherein I had already kept the electrolysis part also you'll be getting this PDF in the telegram. Okay, you'll be getting the PDF of this in the telegram. All right, this PDF I'll be sharing in the telegram. Hello guys, thank you so much. See you. Bye-bye.
