 Hello guys. Welcome to the session. Today we are going to discuss some problems for ITGE and need exams. The topic we picked up today is electrochemistry. We have already had a session already on this electrochemistry. We have discussed around 7-8 problems there, a J-level problem. Subjective questions we are discussing. You see there is no option over here. So these are the questions of ITGE but it is not like it is only useful for those kids who are preparing for ITGE. Concept you must understand. It helps you in other competitive exams also. So look at this question number 9. The standard reduction potential is given for AG. ASP is given for AGI. Evaluate the potential of AG plus to AG electrode in a saturated solution of AGI. Also find the standard reduction potential of this electrode. You can pause the video and you can try this on your own. I am continuing with the solution over here. You see the electrode potential of AG plus to AG it is given. That is 0.799. AGI KSP value is given. When we have this saturated solution of AGI, the half cell reactions are. What are the half cell reactions? We have AG electrode and AGI saturated in the solution. So first reaction is the conversion of AGI into silver plus I minus. This is reduction reaction cathode. And the silver is getting oxidized. This is the reaction at anode. So overall the reaction is what we can add to the two and we can find out the E value of this. This KSP of AGI is given. So we can find out the AG plus concentration. AGI is given. That is the concentration of AG plus into the concentration of I minus. So concentration of AG plus square is equals to because both will be equal. So this is equal to 8.7 into 10 to the power minus 70. When you solve this, you will get AG plus concentration as we calculate this root over of it. That is around 9.3 root over of it and this is 10 to the power minus 9. This is the AG plus concentration we get. Now the standard reduction potential of AG plus to AG is given. We need to find out electrode potential of this. So we can write down the next equation for this, which is E of AG plus to AG is equals to E naught of AG plus to AG minus 0.0591 by 1 log of 1 by AG plus. So now we can substitute the value of E naught of AG plus and AG plus here will get the answer. E naught of AG plus is 0.799 minus 0.0591 by 1 log of 1 by 9.3 into 10 to the power minus 9. You can solve this. You'll get the electrode potential for AG plus to AG is approximately 0.324 volt. This is the first part of this question. AG plus to AG electrode in a saturated solution electrode potential is 0.324 volt. Also find the standard reduction potential of this electrode I minus AG I and AG. The cell that we have right this cell standard reduction potential of this we need to find out. So this means what the reaction if you consider here we have AG I saturated. So for this cell for this cell that E naught cell will be E naught cell will be because E cell is zero. So it is 0.0591 log KSP. KSP of what? AGI. So we know all these value when you substitute you'll get 0.0591 log of KSP is 8.7 into 10 to the power minus 17. So when you solve this you'll get minus 0.95. This is the E naught cell of this. Now you see this E naught cell is equals to what we can write further because we need to find out for. So here you see we need to find out the standard reduction potential of this electrode right I minus to I plus and AG plus to AG right AG plus to AG and I minus to I plus we have. Okay, so the E naught cell for this E naught cell for this one will be the E of E naught of cathode E naught of cathode minus E naught of and both are reduction potential. E naught cell we have calculated and I know this what I know this silver electrode because this is saturated resolution. So this is one electrode and this is another electrode right this is one electrode and this is another electrode. So here you see E naught of cathode and what is AG plus we have so E naught of cathode is equals to minus of 0.95 plus for AG plus it is 0.799 reduction. So when you solve this you'll get negative of 0.151. So this is the answer. So we have to answer one is this the one is this reduction potential E value is this standard reduction potential of E cathode is this right. So we need to find out we have actually two electrode one electrode is AG plus to AG and another electrodes is this I minus to AGI to AG means silver is present in in a solution in which which is saturated with AG. Okay, one behaves as cathode other one behaves as anode the reduction potential of AG plus is given for AG I KSP is given. Right, so we have calculated the reduction potential of standard reduction potential of this one right AG I that is that we have you know overall cell reaction is this overall cell reaction or how do we get when we add these two. Right, so cathode is this it is given here. So cathode is nothing but this which is this is here is the cathode. That's why we'll write this and E naught cathode is nothing but this. Okay, that is the answer for this question. Okay. Next question you see. You see there's a storage cell Edison storage cell is represented by this is given half reactions are also given. Okay, and I two or three such to a gives this. E naught value is given for both and both are reduction reaction. If you observe this, both are reduction reaction. And hence the potential that is given that is the reduction potential. So what we can conclude from this, since the electron is getting consumed here. So both are RPs reduction potential. This is also reduction potential. So the one which has more reduction potential will get reduced. Okay, so cathode is cathode is. This reduction. Okay, I'll write down here. This is cathode. Obviously it is the right side. So it is cathode left side. So it is and right side cathode left side and okay cell reaction will be what will have to add the two reaction. Right, we have to add the two reaction. So what we'll do will inverse the second one and we'll add with two. Right, we'll inverse the second one and we'll add it to the cell reaction is here cell reaction is will inverse the two and add with the first one. So I'm writing it down directly. F e solid. Plus. And I two or three solid. X two or one electron gets cancelled. Right hand side we have two and I oh solid. Plus F E O. Okay, this is the cell reaction. So first one we have done a bees what bees what is the emf the cell. How does it depend on the concentration of keywords emf of the cell is what he said again will use last equation is equals to E naught of the cell. 0.0591 by two log of n I O square F E O divided by F E and and I do. I think this should be right. All these are given solid. So this thing is one. We can write this entire thing as one because all given is solid. Right, so directly we can substitute here. E naught of the cell is what ecathode minus enode. So ecathode is this. That is 0.40 minus E anode is this that is minus of 0.87 minus 0.0591 divided by log. One is zero. So answer for this question is the sum of these two, which is 1.27. Now the OH minus concentration that we have here, it gets cancelled right it is not present in this expression there's no OH present here. Hence what we can write the second part of this, the E cell is independent of KOH concentration. This is the second part of this question. Okay, so this is the another answer we have. Now the last one is C. What is the maximum amount of energy that can be obtained from one mole of ni two or three. Okay, so one mole of the energy is what the energy is nothing but the delta G. But we have to talk about the magnitude here. NF E cell, E cell, right. Since maximum amount of energy has been asked, so it must be E naught cell. You won't take EMF. EMF is this E cell. E naught cell if you take when it has 100% efficiency, all these E naught cell converts into energy, then we'll get the maximum energy, right. That's why we have taken E naught not E cell. So we'll substitute 2 into 96500 into 1.27, the E naught cell, which is 245.11 kilo. So this is the answer for the question. Okay, so I hope all of you understood this. Thank you for listening and see you in the next video.