 Okay, so this is the question we have discussed in the second one. The next one you see, an excess of liquid mercury is added to an acidified solution of this molar Fe3+. It is found that 5% of Fe3+, remains at equilibrium at 25 degree Celsius. We have to calculate E0 for this, assuming that the only reaction that occurs is this. Hosell reaction is given, and E0 of Fe3+, to Fe2+, is given. So you have to use Nernst equation here. This value, it should be E0, it is just a misprint. This should be E0. Now you see this mercury is getting oxidized, and Fe is getting reduced, right? So cathode is what? Cathode is Fe3+, to Fe2+, reduction, and anode is Hg2, Hg2+, okay? E0 of this cell is what? E0 of cathode minus E0 of, right? Here you see this E0 of anode, both are reduction potential. And that is what we need to find out, reduction potential. This E0 is the answer we have, okay? Now when you apply Nernst equation into this, so E cell is equals to this thing, right? E0 cell is nothing but this. E0 cell minus 0.0591 divided by 2, right? Because the N value is 2 log of Hg2+, Hg2, 2+, 102, 102. We have Fe2+, square divided by the ion, we have Fe3+, square. This is what the question we have. It says what? 5% of Fe3+, remains at equilibrium. Means when this reaction takes place and equilibrium achieves, there we have 5% of Fe3+, present in the solution. Obviously the state of equilibrium, E cell is 0, is equals to E0 cell minus 0.0591 by 2. We'll substitute the value of this. The concentration of Fe2+, 2F3+, so it is given 1%. So you see this thing we'll use over here. We have Fe3+, the reaction is this. Concentration we can find out with this value. 5% thing is there. OK, so serious. The reaction is 2Hg plus 2Fe3+, gives Hg2, 2+, plus 2Fe3+. This is 1 into 10 to the power minus 3, present initially. It is 0 and 0. When 5% left with 95% react, right? So this is the Fe3+, concentration which is left here. That will be 0.95 reacts, 1 minus, or we can write simply, 10 to the power minus 3 into 0.05, alpha is this, right? C minus C alpha we can write, so I can write down here. Suppose if I write it in the general term, probably that will be easy for you to understand. If it is C initially and C value is given, so it is C minus C alpha. This will be C alpha. This will be 2C alpha. What is C alpha? C is 10 to the power minus 3 into 1 minus alpha. Alpha is 5%. So it is 1 minus 0.05. C is 10 to the power minus 3. Alpha is 0.05, 2 into 10 to the power minus 3. Alpha is 0.05. So this is 0.95 into 10 to the power minus 3, right? This will be 0.05 into 10 to the power minus 3. This will be 0.1 into 10 to the power minus 3. I will find out this concentration we have already. We'll substitute this and we'll get the answer. One more thing you see here, one thing we must write. This is one thing I'm missing here. Just one small correction here. This should be 2C alpha because 2 we have here. So this will be what? This will be 2 alpha. 2 alpha is 0.1. This is the concentration we have. Now when we substitute all these values here and we'll find out E0 cell, right? So E0 cell here is we can also write this down as E0 cell is equals to. It is E0 of cathode minus E0 of. And that would be equals to. When you solve this, we'll get some value here. E0 of anode we need to find out, which is Hg2 Hg2 plus reduction potential. So E0 of anode is equals to. The value we get is 0.7926. This is the answer we get in this question. So what we are doing here, we are just using Nernst equation. Equilibrium condition will apply here. And then to find out the concentration of the species involved, we use the concept of equilibrium. Ions are in equilibrium. So how it dissociates, how concentration of ion changes. We'll find out that and we'll substitute it in this. We'll get the expression. OK, fine. Now this is the next question. The standard reduction potential of copper is given 0.34. Standard reduction potential of copper is given 0.34. Find the reduction potential at pH 14 for the above couple. Ksp is given. We need to find out the reduction potential of Cu2 plus to Cu, Ksp is given. So Ksp is given for CuOH whole twice. So CuOH whole twice dissociates as Cu2 plus plus 2OH minus. So Ksp expression, if I write it down, that will be concentration of Cu2 plus concentration of OH minus square. So if it is S, this one is 2S. So we can write 4SQ. So concentration of Cu2 plus is equals to S is equals to Ksp by 4 root over of it. And that would be 10 to the power minus 19 divided by 4 root over of it. OK, so this is, one more thing is given, pH is given. So it means PoH is what? PoH is 0, means OH minus concentration. We can do this way also, because why we cannot use this, I'll tell you, because the solution, pH is given. Means it's OH minus concentration is already given. Means in the solution, we have H plus OH minus already present. Hence, we cannot talk about the solubility of this. This solubility we get when there is no common ion. But when common ion is present, the solubility is something different. So this is true when the solution is, does not have any common ion with respect to the salt. Hence, we cannot use this method over here, because the solution has, we do not have any idea about the solubility of this compound, since solution thing is not mentioned. Ksp expression is this, but we know the fact that when pH is 14, it means PoH is 0. This means OH minus concentration, if you find out that is one. And hence, one is substitute here. So concentration of Cu2 plus, we can find out where that is 10 to the power minus 19 with the given data, okay? Now we can use NUST equation and find out the answer, right? So this equation for Cu2 plus to Cu, we can write E Cu2 plus to Cu is equals to E naught is 0.34 minus 0.0591 by 2 log of 1 by Cu2 plus, okay? All the value you can substitute here, 0.34 minus 0.0591 by 2 log of 1 by Cu2 plus is 10 to the power 19. So when you solve this, you'll get it as around minus of 0.221, right? This is the answer for this question, okay? So the point here, we won't use the, because ultimately we have to find out the concentration of Cu2 plus, but we won't use the solubility expression here. Why? Because the pH of the solution is mentioned, right? So there is H plus present into the solution and in presence of H plus OH minus, the solubility may differ, right? That's why we cannot use that KSP is equals to S4S square X4S cube expression here, okay? Hence, we'll find out the OH minus concentration with respect to pH, which is this and then KSP. Okay? Now, the next question you see, how many grams of silver could be plated out on a serving tray by electrolysis of a solution containing silver in plus one oxidation state for a period of eight hours, okay? So obviously when electrolysis is there, the question is based upon Faraday's law of electrolysis. Time is given eight hours, so eight into 60 into 60 second, right? This is the time. I is given. I is also 8.46 ampere. What is the area of the tray if the thickness of the silver is plating is the centimeter, right? Area of the tray. So see density is given. Density is given with density is equals to mass by volume. Volume is equals to what? Area into the thickness. That's the area we need to find out. Thickness, we know. Mass, we can find out the mass is deposited and density also given, right? So first of all, we'll find the idea of behind this question is what? We'll find out the mass deposited, that is M. So mass is what? Mass is equals to density into volume, and volume further, we can write density into, volume further, we can write area into the thickness that is X suppose we have, okay? This area we need to find out. So area is equals to what? Mass divided by density into the thickness X, okay? So density is given, thickness is already given, right? We need to find out mass. Now for mass, what we can write? M is equals to EIT by 96500, right? E is the equivalent mass and that would be 108 for silver. Plus one, so it will 108 only. I is 8.46 and T is 8 into 3600. This whole thing is divided by 96500, okay? So you can calculate this and we'll find out mass that mass would be approximately 270, 2.17 grand. 273 also you can take approximately this. Now this mass will substitute here, 272.17. Density is given 10.5 into the thickness is 0.0025, right? So area is 1.02 into 10 to the power 4 and the unit is centimeter square. This is the answer for this question. Easy one, next question you see. The electrolysis of the solution of MNSO4 in aqueous sulfuric acid is a method of preparation of MNO2, okay? What is said that electrolysis of the solution of MNSO4 in aqueous sulfuric acid is a method of preparation of MNO2 as per the reaction period. Passing the current of 27 ampere, okay? So I is given 27 ampere, T is 24 into 3600 seconds, right? Mass of MNO2 is 1 kg means mass is 1000 grand, okay? We need to find out the value of current efficiency, okay? Now for all these things, you know we have to apply Faraday's law of electrolysis. M is equals to EIT by 96500. We'll find out current for this. What is the current required for this particular mass to get deposited? 1000 into 96500, okay? Divided by E into T. T is 24 into 3600. What is E? E is the equivalent mass. So that would be atomic mass of magnies and the change is what? The change is of, you can see here, the number of electron exchange is two. And this is two, plus two electrons, correct? So if you write down this reaction, I'll tell you. How do we get the number of electron exchange? MN plus two converts into MNO2. So when you balance this in an acidic medium, we'll write two H2O this side, then four H plus this side. When you balance the charge, we have to add two electron this side, okay? So charge will be balanced. Two electron is getting exchanged. This will be atomic mass of magnies divided by two equivalent masses. So when you solve this, the I value that you get here is 25.67 ampere, okay? So current efficiency is what? Current efficiency is the current that has been used divided by the current which is provided into 100. When you solve this, you'll get approximately 90% current efficiency. Next one you see. Calculate the equilibrium constant for the reaction this. Everything is given, right? Equilibrium constant means what? Delta G is zero. So we can write Delta G naught is equals to 2.303 RT log KC. This KC we need to find out, okay? What is the E naught of the cell? Because Delta G, we know it is minus NF E naught of the cell is equals to 2.303 RT log KC. So this we can further write the number of electrons so this we can further write the number of electron exchanges one. So log of KC is equals to minus N divided by 2.303 RT by F. So log of KC is equals to into E naught cell. So log of KC is equals to one. One divided by this value is 0.0591 into E naught cell is CFE is getting oxidized, CE4 plus is getting reduced, right? So CE4 plus to CE3 plus, FE3 plus to FE2 plus. So this is cathode and this is anode. So E cathode minus E anode. E naught cell is 1.44 E cathode minus E anode the reduction potential of this 0.68, okay? This is the E naught cell. That will be 6.76 volt, which we substitute here. KC value is, okay? I have missed a sign here, minus. This minus, minus gets cancelled. This is the answer we get, okay? So answer for this question is 7.6 into 10 to the power 12. The next question is find the solubility product of a saturated solution of this in water at 298 Kelvin. If the EMF of the cell is, this is AG to AG plus and this is this, correct? AG plus to AG. So this is, this side we have cathode and this side we have anode. So for this reaction, if I write down the E cell is equals to E naught cell minus 0.0591 divided by one log of AG plus cathode by AG plus, this is what the reaction we get. Now, AG plus cathode, the concentration is given 0.1, right? Since we need to find out a solubility product. So we have to find out the concentration of AG plus in the anodex side. So if I substitute all the value, E naught cell is 0, it is a concentration cell. E naught cell is 0, 0.0591 by one log of 0.1 by the AG plus concentration on the anodex side. Now, when you solve this, we'll get the concentration of AG plus at anode and that is, that is 1.66 into 10 to the power minus 4 in order, right? So when you look at the dissociation of the salt, which is AG 2 CRO4 here, AG 2 CRO4, CRO4, it dissociates as 2 AG plus plus CRO4 2 minus, okay? The concentration of this is 2S, which is given here. We have already calculated and this is S. So with this relation, you see the concentration of CRO4 2 minus, if you calculate, O4 2 minus, that will be the half of this, divided by 2, half of this. Now, the solubility product KSP is equals to we have 4S cube, right? S value is, S value is this. So answer is 4 into 1.66 into 10 to the power minus 4, divided by 2, 2. So this is 2.287 into 10 to the power minus 12, more per liter, this is the answer, okay? I can understand, guys, there are some calculation into this, some log, anti-log value must require to solve this kind of expression. But these are the questions you can see, these are questions of 1998. There we have this kind of calculation. We have made, we have J is spinning and Vane's exam that point of time. And we have subjective questions also. So we get enough time for this kind of questions there, okay? So that's why they give some weird kind of calculation also in the question. But nowadays, you don't have that much time, right? So you won't get calculation also like this. But the concept you must understand how to do this kind of questions, okay? So fine, guys, thank you. We are done for the session for today, okay? See you in the next session. Take care, bye-bye.