 This is the half reaction, we can apply half reaction as it is, initially the concentration of C u 2 plus is given. Suppose it is given that is final concentration is what? X5. Now we have to apply the next equation for this and this and then we have to subtract this equation. For this if I use e-cell is equal to what? e-cell is equal to minus 0.0591 by 2 log of 1 by X. Now for the other equation, e-cell is equal to minus 0.0591 by 2 log of 1 by 0.1 X. This is final and this is initial. So 2 minus 1 will be what? e-cell final minus e-cell initial and that will be this minus this, we get answer because e0 cell is constant that is why the cell is not given in others. This we get cancel and we have minus 0.0591 by 0.06 because 0.0654 gives you 0.03. That approximation we can make 0.06520 common log of 1 by 0.1 X minus this that will be plus 0.06 by 2 log of 1 by, so if I take common 0.03 log of X divided by 0.1 X. This by this X will go in the numerator and 0.1 X. So it becomes log 10 answer will be what? 0.03. So this minus 0.0591 by 2 log of 1 by 0.1 X by X because minus 0.0591 by 2 log of 1 by 0.1 X. Okay yeah actually that expression is there. It's like log. It's like log. It's like log. Right, right. Right. That's what log e minus log b log e minus log b. Right on B. Right on B. Now we have to find out that these three terms we have to find out. It's not that important but just we'll see the formula. Okay. Only one formula out of the three is important and I'll tell you the temperature coefficient formula. Okay. Now you see for this g we can write it as h minus ds right. And h we can write e plus pd internal energy plus pressure volume. Very small change what we can write dg is equals to de plus pdv plus vdp minus pds minus sdv. Is it? First change in 1 down into de plus p constant change in v plus v constant change in v plus t constant change in s plus sdv. Right understood. Now you see de plus pdv from first law we can write it as what dq from FL 14. First law of thermodynamics. dq is equals to de plus pdv right. And dq is also equals to dvs from second law of thermodynamics. Yes or no? What? dv is equals to dvs second law of thermodynamics. This is term this de plus pdv and tds gets cancelled yes or no? Yes. Right. At constant pressure because usually the cell reaction takes place at constant pressure only at constant pressure dp is equals to what? Zero. Zero. So this term also equals to z. Now this sdt or we can write down s is equals to minus dg by right s is equals to minus dg by dt right. Now change in entropy is what delta s is equals to minus d by dt of delta g. What is delta g? Minus. Minus nf esel right. So when I substitute it here it becomes nf tesel by dt. So the formula of delta s is equals to nf tesel by dt. On this formula they have asked question once out of the three delta s they have asked once. Tesel by dt we call it as temperature coefficient temperature coefficient. Sometimes what happens they will get the relation of esel with temperature like this dt square plus dt plus c something like this then give you the expression. You have to find out they ask what is the temperature coefficient at this temperature. So what do you have to do? You have to differentiate this desel by dt right. This side you differentiate substitute temperature. So this formula on this formula they have asked one question right. Other two del H and delta cp is not that much important but we will see the formula. There are two steps there. Now you see this one with this particular expression we have delta g and we have delta s. And we find out delta h from this right. So delta g is equals to what we can write delta h minus t delta s right. So we have to find out delta h delta h is what? Delta g plus t delta s delta g formula we have minus nf tsel plus t delta s formula we have. So the formula of that minus nf tsel plus nf t tsel by delta d equals to density pt. So for one more for one more what we can write cp is equals to dh by. So delta cp is equals to what? d by dt of delta h delta h you have in substitute okay. The expression you get will be easy and small it won't that big okay. Something you get cancelled when you substitute delta h here and difference here okay. You see d by dt I will write down here d by dt of minus nf tsel plus nf t d tsel by delta d. Can you do the same difference here? First of all what we do u plus v correct. And in the v again we can use u into v right. Here minus nf tsel by dt plus write nf d tsel by dt into dt by dt plus. Did you write this? Del h. Del h we have written down plus we will use t into d2 esel by dt. This is clear. See first of all what we do this nf we take common d esel by dt is out differentiation of t. Plus t is out differentiation of d esel by dt. D esel by dt is what? d esel by dt is what? Right. What? Right. So this dt by dt is equals to what? 1. 1. What? 1. It's differentiation. Sir didn't you already differentiate that first term? What? This one is there nf is out minus d esel by dt. Now differentiation of this again u into v we are applying. Right. If you already don't see this do it on your own. U plus v and then v again u into v you are applying right. Right. Now you see this dt by dt is what? What we have left with nf d esel by dt minus nf d esel by dt. These two terms will also get transferred. Right. If you already doubt into this do it on your own. It's very easy to do it. U plus v and then v also you apply u into v. Right. Okay.