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Question

Given
CuI(s)+eCu(s)+I(aq);E=0.188V ...(i)Cu2+(aq)+I+eCuI(s);E=0.868V ...(ii)Ksp(CuI)=1012M2 ...(iii)
298R ln 10F=0.059
List -1 contains half cells and List-2 contains emf values of half cells

List - IList - II(P)Cu+(aq)+eCu(s);E01=(1)0.34 V(Q)Cu2+(aq)+eCu+(aq);E02=(2)0.16 V(R)Cu2+(aq)+2eCu(s);E03=(3)0.52 V(S)Cu2+(1M)+I(1012M)+eCuI(s);E4=(4)(E01E030.02)V(5)2×E02+0.02
Match the emf of cell in LIST-I with one or more values in LIST-II and choose the correct option.

A
P3;Q1;R2,4;S3,5
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B
P1,5;Q3;R4,5;S2
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C
P2,5;Q2,4;R1;S2
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D
P3;Q2,4;R1,5;S2,4
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Solution

The correct option is D P3;Q2,4;R1,5;S2,4
From (iii) CuI(s)Ksp=1012M2Cu++I
We are going to use (i) and (iii) to get to (P), since (i) has the iodide ion in standard states, we can write:
[I1]=1M
Using (iii) [Cu+]=1012
CuI(s)+eCu(s)+I(aq); E=0.188V
(P) was Cu+(aq)+eCu(s);E1=?;E=0.188V (from (i))
E=E10.0591log110120.188=E10.0591log11012E1=0.52V
So, (P) matches with 3.

(Q)
To get this, we use (ii) and (iii)
Cu2+(aq)+I+eCuI(s);E=0.868V
The idea is the same, copper (I) and copper (II) ions have a concentration of 1 M, the iodide ion now has a concentration of 1012 M from the Ksp value from (iii).
Cu2+(aq)+eCu+(aq);E=?E=0.868V
So E=E20.0591log101210.868=E20.0591log10121E2=0.16V
Q matches with 2.

From just these two, you have the answer to be (d.) already!

(R)
Adding (P) and (Q) we get the equation given in (R)
E3=E1+E22=0.52+0.162=0.34V
And 2×E2+0.02=2×0.16+0.02=0.34
So, (R) matches with 1 and 5.

(S)
E4=E°0.05911log11012
E4=0.8680.0591log1012=0.16 V

Also E1E30.02=0.520.340.02=0.16=E2
And 2×E2+0.02=2×0.16+0.02=0.34
(S) goes with 2 and 4.

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