The correct option is D P→3;Q→2,4;R→1,5;S→2,4
From (iii) CuI(s)Ksp=10−12M2⇌Cu++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:
[I−1]=1M
Using (iii) [Cu+]=10−12
CuI(s)+e−⟶Cu(s)+I−(aq); E∘=−0.188V
(P) was Cu+(aq)+e−⟶Cu(s);E∘1=?;E=−0.188V (from (i))
E=E∘1−0.0591log110−12⇒−0.188=E∘1−0.0591log110−12⇒E∘1=0.52V
So, (P) matches with 3.
(Q)
To get this, we use (ii) and (iii)
Cu2+(aq)+I−+e−⟶CuI(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 10−12 M from the Ksp value from (iii).
Cu2+(aq)+e−⟶Cu+(aq);E∘=?E=0.868V
So E=E∘2−0.0591log10−121⇒0.868=E∘2−0.0591log10−121⇒E∘2=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)
E∘3=E∘1+E∘22=0.52+0.162=0.34V
And 2×E∘2+0.02=2×0.16+0.02=0.34
So, (R) matches with 1 and 5.
(S)
E4=E°−0.05911log110−12
E4=0.868−0.0591log1012=0.16 V
Also E∘1−E∘3−0.02=0.52−0.34−0.02=0.16=E∘2
And 2×E∘2+0.02=2×0.16+0.02=0.34
(S) goes with 2 and 4.