Question

Given
$Cu{I}_{\left(s\right)}+e^{-}⟶C{u}_{\left(s\right)}+I_{\left(aq\right)}^{-};E^{\circ }=-0.188V...\left(i\right)C{u}_{\left(aq\right)}^{2+}+I^{-}+e^{-}⟶Cu{I}_{\left(s\right)};E^{\circ }=0.868V...\left(ii\right)K_{sp}\left(CuI\right)={10}^{-12}{M}^2...\left(iii\right)$
$\frac{298Rln10}{F}=0.059$
List -1 contains half cells and List-2 contains emf values of half cells

$\begin{array}{cccc}& \text{List - I}& & \text{List - II}\\ \text{(P)}& C{u}_{\left(aq\right)}^{+}+e^{-}⟶C{u}_{\left(s\right)};E_1^0=& \text{(1)}& 0.34V\\ \text{(Q)}& C{u}_{\left(aq\right)}^{2+}+e^{-}⟶C{u}_{\left(aq\right)}^{+};E_2^0=& \text{(2)}& 0.16V\\ \text{(R)}& C{u}_{\left(aq\right)}^{2+}+2e^{-}⟶C{u}_{\left(s\right)};E_3^0=& \text{(3)}& 0.52V\\ \text{(S)}& C{u}_{\left(1M\right)}^{2+}+I_{\left({10}^{-12}M\right)}^{-}+e^{-}⟶CuI\left(s\right);E_4=& \left(4\right)& (E_1^0-E_3^0-0.02)V\\ & & \text{(5)}& 2\times E_2^0+0.02\end{array}$
Match the emf of cell in LIST-I with one or more values in LIST-II and choose the correct option.

• A. $P\to 3;Q\to 1;R\to 2,4;S\to 3,5$
• B. $P\to 1,5;Q\to 3;R\to 4,5;S\to 2$
• C. $P\to 2,5;Q\to 2,4;R\to 1;S\to 2$
• D. $P\to 3;Q\to 2,4;R\to 1,5;S\to 2,4$

Answer 1

The correct option is D $P\to 3;Q\to 2,4;R\to 1,5;S\to 2,4$

From (iii) $CuI\left(s\right)\stackrel{K_{sp}={10}^{-12}{M}^2}{\rightleftharpoons }C{u}^{+}+I^{-}$
We are going to use (i) and (iii) to get to (P), since (i) has the iodide ion in standard states, we can write:
$\left[I^{-1}\right]=1M$
Using (iii) $\left[C{u}^{+}\right]={10}^{-12}$
$Cu{I}_{\left(s\right)}+e^{-}⟶C{u}_{\left(s\right)}+I_{\left(aq\right)}^{-}$; $E^{\circ }=-0.188V$
(P) was $C{u}_{\left(aq\right)}^{+}+e^{-}⟶C{u}_{\left(s\right)};E_1^{\circ }=?;E=-0.188V$ (from (i))
$E=E_1^{\circ }-\frac{0.059}{1}log\frac{1}{{10}^{-12}}\Rightarrow -0.188=E_1^{\circ }-\frac{0.059}{1}log\frac{1}{{10}^{-12}}\Rightarrow E_1^{\circ }=0.52V$
So, (P) matches with 3.

(Q)
To get this, we use (ii) and (iii)
$C{u}_{\left(aq\right)}^{2+}+I^{-}+e^{-}⟶Cu{I}_{\left(s\right)};E^{\circ }=0.868V$
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 $K_{sp}$ value from (iii).
$C{u}_{\left(aq\right)}^{2+}+e^{-}⟶C{u}_{\left(aq\right)}^{+};E^{\circ }=?E=0.868V$
So $E=E_2^{\circ }-\frac{0.059}{1}log\frac{{10}^{-12}}{1}\Rightarrow 0.868=E_2^{\circ }-\frac{0.059}{1}log\frac{{10}^{-12}}{1}\Rightarrow E_2^{\circ }=0.16V$
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)
$E_3^{\circ }=\frac{E_1^{\circ }+E_2^{\circ }}{2}=\frac{0.52+0.16}{2}=0.34V$
And $2\times E_2^{\circ }+0.02=2\times 0.16+0.02=0.34$
So, (R) matches with 1 and 5.

(S)
$E_4=E°-\frac{0.0591}{1}\log \frac{1}{{10}^{-12}}$
$E_4=0.868-0.0591\log {10}^{12}$=0.16 V

Also $E_1^{\circ }-E_3^{\circ }-0.02=0.52-0.34-0.02=0.16=E_2^{\circ }$
And $2\times E_2^{\circ }+0.02=2\times 0.16+0.02=0.34$
(S) goes with 2 and 4.