Explanation

(c)
From given data $(from\ {\mathrm{\Delta}G}^{{^\circ}}\ = \ - \ {nE}^{{^\circ}}F\ )$
$Cu\ (s)\ \rightarrow \ {Cu}^{2 +}\ (aq) + \ {2e}^{-}\$
$\mathrm{\Delta}G_{1}^{{^\circ}}\ \ = \ - \ 2\ \times \ ( - \ 0.34)\ \times \ F$
${Cu}^{2 +}\ (aq) + \ e^{-} \rightarrow \ {Cu}^{2 +}\ (aq)$
${\mathrm{\Delta}G}_{2}^{{^\circ}}\ = \ - \ 1\ \times \ (\ 0.15)\ \ F$
On addition,
$Cu\ (s)\ \rightarrow \ {Cu}^{2 +}\ (aq) + \ e^{-}.\ {\mathrm{\Delta}G}_{1}^{{^\circ}}\ = \ - \ 1\ \times \ E^{{^\circ}}\ \times \ F$
${\mathrm{\Delta}G}_{3}^{{^\circ}}\ = \ {\mathrm{\Delta}G}_{1}^{{^\circ}}\ + \ {\mathrm{\Delta}G}_{2}^{{^\circ}}$
$\ \ \ \ \ \ \ \ \ = \ ( - \ 2 \times - \ 0.34 \times F) + (\ - \ 1 \times 0.15 \times F)$
=+ 0.68 F – 0.15 F = 0.53 F
0r $E^{{^\circ}}\ \ = \ - \ 0.53\ V$
Reaction,
${2Cu}^{+}(aq)\ \rightleftharpoons \ {Cu}^{2 +}\ (aq)\ + \ Cu\ (s),\ \ \ E^{{^\circ}}\ = \ ?$
So, ${Cu}^{+}\ (aq) + \ e^{-}\ \rightleftharpoons \ \ Cu\ (s)\ ,\ \ \ \ \ \ \ E^{{^\circ}}\ = \ + \ 0.53\ V$
${Cu}^{+}\ (aq) \rightleftharpoons \ {Cu}^{2 +}\ (aq) + \ e^{-};\ \ \ \ \ \ E^{{^\circ}}\ = \ - \ 0.15\ V\ \ \ \ \$
[table] ${2Cu}^{+}\ (aq) \rightleftharpoons \ {Cu}^{2 +}\ (aq) + \ Cu\ (s);\ \ E^{0} = + 0.38V$ [/table]