Explanation

(a)
The required reaction $\left( {Cu}^{2 +} + Cu \rightarrow 2{Cu}^{+} \right)$ can be obtained by using the following reactions
${Cu}^{2 +} + e^{-} \rightarrow {Cu}^{+},\ E{^\circ}_{{Cu}^{2 +}/\ {Cu}^{+}} = 0.15\ V\ \ \ \$…(i)
${Cu}^{2 +} + 2e^{-} \rightarrow Cu,\ E{^\circ}_{{Cu}^{2 +}/\ Cu} = 0.34\ V\ \ \ \ \ \$…(ii)
Multiplying Eq. (i) by 2, we get
$2{Cu}^{2 +} + 2e^{-} \rightarrow 2{Cu}^{+},$
$\mathrm{\Delta}G_{1} = - nFE = - 2 \times F \times 0.15\ \ \ \ \ \ \ \$…(iii)
${Cu}^{2 +} + 2e^{-} \rightarrow Cu$
$\mathrm{\Delta}G_{2} = - nFE = - 2 \times F \times 0.34\ \ \ \ \ \ \ \$…(iv)
Subtract the Eq. (iv) from Eq. (iii)
${Cu}^{2 +} + Cu \rightarrow 2{Cu}^{+}$
$\mathrm{\Delta}G_{3} = - nFE = - 1 \times F \times E{^\circ}$
Also$\ \ \mathrm{\Delta}G_{3} = \mathrm{\Delta}G_{1} - \mathrm{\Delta}G_{2}$
$\therefore - 1\ FE{^\circ} = ( - 2F \times 0.15) - ( - 2F \times 0.34)$
$E{^\circ} = - 0.38$
This is the value for the reaction
${Cu}^{2 +} + Cu \rightarrow 2{Cu}^{+}$
But the given reaction is just reverse of it
$\therefore E_{cell}$ for given reaction $= + \ 0.38\ V$