Electrochemistry - Result Question 35
####37. The electrode potentials for
$ Cu^{2+}(aq)+e^{-} \longrightarrow Cu^{+}(aq) $
and $Cu^{+}(aq)+e^{-} \longrightarrow Cu(s)$
are $+0.15 V$ and $+0.50 V$, respectively. The value of $E^{\circ} Cu^{2+} / Cu$ will be :
(a) $0.500 V$
(b) $0.325 V$
(c) $0.650 V$
(d) $0.150 V$
Show Answer
Solution:
- (b) $Cu^{2+}+1 e^{-} \to Cu^{+}$
$E_1^{o}=0.15 V ; \Delta G_1^{o}=-n_1 E_1^{o} F$
$Cu^{+}+1 e^{-} \to Cu$
$E_2^{o}=0.50 V ; \Delta G_2^{o}=-n_2 E_2^{o} F$
$Cu^{2+}+2 e^{-} \to Cu$
$\Delta G^{\circ}=\Delta G^{\circ} _1+\Delta G^{\circ} _2$
$-n E^{\circ} F=-1 n_1 E_1{ }^{0} F+(-1) n_2 E_2{ }^{0} F$
$E^{\circ}=\frac{n_1 E_1^{o}+n_2 E_2^{o}}{n}=\frac{0.15 \times 1+0.50 \times 1}{2}$
$\Rightarrow 0.325 V$
When exchange of electrons ( $n$ ) is different for all
\begin{tabular}{|c|c|} \hline $Cu^{2+}+e^{-} \longrightarrow Cu^{+}$ & $(n=1)$ \\ \hline $Cu^{+}+e^{-} \longrightarrow Cu$ & $(n=1)$ \\ \hline $Cu^{2+}+2 e^{-} \longrightarrow Cu$ & $(n=2)$ \\ \hline \end{tabular} three reactions, we can not use the formula for standard electrode potential $(E _{\text{cell }}.$ ). e.g.
