Electrochemistry - Result Question 49
####51. $Cu(aq)$ is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction :
$2 Cu^{+}(aq) \rightarrow Cu^{2+}(aq)+Cu(s)$
[2000] choose correct $E^{o}$ for above reaction if $E^{o}$ $Cu^{2+} / Cu=0.34 V$ and $E^{0} Cu^{2+} / Cu^{+}=0.15 V$
(a) $-0.38 V$
(b) $+0.49 V$
(c) $+0.38 V$
(d) $-0.19 V$
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Solution:
- (c) $2 Cu^{+} \longrightarrow Cu^{+2}+Cu$
$ \begin{align*} & 2 e^{-}+Cu^{+2} \longrightarrow Cu ; E_1^{\circ}=0.34 V ; \tag{i}\\ & e^{-}+Cu^{+2} \longrightarrow Cu^{+} ; E_2^{\circ}=0.15 V ; \tag{ii}\\ & Cu^{+}+e^{-} \longrightarrow Cu ; E_3^{\circ}=? \tag{iii} \end{align*} $
Now, $\Delta G_1^{\circ}=-n F E_1^{\circ}=-2 \times 0.34 F$
$\Delta G_2^{\circ}=-1 \times 0.15 F, \Delta G_3^{\circ}=-1 \times E_3^{\circ} F$
Again, $\Delta G_1^{\circ}=\Delta G_2^{\circ}+\Delta G_3^{\circ}$
$\Rightarrow-0.68 F=-0.15 F-E_3^{\circ} F$
$\Rightarrow E_3^{\circ}=0.68-0.15=0.53 V$
$E _{\text{cell }}^{\circ}=E _{\text{cathode }}^{\circ}(Cu^{+} / Cu)$
$ -E _{\text{anode }}^{\circ}(Cu^{+2} / Cu^{+}) $
$ =0.53-0.15=0.38 V \text{. } $
