The minimum voltage (in $\mathrm{V}$ ) required to bring about the electrolysis of $1 \mathrm{M}$ copper…
The minimum voltage (in $\mathrm{V}$ ) required to bring about the electrolysis of $1 \mathrm{M}$ copper (II) sulphate solution at $298 \mathrm{~K}$ is
$$
\text { (Given } \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{H}_2 \mathrm{O} / \mathrm{H}^{+}}^{\circ}=-1.23 \mathrm{~V} \text { ) }
$$
+1.57
+0.89
-0.89
-1.57
Solution
Oxidation : (anode)
$
2 \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+} \text {(aq.) } ; \mathrm{E}^{\circ}=+1.23 \mathrm{~V}
$
Reduction : (cathode)
$
\begin{aligned}
& \mathrm{Cu}^{2+}(\mathrm{aq} .)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) ; \mathrm{E}^{\circ}=+0.34 \mathrm{~V} \\
& \mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ}=(0.34)-(+1.23)=-0.89 \mathrm{~V} \text {. }
\end{aligned}
$
Thus, a potential of minimum of $+0.89 \mathrm{~V}$ would be required to carry out the given reaction