Chemistry - Electrochemistry Question with Solution | TestHub

For the reaction:
${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Cu}\left(\mathrm{s}\right)$

The reduction potential is

$\mathrm{E}={\mathrm{E}}^{\mathrm{o}}-\frac{\mathrm{RT}}{2\mathrm{F}}\ln \frac{1}{\left[{\mathrm{Cu}}^{2+}\right]}$

From the given data$\mathrm{pH} = 14$and

${\mathrm{K}}_{\mathrm{sp}}=\left(\mathrm{Cu}{\left(\mathrm{OH}\right)}_2\right)=1.0\times {\mathrm{10}}^{-20}$, we get

$\left[{\mathrm{H}}^{+}\right]={10}^{-14}\mathrm{M},\left[{\mathrm{OH}}^{-}\right]=\frac{{\mathrm{K}}_{\mathrm{W}}}{\left[{\mathrm{H}}^{+}\right]}=\frac{{10}^{-14}{\mathrm{M}}^2}{{10}^{-14}\mathrm{M}}=1\mathrm{M}$

$\left[{\mathrm{Cu}}^{2+}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{{\left[{\mathrm{OH}}^{-}\right]}^2}=\frac{1.0\times {\mathrm{10}}^{-20}{\mathrm{M}}^2}{1\mathrm{M}}=1.0\times {\mathrm{10}}^{-20}\mathrm{M}$

$$\begin{gathered} \mathrm{E}=0.34\mathrm{V}-\left(\frac{0.059\mathrm{V}}{2}\right)\log \frac{1}{1.0\times {\mathrm{10}}^{-20}}=0.34\mathrm{V}-\frac{0.059\times 20\mathrm{V}}{2} \\ \mathrm{E}=0.34-0.59 =-0.25 \mathrm{V} \end{gathered}$$