The reduction potential at P^H=14 for the ${\mathrm{Cu}}^{2+}/\mathrm{Cu}$ couple is [given $\left.{\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^0=0.34\mathrm{V},{\mathrm{K}}_{\mathrm{sp}}\left[\mathrm{Cu}(\mathrm{OH}{)}_2\right]=1\times {10}^{-19}\right]$ 

Given pH = 14 

So pOH = 0 

Therefore [OH^-]=1.

$$\begin{gathered} Cu{\left(OH\right)}_2\iff C{u}^{2+}+\hspace{0.17em}2O{H}^{-} \\ {K}_{sp}=\left[C{u}^{2+}\right] . {\left[O{H}^{-}\right]}^2=1\times {10}^{-19} \\ \left[C{u}^{2+}\right]={10}^{-19} \end{gathered}$$

For reduction of Cu²⁺ to Cu  , Nernst equation is written as 

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.0591}{2}\log \frac{1}{\left[{\mathrm{Cu}}^{2+}\right]} \\ = 0.34-\frac{0.0591}{2}\log \frac{1}{{10}^{-19}} \\ \mathrm{on} \mathrm{solving} \\ {\mathrm{E}}_{\mathrm{cell}}=-0.22\mathrm{V} \end{gathered}$$