Thank you for reporting, we will resolve it shortly
Q.
The standard reduction potential for $Cu ^{2+} / Cu$ is $+0.34$. Calculate the reduction potential at $pH =14$ for the above couple. $\left( K _{\text {sp }} Cu ( OH )_2=1 \times 10^{-19}\right)$
When $pH =14,\left[ H ^{+}\right]=10^{-4}$ and $\left[ OH ^{-}\right]=1 M$
$ K_{s p}=\left[ Cu ^{2+}\right]\left[ OH ^{-}\right]^2=10^{-19} $
$ \therefore\left[ Cu ^{2+}\right]=\frac{10^{-19}}{\left[ OH ^{-}\right]^2}=10^{-19}$
The half-cell reaction
$Cu ^{2+}+2 e ^{-} \rightarrow Cu $
$ E = E ^{\circ}-\frac{0.059}{2} \log \frac{1}{\left[ Cu ^{2+}\right]} $
$ =0.34-\frac{0.59}{2} \log \frac{1}{10^{-19}}=-0.22\,V$