Chemistry - Electrochemistry Question with Solution | TestHub

When $\mathrm{pH}=14\left[{\mathrm{H}}^{+}\right]=10^{-14}$ and $\left[{\mathrm{OH}}^{-}\right]=1\mathrm{M}$ $\begin{array}{l}{\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Cu}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2=10^{-19}\\ \therefore \left[{\mathrm{Cu}}^{2+}\right]=\frac{10^{-19}}{{\left[{\mathrm{OH}}^{-}\right]}^2}=10^{-19}\end{array}$ The half cell reaction $\begin{array}{l}{\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}⟶\mathrm{Cu}\\ \mathrm{E}={\mathrm{E}}^{\circ }-\frac{0.059}{2}\log \frac{1}{\left[{\mathrm{Cu}}^{2+}\right]}\\ =0.34-\frac{0.059}{2}\log \frac{1}{10^{-19}}=-0.22\mathrm{V}\end{array}$