Chemistry - Electrochemistry Question with Solution | TestHub

When$\text{p}\text{H}=\mathrm{14, }\left[ {\text{H}}^{+}\right]={10}^{-14}\text{and}\left[\text{O}{\text{H}}^{-}\right]=1\mathrm{ M}$

${\text{K}}_{\text{s}\text{p}}=\left[\text{C}{\text{u}}^{2+}\right]{\left[\text{O}{\text{H}}^{-}\right]}^2={10}^{-19}$

$\therefore \left[\text{C}{\text{u}}^{2+}\right]=\frac{{10}^{-19}}{{\left[\text{O}{\text{H}}^{-}\right]}^2}={10}^{-19}$

The half cell reaction

$\text{C}{\text{u}}^{2+}+2{\text{e}}^{-}\to \text{C}\text{u}$

$\text{E}={\text{E}}^{\text{O}}-\frac{0.059}{2}\log \frac{1}{\left[\text{C}{\text{u}}^{2+}\right]}$

$=0.34-\frac{0.059}{2}\log \frac{1}{{10}^{-19}}=-\mathrm{0.22 }\text{V}$