Chemistry - Electrochemistry Question with Solution | TestHub

$\mathrm{A}{\mathrm{l}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{A}\mathrm{l}$
$\mathrm{E}={\mathrm{E}}^{\mathrm{o}}-\frac{0.059}{3}\log \frac{1}{\left[\mathrm{A}{\mathrm{l}}^{3+}\right]}$
at$\text{pH}\text{=}\text{12}\Rightarrow \text{pOH}\text{=}\text{14}-\text{12}\text{=}\text{2}\Rightarrow \left[{\text{OH}}^{-}\right]\text{=}\text{0}\text{.01}\text{M}$
$\left[\mathrm{A}{\mathrm{l}}^{+3}\right]=\frac{10^{-36}}{[\mathrm{O}{\mathrm{H}}^{-}{]}^3}=\frac{10^{-36}}{(0.01{)}^3}=10^{-30}$
$\therefore \mathrm{E}=-1.66-\frac{0.059}{3}\log \left(\frac{1}{10^{-30}}\right)$
$\text{=}-\text{1}\text{.66}-\frac{\text{0}\text{.059}}{\text{3}}\text{×}\text{30}\text{=}-\text{1}\text{.66}-\text{0}\text{.59}\text{=}-\text{2}\text{.25}\text{V}$