Chemistry - Electrochemistry Question with Solution | TestHub

For hydrogen electrode, oxidation half reaction is $\underset{(1\mathrm{atm})}{{\mathrm{H}}_2}⟶\underset{(\text{At }\mathrm{pH}10)}{2{\mathrm{H}}^{+}}+2{\mathrm{e}}^{-}$ If $\mathrm{pH}=10$ ${\mathrm{H}}^{+}=1\times 10^{-\mathrm{pH}}=1\times 10^{-10}$ From Nernst equation, $E_{\text{cell }}=E_{\text{cell }}^0-\frac{0.0591}{2}\log \frac{{\left[{\mathrm{H}}^{+}\right]}^2}{P_{{\mathrm{H}}_2}}$ For hydrogen electrode, $E_{\text{cell }}^{\circ }=0$ $\begin{array}{rl}{E}_{\text{cell }}& =-\frac{0.0591}{2}\log \frac{{\left(10^{-10}\right)}^2}{1}\\ & =+\frac{0.0591\times 2}{2}\log \frac{1}{10^{-10}}\\ & =0.0591\log 10^{10}\\ & =0.0591\times 10=0.591\mathrm{V}\end{array}$