Chemistry - Electrochemistry Question with Solution | TestHub

Hints : ${\mathrm{H}}^{+}(\mathrm{pH}=10)\left|{\mathrm{H}}_2(1\mathrm{atm})\right|\mathrm{Pt}(\mathrm{s})$ Reaction: $2{\mathrm{H}}^{+}\left({\mathrm{p}}^{\mathrm{H}}=10\right)+2\mathrm{e}\to {\mathrm{H}}_2(1\mathrm{atm})$ $$\begin{array}{rl}& \mathrm{E}={\mathrm{E}}^0-\frac{0.0591}{2}\log \left(\frac{{\mathrm{P}}_{{\mathrm{H}}_2}}{{\left[{\mathrm{H}}^{+}\right]}^2}\right)\\ & =0-\frac{0.0591}{2}\log \frac{1}{{\left(10^{-10}\right)}^2}=-\frac{0.0591}{2}\times 2\log \frac{1}{10^{-10}}=-0.0591\times 10=-0.591\\ & \text{ i.e. }E=-0.591\mathrm{V}\\ & \end{array}$$