d
\(\mathrm{H}_{2} \quad \rightarrow \quad 2\mathrm{H}^{+}+2 \mathrm{e}^{-}\)

\(1 atm\) \(\quad\) (at \(pH\; 10\))

If \(\mathrm{pH}=10\)

\(\mathrm{H}^{+}=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-10}\)

From nernst equation, \(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{0}-\frac{0.0591}{2} \log \frac{\left[\mathrm{H}^{+}\right]^{2}}{\mathrm{p}_{\mathrm{H}_{2}}}\)

For hydrogen electrode, \(E_{\text {cell }}^{0}=0\)

\(E_{\text {cell }}=-\frac{0.0591}{2} \log \frac{\left(10^{-10}\right)^{2}}{1}\)

\(0.0591 \times \log 10^{10}\)

\(0.0591 \times 10=0.591\; V\)