Question
Calculate the potential of hydrogen electrode in contact with a solution whose $pH$ is $10.$

Answer

For hydrogen electrode, $\text{H}^++\text{e}^-\rightarrow\frac{1}{2}\text{H}_2$ it is given that $pH = 10$
$\therefore\ [\text{H}_+]=10\ \text{M}$
Now, using Nernst equation:
$\text{H}_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}=\text{E}^\ominus_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}-\frac{\text{RT}}{\text{nF}}\text{In}\frac{1}{[\text{H}^+]}$
$\text{E}^\ominus_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}-\frac{0.0591}{1}\text{log}\frac{1}{[\text{H}^+]}$
$=0-\frac{0.0591}{1}\text{log}\frac{1}{[10^{-10}]}$
$= -0.0591\ \log\ 10^{10}$
$= -0.591\ V$

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