If the hydrogen electrode is dipped in a solution of pH = 2 at 25° C, then the reduction potential of the electrode would be

$$\begin{gathered} {\mathrm{H}}^{+}+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right) \\ {\mathrm{E}}_{{\mathrm{H}}^{+}/\frac{1}{2}{\mathrm{H}}_2}={{\mathrm{E}}^0}_{{\mathrm{H}}^{+}/\frac{1}{2}{\mathrm{H}}_2}-\frac{0.0591}{1}\log \frac{1}{\left[{\mathrm{H}}^{+}\right]} \\ =0-\frac{0.0591}{1}\log \left[\frac{1}{{10}^{-2}}\right] \\ =-0.118 \mathrm{V} ; \\ \mathrm{Alternate} \mathrm{method} : \\ {\mathrm{E}}_{\mathrm{RP}}=-0.0591{\mathrm{p}}^{\mathrm{H}} ; \end{gathered}$$