Reduction potential of hydrogen electrode can be calculated Using Nernst equation for ${2\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2$

${\mathrm{E}}_{\mathrm{red}}={\mathrm{E}}_{\mathrm{red}}^0-\frac{0.059}{\mathrm{n}}\log \frac{{\mathrm{p}}_{{\mathrm{H}}_2}}{{\left[{\mathrm{H}}^{+}\right]}^2}$

For  a neutral solution H⁺ ion concentration is ${10}^{-7}$ as $\mathrm{pH}=7$

$$\begin{gathered} {\mathrm{E}}_{{\mathrm{H}}^{+}|{\mathrm{H}}_2}^0=0-\frac{0.0591}{2}\log \frac{1}{{\left({10}^{-7}\right)}^2} \\ =0-\frac{0.0591}{2}\log \frac{{10}^{14}}{} \\ =-\frac{0.0591\times 14}{2}\log 10 \\ =-0.413V \end{gathered}$$