Apply Nernst Equation on the given reduction reaction of hydrogen.

${\mathrm{H}}^{+}\left(\mathrm{aq}\right) + {\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)$

${\mathrm{E}}_{\mathrm{cell}}= {\mathrm{E}}_{\mathrm{cell}}^{\circ } -\frac{0.059}{\mathrm{n}}\log \frac{1}{\left[{\mathrm{H}}^{+}\right]}$

Where ${\mathrm{E}}_{{\mathrm{H}}^{+}/{\mathrm{H}}_2}^{\circ }=0 \mathrm{V}$ and $\mathrm{n}=1$

$-0.413=-0.059\log \frac{1}{\left[{\mathrm{H}}^{+}\right]}$ or $\frac{0.414}{0.059}=-\log \left[{\mathrm{H}}^{+}\right]=\mathrm{pH}$ 

$\mathrm{pH}=7$