Given that $E_{{\mathrm{H}}_3\mathrm{O}/{\mathrm{H}}_2\mid \mathrm{Pt}}=0\text{ at }298\mathrm{K}.$. The pressure of ${\mathrm{H}}_2$ gas would be

For water, $\left[{\mathrm{H}}^{+}\right]={10}^{-7}\mathrm{M}\text{ at }298\mathrm{K}$

${\mathrm{H}}^{+}+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2;E=-\frac{RT}{F}\ln \frac{p_{{\mathrm{H}}_2}^{1/2}}{\left[{\mathrm{H}}^{+}\right]}$

Obviously, when E = 0

${\left(\frac{p_{{\mathrm{H}}_2}}{\mathrm{atm}}\right)}^{1/2}=\frac{\left[{\mathrm{H}}^{+}\right]}{{\mathrm{moldm}}^{-3}}={10}^{-7}\text{ or }{p}_{{\mathrm{H}}_2}={10}^{-14}\mathrm{atm}$