The pressure of ${\mathrm{H}}_2$ required to make the potential of ${\mathrm{H}}_2$ electrode zero in pure water at 298 K

$\mathrm{pH}=7 \mathrm{for} \mathrm{pure} \mathrm{water} \mathrm{because} \mathrm{it} \mathrm{is} \mathrm{neutral}.$

$-\log \left[{\mathrm{H}}^{+}\right]=7\Rightarrow \left[{\mathrm{H}}^{+}\right]={10}^{-7}$

$2{\mathrm{H}}_{\left(aq\right)}^{+}+2e^{-}⟶{\mathrm{H}}_{2( \mathrm{g})}$

${\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{cell }}°-\frac{0.0591}{2}\log \frac{{\mathrm{p}}_{{\mathrm{H}}_2}}{{\left[{\mathrm{H}}^{+}\right]}^2}$

$0=0-\frac{0.0591}{2}\log \frac{{\mathrm{p}}_{{\mathrm{H}}_2}}{{\left({10}^{-7}\right)}^2}$

$\log \frac{{\mathrm{p}}_{{\mathrm{H}}_2}}{{\left({10}^{-7}\right)}^2}=0\Rightarrow \frac{{\mathrm{p}}_{{\mathrm{H}}_2}}{{\left({10}^{-7}\right)}^2}=1 [\because \log 1=0]$

${\mathrm{p}}_{{\mathrm{H}}_2}={10}^{-14} \mathrm{atm}$