The standard cell potential for the electrode reaction ${\mathrm{H}}_2\mathrm{O}+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2+{\mathrm{OH}}^{-}\text{at }{25}^{\circ }\mathrm{C}$ is.

The potential of the given electrode potential is

$E=E^{\circ }-\frac{RT}{F}\ln \left\{{\left(p_{{\mathrm{H}}_2}/\mathrm{bar}\right)}^{1/2}\left(\left[{\mathrm{OH}}^{-}\right]/\mathrm{M}\right)\right\}$

Multiplying and dividing the term witain the logarithm by $\left[{\mathrm{H}}^{+}\right]/\mathrm{M}$, we get

$E=E^{\circ }-\frac{RT}{F}\ln \left\{\frac{\left(\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]/{\mathrm{M}}^2\right){\left(p_{{\mathrm{H}}_2}/\mathrm{bar}\right)}^{1/2}}{\left[{\mathrm{H}}^{+}\right]/\mathrm{M}}\right\}=\left(E^{\circ }-\frac{RT}{F}\ln K_{\mathrm{w}}^{\circ }\right)-\frac{RT}{F}\ln \left\{\frac{{\left(p_{{\mathrm{H}}_2}/\mathrm{bar}\right)}^{1/2}}{\left[{\mathrm{H}}^{+}\right]/\mathrm{M}}\right\}$The electrode in which the given reaction occurs may be considered as ${\mathrm{H}}^{+}\left|{\mathrm{H}}_2\right|\mathrm{Pt}$ with electrode reaction ${\mathrm{H}}^{+}+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2$ for which we have

$E=-\frac{RT}{F}\ln \left\{\frac{{\left(p_{{\mathrm{H}}_2}/\mathrm{bar}\right)}^{1/2}}{\left[{\mathrm{H}}^{+}\right]/\mathrm{M}}\right\}$

Equating Eqs (1) and (2), we get

$E^{\circ }=\frac{RT}{F}\ln K_{\mathrm{w}}^{\circ }=\frac{2.303RT}{F}\log K_{\mathrm{w}}^{\circ }=\left(0.059\mathrm{V}\right)(-14)=-0.826\mathrm{V}$