Given: $\frac{1}{2}\mathrm{HgO}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{e}}^{-}\to \frac{1}{2}\mathrm{Hg}\left(1\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$ If $E^{\circ }$ is the standard potential of the half-cell producing this reaction, then the standard potential of the cell producing the reaction $\mathrm{HgO}\left(\mathrm{s}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to \mathrm{Hg}\left(\mathrm{l}\right)+\frac{1}{2}{\mathrm{H}}_2\mathrm{O}\left(1\right)$ will be

For $\frac{1}{2}\mathrm{HgO}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{e}}^{-}\to \frac{1}{2}\mathrm{Hg}\left(\mathrm{l}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right).E=E^{\circ }-\frac{RT}{F}\ln \left(\left[{\mathrm{OH}}^{-}\right]/\mathrm{M}\right)$

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

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

$=\left(E^{\circ }-\frac{RT}{F}\ln K_w^{\circ }\right)+\frac{RT}{F}\ln \left(\left[{\mathrm{H}}^{+}\right]/\mathrm{M}\right)$

This is the Nernst equation for the reaction $\mathrm{HgO}\left(\mathrm{s}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to \mathrm{Hg}\left(\mathrm{l}\right)+\frac{1}{2}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Hence, its standard potential is $E^{\circ }-(RT/F)\ln K_{{\mathrm{w}}^{\circ }}^{\circ }$