Given: 12HgO(s)+12H2O(l)+e−→12Hg(1)+OH−(aq) If E∘ is the standard potential of the half-cell producing this reaction, then the standard potential of the cell producing the reaction HgO(s)+H+(aq)+e−→Hg(l)+12H2O(1) will be

# 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

1. A

${E}^{\circ }+\left(RT/F\right)\mathrm{ln}{K}_{\mathrm{w}}^{\circ }$

2. B

${E}^{\circ }-\left(RT/F\right)\mathrm{ln}{K}_{\mathrm{w}}^{\circ }$

3. C

${E}^{\circ }+\left(2RT/F\right)\mathrm{ln}{K}_{\mathrm{w}}^{\circ }$

4. D

${\mathrm{E}}^{\circ }-2\left(\mathrm{RT}/\mathrm{F}\right)\mathrm{ln}{K}_{\mathrm{w}}^{\circ }$

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### Solution:

For

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

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

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

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 ${E}^{\circ }-\left(RT/F\right)\mathrm{ln}{K}_{{\mathrm{w}}^{\circ }}^{\circ }$

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