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

A
$E^{\circ} + (R T / F) \ln K_{w}^{\circ}$
B
$E^{\circ} - (R T / F) \ln K_{w}^{\circ}$
C
$E^{\circ} + (2 R T / F) \ln K_{w}^{\circ}$
D
$E^{\circ} - 2 (RT / F) \ln K_{w}^{\circ}$

Solution:

For $\frac{1}{2} HgO (s) + \frac{1}{2} H_{2} O (l) + e^{-} \to \frac{1}{2} Hg (l) + {OH}^{-} (aq) . E = E^{\circ} - \frac{R T}{F} \ln ([{OH}^{-}] / M)$

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

$E = E^{\circ} - \frac{R T}{F} \ln (\frac{[H^{+}] [{OH}^{-}] / M^{2}}{[H^{+}] / M}) = E^{\circ} - \frac{R T}{F} \ln [([H^{+}] [{OH}^{-}] / M^{2 +}) + \ln ([H^{+}] / M)]$

$= (E^{\circ} - \frac{R T}{F} \ln K_{w}^{\circ}) + \frac{R T}{F} \ln ([H^{+}] / M)$

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

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

Solution:

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