Consider the following unbalanced redox reaction
$\underset{(1 M)}{{Hg}_{2}^{2 +}} + \underset{(1 M)}{{NO}_{3}^{-}} + H^{+} ⇌ \underset{(1 M)}{{Hg}^{2 +}} + \underset{(1 M)}{{HNO}_{2}} + H_{2} O$
[Concentration of ionic species are given at equilibrium]
If $E_{cell}^{o}$ of the given reaction is 0.09 V, calculate the minimum pH required to reach equilibrium.
$[Take \frac{2.303 RT}{F} = 0.06, \log 2 = 0.3]$

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Detailed Solution

Balanced equation is

$\begin{array}{l} {Hg}_{2}^{2 +} + {NO}_{3}^{-} + 3 H^{+} \to 2 {Hg}^{2 +} + {HNO}_{2} + H_{2} O \\ E_{cell} = 0 = E^{0} - \frac{2.303 RT}{nF} \log \frac{{[{Hg}^{2 +}]}^{2}}{{[{Hg}_{2}^{2 +}]}^{2} [{NO}_{3}^{-}] {[H^{+}]}^{3}} \\ 0.09 = \frac{0.06}{2} \log \frac{(1)^{2}}{(1)^{2} (1)^{2} {(H^{+})}^{3}} \\ pH = 1 \end{array}$