$E_{r e d}^{o}$ of the half-cell $H g_{2} C l_{2} (s) + 2 e^{-} \to 2 H g (s) + 2 C l^{-} (a q)$ is $0.27 V$ . If $p K_{s p}$ corresponding to solubility product for $H g_{2} C l_{2} (s)$ is 17 then find the $E_{r e d}^{\circ}$ for the half cell
$H g_{2}^{2 +} (a q) + 2 e \to 2 H g (s) .$ (Given $\frac{2.303 R T}{F} = 0.06)$

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Solution

Given, $E_{r e d}^{o}$ for the half cell
$H g_{2} C l_{2} (s) + 2 e^{-} \to 2 H g (s) + 2 C l^{-} (a q) E_{r e d}^{o} = 0.27 V$ is $0.27 V$
Which is defined when $[C l^{-}] = 1 M$
So, from $H g_{2} C l_{2} (s)_{s p} H g_{2}^{2 +} (a q) + 2 C l^{-} (a q)$
$K_{s p} = [H g_{2}^{2 +}] [C l^{-}]^{2} \Rightarrow K_{s p} = [H g_{2}^{2 +}] w h e n [C l^{-}] = 1 M$

Hence for $H g_{2}^{2 +} (a q) + 2 e^{-} \to 2 H g (s)$
$E = E^{o} - \frac{0.06}{n} l o g \frac{1}{[H g_{2}^{2 +}]}$ $\Rightarrow 0.27 = E^{o} - \frac{0.06}{2} l o g \frac{1}{K_{s p}}$ $\Rightarrow 0.27 = E^{o} - 0.03 \times p K_{s p}$
$\Rightarrow 0.27 = E^{o} - 0.03 \times 17 \Rightarrow E^{o} = 0.78 V$

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