EMF of the following cell is $0.634 v o l t$ at $298 K$ $P t | H_{2} (1 a t m) | H^{+} (a q) | | H g_{2}^{2 +} (a q . 1 N) | H g (l)$ . The pH of anode compartment is :

(Given, $E_{H g_{2}^{2 +} | H g}^{o} = 0.28 and \frac{2.303 R T}{F} = 0.059$ )

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Solution

$At cathode : \frac{1}{2} H g_{2}^{2 +} + e^{-} \to H g (l)$
$At anode : \frac{1}{2} H_{2} (g) \to H^{+} (a q) + e^{-}$
$\Rightarrow \frac{1}{2} H g_{2}^{2 +} + \frac{1}{2} H_{2} (g) \to H g (l) + H^{+} (a q)$
$E_{c e l l} = E_{c e l l}^{o} - \frac{0.059}{1} log [H^{+}]$
$0.634 = (0.28 - 0) + 0.059 p H$
$\Rightarrow p H = \frac{0.634 - 0.28}{0.059} = 6$

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P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q, 1 M) | | M^{4 +} (a q), M^{2 +} (a q) | P t (s) E_{c e l l} = 0.092 V w h e n \frac{[M^{2 +} (a q)]}{[M^{4 +} (a q)]} = 10^{x} G i v e n : E_{M^{4 +} / M^{2 +}}^{0} = 0.151 V; 2.303 \frac{R T}{F} = 0.059 V

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P t [H_{2} (1 a t m) | H^{+} (p H = x) ∥ K C l (1 N) | H g_{2} C l_{2} (s) | H g (l)

E^{o}

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E_{c e l l} = 0.092 V w h e n \frac{[M^{2 +} + (a q)]}{[M^{4 +} (a q)]} = 10^{x}

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E_{Z n^{2 +} / Z n}^{0} = - 0.76 V

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