Calculate the potential $(e m f)$ of the cell $C d | C d^{2 +} (0.10 M) ∥ H^{+} (0.20 M) | P t, H_{2} (0.5 a t m)$ .

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Solution

$C d + 2 H^{+} \to C d^{+ 2} + H_{2}$ , $n = 2$

$E_{c e l l}^{0} = E_{r}^{0} - E_{l e f t}^{0} = 0 - (0.403) = 0.403 V$

$E_{c e l l} = E_{c e l l}^{0} + \frac{2.303 R T}{n F} log \frac{[H^{+}]^{2} [C d]}{P_{H_{2}} [C d^{2 +}]}$

$= 0.403 + \frac{8.314 \times 298 \times 2.303}{2 \times 96500} log \frac{(0.2)^{2}}{(0.5) (0.1)}$

$= 0.403 + 0.0295 log 0.8$

$0.403 - 0.0028 = 0.400 V$

Hence, Emf is $0.4 V$

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