Calculate the cell potential and Gibb's energy change for the following galvanic cell at $298 K$ .
$C d (s) | C d_{(a q)} {(0.001 M) | | H^{+}}_{(a q)} (0.01 M) | H_{2} (g), 1 a t m | P t$

0.38

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-0.38

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0.36

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Solution

The correct option is A 0.38
${C d}^{2 +} + 2 e^{-} \to C d E^{0} = - 0.41 V$
$H^{+} + 2 e^{-} \to H_{2} E^{0} = 0.0 V$
$∴ C d + 2 H^{+} \to {C d}^{2 +} + H_{2}$
${E^{0}}_{c e l l} = 0.0 V - (- 0.41) V$
$= 0.41 V$
Now, nernst equation:
$E_{c e l l} = E^{0} - \frac{0.059}{n} l o g ⎡ ⎣ \frac{{C d}^{2 +}}{{(H^{+})}^{2}} ⎤ ⎦$
$E_{c e l l} = 0.41 - \frac{0.059}{2} l o g [\frac{0.001}{{(0.01)}^{2}}]$
$E_{c e l l} = 0.3805 V$

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