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Application of Electrolysis

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Question

Calculate using Nernst equation cell potential of the following electrochemical cell at $298 K$ .
$Z n (s) | Z n_{(e q)}^{2 +} (0.04 M) | | S n_{(e q)}^{2 +} | S n_{(s)} (0.03 M)$
(Given $E^{\circ} = - 0.76 V, E^{\circ} = - 0.14 V)$ .

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Solution

Nernst equation for the electrochemical process at 298 K:
$Z n (s) | Z n_{(e q)}^{2 +} (0.04 M) | | S n_{(e q)}^{2 +} | S n_{(s)} (0.03 M)$
$E = E^{\circ} - \frac{0.0592}{2} {log}_{10} (\frac{[Z n^{2 +}]}{[S n^{2 +}]})$
$E^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ}$
Given: $E_{Z n (s) | Z n_{(e q)}^{2 +}}^{\circ} = - 0.76 V, E_{S n_{(e q)}^{2 +} | S n_{(s)}}^{\circ} = - 0.14 V$
$E^{\circ} = E_{S n_{(e q)}^{2 +} | S n_{(s)}}^{\circ} - E_{Z n (s) | Z n_{(e q)}^{2 +}}^{\circ}$
$E^{\circ} = - 0.14 - (- 0.76) = 0.62 V$
Also, $[Z n^{2 +}] = 0.04 M$ and $[S n^{2 +}] = 0.03 M$
Upon substitution in Nernst equation we get
$E = 0.62 - \frac{0.0592}{2} {log}_{10} (\frac{0.04}{0.03})$
$E = 0.616 V$

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