E° cell for the given redox reaction is 2.71 V
$M g_{(s)} + C u_{(0.01 M)}^{2 +} \to M g_{(0.001 M)}^{2 +} + C u_{(s)}$
Calculate $E_{c e l l}$ for the reaction. Write the direction of flow of current when an external opposite potential applied is (i) less than 2.71 V and (ii) greater than 2.71 V

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Solution

$M g_{(s)} + C u_{(0.01 M)}^{2 +} \to M g_{(0.001 M)}^{2 +} + C u_{(s)}$
$∴ Q = \frac{[M g^{2 +}] \times [C u]}{[C u^{2 +}] \times [M g]} = \frac{0.001 \times 1}{0.01 \times 1} = 0.1$
using Nerst equation
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n} l o g Q$
$= 2.71 - \frac{0.0591}{2} l o g 0.1$
$= 2.74 V$
(i) In the first case voltage applied is less than 2.71 V so current will be from from cathode to anode
(ii) In the second case voltage applied is more than 2.71 V so current will be from from anode to cathode

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