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Question

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Calculate the maximum possible electric work that can be obtained from the following cell under the standard conditions at 25°C:
$Fe | {Fe}^{2 +} (aq) | | {Cu}^{2 +} (aq) | Cu$
$E °_{{Fe}^{2 +} (aq) / Fe} = - 0.44 V E °_{{Cu}^{2 +} (aq) / Cu} = + 0.34 V and F = 96, 500 C .$

Solution

W_{\max} = nFE °_{cell}

The cell is

Fe | {Fe}^{2 +} (aq) | | {Cu}^{2 +} (aq) | Cu

Fe \to {Fe}^{2 +} + 2 e

{Cu}^{2 +} + 2 e \to Cu

__________________________

Fe + {Cu}^{2 +} \to {Fe}^{2 +} + Cu

___________________________

Here, n = 2
Therefore,

E °_{cell} = E °_{cathode} - E °_{anode} = E °_{{Cu}^{2 +} / Cu} - E °_{{Fe}^{2 +} / Fe} = + 0.34 V - (- 0.44 V) = 0.78 V

W_{\max} = 2 \times 96500 C \times 0.78 V = 150540 J = 150.5 kJ .

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