Electrical work done in one second is equal to electrical potential multiplied by total charge passed. If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy and therefore, if the emf of the cell is E and nF is the amount of charge passed and $Δ_{r} G$ is the Gibbs energy of the reaction, then
$Δ_{r} G = {n F E}_{(c e l l)}$
It may be remembered that $E_{(c e l l)}$ is an intensive parameter but $Δ_{r} G$ is an extensive thermodynamic property and the value depends on n. Thus, if we write the reaction.

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Solution

$Z n (s) + {C u}^{2 +} (a q) ⟶ {Z n}^{2 +} (a q) + C u (s)$
$Δ_{r} G = - 2 F E (c e l l)$
But when we write the reaction
$2 Z n (s) + 2 {C u}^{2 +} (a q) ⟶ 2 {Z n}^{2 +} (a q) + 2 C u (s)$
$Δ_{r} G = - 4 F E (c e l l)$
If the concentration of all the reacting species in unity, then $E_{(c e l l)} = (c e l l) E v$ and we have
$Δ_{r} G = {n F E}^{0} (c e l l) ⟶ (I)$
We have the relation
$Δ_{r} G = R T I n K ⟶ (I I)$
From the equation $(I)$ and $(I I)$
${n F E}^{0} (c e l l) = R T l n K$
$E_{(c e l l)}^{0} = \frac{R T}{n F l n K}$

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