Question

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 ${\Delta }_{r}G$ is the Gibbs energy of the reaction, then
${\Delta }_{r}G={nFE}_{\left(cell\right)}$
It may be remembered that $E_{\left(cell\right)}$ is an intensive parameter but ${\Delta }_{r}G$ is an extensive thermodynamic property and the value depends on n. Thus, if we write the reaction.

Answer 1

$Zn\left(s\right)+{Cu}^{2+}\left(aq\right)⟶{Zn}^{2+}\left(aq\right)+Cu\left(s\right)$

${\Delta }_{r}G=-2FE\left(cell\right)$

But when we write the reaction

$2Zn\left(s\right)+2{Cu}^{2+}\left(aq\right)⟶2{Zn}^{2+}\left(aq\right)+2Cu\left(s\right)$

${\Delta }_{r}G=-4FE\left(cell\right)$

If the concentration of all the reacting species in unity, then $E_{\left(cell\right)}=\left(cell\right)Ev$ and we have

${\Delta }_{r}G={nFE}^0\left(cell\right)⟶\left(I\right)$

We have the relation

${\Delta }_{r}G=RTInK⟶\left(II\right)$

From the equation $\left(I\right)$ and $\left(II\right)$

${nFE}^0\left(cell\right)=RTlnK$

$E_{\left(cell\right)}^0=\frac{RT}{nFlnK}$