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What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?

Explanation

If concentration of all reacting species is unity, then $E_{\text {cell }}=E_{\text {cell }}^{\circ}$ and $\Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}$ where, $\Delta_r G^{\circ}$ is standard Gibbs energy of the reaction

$$\begin{aligned} E_{\text {cell }}^{\circ} & =\text { emf of the cell } \\ n F & =\text { charge passed } \end{aligned}$$

If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly.

The reversibly work done by a galvanic cell is equal to decrease in its Gibbs energy.

$$\Delta_r G=-n F E_{\text {cell }}$$

As $E_{\text {cell }}$ is an intensive parameter but $\Delta_r G$ is an extensive thermodynamic property and the value depends on $n$.

For reaction, $\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(\mathrm{s})$ in a galvanic cell.

$$\Delta_r G=-2 F E_{\text {cell }} \quad[\text { Here, } n=2]$$