Write a note on the relation between Gibbs free energy and cell potential for a cell reaction.

(N/A) The electrical work $(\omega_{\text{ele}})$ done in one second is equal to the electrical potential multiplied by the total charge passed: $\omega_{\text{ele}} = E \times Q$.
To obtain maximum work from a galvanic cell, the charge must be passed reversibly. The reversible work done by a galvanic cell is equal to the decrease in its Gibbs energy: $\omega_{\text{ele}} = -\Delta_{r}G$.
If the electromotive force $(EMF)$ of the cell is $E_{\text{cell}}$ and the total charge passed is $nF$ (where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant), then the Gibbs energy change for the reaction is given by: $\Delta_{r}G = -nF E_{\text{cell}}$.
Here, $E_{\text{cell}}$ is an intensive property, while $\Delta_{r}G$ is an extensive thermodynamic property, and its value depends on the value of $n$.
If all species taking part in the reaction are at unit activity, then $E_{\text{cell}} = E_{\text{cell}}^{o}$ and $\Delta_{r}G^{o} = -nF E_{\text{cell}}^{o}$.
Thus, the standard Gibbs energy change $\Delta_{r}G^{o}$ can be determined by measuring the standard cell potential $E_{\text{cell}}^{o}$.
Using the value of $\Delta_{r}G^{o}$, the equilibrium constant $(K)$ can be calculated using the formula: $\Delta_{r}G^{o} = -RT \ln K = -2.303 RT \log K$.