Q.For the cell $\text{Mg(s)} \mid \text{Mg}^{2+}\text{(aq)} \ \| \ \text{Ag}^{+}\text{(aq)} \mid \text{Ag(s)}$, calculate the equilibrium constant at 25°C and the maximum work that can be obtained during operation of the cell. Given: $E^{o}_{\text{Mg}^{2+}/\text{Mg}} = -2.37$ V and $E^{o}_{\text{Ag}^{+}/\text{Ag}} = 0.80$ V.

The electrical energy a galvanic cell produces equals the total charge moved (n moles of electrons, i.e. nF coulombs, where 1 faraday $F \approx 96500$ C) multiplied by the cell's emf: $\text{Electrical energy} = nFE_{cell}$, so the maximum electrical work obtainable is $W_{max} = -nFE_{cell}$. Since the maximum work of a process equals its Gibbs free energy change, this gives one of electrochemistry's central equations, $\Delta G = -nFE_{cell}$ (and $\Delta G^{o} = -nFE^{o}_{cell}$ under standard conditions) — showing directly that a spontaneous reaction ($\Delta G < 0$) requires a positive $E_{cell}$. Comparing this with $\Delta G^{o} = -RT\ln K_{eq}$ links cell potential to the equilibrium constant: $E^{o}_{cell} = \dfrac{2.303RT}{nF}\log K_{eq}$.

The Nernst equation extends this to non-standard concentrations. Starting from $\Delta G = \Delta G^{o} + RT\ln Q$ and substituting the electrochemical relations above gives, after simplification at 25°C,

$E_{cell} = E^{o}_{cell} - \dfrac{0.0591}{n}\log Q$ …