work done by electrons through galvanic cell

The work done by electrons following through a galvanic cell can be calculated with:

$$w=-nF\xi$$

where $w$ is work, $n$ is the number of moles of $\mathrm{e}{}^{-}$, $F=96485\mathrm{C}/\mathrm{mol}\mathrm{e}{}^{-}$.

Derivation

Since we know cell potential is work divided by charge ($\xi =\frac{w}{q}$), we can rearrange the relationship to get $w=q\xi$. However, when the work is done by the system, work is negative, so $w=-q\xi$. We know that the charge of $n$ moles of $\mathrm{e}{}^{-}$ is $nF$ (where $F$ is Faraday, the charge carried by one mole of electrons). Substituting the charge for $nF$, we get $w=-nF\xi$.

Works equals change in Gibbs free energy

Note that since the change in Gibbs free energy is defined as the maximum amount of work done by a reaction, we have $\Delta G=w=-nF\xi$