# Gibbs free energy and electrical potential

The relation between the Gibbs free energy and the electrical potential is: $$\Delta G=-zFE$$ $z$ is the valence of the electrochemical reaction and $F$ is the Faraday constant.

If at equilibrium the $\Delta G$ is zero this means that also $E$ is zero but this is not true because $E$ at equilibrium is the equilibrium potential of the electrochemical reaction considered.

This makes some confusion in my mind and for sure I'm not properly consider something. Can someone help to clarify my mind?

• Look in your textbook for the distinction between $\Delta G$ and $\Delta G^o$. – porphyrin Jul 2 '17 at 10:26
• That $\Delta G0$ is evaluated at standard conditions. I don't get your answer – user45328 Jul 2 '17 at 10:36
• The equilibrium potential $E_\text{eq}$ is always zero -- at equilibrium we have no net flux and hence no potential difference. – a-cyclohexane-molecule Jul 2 '17 at 11:01
• In case of a galvanic cell it is not true that when the two electrodes are not connected the two half cells are at equilibrium? If I connect the electrodes to a voltmeter I measure a potential difference. – user45328 Jul 2 '17 at 12:02

Your $\Delta G$ is really $\Delta G_{\mathrm{reaction}}$, and the reaction does not happen alone: it is accompanied by charging of electrodes. The Gibbs free energy of charging is just the associated change in the electroststic energy. That is, it equals $zF$ (the amount of charge moving against the electric potential for 1 mole of reaction) times $E$. Therefore, at equilibrium, the total change of the Gibbs free energy satisfies $$\begin{split} \Delta G_{\mathrm{total}} &= \Delta G_{\mathrm{reaction}} + \Delta G_{\mathrm{charging}}\\ &= 0 \end{split}$$ just as it should.
In fact, when deriving the expression for the cell potential $E$, the logic goes backwards. We know that the electrodes are charged until equilibrium is reached, in which case the sum of $\Delta G_{\mathrm{reaction}}$ and $\Delta G_{\mathrm{charging}} = zFE$ should vanish. Thus, the cell potential is given by $$E = -\frac{\Delta G_{\mathrm{reaction}}}{zF}.$$