Why is $\Delta G = -nFE$?

I don't understand what the motivation is behind this definition. Was it derived or just given? The textbook provides no justification for this equation. In fact, much of the book associated with the Gibbs free energy provides no justification and just says, 'This is how it is. Now go and solve some problems.'


There are two ways to understand this equation.

One is to realise that (reversible ideal case) $\Delta G = W_\text{non-exp}$ (non-expansion). Therefore, in an ideal chemical cell, if the potential difference between the electrodes is $E$, to move one mole electrons across the external circuit will be $FE$, which must be equal to the decrease in gibbs free energy of the system. Hence for $n$ mole electrons transferred at the same potential, $W_\text{non-exp} = \Delta G = -nFE$.

The fact that $\Delta G = W_\text{non-exp}$ can be derived as under: \begin{align} \mathrm{d}S &= \frac{\delta q}{T} && \text{(reversible case)}\\ \mathrm{d}U &= \delta q + W_\text{non-exp} + W_\text{exp}\\ W_\text{non-exp} &= \mathrm{d}U - W_\text{exp} - \delta q \\ &= \mathrm{d}U + p\,\mathrm{d}V - T\,\mathrm{d}S && \text{(const. $p$ and $T$)} \\ &= \mathrm{d}H - T\,\mathrm{d}S \\ &= \mathrm{d}G \end{align}

Another approach is to use $$\mu_\mathrm{electrochemical} = \mu_\mathrm{chemical} + zF\phi,$$ where $\phi$ is the electric potential at the point, $z$ is the charge on the species. Using this and $$\mathrm{d}G_{T,P} = \sum_{i=1}^n \mu_i\,\mathrm{d}n_i,$$ we get the required equation for $z = 1$ for an electron, and considering the electrochemical potential of both the electrodes. I am unsure of a rigorous derivation for the electrochemical potential but Wikipedia has some basics on it.

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I'm surprised your textbook did not derive this equation from the reaction isotherm relationship between $\Delta G$ and the reaction quotient $Q$ and the Nernst equation. The derivation is not hard.

Reaction isotherm equation:

$$\Delta_\mathrm{r} G =\Delta_\mathrm{r} G^\circ + RT\ln Q$$

Nernst equation:

$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{RT}{nF}\ln Q$$

If we solve both equations for $RT\ln Q$, we get your equation (almost).

\begin{align} RT\ln Q &= \Delta_\mathrm{r} G -\Delta_\mathrm{r} G^\circ\\ RT\ln Q &= nFE^\circ_\mathrm{cell} - nFE_\mathrm{cell}\\ \Delta_\mathrm{r} G -\Delta_\mathrm{r} G^\circ &= nFE^\circ_\mathrm{cell} - nFE_\mathrm{cell} \end{align}

Why is my equation not as simple as the one you started with? Your equation is at equilibrium, and I assumed that we might not be at equilibrium. At equilibrium, the following are true, which simplify the relationship.

\begin{align} Q &= K\\ \Delta_\mathrm{r} G &= 0\\ E_\mathrm{cell} &= 0 \end{align}

At equilibrium, $\Delta_\mathrm{r} G = 0$ because the reaction has achieved a minimum energy state — the chemical potential $\mu_i = \left(\frac{\partial G}{\partial N_i}\right)_{T,P}$ is also $0$ because there is no net change of state at equilibrium. Similarly $E_\mathrm{cell} = 0$ at equilibrium. There is no change of state, and thus the redox reaction has ground to a halt.

At equilibrium, the final relationship is

$$\Delta_RG^\circ = nFE^\circ_\mathrm{cell}$$

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  • $\begingroup$ I guess his textbook is going the other way, and derives the Nernst from the one above (since the first is a more or less trivial relationship between the electric work and enthalpy). $\endgroup$ – Greg Feb 6 '16 at 15:50
  • $\begingroup$ The chemical potential is not zero. I would remove that statement since it is incorrect and doesn't add value to your explanation. $\endgroup$ – Buck Thorn Jun 1 at 6:28

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