# Chemistry - Derivation of relationship between Gibbs free energy and electrochemical cell potential

## Solution 1:

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.

## Solution 2:

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

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