Combining electrode potentials

Standard electrode potentials look like they should add up like enthalpies, but they don't. A potential is an energy per electron, which makes it intensive. Gibbs energies are extensive, and they are the quantities that add.

The rule

For each half-reaction, convert to Gibbs energy with :

\Delta G^\circ = -nFE^\circ

Add the $\Delta G^\circ$ values along any path, then convert back using the electron count of the overall half-reaction:

E^\circ_\text{overall} = \frac{\sum_i n_iE^\circ_i}{\sum_i n_i}

Worked example: iron

Find $E^\circ(\mathrm{Fe^{3+}/Fe})$ from the two tabulated couples:

Couple	$n$	$E^\circ$ / V	$nE^\circ$ / V
$\mathrm{Fe^{3+} + e^- \to Fe^{2+}}$	1	+0.77	+0.77
$\mathrm{Fe^{2+} + 2e^- \to Fe}$	2	−0.44	−0.88
$\mathrm{Fe^{3+} + 3e^- \to Fe}$	3	?	−0.11

So $E^\circ = -0.11/3 \approx -0.04$ V. That is nowhere near the plain sum (+0.33 V) or the average (+0.17 V).

Full cells are the exception

When two half-reactions make a balanced cell, the electrons cancel and both halves carry the same $n$ . That $n$ divides out, which is why the , $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$ , works with raw potentials.

Latimer diagrams and disproportionation

A Latimer diagram lists one element's oxidation states left to right, from most to least oxidised, with $E^\circ$ for each step. A species in the middle disproportionates if the potential on its right is higher than the one on its left. That means it is a better oxidant than its higher oxidation state is.

Copper is the classic case. Cu²⁺ —(+0.15 V)→ Cu⁺ —(+0.52 V)→ Cu, so aqueous Cu⁺ falls apart into Cu²⁺ and copper metal.

These are standard potentials. For real concentrations, correct each couple with the before comparing.