[Meter]

Given a "Latimer diagram":

A --E₁--> B --E₂--> C

We are told that B disproportionates into A and C. If I want to find the cell potential for that reaction, can I generally simply add E₁ and E₂?

I recall my professor saying that you can't always add cell potentials and get a correct result. Any ideas what he meant?

[Babcock_Hall]

This is going to be the blind leading the blind.  IIRC reduction potentials are not additive when the number of electrons is not the same in the two reactions.  Gibbs' free energies are additive.  It might be better to wait until a physical or analytical chemist answers.

[Meter]

I looked at some notes I had forgotten about and only Gibbs free energies are generally summable whereas you should not add standard potentials. The reason why you can add standard potentials when only one electron is being transferred is because it checks out with the math for Gibbs free energy.

[mjc123]

The potential for disproportionation is E₂ - E₁. You can do the subtraction because the reaction must be balanced to give the same number of electrons in the reduction as the oxidation. E.g. if A B was a 2-electron reduction, and B C a 1-electron reduction, the balanced half-reactions would be
B A + 2e     ΔG° = +2FE°₁  (+ because the reaction is reversed)
2B + 2e 2C    ΔG° = -2FE°₂
Overall 3B A + 2C     ΔG° = -2F(E°₂ - E°₁) = -2FE°_cell

On the other hand, if you wanted E° for the process A C, it would be
A + 2e B     ΔG° = -2FE°₁ 
B + e C    ΔG° = -FE°₂     (same number of B created and destroyed)
Overall A + 3e C     ΔG° = -F(E°₂ + 2E°₁) = -3FE°_AC
so E°_AC is the weighted average of E°₁ and E°₂.

These things become clearer on a Frost diagram, where G (or G/F, in V) is plotted vs. oxidation number. The line joining two oxidation states has a slope of E for that couple. If B lies above the line joining A and C, B can disproportionate to A and C.