[xenonp]

Say I want to find the E value for Mn3+ + 3e- > Mn
I have been given E values for Mn3+/Mn2+ and Mn2+/Mn
Why can't I just construct two half (reduction) equations, combine them, add the E values and get the correct E value for the first reaction I mentioned?

Instead I have to calculate -nFE for both of the latter half reactions, add them and solve for the overall E value.

When do you have to use this? Only when you are combining two oxidation / reduction reactions as E is not equal to E(reduction) - E(oxidation)?

Thanks :-)

[MrTeo]

I actually never thought about this but it looks like the right thing to do: when you work out electrochemical problems you are actually using Gibbs' free energies (which are additive) and the usual relationship ∆G=-nF∆E is applied.

No one assures you that Es are additive too (and apparently they're not) so what you have to do is convert them to free energies and then go back to the E or ∆E value needed.

When you have a redox reaction the only reason to use ∆E is that the reactions are coupled so n_Red=n_Ox and the two methods give you the same answer (you can easily check it by yourself).