[Prestomatic]

Here are two standard reduction potentials:
Fe3+ + 3e-  Fe   E° = +.04V
Fe2+ + 2e-  Fe   E° = -.41V
And I intended to find the standard reduction potential:
Fe3+ + 3e-  Fe2+

I assumed to flip the second equation to oxidation, thus flipping the sign, then E°red + E°ox = (.04 V) + (.41 V) = .45V.

The actual value is .94 V, and i noticed that this is 3(.04 V) + 2(.41 V). I thought that reduction potential was intensive, thus potentials can be flipped but not multiplied when combined with other equations? Why are these coefficients corresponding to the electron counts in their respective equations?

Any input is highly appreciated, thanks.

[Babcock_Hall]

There are pitfalls in trying to add or subtract standard reduction potentials, although there are others here with far more knowledge of thermodynamics than I have.  In the example you gave the number of electrons is not the same in all three equations.  Try converting the reduction potentials to ΔG° values, and the coefficients may make more sense.

[Corribus]

Just to elaborate on Babcock Hall's post: Reduction potentials aren't strictly additive - removing two electrons in concerted fashion, say, takes a different amount of energy than removing them sequentially.  You have to do a specific type of conversion to get the right answer.

For example, consider the following two half reactions:

Co³⁺(aq) + e^- -->  Co²⁺(aq)   E° = 1.82 V
Co²⁺(aq) + 2e^-  -->  Co(s)   E° = -0.28 V

If you add these "Hess law style" to get the following balanced reaction for reduction of cobaltic form to zero valent cobalt:

Co³⁺(aq) + 3e^- -->  Co(s)   E° = 1.54 V

But the correct answer is 0.42 V.

To solve this kind of problem, you have to convert the potentials into quantities that are additive - that is, Gibbs energies. The equation that allows you to do this is

ΔG = -nFE

Since F is just a constant, you will find that converting between E (in V) and G (in kJ/mol, say) only involves factors of the number of electrons involved in each step. If you do these conversions you will get the right answer of 0.42 V for the reduction of Co³⁺ to Co⁰, just as you will arrive at the solution you mentioned in your first post.

[mjc123]

Have you come across oxidation state diagrams (Frost diagrams)? In such a diagram, oxidation state is plotted on the x-axis and Gibbs energy (or Gibbs energy divided by F, to give a number in volts for convenience) on the y-axis. The difference in Gibbs energy between two states is the vertical distance between them. The electrode potential is the slope of the line between them. Try plotting it for Fe, Fe²⁺ and Fe³⁺ (take G(Fe⁰) as 0 for convenience). You will see how the Gibbs energy is additive, but the slopes (electrode potentials) are not. Then try it for different oxidation states - e.g. M⁰, M⁺ and M²⁺, or M⁰, M⁵⁺ and M⁶⁺. Keep the relative Gibbs energies the same; you will see that the electrode potentials (slopes) are different.
See e.g. https://en.wikipedia.org/wiki/Frost_diagram