[galpinj]

Hey guys,

I think I'm misunderstanding something regarding reduction potentials. When looking at standard potential (E°), it is necessary to be at 25°C, 1atm, and aqueous solution of 1M. Moreover, the value for E is intrinsic and won't change even if the stoichiometric values of half reactions are changed.

However, in concentration cells the molarity of the half-cells is different, and this will impact the potential. If changes to the amount of moles (i.e. stoichiometric coefficient) doesn't affect E, why does changing the molarity affect the value of E? I understand that it is no longer held at standard conditions, but doesn't both concentration cells and stoichiometric changes involve a change in the amount of moles?

Any help is much appreciated.

[mjc123]

What do you mean by "stoichiometric values"? Do you mean e.g.
H⁺ + e  1/2 H₂ vs.
2H⁺ + 2e  H₂ ?
The value of E° is the same for these two half reactions. ΔG = -nFE; ΔG for the second reaction is twice that of the first, but there are two electrons consumed, so E is the same. E is an intensive property, ΔG extensive. I could write what coefficients I like, as long as they balance. I could write 87H⁺ + 87e  43.5 H₂ if I wanted; E would still be the same.
The molarity of the solution affects E via the Nernst equation. Note "amount of moles" is not the same as "stoichiometric coefficient", notr is it the same as concentration. If you have a 1L solution of 1M H⁺, you can change the number of moles in two ways. You could have, say, 2L of 1M solution, or 1L of 2M solution. How will each of these changes affect E? (hint: no of moles extensive, concentration intensive)