Reduction Potential - Intensive, but does it change with molarity?

[Cundiff1080]

So I keep hearing that reduction potential is an intensive property. Thus the stoichiometry does not change the E value. However that is at 1M, if the concentration is bumped up to 2M does the E value increase by a factor of 2 as well?

Thanks!

---

Members don't see this ad.

 

[Raiden2012]

Ok take my answer with a pinch of salt. I am 70% sure of my answer.

Since molarity is an intensive property as well, the reduction potential would change as well. However, I don't think it's just a multiple. It depends on what the concentration you're referring to is. Is it just the reactant/product or both.

To help, use the equation
delta E = delta E (standard) - RT/(zF) ln Q

where Q is the ration of products over reactants raised to respective stoichometric ratios

So if the products are twice the reactants, the delta E would go down but not my a multiple of 2

 

[LoLCareerGoals]

Not only that even if all concentrations of all reactants and products are increased by factor of 2, delta E doesn't change only if all the coefs are equal. If one or of the reactants/products coef is different from all others doubling will have a different (non canceling) effect on Q and in turn on delta E.

There is a reason why standard E is defined as one computed at 1M concentrations (since lnQ=0 for any reaction for such concentrations).