Voltages, EMF, electrochemistry.

[Rhino1000]

Excerpt from TBR electrochemistry: When you multiply a (half) reaction by an integer, you do not multiply its contribution to the overall reactions' emf by an integer. This is because reduction potential is an intensive property that does not change with conditions. * I added the (half) because I assume they are referring to half reactions that are components of a larger redox reaction.

I'm trying to understand why a system of reactions that requires the oxidation of 2 molar equivalences of A for the reduction one molar equivalences of B shouldn't multiply the oxidation potential of A by two in order to find the oxidation/reduction potential (emf) of the reaction. If two A's are oxidized in a given reaction, and one B is reduced, shouldn't that mean that two As' worth of energy is released, and one B's worth of energy is required? I realize energy is different from voltage, but I don't see how voltage shouldn't follow the same rule. An explanation is appreciated.

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[Czarcasm]

edit: Sorry, didn't want to confuse you or anyone reading this. I provided my reasoning, but wasn't 100% with my reasoning, so I removed my reply.

 

[Rhino1000]

edit: Sorry, didn't want to confuse you or anyone reading this. I provided my reasoning, but wasn't 100% with my reasoning, so I removed my reply.

Ok, next time maybe

 

[Cawolf]

The easiest way to think of the potential as an intensive property is to consider the units (for me). A volt is a Joule per Coulomb. It is a measure of intensity and shows how the energy will increase with charge. It is not defining anything based on the current amount of material in the reaction.

So the half reaction produces some energy per charge whether the reaction has 0.5 moles or 1 mole.