# Redox reactions and potentials/potential differences

1. Oct 11, 2016

### s3a

Hello to everyone who reads this. :)

1. The problem statement, all variables and given/known data
I DO understand the two following points.:

1. If the redox potential of a redox reaction is positive, then the redox reaction is spontaneous.

2. If the redox potential of a redox reaction is negative, then the redox reaction is NOT spontaneous/non-spontaneous.

What I am NOT 100 percent sure I understand is what a redox potential of 0 (so neither positive nor negative) means (since most, if not all, sources don't seem to mention that). I've been told that it means that the system is at equilibrium, which does make sense to me, but being told that, I have another (sub)question to inquire about this further. (See part 3, below.)

2. Relevant equations
http://chem.libretexts.org/Core/Phy...es_of_Chemical_Equilibria/Dynamic_equilibrium

3. The attempt at a solution
Basically, a redox reaction with a potential of 0 means the system in question is ALWAYS in DYNAMIC equilibrium (and never static equilibrium), since redox reactions ALWAYS have motion between reactants and products, albeit at an equal and constant rate, when the redox potential of a redox reaction is 0, right?

If I'm wrong, please correct me.

Any input would be GREATLY appreciated!

P.S.
"Potentials" = "Potential differences", right?

2. Oct 11, 2016

### Staff: Mentor

I am not sure I understand what it is supposed to mean.

What do you mean by "redox reaction"? The one in which two half cells are combined to react?

If so, whether the cell potential is positive or negative is just a matter of how the half cells are connected to the voltmeter and the negative potential can be made positive just by switching the cells. But you can't make the system spontaneous just by shifting flasks, so there is something wrong.

And if by "redox reaction" you mean a half cell reaction, it is never spontaneous, as it won't work without an electron source (or sink) which is typically a second half cell. Which moves us back to the first case (which is already wrong).