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Topic: Which values of the water half-cell potentials are correct?  (Read 3885 times)

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Offline o1ocups

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Which values of the water half-cell potentials are correct?
« on: April 08, 2009, 04:39:41 AM »
The redox reactions for water are these right?
oxidation: 2H2O(aq) --> O2(g) + 4H+(aq) + 4e-
reductoin: 2H2O(l) + 2e- --> H2(g) + 2OH-(aq)

So, my textbook has a list of the Standar Reduction Potentials at 25 C for many different half reactions. According to that table, the SRP value for the water oxidation reaction is 1.23 (or -1.23 in terms of Ecell,ox) and the reduction one is -0.83.

However, in the solution manual, it gives the SRP values 0.82 and -0.41, accordingly. Why? And which ones are correct?

Offline o1ocups

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Re: Which values of the water half-cell potentials are correct?
« Reply #1 on: April 08, 2009, 10:29:49 AM »
Never mind! My friend told me why. It's because 0.82 and -0.41 are the potentials at nonstandard conditions, with the concentration being 1*10^-7 M instead of 1 M.

Offline Borek

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Re: Which values of the water half-cell potentials are correct?
« Reply #2 on: April 08, 2009, 11:13:44 AM »
So these are not standard potentials, but so called formal potentials.
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Offline o1ocups

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Re: Which values of the water half-cell potentials are correct?
« Reply #3 on: April 08, 2009, 03:25:20 PM »
So these are not standard potentials, but so called formal potentials.

"formal potentials"? I don't think we learned about them. are they basically just nonstandard potentials?

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Re: Which values of the water half-cell potentials are correct?
« Reply #4 on: April 08, 2009, 04:27:38 PM »
"Formal potential" is just translation from Polish - I think the same name is used in English, although I am not 100% sure.

Formal potential is the potential observed in specific circumstances (solution).

For example for permanganate solution, potential depends on the pH:

E = E0 + RT/5F ln ([MnO4-][H+]8/[Mn+2])

If you know that pH will be constant, you can write the same equation as

E = E0' + RT/5F ln ([MnO4-]/[Mn+2])

moving pH into E0' - now it is a formal potential of the solution, equal

E0' = E0 + 8RT/5F ln ([H+])

That's just a mathematical trick, but very convenient one.

So, in your case, potential was given not for standard solution (where activities of all ions are 1) but for most likely to be used solution (neutral one) - that's application of the same approach.
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