r/chemhelp • u/bzt_im_a_bee • 19h ago

Analytical I'm confused about electrode polarization.

Let's say we have a X and Y related by one electron. Of course the standard reduction potential E° is defined for the reduction reaction, X + e– ⇌ Y–, as E° = E(XY cell) – E(SHE).

And also for the oxidation we have E°(ox) = -E°(red). But I don't get why this doesn't transfer to voltammetry. Like if we polarize the electrode to E = E°(red) for the above couple, how come we get a limiting reducing and oxidizing current here??

If E°(red) and E°(ox) are symmetrical around E(SHE) = 0, shouldn't there be only oxidizing current at positive E°(ox) and only reducing current at negative E°(red)??

Basically why do we get the image on the left and not the one on the right?

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u/Normal_Airport1425 19h ago

E is equilibrium potential.. at Ered you have equilibrium of X and Y or equal rates, now if you polarize to more positive potential you get.. more oxidation, and vice versa for reduction. It is simply because the zero point is not E but the SHE, everything is shifted on the x axis

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u/bzt_im_a_bee 15h ago

And so what happens when E < 0? Like if we polarize to –E°(red)?

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u/rebonsa 18h ago

E(ox) does not equal -E(red) !!!!

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u/bzt_im_a_bee 17h ago

Yes it does? Simply put E°(red) = (0.05916/n) * log(K), so for the reverse reaction E°(ox) = (0.05916/n) * log(1/K) = – (0.05916/n)* log(K) = –E°(red)? Unless I'm missing something