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How is the redox couple as it is written for the oxidation of $\ce{MnO4-}$, as $\ce{MnO4-,H+/Mn2+,H2O}$ correct?

I agree that $\ce{MnO4}$ is being oxidized as it loses electrons (going from -1 charge to +2) but I don't see why $\ce{H+}$ is also being oxidized. Could someone explain?

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    $\begingroup$ H+ isn't and can't be oxidized. It is already in the highest possible oxidation state. $\endgroup$ Nov 17 '19 at 16:07
  • $\begingroup$ Can you help me understand why it should be on the left of the /, meaning how do I see that H+ is being oxidized to H2O? Because oxidation is loss of electrons but H+ goes from positive to neutral, in my head I see that perhaps too simplified as being reduced @IvanNeretin $\endgroup$
    – J.Se
    Nov 17 '19 at 16:15
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    $\begingroup$ The charge on ions is irrelevant. It is oxidation state of elements that matters. $\endgroup$ Nov 17 '19 at 16:30
  • $\begingroup$ My take is: $\ce{MnO4-}$ can be reduced in two media, namely acid or base/neutral with different electrode potentials. The equation given here is for acidic medium half-reaction. So, $\ce{H+/H2O}$ represent the medium. $\endgroup$ Nov 18 '19 at 5:05
  • $\begingroup$ @ Mathew, It is still a poor notation, if several PhDs have to make a guess or an deliver an opinion on the meaning of certain notation which is not worth the complication, that notation is indeed useless. The author is unnecessarily making the redox notation complicated. $\endgroup$
    – M. Farooq
    Nov 18 '19 at 14:56

The H atom is neither oxidized nor reduced in the redox semi-equation. It remains at +1. Only Manganese reacts, as it is reduced from +7 to +2. The redox couple should be written : MnO4-/Mn2+, without mentioning Hydrogen

And MnO4 does not lose electrons, as you say. On the contrary, MnO4 fixes or consumes 5 electrons in its reduction to Mn2+.


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