For oxidant/reductants and conjugate pairs, for example $\ce{MnO2 -> Mn^2+}$, do we write $\ce{Mn^4+}$ or $\ce{MnO2}$ as the oxidant? Websites conflicts what my text book says, because it appears that my text book would write the conjugate pair as $\ce{Mn^4+/Mn^2+}$ which implies that it considers $\ce{Mn^4+}$ is the oxidant not $\ce{MnO2}$.

I just realised my example wouldn't work for what I was trying to say since $\ce{MnO2}$ does not exist in ionic form. But using a compound that did exist in the ionic form, would I be saying that the oxidant is the entire compound or the ion that has a change in oxidation number?


1 Answer 1


You generally use whichever compound is actually present in solution.

For manganese(IV), it will practically always be $\ce{MnO2}$ so use that. Manganese(VII) will always be $\ce{MnO4-}$ so use that. Manganese(II) salts dissolved in water dissociate and coordination complexes of manganese(II) are very labile. Thus, it doesn’t make sense to assume any $\ce{[MnL_n]^2+}$ complex to predominate and you should simply use $\ce{Mn^2+}$.

It does, for example, make a difference if your iron redox pair is $\ce{Fe^2+/Fe^3+}$ or $\ce{[Fe(CN)6]^3-/[Fe(CN)6]^4-}$. Likewise, a reduction using tin(II) should better be written $\ce{Sn^2+/[SnCl6]^2-}$.

If you are in doubt whether something dissociates or whether it should be considered a single entity, the cleaner solution is probably to write the fully dissociated salt, so you can get away with using $\ce{Sn^2+/Sn^4+}$. However, as soon as molecules or oxyanions are involved you cannot resort to this; oxalate reductions have to be $\ce{C2O4^2-/CO2}$ and permanganate has to be $\ce{MnO4-/product}$.

  • $\begingroup$ Some data for different Co(III)/Co(II) and Fe(III)/Fe(II) couples: i.stack.imgur.com/nCYyg.jpg (Yeah, that's my handwriting haha) $\endgroup$
    – orthocresol
    Mar 20, 2017 at 14:03

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