$\ce{KMnO4}$ should be a better oxidizing agent in acidic medium, than in alkaline medium, as per my knowledge. However, it oxidizes iodide (oxidation state = -1) to iodine (0) in acidic medium, whereas iodide (-1) to iodate (+5) in alkaline medium. The latter has a higher change in oxidation number, which we would have instead expected for the former. Why is this so?
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asked Feb 3, 2017 at 7:51
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1$\begingroup$ My book (NCERT India) mentions the same oxidation (iodide to iodate) for even neutral/faintly alkaline medium ($\ce{2MnO4- + H2O + I2 -> 2MnO2 + 2OH- + IO3-}$), which should be even weaker than the alkaline $\ce{KMnO4}$. It'd be interesting if an answer elaborated upon this as well. $\endgroup$– Gaurang TandonMay 7, 2018 at 5:17
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$\begingroup$ @GaurangTandon But alkaline $\ce{KMnO4}$ gets reduced to $\ce{MnO4^2-}$, which seems to imply that alkaline $\ce{KMnO4}$ is weaker than neutral/faintly alkaline medium $\ce{KMnO4}$... so if the answer explains it for alkaline medium, I think the same should apply for neutral/faintly alkaline medium as well. $\endgroup$– Box Box Box BoxFeb 17, 2022 at 16:12
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1 Answer
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A simple way to explain this is to illustrate with the following Latimer diagrams:
If you calculate the standard electrode potential of the highlighted portions it will be positive, so the reaction is spontaneous under standard conditions.
For acidic conditions, iodine is not converted to iodate because the electrode potential of the reaction would become negative, whereas for basic conditions it is possible.
Note that both the manganate and the iodide ions are both affected by acidic and basic conditions; thus it is important to consider their respective electrode potentials.
(Data was taken from Inorganic Chemistry, Shriver et al., Sixth Edition, Oxford University Press)
