How do ${MnO}_{2}$ $"MnO"_2$ and ${KMnO}_{4}$ $"KMnO"_4$ behave? Oxidizing or reducing agents?

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How do ${MnO}_{2}$ $"MnO"_2$ and ${KMnO}_{4}$ $"KMnO"_4$ behave? Oxidizing or reducing agents?

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Answer 1 · 2016-12-14T08:54:03

$Manganese (IV) oxide$ $"Manganese (IV) oxide"$ and $K M n O_{4}$ $KMnO_4$ are $oxidizing agents.$ $"oxidizing agents."$

Explanation:

$M n^{2 +}$ $Mn^(2+)$ is the reduction product. Because this is a $d^{5}$ $d^5$ high spin metal centre, its electronic transitions are spin-forbidden, and a solution of $M n^{2 +}$ $Mn^(2+)$ is (almost!) completely colourless (very conc. solutions are a pale rose).

So, when we take strongly oxidizing permanganate ion, which is very strongly coloured, the redox reaction has a self-indicating end-point, and decolorizes:

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ $MnO_4^(-) +8H^(+) + 5e^(-) rarr Mn^(2+) +4H_2O$

We habitually use this reaction in redox titrations.

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How do ${MnO}_{2}$ $"MnO"_2$ and ${KMnO}_{4}$ $"KMnO"_4$ behave? Oxidizing or reducing agents?

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How do MnO2"MnO"_2 and KMnO4"KMnO"_4 behave? Oxidizing or reducing agents?

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How do ${MnO}_{2}$ $"MnO"_2$ and ${KMnO}_{4}$ $"KMnO"_4$ behave? Oxidizing or reducing agents?