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Ion Electron Method

When KMnO4 ...

Question

When $K M n O_{4}$ acts as an oxidising agent and forms $M n_{2} O_{3}$ , the number of electrons transferred per molecule of $K M n O_{4}$ is :

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Solution

Reaction when $K M n O_{4}$ acts as an oxidising agent and forms $M n_{2} O_{3}$ is given below:
$+ 7 M n O_{4}^{-} + 4 e \to {+ 3 M n}_{2} O_{3}$
The oxidation state of $M n$ in $M n O_{4}^{-}$ and $M n_{2} O_{3}$ are $+ 7$ and $+ 3$ respectively.
Therefore, change in oxidation is $4$ which is the number of electrons transferred per molecule.

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Similar questions

K M n O_{4}

acts as an oxidising agent and forms

M n O_{4}^{2 -}

, the number of electrons transferred per molecule of

K M n O_{4}

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acts as an oxidising agent and forms

M n^{+ 2}

, the number of electrons transferred per molecule of

K M n O_{4}

K M n O_{4}

acts as an oxidising agent and ultimately forms

M n O_{4}^{2 -}

M n O_{2}

M n_{2} O_{3}

M n^{2 +}

, then the number of electrons transferred per molecule of

K M n O_{4}

in each case respectively, is:

k M n O_{4}

acts as an oxidising agent and ultimately forms

M n O_{4}^{2 -}, M n O_{2}, M n_{2} O_{3},

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, determine the number of electrons transferred in each case.

K M n O_{4}

acts as an oxidising agent and ultimately forms

[M n O_{4}]^{2 -}

M n O_{2}

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, then the number of electrons transferred in each case respectively is:

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