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Ion Electron Method

When KMnO4 ...

Question

When $K M n O_{4}$ acts as an oxidising agent and forms $M n O_{4}^{2 -}$ , 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 $M n O_{4}^{2 -}$ is given below:
$+ 7 M n O_{4}^{-} + e \to + 6 M n O_{4}^{2 -}$
The oxidation state of $M n$ in $M n O_{4}^{-}$ and $M n O_{4}^{-}$ are $+ 7$ and $+ 6$ respectively.
Therefore, change in oxidation state of $M n$ is $1$ 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_{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:

Q. Number of electrons transfer in each case when

K M n O_{4}

acts as an oxidising agent to given:

M n O_{2}, {M n}^{2 +}, M n {(O H)}_{3}

M n O_{4}^{2 -}

are respectively:

K M n O_{4}

acts as an oxidising agent, it can form

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

{M n}^{2 +}

, the numbers of electrons transferred in each case respectively, are :

K M n O_{4}

acts as an oxidizing 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 in each case respectively are:

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