MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Ion Electron Method

When KMnO4 ...

Question

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

Open in App

Solution

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

Suggest Corrections

thumbs-up

0

Similar questions

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 forms

M n O_{4}^{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 +}

, determine the number of electrons transferred in each case.

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 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 is:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Balancing_ion elec

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Ion Electron Method

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon