The reaction of aqueous $K M n O_{4}$ with $H_{2} O_{2}$ in acidic conditions gives which of the following products?
A. $M n^{4 +}$ and $O_{2}$
B. $M n^{2 +}$ and $O_{2}$
C. $M n^{2 +}$ and $O_{3}^{}$
D. $M n^{4 +}$ and $M n O_{2}$

Answer

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Hint:Check for the oxidizing and reducing properties of the given reactants. If one of the reactants is an oxidizing agent then it oxidizes the reactant to a higher value of charge and reduces itself by taking up electrons. Find the charge on both the central atoms of the given reactants.

Complete step-by-step answer:Our given reactants are

K M n O_{4}

and

H_{2} O_{2}

.
Here

K M n O_{4}

is a strong oxidizing agent so it oxidizes the other reactant given to us i.e.

H_{2} O_{2}

.
This

H_{2} O_{2}

when oxidized, loses its Hydrogen atoms and forms

O_{2}

. So the charge on both the oxygen changes from

- 1

0

. We know that there are two oxygen atoms so each oxygen atom gives away one electron.
These electrons are taken up by

K M n O_{4}

and since there are two electrons,

M n

changes to

M n^{2 +}

2 K M n O_{4} + 5 H_{2} O_{2} + 3 H_{2} S O_{4} \to 2 M n S O_{4} + 5 O_{2} + 8 H_{2} O + K_{2} S O_{4}

Therefore the products of the given reaction in acidic conditions are $M n^{2 +}$ and $O_{2}$ i.e. option B.

Additional information: Usually, molecules/atoms/ions that have high affinity for electrons or unusually large oxidation states tend to be strong oxidizing agents. Conversely, molecules that have low ionization energies and low electro-negativities account for strong reducing agents.

Note:

K M n O_{4}

is a strong oxidizing agent which means that it takes up electrons from

H_{2} O_{2}

to reduce its charge and increase the other reactant’s charge and hence oxidizing it. This reaction is a redox reaction where reduction on one compound and the oxidization of another compound occur simultaneously in the same reaction.