Question

The reaction of aqueous $KMn{O_4}$ 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 $Mn{O_2}$

Answer 1

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 $KMn{O_4}$ and ${H_2}{O_2}$.
Here $KMn{O_4}$ 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 $- 1$ to $0$ . We know that there are two oxygen atoms so each oxygen atom gives away one electron.
These electrons are taken up by $KMn{O_4}$ and since there are two electrons, $Mn$ changes to $M{n^{2 + }}$ .

$2KMn{O_4} + 5{H_2}{O_2} + 3{H_2}S{O_4} \to 2MnS{O_4} + 5{O_2} + 8{H_2}O + {K_2}S{O_4}$

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: $KMn{O_4}$ 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.