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

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 $KMn{O}_4$ 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 $M{n}^{2+}$ .

$$2KMn{O}_4+5H_2{O}_2+3H_{2}S{O}_4\to 2MnS{O}_4+5O_2+8H_{2}O+K_{2}S{O}_4$$

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.