Question

What is the change in the oxidation state of ${\text{Mn}}$, in the reaction of ${\text{MnO}}_4^ - $ with ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ in acidic medium?
A. $7 \to 4$
B. $6 \to 4$
C. $7 \to 2$
D. $6 \to 2$

Answer 1

Hint:We know that oxidation number is the charge acquired by the species may be positive, negative or zero by gain or loss of electrons. To solve this we must first write the reaction of ${\text{MnO}}_4^ - $ with ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ in acidic medium.

Complete answer:
The reaction of ${\text{MnO}}_4^ - $ with ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ in acidic medium means that ${\text{MnO}}_4^ - $ reacts with ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ in presence of hydrogen ions. The reaction is as follows:
${\text{2MnO}}_4^ - + 5{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} + 6{{\text{H}}^ + } \to 2{\text{M}}{{\text{n}}^{2 + }} + 8{{\text{H}}_2}{\text{O}} + {\text{5}}{{\text{O}}_2}$
Thus, the reaction of ${\text{MnO}}_4^ - $ with ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ in acidic medium produces ${\text{M}}{{\text{n}}^{2 + }}$ ion.
First we will calculate the oxidation number (ON) of ${\text{Mn}}$ in ${\text{MnO}}_4^ - $ as follows:
${\text{ON of Mn + 4}} \times {\text{ON of O}} = {\text{Charge of the ion}}$
Substitute $\left( { - 2} \right)$ for the oxidation number of oxygen, $ - 1$ for the charge on the ion. Thus,
${\text{ON of Mn: + 4}} \times \left( { - 2} \right) = - 1$
${\text{ON of Mn}} = - 1 + 8$
${\text{ON of Mn}} = + {\text{7}}$
Thus, the oxidation number (ON) of ${\text{Mn}}$ in ${\text{MnO}}_4^ - $ is $ + {\text{7}}$.
When ${\text{MnO}}_4^ - $ reacts with ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ in acidic medium, ${\text{MnO}}_4^ - $ is converted to ${\text{M}}{{\text{n}}^{2 + }}$ ion. The oxidation number of ${\text{M}}{{\text{n}}^{2 + }}$ ion is $ + 2$. Thus, the oxidation state of ${\text{Mn}}$, in the reaction of ${\text{MnO}}_4^ - $ with ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ in acidic medium is decreases from $ + {\text{7}}$ to $ + 2$.
Thus, the change in the oxidation state of ${\text{Mn}}$, in the reaction of ${\text{MnO}}_4^ - $ with ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ in acidic medium is $7 \to 2$.

Thus, the correct option is (C) $7 \to 2$.

Note:
1. The decrease in the oxidation number during a chemical reaction is known as reduction. Thus, in the reaction of ${\text{MnO}}_4^ - $ with ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ in acidic medium, the oxidation number of ${\text{Mn}}$ decreases which shows that ${\text{Mn}}$ is getting reduced. The oxidation number decreases because the ${\text{Mn}}$ atom has gained electrons.
2. The increase in oxidation number during a chemical reaction is known as oxidation. Electrons are lost in an oxidation reaction.
3. The reaction in which both oxidation and reduction occurs is known as a redox reaction.