Question

Find the number of electrons that are involved in the reduction of permanganate to manganese (II) salt, manganate and manganese dioxide respectively:
(A) 5,1,3
(B) 5,3,1
(C) 2,7,1
(D) 5,2,3
(E) 2,3,1

Answer 1

Hint: (1) By the term reduction, we mean the gain of electrons or the decrease in the oxidation state of an ion or an atom in a molecule. It is a part of the redox reaction process and simultaneously takes place with oxidation.
(2) When a species undergoes reduction, it is called an oxidizing agent and is said to be reduced.
 Complete step-by-step answer: ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ is the permanganate ion. We need to find out the number of electrons that are involved in the reduction of the permanganate ion to manganese (II) salt $\left( {{\text{M}}{{\text{n}}^{{\text{2 + }}}}} \right)$ , manganate $\left( {{\text{Mn}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}} \right)$ and manganese dioxide $\left( {{\text{Mn}}{{\text{O}}_{\text{2}}}} \right)$ respectively.
First, let us consider the conversion of the permanganate ion to manganese (II) salt.
${\text{Mn}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{ + 8}}{{\text{H}}^{\text{ + }}}{\text{ + 5}}{{\text{e}}^{\text{ - }}} \to {\text{M}}{{\text{n}}^{{\text{2 + }}}}{\text{ + 4}}{{\text{H}}_{\text{2}}}{\text{O}}$
Here, ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ is the species being reduced and so it is the oxidant while ${\text{M}}{{\text{n}}^{{\text{2 + }}}}$ is the species being oxidized and so it is the reductant.
For polyatomic ions, we need to consider the oxidation state of each element in the chemical species and the oxidation number or oxidation state can be thought of as the charge that atoms will have if the chemical species were broken up.
We know, the sum of all the oxidation numbers in a chemical species must equal the charge on the chemical species. We also know that the oxidation number of a monatomic ion is its charge.
Let us consider the oxidation number of Mn be x. Now, for the oxidant ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$, we will find out the value of x: $x + \left( { - 2} \right) \times 4 = - 1 \Rightarrow x = + 7$
For the reductant ${\text{M}}{{\text{n}}^{{\text{2 + }}}}$, oxidation state of Mn is equal to its charge 2. Therefore, the change in the number of electrons involved $ = 7 - 2 = 5$
Thus, the number of electrons that are involved in the reduction of the permanganate ion to manganese (II) salt is 5.
Now, let us consider the conversion of the permanganate ion to manganate$\left( {{\text{Mn}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}} \right)$.
${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + }}{{\text{e}}^{\text{ - }}}{\text{ + O}}{{\text{H}}^{\text{ - }}} \to {\text{Mn}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$
Here, ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ is the species being reduced and so it is the oxidant while ${\text{Mn}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$ is the species being oxidized and so it is the reductant.
We know, the sum of all the oxidation numbers in a chemical species must equal the charge on the chemical species. Let us consider the oxidation number of Mn be x. Now, for the oxidant${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$, we will find out the value of x: $x + \left( { - 2} \right) \times 4 = - 1 \Rightarrow x = + 7$
For the reductant ${\text{Mn}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$, we will have: $x + \left( { - 2} \right) \times 4 = - 2 \Rightarrow x = + 6$
Therefore, the change in the number of electrons involved $ = 7 - 6 = 1$
Thus, the number of electrons that are involved in the reduction of the permanganate ion to manganese, ${\text{Mn}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$is 1.
Lastly, let us consider the conversion of the permanganate ion to manganese dioxide$\left( {{\text{Mn}}{{\text{O}}_{\text{2}}}} \right)$.
${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + 3}}{{\text{e}}^{\text{ - }}} \to {\text{Mn}}{{\text{O}}_{\text{2}}}$
Here, ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ is the species being reduced and so it is the oxidant while ${\text{Mn}}{{\text{O}}_{\text{2}}}$ is the species being oxidized and so it is the reductant.
We know, the sum of all the oxidation numbers in a chemical species must equal the charge on the chemical species. Let us consider the oxidation number of Mn be x. Now, for the oxidant ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$, we will find out the value of x: $x + \left( { - 2} \right) \times 4 = - 1 \Rightarrow x = + 7$
For the reductant ${\text{Mn}}{{\text{O}}_{\text{2}}}$, we will have: $x + \left( { - 2} \right) \times 2 = 0 \Rightarrow x = + 4$
Therefore, the change in the number of electrons involved $ = 7 - 4 = 3$
Thus, the number of electrons that are involved in the reduction of the permanganate ion to manganese dioxide $\left( {{\text{Mn}}{{\text{O}}_{\text{2}}}} \right)$ is 3.
Thus, 5, 1, 3 are the number of electrons which are involved in the reduction of the permanganate ion to manganese (II) salt $\left( {{\text{M}}{{\text{n}}^{{\text{2 + }}}}} \right)$, manganate $\left( {{\text{Mn}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}} \right)$ and manganese dioxide $\left( {{\text{Mn}}{{\text{O}}_{\text{2}}}} \right)$ respectively. So, the correct option is (A).
Note: In redox reactions, the number of electrons lost by one species is the same as the number gained by the other species, i.e., the number of electrons involved in this process are balanced.