Question

In the equation $Mn{{O}_{4}}^{-}+{{H}^{+}}+n{{e}^{-}}\to M{{n}^{2+}}+{{H}_{2}}O$,the value of n is:

Answer 1

Hint: The equation where there is an increase and a decrease in oxidation states of reactants is a redox equation. Here oxidation and reduction happens simultaneously. The reduction half is the half where a species gains electrons and decreases in oxidation number and itself gets oxidized. The given equation is a reduction half. We have to balance it to find the value of n.

Complete answer:
In a reduction equation there is a gain of electrons by the reactant and there is removal of oxygen or addition of hydrogen, so the oxidation state decreases of the oxidizing agent in the reaction.
The given equation consists of permanganate ions that get reduced to form manganese ions. The equation is a reduction reaction as the oxidation number of manganese ions gets decreased in the reaction. On balancing this equation, we will get the number of ‘n’.
So, the reduction happens with permanganate ion having manganese with +7 oxidation state that gets reduced to +2 oxidation state as:
$\overset{+7}{\mathop{Mn}}\,{{\overset{-2}{\mathop{{{O}_{4}}}}\,}^{-}}+{{H}^{+}}+n{{e}^{-}}\to M{{n}^{2+}}+{{H}_{2}}O$
As we can see there is a decrease in 5 units in the oxidation number of Mn, so there must be a gain of 5 electrons. Also the equation will be balanced as the same number of oxygen and hydrogen should be there in the reactants and products. So 5 electrons are added in this equation, and 4 O atoms are balanced, as:
$Mn{{O}_{4}}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O$
Hence, the value of n is 5.

Note:
The given equation is an example of a reduction half; there may be an oxidation half where oxidation occurs. The suitable reagent used may be sulfur dioxide. Oxidation occurs where there is an increase in the oxidation number and the species loses electrons. While balancing, hydrogen ions are added to balance the positive charges and water is added to balance the negative charges.