Question

On reduction of $KMn{{O}_{4}}$ by oxalic acid in the acidic medium, the oxidation number of Mn changes. What is the magnitude of this change?
(a)- 7 to 2
(b)- 6 to 2
(c)- 5 to 2
(d)- 7 to 4

Answer 1

Oxidation number is the charge on the atom in which it appears in the combined state. Write the correct reaction of potassium permanganate with oxalic acid in an acidic medium. It forms manganese sulfate, potassium sulfate, carbon dioxide, and water.

Complete step by step answer:
The oxidation number of an element may be defined as the charge in which an atom of the element has its ion or appears to have when present in the combined state with other atoms. The oxidation number is also called the oxidation state.
For finding the change in the oxidation number of Mn first we have to write the balanced equation of potassium permanganate with oxalic acid in the acidic medium. This will form manganese sulfate, potassium sulfate, carbon dioxide, and water. The equation is
$KMn{{O}_{4}}+{{H}_{2}}{{C}_{2}}{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+MnS{{O}_{4}}+{{H}_{2}}O+C{{O}_{2}}$
So, on the reactant side, the Mn is present in $KMn{{O}_{4}}$
The oxidation number of K is +1 because it has 1 electron in its outermost shell and the oxidation number of oxygen is -2. Since this is a neutral compound the sum of oxidation states of all the atoms is zero. Hence, we can calculate:
$KMn{{O}_{4}}$ = +1 + x +4(-2) = 0
x = +7.
So, the oxidation state of Mn in $KMn{{O}_{4}}$ is +7.
Now, in the product side the Mn is present in $MnS{{O}_{4}}$
We can take the combined oxidation state of $S{{O}_{2}}^{2-}$ . The oxidation number of $S{{O}_{2}}^{2-}$ is -2. This is also a neutral compound. Hence, we can calculate:
$MnS{{O}_{4}}$ = x + (-2) = 0
x = +2.
So, the oxidation state of Mn in $MnS{{O}_{4}}$ is +2.
Hence, the oxidation state changes from +7 to +2.

Hence the correct option is option (a)- 7 to 2.

Note:
The compound is neutral then the sum of the oxidation states is equal to zero, but if it has some charge, then the sum of the oxidation states is equal to the charge.