Question

When ${ KMnO }_{ 4 }$ is reduced with oxalic acid in an acidic solution, the oxidation number of Mn changes from:
(A) +2 to +7
(B) +4 to +7
(C) +7 to +2
(D) +6 and +2

Answer 1

Hint: Oxidation number is the residual charge which a molecule has or seems to have when different elements are taken out as ions by counting the number of shared electrons.

Complete step-by-step answer:
The reaction involved between potassium permanganate and oxalic acid is given below;
${ 2KMnO }_{ 4 }{ +5H }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }{ +3H }_{ 2 }{ SO }_{ 4 }{ \rightarrow K }_{ 2 }{ SO }_{ 4 }{ +2MnSO }_{ 4 }{ +10CO }_{ 2 }{ +8H }_{ 2 }{ O }$
Firstly, assign the oxidation number of Manganese:
Let, the oxidation number of Mn is ‘x’.
The oxidation number of ${ SO }_{ 4 }^{ 2- }$ = ${ x+4\times (-2)=-2 }$
x = ${ +6 }$
and K = ${ +1 }$
Now, for ${ KMnO }_{ 4 }$, we can calculate;
${ 1+ x + 4\times(-2) = 0 }$
${ x +1 - 8 = 0 }$
${ x -7 = 0 }$
x = ${ +7 }$
In ${ MnSO }_{ 4 }$, x + (-2) = 0
x = ${ +2 }$

Therefore, When ${ KMnO }_{ 4 }$ is reduced with oxalic acid in acidic solution, the oxidation number of Mn changes from ${ +7 }$ to ${ +2 }$.

Hence, the correct option is C.

Additional Information:
(i) ${ KMnO }_{ 4 }$ is a deep purple crystalline solid which gives deep purple color to the solution when dissolved in water.
(ii) It is also a strong oxidizing agent, and hence widely used as an oxidizing agent in organic chemistry.
(iii) It is capable of destroying organic impurities and hence also used to purify impure water in well and ponds.
(iv) Oxalic acid: It is a colorless crystalline solid that dissolves in water to give colorless solutions. It is a weak acid and is used as a bleaching agent, cleanser rust remover, etc.

Note: The possibility to make a mistake is that you may choose option D. But the oxidation state of S in ${ SO }_{ 4 }^{ 2- }$ is +6 while the oxidation state of Mn in ${ KMnO }_{ 4 }$ is ${ +7 }$, not ${ +6 }$.