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# 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

Last updated date: 13th Jun 2024
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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.

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.

(i) ${ KMnO }_{ 4 }$ is a deep purple crystalline solid which gives deep purple color to the solution when dissolved in water.
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 ${ +6 }$.