Question

When kmno4 is reduced with oxalic acid in acidic medium the oxidation number of mn changes from?

Answer 1

Mn +7 in KMnO4 changes to Mn+2
There's gain of 5 electron by Manganese

Answer 2

Explanation:

$\mathrm{Kmno4}$ can be reduced with the “oxalic acid” in the medium of acidic and the “oxidation number” of mn changes from +7 to +2.

The chemical reaction is as follows:

$2 \mathrm{K}_{2}+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}$

Let give oxidation states to the “mn “ in both “reactants and products side”.

The reactant is ‘left side’ of the arrow mark in the equation of the reaction and the product is the ‘right side’ of the arrow mark.

Therefore, with the “oxalic acid” in the medium of acidic and the “oxidation number” of mn changes from +7 to +2.