Question

What element is being reduced in the following redox reaction?

C₃H₈O₂(aq) + KMnO₄(aq) → C₃H₂O₄K₂(aq) + MnO₂(aq)

A.C

B.K

C.H

D.O

E.Mn

Answer 1

In KMnO4:

Oxidation number of K = +1

Oxidation number of O = -2

lets the oxidation number of Mn be x

use:

1* oxidation number (K) + 4* oxidation number (O) + 1* oxidation number (Mn) = net charge

1*(+1)+4*(-2)+1* x = 0

-7 + 1 * x = 0

x = 7

So oxidation number of Mn = +7

In MnO2, oxidation state of Mn is +4

Oxidation state of Mn is decreasing from +7 to +2.

So, Mn is being reduced.

Answer: E