Question

(a) Assign oxidation number to the underlined elements in each of the following species:

(i) KMnO₄   (ii) CaO₂

(b) Identify the oxidising agent and reducing agent in the following reaction:

2MnO₄ + 6I^– + 4H₂O → MnO₂ + 3I₂ + 8OH^–

(c) What do you understand by disproportionation reaction?

Answer 1

(a) (i) KMnO₄

KMnO₄ = 1 + x + 4 (–2) = 0

= 1 + x – 8 = 0

x = +7

\(\therefore\) Oxidation number of Mn in KMnO₄ is +7.

(ii) CaO₂

If oxidation number of O is x,

the CaO₂ + 2 + 2 (x) = 0

+2 + 2x = 0 

2x = – 2

\(\therefore\) x = -1

\(\therefore\) Oxidation number O in CaO₂ is –1.

(b) 

Here, MnO₄^– is an oxidising reagent as itself gets reduced and I^– is a reducing reagent as it itself oxidised.