Whenever a reaction between an oxidizing agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidizing agent is in excess. This can be illustrated as follows:

(a) C is a reducing agent, while O₂acts as an oxidizing agent.

If an excess of C is burnt in the presence of an insufficient amount of O₂, then CO will be produced, wherein the Oxidation number of C is + 2.

C(excess) + O₂→ CO

Let the oxidation number of C be x in CO.

X + 1(-2) = 0 {oxidation number of O = -2}

X = 2

On the other hand, if C is burnt in an excess of O₂, then CO₂ will be produced wherein the oxidation number of C is + 4.

C + O₂(excess) → CO₂

Let the oxidation number of C be x in CO₂.

X + 2(-2) = 0 {oxidation number of O = -2}

X = 4

(b) P₄ and F₂ are reducing and oxidizing agents respectively.

If an excess of P₄ is treated with F₂, then PF₃ will be produced, wherein the O.N. of is + 3.

P₄(excess) + F₂→ PF₃

Let the oxidation number of P be x in PF₃.

X + 3(-1) = 0 {oxidation number of F = -1}

X = 3

However, of P is treated with an excess of F₂, then PF₅ will be produced wherein the O.N. of P is + 5.

P₄ + F₂(excess) → PF₅

Let the oxidation number of P be x in PF₅.

X + 5(-1) = 0 {oxidation number of F = -1}

X = 5

(c) K acts as a reducing agent, whereas O₂ is an oxidizing agent.

If an excess of K reacts with O₂ then K₂O will be formed wherein the O.N. of O is -2.

4K(excess) + O₂→ 2K₂O

Let the oxidation number of O be x in K₂O.

2(+ 1) + x = 0 {oxidation number of K = + 1}

X = -2

However, if K reacts with an excess of O₂,then K₂O₂ will be formed wherein the O.N. of O is -1.

2K + O₂(excess) → K₂O₂

let the oxidation number of O be x in K₂O₂.

2(+ 1) + 2x = 0 {oxidation number of K = + 1}

X = -1