Answer :

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 O_{2}acts as an oxidizing agent.

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

C(excess) + O_{2}→ 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 O_{2}, then CO_{2} will be produced wherein the oxidation number of C is + 4.

C + O_{2}(excess) → CO_{2}

Let the oxidation number of C be x in CO_{2}.

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

X = 4

(b) P_{4} and F_{2} are reducing and oxidizing agents respectively.

If an excess of P_{4} is treated with F_{2}, then PF_{3} will be produced, wherein the O.N. of is + 3.

P_{4}(excess) + F_{2}→ PF_{3}

Let the oxidation number of P be x in PF_{3}.

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

X = 3

However, of P is treated with an excess of F_{2}, then PF_{5} will be produced wherein the O.N. of P is + 5.

P_{4} + F_{2}(excess) → PF_{5}

Let the oxidation number of P be x in PF_{5}.

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

X = 5

(c) K acts as a reducing agent, whereas O_{2} is an oxidizing agent.

If an excess of K reacts with O_{2} then K_{2}O will be formed wherein the O.N. of O is -2.

4K(excess) + O_{2}→ 2K_{2}O

Let the oxidation number of O be x in K_{2}O.

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

X = -2

However, if K reacts with an excess of O_{2},then K_{2}O_{2} will be formed wherein the O.N. of O is -1.

2K + O_{2}(excess) → K_{2}O_{2}

let the oxidation number of O be x in K_{2}O_{2}.

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

X = -1

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