Q. 114.2( 13 Votes )

# Whenever a reacti

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 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

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