Q. 113.9( 9 Votes )

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. Justify this statement giving three illustrations.

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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