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Question 11

Whenever a reaction between an oxidising 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 oxidising agent is in excess. Justify this statement giving three illustrations.

Answer

Whenever a reaction between an oxidising 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 oxidising agent is in excess. This can be illustrated as follows:

 

(i) P4 and F2 are reducing and oxidising agents respectively.

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

P4 (excess) + F2 → P+3F3

However, if P4 is treated with an excess of F2, then PF5 will be produced, wherein the O.N. of P is +5.



P4  + F2 (excess) → P+5F5

 

(ii) K acts as a reducing agent, whereas O2 is an oxidising 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-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)  →  K2O-12

 

(iii) C is a reducing agent, while O2 acts as an oxidising agent.

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

C (excess) + O2 →  C+2O

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

C + O2 (excess)  →  C+4O2

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