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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 - Chemistry

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प्रश्न

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.

दीर्घउत्तर
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उत्तर

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.

\[\ce{P4 (excess) + F2 -> ^{+3}PF3}\]

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

\[\ce{P4 + F2 (excess) -> ^{+5}PF5}\]

(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.

\[\ce{4K (excess) + O2 -> 2K2 ^{-2}O}\]

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

\[\ce{2K + O2 (excess) -> K2 ^{-1}O2}\]

(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.

\[\ce{C (excess) + O2 -> ^{+2}CO}\]

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.

\[\ce{C + O2 (excess) -> ^{+4}CO2}\]

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