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Class 11
Chemistry
Redox Reactions
whenever a reaction between an oxidising agent and...

Question:
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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