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 1

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)\(P_4\) and \(F_2\) are reducing and oxidising agents respectively. 
If an excess of \(P_4\) is treated with \(F_2\), then \(PF_3\) will be produced, wherein the oxidation number (O.N.) of \(P\) is \(+3\).
\(P_4\,(excess)+F_2\rightarrow \overset{+3}PF_3\)
However, if \(P_4\) is treated with an excess of \(F_2\) , then \(PF_5\) will be produced, wherein the O.N. of \(P\) is \(+5\).
\(P_4+F_ 2\,(excess)\rightarrow \overset{+5}PF_5\)

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(ii) \(K\) acts as a reducing agent, whereas \(O_2\) is an oxidising agent.
If an excess of \(K\) reacts with \(O_2\), then \(K_2O\) will be formed, wherein the O.N. of \(O\) is \(-2\).
\(4K\,(excess)+O_2\rightarrow2K_2\overset{+2}O\)
However, if \(K\) reacts with an excess of \(O_2\), then \(K_2O_2\) will be formed, wherein the O.N. of \(O\) is \(-1\).

\(2K+O_2\,(excess)\rightarrow K_2\overset{-1}O_2\)

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(iii) \(C\) is a reducing agent, while \(O_2\) acts as an oxidising agent.
If an excess of \(C\) is burnt in the presence of insufficient amount of \(O_2\), then \(CO\) will be produced, wherein the O.N. of \(C\) is \(+2\).

\(C\,(excess)+O_2\rightarrow \overset{+2}CO\)

On the other hand, if \(C\) is burnt in an excess of \(O_2\), then \(CO_2\) will be produced, wherein the O.N. of \(C\) is \(+4\).
\(C+O_2\,(excess)\rightarrow \overset{+4}CO_2\)