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 $+4$.
$C+O_2\,(excess)\rightarrow \overset{+4}CO_2$