8.11 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.

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)   are reducing and oxidising agents respectively.
If an excess of ${\mathrm{P}}_{4}$ is treated with ${\mathrm{F}}_{2}$, then ${\mathrm{PF}}_{3}$ will be produced, wherein the oxidation number (O.N.) of P is +3.

(ii)  K acts as a reducing agent, whereas ${\mathrm{O}}_{2}$ is an oxidising agent.

if an excess of K reacts with ${\mathrm{O}}_{2}$, then ${\mathrm{K}}_{2}\mathrm{O}$ will be formed,

wherein the O.N. of O is -2.

4K(excess)+${\mathrm{O}}_{2}$$\to 2{K}_{2}{O}^{-2}$

However, if K reacts with an excess of ${\mathrm{O}}_{2}$,

then ${K}_{2}{O}_{2}$ will be formed, wherein the O.N. of O is -1.

$2K+{O}_{2}\left(excess\right)\to {K}_{2}{{O}_{2}}^{-2}$

(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 an insufficient amount of ${O}_{2}$,

then CO will be produced, wherein the O.N. of C is +2.

$\mathrm{C}\left(\mathrm{excess}\right)+{\mathrm{O}}_{2}\to \stackrel{+2}{\mathrm{C}}\mathrm{O}$

On the other hand, if C is burnt in an excess of ${O}_{2}$, then C${O}_{2}$ will be produced, wherein the O.N. of C is +4.

$\mathrm{C}+{\mathrm{O}}_{2}\left(\mathrm{excess}\right)\to \stackrel{+4}{\mathrm{C}}{\mathrm{O}}_{2}$