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.

(N/A) 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 $PF_3$ is produced,wherein the oxidation number $(O.N.)$ of $P$ is $+3$: $P_4 (\text{excess}) + 6F_2 \to 4PF_3$.
However,if $P_4$ is treated with an excess of $F_2$,then $PF_5$ is produced,wherein the $O.N.$ of $P$ is $+5$: $P_4 + 10F_2 (\text{excess}) \to 4PF_5$.
$(ii)$ $K$ acts as a reducing agent,whereas $O_2$ is an oxidising agent.
If an excess of $K$ reacts with $O_2$,then $K_2O$ is formed,wherein the $O.N.$ of $O$ is $-2$: $4K (\text{excess}) + O_2 \to 2K_2O$.
However,if $K$ reacts with an excess of $O_2$,then $K_2O_2$ is formed,wherein the $O.N.$ of $O$ is $-1$: $2K + O_2 (\text{excess}) \to K_2O_2$.
$(iii)$ $C$ 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 $CO$ is produced,wherein the $O.N.$ of $C$ is $+2$: $2C (\text{excess}) + O_2 \to 2CO$.
On the other hand,if $C$ is burnt in an excess of $O_2$,then $CO_2$ is produced,wherein the $O.N.$ of $C$ is $+4$: $C + O_2 (\text{excess}) \to CO_2$.