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)  ${\mathrm{P}}_4 \mathrm{and} {\mathrm{F}}_2$ 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.
${\mathrm{P}}_4 + {\mathrm{F}}_2\left(\mathrm{excess}\right)\to \stackrel{+5}{\mathrm{P}}{\mathrm{F}}_5$

(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 2K_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.

$\mathrm{C}+{\mathrm{O}}_2\left(\mathrm{excess}\right)\to \stackrel{+4}{\mathrm{C}}{\mathrm{O}}_2$