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) Carbon $(C)$ is a reducing agent and $O_2$ is an oxidizing agent. When $C$ is in excess,$CO$ is formed (oxidation state of $C = +2$). When $O_2$ is in excess,$CO_2$ is formed (oxidation state of $C = +4$).
$2C(s) + O_2(g) \rightarrow 2CO(g)$ (Excess $C$)
$C(s) + O_2(g) \rightarrow CO_2(g)$ (Excess $O_2$)
$(b)$ $P_4$ is a reducing agent and $Cl_2$ is an oxidizing agent. When $P_4$ is in excess,$PCl_3$ is formed (oxidation state of $P = +3$). When $Cl_2$ is in excess,$PCl_5$ is formed (oxidation state of $P = +5$).
$P_4(s) + 6Cl_2(g) \rightarrow 4PCl_3(l)$ (Excess $P_4$)
$P_4(s) + 10Cl_2(g) \rightarrow 4PCl_5(s)$ (Excess $Cl_2$)
$(c)$ $Na$ is a reducing agent and $O_2$ is an oxidizing agent. When $Na$ is in excess,$Na_2O$ is formed (oxidation state of $O = -2$). When $O_2$ is in excess,$Na_2O_2$ is formed (oxidation state of $O = -1$).
$4Na(s) + O_2(g) \rightarrow 2Na_2O(s)$ (Excess $Na$)
$2Na(s) + O_2(g) \rightarrow Na_2O_2(s)$ (Excess $O_2$)