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) P 4 and F 2 are reducing and oxidising agents respectively. If an excess of P 4 is treated with F 2 , then PF₃ will be produced, wherein the oxidation number (O.N.) of P is+3.
+3 P, (excess) + P, —> P F3
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.
+5 P4 + F2 (excess) —> P F5
(ii) K acts as a reducing agent, whereas 02 is an oxidising agent. If an excess of K reacts with 02 , then K 2 0 will be formed, wherein the O.N. of 0 is —2.
4K (excess) + 0, —>2K2 0
However, if K reacts with an excess of 02 , then K 2 02 will be formed, wherein the O.N. of 0 is —1.
2K + 0, (excess) -> K, 02
(iii) C is a reducing agent, while 02 acts as an oxidising agent. If an excess of C is burnt in the presence of insufficient amount of 02, then CO will be produced, wherein the O.N. of C is +2.