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.

Explanation

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)` will be produced , wherein the oxidation number (O.N.) of P is +3 .
`P_(4)` (excess ) + `F_(2) to overset(+3)(P)F_(3)`
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.
`P_(4) + F_(2) ("excess") to overset(+5)(P)F_(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_(2)O` will be formed, wherein the O.N. of O is –2.
4K (excess ) `+ O_(2) to 2K_(2) overset(-2)(O)`
However , if K reacts with an excess of `O_(2)` , then `K_(2)O_(2)` will be formed , wherein the O.N. of O is `-1` .
2K + `O_(2) ("excess") to K_(2) overset(-1)(O_(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 insufficient amount of `O_(2)` then CO will be produced . wherein the O.N. of C is `+2` .
C (excess ) + `O_(2) to overset(+2)(C)O`
On the other hand , if C is burnt in an excess of `O_(2)` , then `CO_(2)` will be produced , wherein the O.N. of C is +4 .
`C + O_(2) ("excess") to overset(+4)(C) O_(2)`