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

a. `C` is a reducing agent while `O_(2)` is an oxidising agent. If excess of carbon is burnt in a limited supply of `O_(2), CO` is formed in which the oxidation state of `C` is `+2`. If, however, excess of `O_(2)` is used, the initially formed `CO` get oxidised to `CO_(2)` in which oxidation state of `C` is `+4`.
`underset(("Excess"))(2C(s))+O_(2)(g) rarr 2overset(+2)(C )O(g)`
`C(s)+underset(("Excess"))(O_(2)(g))rarr overset(+4)(CO_(2))(g)`
b. `P_(4)` is a reducing agent while `Cl_(2)` is an oxidising agent. When excess of `P_(4)` is used, `PCl_(3)` is formed in which the oxidation state of `P` is `+3`. If, however, excess of `Cl_(2)` is used, the initially formed `PCl_(3)` reacts further to from `PCl_(5)` in which the oxidation state of `P` is `+5`.
`underset(("Excess"))(P_(4)(s))+6Cl_(2)(g) rarr overset(+3)(4 PCl_(3)),`
`P_(4)(s)+underset(("Excess"))(10Cl_(2))(g) rarr overset(+5)(4PCl_(5))`
c. `Na` is a reducing agent while `O_(2)` is an oxidising agent. When excess of `Na` is used, sodium oxide is formed in which the oxidation state of `O` is `-2`. If however, excess of `O_(2)` is used, `Na_(2)O_(2)` is formed in which the oxidation state of `O` is `-1` which is higher than `-2`.
`underset(("Excess"))(4Na(s))+O_(2)(g) rarr Na_(2)overset(-2)(O)(s)`
`4Na(s)+underset(("Excess"))(2O_(2)(g))rarr 2Na_(2)overset(-1)(O_(2))(s)`