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

(i) C is a reducing agent while `O_2` is an oxidisin agent. If excess of C is burnt in a limited supply of `O_2` , CO is formed in which the oxidation state of C is +2. However , if `O_2` is excess , the initially formed CO gets oxidised to CO in which the oxidation state of C is +4.
`underset("(excess)")(2C(s))+O_2(g) rarroverset(+2)(2CO)`
`C(s)+underset("(excess)")(O_2(g))rarroverset(+4)(CO_2)`
(ii) Sodium is a reducing agent while `O_2` is an oxidising agent . When excess of Na is used , sodium oxide is formed in which the oxidaiton state of O is -2 . If howere , excess of `O_2` is used , `Na_2O_2` is formed in which the oxidation state of O is - 1, which is higher than -2 .
`underset(("excess"))(4Na(s))+O_2(g)rarr2Na_2overset(-2)O(s)`
`2Na(s) +underset(("excess"))(O_2(g))rarrNa_2overset(-1)O_2(s)`
(iii) `P_4` is a reducing agent while `Cl_2` is an oxidising agnet . When excess of `P_4` is used , `PCl_3` is formed in which the oxidaiton state of P is +3. However, if excess of `Cl_2` is used , then initially formed `PCl_3` reacts further to form `PCl_5` in which the oxidation state of P is +5.
`underset(("excess"))(P_4(S))rarr6Cl_2rarroverset(+3)(4PCl_3)`
`P_4(s) +underset(("excess"))(10Cl_2)rarrNa_2overset(+5)4PCl_5`