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) Carbon is an reducing agent. `O_(2)` is an oxidizing agent. C combines with half of `O_(2)` molecules in which oxidation number of C is +2. As more amount of carbon reacts with oxygen `(O_(2))` forms `CO_(2)` in which oxidation number of C in +4.
`underset(("More amount"))(2C_((s)))+O_(2(g))tooverset(+2)(2CO_((g)))`
`underset(("More amount"))(C_((s))+O_(2(g)))tooverset(+4)(CO_(2(g)))`
(b) `P_(4)` is a reducing agent while `Cl_(2)` is oxidizing agent. When more amount of `P_(4)` is used then form `PCl_(3)` in which the oxidation number of P is +3.
Same way when more `Cl_(2)` is used at initial stage it forms `PCl_(3)`, which reacts again and form `PCl_(5)` in which oxidation number of P is +5.
`underset(("More amount"))(P_(4(s)))+6Cl_(2(g))tooverset(+3)(4PCl_(3))`
`underset(("More amount"))(P_(4(s))+10Cl_(2))tooverset(+5)(4PCl_(5))`
( c) Na is reducing agnet, while `O_(2)` is an oxidizing agent. More amount of Na used forms `Na_(2)O` in which oxidation number of oxygen is -2. Same as more amount of `O_(2)` used forms `Na_(2)O_(2)`, in which, oxidation number of oxygen is -1.
`underset(("More amount"))(4Na_((s)))+O_(2(s))tooverset(-2)(Na_(2)O_((s)))`
`underset(("More amount"))(2Na_((s))+2O_(2(s)))tooverset(-1)(Na_(2)O_(2(s)))`