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 reducing agent while `O_(2)` is n oxidising agent if excess of carbon is burnt in a limited support of `O_(2)` CO formed in which the oxidation state of C is +2 if however excess of `O_(2)` is used the initial formed CO gets oxidised to `CO_(2)` in oxidation state of C is +4
`2C(s) +O_(2)(g)rarr 2CO(g), C(s)+O_(2)(g)rarrCO_(2)(g)`
(ii) `P_(4)` is a reducing agent while `CI_(2)` is used the initally formed `PCI_(3)` react further to form `PCI_(5)` in which the oxidation state of p is +5
`P_(4)(s)+6CI_(2)(g)rarr4overset(+3)PCI_(3), P_(4)(s)10 CI_(2)rarr4 overset(+5)PCI_(5)`
(iii) Na is a reducing agent while `O_(2)` is an oxidising agent when excess of Na is used sodium oxide formed in the oxidation state of O is -1 which is higher than -2
`4Na(s)+O_(2)(g)rarrNa_(2)O(s), 2Na(s)+2O_(2)(g)rarrNa_(2)O_(2)(s)`