Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the oxidising agent is in excess justify this statement giving three illustrations .

Explanation

The given statement can be justified form the examples of ractions below
1 The reactions of C in (reductant) with `O_(2) `(oxidant) may result in Co or `CO_(2)` or a misture of CO and `CO _(2)` . However ,if this reaction is initilated with excess amount of C, the only producer that formed is CO. On the other hand, if `O_(2)` is taken in excress in the reaction ,only `CO_(2)` is formed in CO, the oxidation state of C is +2 and in `CO_(2)`, it is +4 Thus we see that taking excess amount of reducant leads to the formation of a compound of lower oxidation state. Conversely, a compound of higher oxidation state is formed when the oxidation is taken to excess.
`underset(("Excess"))(C(s))+O_(2)(g)toCO(g):C(s)+underset(("Excess"))(O_(2)(g))toCO_(2)(g)`
2 The reaction of `P_(4)` (reductant) with `CI_(2)` (oxidant) results in `PCI_(3)` when `P_(4)` is taken in excess ,while it results in `PCI_(5)` when `CI_(2)` is takan
`underset(("Excess"))(P_(4)(s)+6)CI_(2)(g)to4CI_(3)g,P_(4)(s)+underset(("Excess"))(10CI_(2)(g))to4PCI_(5)(g)`
The oxidation state P in `PCI_(3)` is +3 and that `PCI_(3)` is +5 Thus an excess amount of reductant produces a compound of lower oxidation state and an excess amount of oxidation produces a compound of higher oxdation state.
3 The same thing happens when na (reductant) is reacted with `O_(2)` (oxidant) .In presence of excess Na, the resulting compound is `Na_(2)O`, in which the oxidation state of oxygen is -2 and in pressence of excess `O_(2)` the resulting compound is `Na_(2)O_(2)`. in which the oxidation state of oxygen is -1.
`underset(("Excess"))(Na(s))+(1)/(2)O_(2)(g)toNa_(2)O(s):2Na(s)+underset(("Excess"))(2O_(2)(g)) to Na_(2)O_(2)(s)`
In the reaction ,the mass of Na is 46g and that of oxygen is 64gl