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 agen

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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.

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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)