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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Solution

Whenever a reaction between an oxidizing agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing ent is in excess and a compound of higher oxidation state is formed if the oxidizing agent is in excess. Following illustrations justify this.
(i) Oxidizing agent is $F_{2}$ and reducing agent is $P_{4}$ . When excess $P_{4}$ reacts with $F_{2}$ , $P F_{3}$ is produced in which P has +3 oxidation number.
$P_{4} ( excess ) + F_{2} \to P F_{3}$
But if fluorine is in excess, $P F_{5}$ is formed in which P has oxidation number of +5.
$P_{4} + F_{2} ( excess ) \to P F_{5}$
(ii) Oxidizing agent is oxygen and reducing agent is K. When excess K reacts with oxygen, $K_{2} O$ is formed in which oxygen has oxidation number of -2.
$4 K ( excess ) + O_{2} \to 2 K_{2} O$
But if oxygen is in excess, then $K_{2} O_{2}$ is formed in which O has oxidation number of -1.
$2 K + O_{2} ( excess ) \to K_{2} O_{2}$
(iii) The oxidizing agent is oxygen and the reducing agent is C. When an excess of C reacts with oxygen, CO is formed in which C has +2 oxidation number.
$C ( excess ) + O_{2} \to C O$
When excess of oxygen is used, $C O_{2}$ is formed in which C has +4 oxidation number.
$C + O_{2} ( excess ) \to C O_{2}$

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