8.11 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?

Answers (1)


These can be understood by the following examples-

  • P_{4}  is reducing agent and  Cl_{2}  is an oxidizing agent

P_{4}+10Cl_{2}(Excess)\rightarrow PCl_{5}        [O.N of phosphorus +5]    \Rightarrow  Higher O.S of P

P_{4}(Excess)+6Cl_{2}\rightarrow 4PCl_{3}        [O.N of phosphorus +3]     \Rightarrow Lower O.S of P

  •   is an O_{2} is a reducing agent and  C oxidising agent

C+O_{2}\rightarrow CO_{2}   (O is in excess)    [O.N of C +4]

C+O_{2}\rightarrow CO     (C is in excess)       [O.N of C +2]

  • K is a reducing agent and O_{2} is an oxidizing agent

K+O_{2}\rightarrow K_{2}O (K is in excess)  [O.N of O -2] (lower O.S.)

K+O_{2}\rightarrow K_{2}O_{2}  (O is in excess) [O.N of O -1] (lower O.S.)