question_answer 21)

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 oxidising agent is in excess. Justify this statement giving three illustrations.

Answer:

(i) Let us consider reaction between carbon and oxygen. Reducing Agent + Oxidising Agent \[\underset{(Excess)}{\mathop{C(s)}}\,+\frac{1}{2}{{O}_{2}}(g)\to \overset{+2}{\mathop{C}}\,\overset{-2}{\mathop{O}}\,\] Lower state of carbon \[C(s)+\underset{(Excess)}{\mathop{{{O}_{2}}(g)}}\,\to \overset{+4}{\mathop{C}}\,{{\overset{-4}{\mathop{O}}\,}_{2}}\] Higher state of carbon (ii) Let us consider the reaction between white phosphorous (\[{{P}_{4}}\]) and\[C{{l}_{2}}(g)\]. Reducing agent + Oxidising agent \[\underset{(Excess)}{\mathop{{{P}_{4}}(s)}}\,+3C{{l}_{2}}(g)\to 4\overset{+3-3}{\mathop{PC{{l}_{3}}}}\,\] Lower oxidation state of phosphorous \[{{P}_{4}}(s)+\underset{(Excess)}{\mathop{10C{{l}_{2}}(g)}}\,\to 4\overset{+5-5}{\mathop{PC{{l}_{5}}(g)}}\,\] Higher oxidation of phosphorous (iii) Let us consider the reaction between sulphur and oxygen. Reducing agent + Oxidising agent \[\underset{(Excess)}{\mathop{S}}\,+{{O}_{2}}\to \overset{+4-4}{\mathop{S{{O}_{2}}}}\,\] Lower oxidation state of sulphur \[2S+3{{O}_{2}}\to 2\overset{+6-6}{\mathop{S{{O}_{3}}}}\,\] Higher oxidation state of sulphur