• # question_answer 22) 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.

(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