Question

Question 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.

Class XI Redox Reactions Page 273

Answer 1

(i) C is oxidising agent while O2 is reducing agent. if excess of C is burnt with a limited supply of O2. CO is formed in which oxidation number of C is +2 .
2C(excess) + O2 ------>CO
and if excess of O2 is used , the intially CO formed and then get oxidised and formed CO2 in which oxidation number of C is +4.
C + O2 (excess)-------> CO2


(ii) P4 is reducing agent while Cl2 is an oxidizing agent. when we use excess of P4, PCl3 is formed in which oxidation number of P is +3.
e.g., P4 (excess)+ 6Cl2 -----> 4PCl3

and if we use excess of Cl2, intially PCl3 is formed which react further to form PCl5 in which oxidation number of P is +5 .
e.g., P4 + 10Cl2(excess) ------> 4PCl5


(iii) Na is reducing agent while O2 is oxidising agent. when we use excess of Na , Na2O is formed in which oxidation number of O is -2.
e.g., 4Na (excess)+ O2 ------>2Na2O

and if we use excess of O2 , Na2O2 is formed in which oxidation number of O is -1
e.g., 2Na + 2O2(excess) -------> 2Na2O2

Answer 2

Hi

Please see the attached file !

Hope it helps you !