149. Balance the following equation in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent.

(A) Ion electron method:
1. Write the oxidation and reduction half reaction by observing the changes in oxidation number.
           Oxidation half reaction:
              -2           +2
              
                                    +5             -1
Reduction half reaction:    
2. Balancing the oxidation half reaction:
  (i) Balancing N atoms by multiplying NO by 2

                       
(ii)  Add eight electron towards R.H.S. to balance the charges
                              
(iii) Add eight OH^- ions towards LHS to balance the charge. 
                
(iv) Balance O atoms by adding six H2O molecules towards R.H.S.
                 
                                   [Balanced oxidation half reactions]
3. Balancing the reduction half reaction:
(i) Add 6 electrons towards LHS to balance the charges on Cl atom
               
(ii) Add six OH^- ions towards RHS to balance the charges
                     
(iii) Balance O atoms by adding three H₂O molecules towards LHS.

(iv) Multiplying balanced oxidation half reaction by 3 and reduction half reaction by 4 to equate the electrons and add both the half reactions.


This is the balanced redox equation.

(b) Oxidation number method
(i) The skeleton equation along with oxidation number of each atom is
-2 + 1             +5                +2 -2           -1

(ii) The oxidation number of N increase by 4 per N atom while that of CI decreases by 6 per atom.



Therefore,  acts as the reducing agent while  acts as the oxidising agent.
(iii) Equalise the increase/decrease in O.N. by multiplying  and 
(iv) To balance N and Cl atoms, multiplying NO by 6 and Cl^- by 4.

(v) Balance O atoms, by adding six H₂O molecules towards R.H.S.
           
which is the balanced equation.