Balance the following equation by half reaction method:

 

(1) Write the oxidation and reduction half-reactions by observing the changes in oxidation numbers and write these separately.
                                      -2             0
Oxidation half-reaction:   H₂S       S
                                     
                                      +5           +2
Reduction half-reaction:     HNO₃    NO
(2) To balance the oxidation half reaction. (i) Balancing of S is not required as the number of each S is one on both the sides.
               2-             0
           H₂S         S
(ii) Add two electrons towards R.H.S. to balance the charges on sulphur atom.
                     2-           0
                    
(iii) Balance H atoms by adding two H⁺ towards 
R.H.S.
           2-        0
              + 
   (Balanced oxidation half reduction)
(3) To balance the reduction half reaction
(i) Balancing of N is not required as the number of each N is one on both the sides.
                          + 5           +2
                         HNO₃      NO
(ii) Add 3 electrons towards L.H.S. in order to balance the charges on the nitrogen atom. 
            +5                  +2
        HNO3   + 3e^-    NO
(iii) Balance the O atoms by adding two H2O molecules on R.H.S.
                                       +2
           HNO3 + 3e^-       NO + 2H₂O
(iv) Balancing H atoms by adding 3H⁺ on L.H.S.
                             (Balancing reduction half reaction)
4. Multiplying oxidation half-reaction by 3 and reductive half-reaction by 2 to equate the electrons and adding both the half reactions. 
      
This is a balanced redox reaction.