Oxidation state of elements in reactant side are as follow, Zn = +2 S = -2 N = +5 O = -2 Oxidation state of elements in product side are as follow, Zn = +2 S = 0 N = +2 O = -2 So, In this reaction Nitrogen is Reduced while Sulfur is Oxidized.
Now Split the reaction into two half cell reactions,
Reduction Reaction, ZnS → S + 2e⁻ Oxidation Reaction, NO₃⁻ + 3e⁻ → NO
As the oxygen atoms are not balance, So, in acidic medium add H⁺ on the side having greater number of Oxygen atom and H₂O on the side having less number of Oxygen atoms, Hence, H⁺ + NO₃⁻ + 3e⁻ → NO + H₂O Now Balance the reaction,
4H⁺ + NO₃⁻ + 3e⁻ → NO + 2H₂O
So, write both half cell equations as,
ZnS → S + 2e⁻ --------- (1) 4H⁺ + NO₃⁻ + 3e⁻ → NO + 2H₂O -------(2)
Multiply eq. 1 with 3 and eq. 2 with 2 to equalize the electron. So, 3ZnS → 3S + 6e⁻