The equation given in question is as follow,
ZnS + NO₃⁻ → Zn²⁺ + S + NO
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⁻
8H⁺ + 2NO₃⁻ + 6e⁻ → 2NO + 4H₂O
_____________________________________(e⁻ cancelled)
3ZnS + 2NO₃⁻ + 8H⁺ → 3Zn + 3S + 2NO + 4H₂O
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