HNO₃ + H₂S → S + NO + H₂<span>O
Assign Oxidation Number:
L.H.S R.H.S
N in HNO</span>₃ = +5 +2 = N in NO
S in H₂S = -2 0 = S in S
Write Half cell Reactions:
Reduction Reaction:
3e⁻ + HNO₃ → NO -------(1)
Oxidation Reaction:
H₂S → S + 2e⁻ -------(2)
Multiply eq. 1 with 2 and eq. 2 with 3 to balance electrons.
6e⁻ + 2 HNO₃ → 2 NO
3 H₂S → 3 S + 6e⁻
Cancel e⁻s,
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2 HNO₃ + 3 H₂S → 2 NO + 3 S + H₂O
Balance Oxygen Atoms by multiplying H₂O with 4, Hydrogen atoms will automatically get balance.
2 HNO₃ + 3 H₂S → 2 NO + 3 S + 4H₂O
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If its sodium it would have 11 electrons
