Question

Balance the reaction by oxidation method HNO3 + H2S gives no + S + H2O

Answer 1

3H₂S + 2HNO₃ → 2NO + 4H₂O + 3S

Answer 2

Solution Explanation:

1. Assign oxidation numbers:
    • In HNO₃, N is +5; in NO, N is +2 (reduction: gain 3e⁻).
    • In H₂S, S is −2; in elemental S, S is 0 (oxidation: loss of 2e⁻).

2. Write half‐reactions:
    Reduction:
     NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O
    Oxidation:
     H₂S → S + 2H⁺ + 2e⁻

3. Balance electrons:
    LCM of electrons is 6. Multiply the reduction half-reaction by 2 and the oxidation half-reaction by 3:
     Reduction: 2NO₃⁻ + 8H⁺ + 6e⁻ → 2NO + 4H₂O
     Oxidation: 3H₂S → 3S + 6H⁺ + 6e⁻

4. Combine and simplify:
    Adding both:
     2NO₃⁻ + 8H⁺ + 3H₂S → 2NO + 4H₂O + 3S + 6H⁺
    Cancel 6H⁺ from both sides:
     2NO₃⁻ + 2H⁺ + 3H₂S → 2NO + 4H₂O + 3S

5. Convert back to molecular form:
    Since HNO₃ dissociates into NO₃⁻ and H⁺, the two NO₃⁻ and 2H⁺ come from 2 HNO₃.
    Final balanced equation:
     3H₂S + 2HNO₃ → 2NO + 4H₂O + 3S