What is the coefficient of H2S in the balanced equation obtained by using oxidation numbers?

Question

HNO₃(aq) + H₂S(aq) → NO(g) + S(s) + H₂O(l)

Answer 1

Oxidation: H₂S --> S + 2e + 2H⁺

Reduction: 3H⁺ + HNO₃ + 3e --> NO + 2H₂O

Multiply the oxidation equation by 3 and the reduction reaction by 2 so both equation exchange equal number of electrons.

Oxidation: 3x( H₂S --> S + 2e + 2H⁺) = 3H₂S --> 3S + 6e + 6H⁺

Reduction: 2x (3H⁺ + HNO₃ + 3e --> NO + 2H₂O) = 6H⁺ + 2HNO₃ + 6e --> 2NO + 4H₂O

Now add both equations: 3H₂S + 6H⁺ + 2HNO₃ + 6e --> 3S + 6e + 6H⁺ + 2NO + 4H₂O

Cancel on the reactant and product sides what can be canceled:

3H₂S + 6H⁺ + 2HNO₃ + 6e --> 3S + 6e + 6H⁺ + 2NO + 4H₂O

Final equation is:

3H₂S + 2HNO₃ --> 3S + 2NO + 4H₂O

Answer 2

Since O has a usual standard oxidation number of -2, and H is +1 (refer to texts for the few exceptions), and the sum of oxidation numbers for a molecule must be 0, the oxidation numbers of N are +5 and +2 in the reactants and products, respectively, and S is -2 in reactants and 0 as an element in the products. Therefore, N is being reduced by 3 charges, and S is gaining 2 by oxidation. In order to balance these changes, the N coefficients will be 2, and S coefficients will be 3 (giving 6 plus for oxidation and 6 minus for reduction). Also, water will have a coefficient of 4, since the reactants will have 2H and 6H. The O will also be balanced, since there are 6 in the reactants, and 2 + 4 in the products. H₂S will have coefficient of 3.