How do we represent the oxidation of hydrogen sulfide by nitric acid to give sulfur and #NO(g)#?

1 Answer
Nov 14, 2016

Answer:

This is a redox reaction: sulfur is oxidized, and nitrogen is reduced.

Explanation:

#"Oxidation (of sulfur) (i)":#

#H_2S^(-II) rarr 1/8S""_8^(0) +2H^(+)+2e^(-)#

#"Reduction (of nitrogen) (ii)":#

#HN^(+V)O_3 +3H^(+)+3e^(-) rarr N^(+II)O +2H_2O#

In both cases, mass is balanced, and charge is balanced, as is required for a chemical reaction.

For the overall redox reaction, we simply cross-multiply each equation in order to eliminate the electrons that appear in each half equation:

i.e. #3xx(i)+2xx(ii):#

#3H_2S^(-II) +2HN^(+V)O_3 rarr 3/8S_8 +2N^(+II)O+4H_2O#

Are charge and mass balanced here? What would be observed in this reaction is the appearance of a fine white precipitate of sulfur. The evolved #NO# would be hard to observe.

You are certainly free to multiply each stoichiometric coefficient by #8# in order to remove the non-integral coefficients.