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

Nov 14, 2016

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

#### Explanation:

$\text{Oxidation (of sulfur) (i)} :$

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

$\text{Reduction (of nitrogen) (ii)} :$

$H {N}^{+ V} {O}_{3} + 3 {H}^{+} + 3 {e}^{-} \rightarrow {N}^{+ I I} O + 2 {H}_{2} O$

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. $3 \times \left(i\right) + 2 \times \left(i i\right) :$

$3 {H}_{2} {S}^{- I I} + 2 H {N}^{+ V} {O}_{3} \rightarrow \frac{3}{8} {S}_{8} + 2 {N}^{+ I I} O + 4 {H}_{2} O$

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 $N O$ would be hard to observe.

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