# Question #8b77f

Jun 7, 2017

You get a precipitate of octasulfur.

#### Explanation:

Sulfur dioxide and hydrogen sulfide react in aqueous solution to produce sulfur, which precipitates out of the solution, and water.

The balanced chemical equation that describes this redox reaction looks like this

$8 {\text{SO"_ (2(aq)) + 16"H"_ 2"S"_ ((aq)) -> 3"S"_ (8(s)) darr + 16"H"_ 2"O}}_{\left(l\right)}$

The interesting thing to notice here is that this is actually a disproportionation reaction, i.e. a redox reaction in which the same species is being oxidized and reduced.

Here sulfur is being reduced from a $+ 4$ oxidation state in sulfur dioxide to a $0$ oxidation state in octasulfur, ${\text{S}}_{8}$. Sulfur is also being oxidized from a $- 2$ oxidation state in hydrogen sulfide to a $0$ oxidation state in octasulfur, hence why this reaction is classified as a disproportionation reaction.