Question

Thiosulfate ion $\left(S_2{O}_3^{2-}\right)$ disproportionates in acidic solution to give solid sulfur and aqueous hydrogen sulfite ion $\left({\mathrm{HSO}}_3^{-}\right)$. Write a balanced equation for this reaction.

Short answer

The balanced equation is:${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2 -}}{\text{+ 2H}}^{\text{+}}\to {\text{S + SO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}$

Step-by-step solution

Disproportionation reaction

Disproportionation reaction constitutes a reaction in whichthe compound gets subjected to oxidation and reduction. It comes under the category of redox reactions.

Writing a balanced equation for the reaction

Thiosulfate,${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2 -}}$has a formal${\text{S}}^{\text{2 -}}$and${\text{S}}^{\text{+ 6}}$oxidation states, i.e the average oxidation state is:

$$\begin{gathered} \text{Average oxidation state =}\frac{\text{- 2 + 6}}{\text{2}} \\ \text{= + 2} \end{gathered}$$

It is reduced to S with a zero oxidation state, i.e zerovalent sulfur, and oxidized to ${\text{SO}}_{\text{2}}$.

The reduction reaction is:

${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2 -}}{\text{+ 6H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2 S + 3H}}_{\text{2}}\text{O}$

The oxidation reaction is:

${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2 -}}{\text{+H}}_{\text{2}}\text{O}\to {\text{2 SO}}_{\text{2}}{\text{+ 2H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}$

As${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2 -}}$ is both oxidized and reduced in this reaction, this reaction is termed disproportionation reaction.

The balanced equation of the reaction is:

${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2 -}}{\text{+ 2H}}^{\text{+}}\to {\text{S + SO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}$