# Question #21c09

Jan 7, 2016

Thiosulfate ion, ${S}_{2} {O}_{3}^{2 -}$ is an interesting customer in the context of redox chemistry. The average oxidation number of sulfur in the anion is $I {I}^{+}$. Do you agree? Why?

#### Explanation:

In fact, one of the sulfurs has replaced an oxygen. In $S {O}_{4}^{2 -}$, sulfate anion, the oxidation number of sulfur is $V {I}^{+}$. We could designate the central sulfur in thiosulfate as $V {I}^{+}$ also, and the terminal sulfur has the SAME oxidation state as oxygen, i.e. $I {I}^{-}$. The average oxidation state of the sulfurs in thiosulfate is of course still $I {I}^{+}$.

Oxidation:

${S}_{2} {O}_{3}^{2 -} + 3 {H}_{2} O \left(l\right) \rightarrow 2 S {O}_{3}^{2 -} + 6 {H}^{+} + 4 {e}^{-}$ $\left(i\right)$

Reduction:

${S}_{2} {O}_{3}^{2 -} + 6 {H}^{+} + 4 {e}^{-} \rightarrow 2 S \downarrow + 3 {H}_{2} O$ $\left(i i\right)$

Both (i) and (ii) are balanced with respect to mass and charge (as they must be!), and, conveniently, we don't have to cross multiply because 4 electrons are involved in each reaction.)

Overall $\left(i\right) + \left(i i\right)$:

$2 {S}_{2} {O}_{3}^{2 -} \rightarrow 2 S {O}_{3}^{2 -} + 2 S \downarrow$

So, what do you observe? Treatment of thiosulfate with strong acid should result in a fine white precipitate of elemental sulfur (reduction), and also the formation of (acrid) ${H}_{2} S {O}_{3}$ (oxidation); clearly a disproportionation reaction has occurred.

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