# Why does thiosulfate oxidize to sulfate in an acidic medium?

Feb 9, 2017

Presumably because oxidation to sulfate ($S \left(V I +\right)$) is the thermodynamic product under these conditions.

#### Explanation:

Sodium thiosulfate is known to undergo disproportionation in acidified aqueous solution:

$\text{Oxidation:}$
${S}_{2} {O}_{3}^{2 -} + {H}_{2} O \left(l\right) \rightarrow 2 S {O}_{2} \left(g\right) + 2 {H}^{+} + 4 {e}^{-}$

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

$\text{Overall:}$
$2 {S}_{2} {O}_{3}^{2 -} + 4 {H}^{+} \rightarrow 2 S {O}_{2} \left(g\right) + 2 S \left(s\right) + 2 {H}_{2} O$

Now this is a balanced chemical equation, but the equation should conform to reality, not vice versa. Under these conditions, I presume, that oxidation to $\text{sulfate ion}$, would be the preferred outcome, i.e.

$\text{Oxidation:}$
${S}_{2} {O}_{3}^{2 -} + 5 {H}_{2} O \left(l\right) \rightarrow 2 S {O}_{4}^{2 -} + 10 {H}^{+} + 8 {e}^{-}$

Note that I presume that sulphur dioxide $\left(S \left(I V +\right)\right)$ is fully oxidized to $S \left(V I +\right)$ under the given conditions.

$\text{Overall:}$
$3 {S}_{2} {O}_{3}^{2 -} + 4 {H}^{+} \rightarrow 2 S {O}_{4}^{2 -} + 2 S \left(s\right) + 2 {H}_{2} O$

So I would presume that the answer your question is that under these conditions oxidation of sulfur to sulfate is the thermodynamic outcome.