Question

Why does thiosulfate oxidize to sulfate in an acidic medium?

Answer 1

Presumably because oxidation to sulfate (`S(VI+)`) is the thermodynamic product under these conditions.

Explanation:

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

`"Oxidation:"`
`S_2O_3^(2-) + H_2O(l) rarr 2SO_2(g)+2H^(+) + 4e^-`

`"Reduction:"`
`S_2O_3^(2-) +6H^(+) + 4e^(-)rarr 2S(s)darr+3H_2O`

`"Overall:"`
`2S_2O_3^(2-) + 4H^(+) rarr 2SO_2(g) + 2S(s) + 2H_2O`

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 `"sulfate ion"`, would be the preferred outcome, i.e.

`"Oxidation:"`
`S_2O_3^(2-) + 5H_2O(l) rarr 2SO_4^(2-)+10H^(+) + 8e^-`

Note that I presume that sulphur dioxide `(S(IV+))` is fully oxidized to `S(VI+)` under the given conditions.

`"Overall:"`
`3S_2O_3^(2-) + 4H^(+) rarr 2SO_4^(2-) + 2S(s) + 2H_2O`

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