Question

Question #21c09

Answer 1

Thiosulfate ion, `S_2O_3^(2-)` is an interesting customer in the context of redox chemistry. The average oxidation number of sulfur in the anion is `II^+`. Do you agree? Why?

Explanation:

In fact, one of the sulfurs has replaced an oxygen. In `SO_4^(2-)`, sulfate anion, the oxidation number of sulfur is `VI^+`. We could designate the central sulfur in thiosulfate as `VI^+` also, and the terminal sulfur has the SAME oxidation state as oxygen, i.e. `II^-`. The average oxidation state of the sulfurs in thiosulfate is of course still `II^+`.

Anyway, to your problem:

Oxidation:

`S_2O_3^(2-) + 3H_2O(l) rarr 2SO_3^(2-) + 6H^+ + 4e^-` `(i)`

Reduction:

`S_2O_3^(2-) + 6H^+ +4 e^(-) rarr 2Sdarr + 3H_2O` `(ii)`

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 `(i)+(ii)`:

`2S_2O_3^(2-) rarr 2SO_3^(2-) + 2Sdarr`

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_2SO_3` (oxidation); clearly a disproportionation reaction has occurred.

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