Question

How does thiosulfate ion disproportionate to give sulfur and sulfur dioxide?

Answer 1

Well, we gots a formal disproportionation reaction....the protium ion is a reactant not a catalyst....

Explanation:

We got (I think) `Na_2S_2O_3`...i,e, sodium thiosulfate. Formally, we have `S_2O_3^(2-)`, for which the average sulfur oxidation state is `+II`. I like to think that one sulfide has replaced an oxide ion in `SO_4^(2-)` and we gots `stackrel(VI+)S`, and `stackrel(-II)S`...of course the average sulfur oxidation state is still `+II`.

So we write the oxidation reaction. `S(II+)rarrS(IV+)`:

`S_2O_3^(2-) +H_2O rarr 2SO_2(g)uarr+2H^+ +4e^(-)` `(i)`

Charge and mass are balanced so this is kosher. And now we write the reduction reaction of `S(II+)rarrS(0)`:

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

And we add `(i)+(ii)` together to eliminate the electrons....

`2S_2O_3^(2-) +4H^+ rarr 2S(s)darr+2SO_2(g)uarr+2H_2O`

...or more simply...

`2H_2S_2O_3 rarr 2S(s)darr+2SO_2(g)uarr+2H_2O`

...and this is balanced with respect to mass and charge...as required.

I will have to think on the latter question you asked...