Question

How does `"thiosulfate ion"`, `S_2O_3^(2-)` give rise to sulfur and sulfur dioxide upon disproportionation?

Answer 1

This is a disproportionation reaction........

`S_2O_3^(2-) +2H^(+) rarr S +SO_2+H_2O`

Explanation:

Thiosulfate, `S_2O_3^(2-)` has formal `stackrel(-II)S` and `stackrel(+VI)S` oxidation states, i.e. an average oxidation state of `(-II+VI)/2=+II`.

It is reduced to `stackrel(0)S`, i.e. zerovalent sulfur, and oxidized to `stackrel(+IV)"SO"_2`.

And thus reduction............:

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

And oxidation............:

`S_2O_3^(2-) +H_2O rarr 2SO_2 +2H^(+)+4e^(-)` `(ii)`

For both `(i)` and `(ii)`, charge and mass are balanced as is absolutely required. Are they balanced? Don't trust my arithmetic.

We add `(i) + (ii)` to remove the electrons...........

`2S_2O_3^(2-) +4H^(+) rarr 2S +2H_2O +2SO_2`

And of course we could halve this, as per Nam D.'s suggestion..

`underbrace(S_2O_3^(2-) +2H^(+))_("sulfurous acid"*H_2SO_3) rarr S +H_2O +SO_2`

Again, if mass and charge ARE NOT BALANCED, then we CANNOT accept it as a representation of chemical reality. Looking at it again, we gots both LHS and RHS neutral, LHS and RHS, `2xxS`, `3xxO`, and `2xxH`.........so balanced with respect to mass and charge.

What would you observe in this reaction? Well, probably (i) the precipitate of a fine white powder of elemental sulfur, and (ii) maybe the foul odour of `SO_2`.

Just as a comment, I like to think of sulfurous acid as the SULFUR analogue of sulfuric acid, where ONE sulfur has replaced ONE oxygen to give `H_2S_2O_3`. And this sulfur assumes THE SAME oxidation state as the oxygen it replaces, i.e. `S(-II)`. And so we got `S(VI+)` and `S(-II)`...the average oxidation state is still `S(+II)`. And we compare this with `SO_4^(2-)`, i.e. `S(VI+)+4xxO(-II)`...