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

1 Answer
Jun 4, 2017

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)...