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

1 Answer
Jun 4, 2017

Answer:

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)#...