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. $S(VI+)+4xxO(-II)$...
