Need help with a half equation?

Question

I need the half equations (oxidation & reduction) for the reaction between the thiosulfate anion and H+ protons:

S2O32−(aq) + 2H+(aq) → SO2(g) + S(s) + H2O

I need a separate oxidation and reduction equation please

Answer 1

S2O32-  (S +2)   + 4e-  ------> 2S (S  0)

S2O32-  (S +2)  ------------> 2SO2 ( S  +4)  +  4e-

1° half equation

S2O32-   +  4e-  +  6H+  ------->  2S  +  3H2O

2° half equation  

S2O32-  +  H2O  --------->  2SO2   +  2H+

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1°+2°  =

2S2O32-   +  4H+  ------------>  2SO2  +  2S  +  2H2O

then  S2O32-  +  2H+  ----------> SO2  +  S  +  H2O

Answer 2

Actually, this is not an oxidation reduction involving H+ at all. H is in the +1 oxidation state on both sides of the equation.

If you subtract S from S2O3 2-, you have SO3 2-. This is the anion of a weak and unstable acid, which reacts with H+ to give H2SO3 and hence SO2 + H2O (a bit like what happens when you acidify a carbonate).

It is usual to assign oxidation state +6 to the central S in S2O3 2-, by analogy with sulphate, and oxidation state -2 to the peripheral S, again just like the peripheral oxygens in sulphate. If you follow this convention, you will see that there has indeed been an oxidation reduction process taking place, but one involving the two sulfur atoms only.