Question

How is `"thiosulfate"`, `S_2O_3^(2-)`, oxidized to `"dithionate"`, `S_4O_6^(2-)`?

Answer 1

`2S_2O_3^(2-) rarr S_4O_6^(2-) + 2e^-`

Explanation:

Is the reaction balanced with respect to mass and charge. If not, then we know that it is wrong.

In thiosulfate the average sulfur oxidation state is `+II`. In dithionate, the average oxidation state is `+II1/2` (the central sulfur linkage has oxidation state `0`, and the termini are `S(V)`.

We could assign individual sulfur oxidation numbers in dithionate as...`""^(-)O(O=)stackrel(+V)S-stackrel(0)S-stackrel(0)S-stackrel(+V)S(=O)O^(-)`. The internal sulfurs are conceived to share the electrons in the `S-S` bonds equally...and are thus formally zerovalent.