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

1 Answer

Answer:

#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.