What is the change in oxidation number of sulfur when the dithionite anion, s2o42-, is converted to so2 in an oxidation-reduction reaction? does each sulfur atom gain or lose electrons in the reaction? is sulfur oxidized or reduced in the reaction?

1) In dithionite anion (S ₂O₄²⁻) sulfur has oxidation number + 3, because oxygen has oxidation number - 2, so:

2 · x + 4 · (-2) = - 2.

2x = + 6.

x = + 3; oxidation number of sulfur.

In sulfur (IV) oxide (SO₂), sulfur has oxidation number + 4, because oxygen is again - 2 and compound has neutral charge:

x + 2 · (-2) = 0.

x = + 4.

Change in sulfur is from + 3 to + 4.

Sulfur atoms lose electrons in the reaction. Sulfur is oxidized in the reaction, because his oxidation number raise from + 3 to + 4.