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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?

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1) In dithionite anion (S₂O₄²⁻) 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.

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