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 (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.
