In an experiment, 50.0 cm3 of a 0.10 mol dm–3 solution of a metallic salt reacted exactly with
25.0cm3 of 0.10moldm–3 aqueous sodium sulphite. The half-equation for oxidation of sulphite ion is shown below.
SO3^2− (aq) + H2O(I) → SO4^2− (aq) + 2H+(aq) + 2e–
If the original oxidation number of the metal in the salt was +3, what would be the new oxidation
number of the metal?
25.0cm3 of 0.10moldm–3 aqueous sodium sulphite. The half-equation for oxidation of sulphite ion is shown below.
SO3^2− (aq) + H2O(I) → SO4^2− (aq) + 2H+(aq) + 2e–
If the original oxidation number of the metal in the salt was +3, what would be the new oxidation
number of the metal?