In an experiment, 50 mL of 0.1 M solution of a metallic salt reacted exactly with 25 mL of 0.1 M solution of sodium sulphite. In the reaction, SO3− is oxidised to SO42−. If the original oxidation number of the metal in the salt was 3, what would be the new oxidation number of the metal?
SO32− is oxidised to SO42− (change in O.N =2)
25 mL of 0.1 M SO32− =2.5 millimoles =5.0 milliequivalents of SO32−=5.0 milliequivalents of M3+
50 mL of 0.1 M M3+ should be 1.
So, 5 millimoles = 5 milliequivalents. Thus, new O.N of metal = 2 as the metal is reduced by 1.