In the reaction of KMnO4and potassium iodide (KI) in an acidic medium, the KMnO4 acts as an oxidizing agent, and the iodide ions (I−) in KI act as reducing agents.
KMnO4+KI+H+⟶Mn+2+I2+H2O
In KMnO4, the oxidation state of manganese is +7, which is the highest possible oxidation state for manganese in this compound.
In the final product, MnSO4, the oxidation state of manganese is +2.
Therefore, the total change in the oxidation state of manganese is:
+7 (initial)−+2 (final) =5
Hence the change in O.S. of Mn is (5).