If the reactant is MnO 2 and the product is MnO 4 - , calculate the change in th
ID: 825315 • Letter: I
Question
If the reactant is MnO2 and the product is MnO4-, calculate the change in the oxidation state of manganese (oxidation state of manganese in product minus the oxidation state of manganese in the reactant). Enter the sign and magnitude of the result, and separated by a comma, also enter O or R to indicate whether manganese is oxidized (O) or reduced (R). Example: -2,O.
If the product is S and the reactant is H2SO3, calculate the change in the oxidation state of sulfur (oxidation state of sulfur in product minus the oxidation state of sulfur in the reactant). Enter the sign and magnitude of the result, and separated by a comma, also enter the number of electrons gained or lost (no signs) by an atom of sulfur. Example: -2,2.
If the reactant is Ce4+ and the product is Ce3+, calculate the change in the oxidation state of cerium (oxidation state of cerium in product minus the oxidation state of cerium in the reactant). Enter the sign and magnitude of the result, and separated by a comma, also enter the number of electrons gained or lost (no signs) by an atom of cerium. Example: -2,2.
Explanation / Answer
MnO2 ---------> MnO4-
oxidation number of Mn in MnO2 = +4
oxidation number of Mn in MnO4- = +7
therefore change in oxidation state = 7-4 = +3
definitely mangenese is oxidized as its oxidation state increases
S --------------> H2SO3
oxidation number of S in S = 0
oxidation number of S in H2SO3 = +4
therefore change in oxidation state = 4-0 = +4
definitely sulfur is oxidized as its oxidation state increases
Ce4+ -------------> Ce3+
oxidation number of Ce in Ce4+ = +4
oxidation number of Ce in Ce3+ = +3
therefore change in oxidation state = 3-4 = -1
definitely cerium is reduced as its oxidation state decreases