Question

What is the oxidation number change for the manganese atom in the following unbalanced reduction half reaction?

MnO₄⁻(aq) + H⁺(aq) → Mn²⁺(aq) + H₂O(l)


A) +5
B) -7
C) +7
D) -5

Answer 1

The oxidation number of manganese changes from +7 in MnO₄⁻ to +2 in Mn²⁺, which is a change of -5.

Step-by-step explanation:

The oxidation number change for the manganese atom in the unbalanced reduction half reaction MnO₄⁻(aq) + H⁺(aq) → Mn²⁺(aq) + H₂O(l) can be determined by examining the oxidation states of manganese in the reactant and product. In MnO₄⁻, each oxygen atom has an oxidation number of -2, and thus for the four oxygen atoms, the total is -8. Since the overall charge of permanganate (MnO₄⁻) is -1, the oxidation number of Mn must be +7 to balance the charge (-8 from oxygen and +1 to balance the -1 charge). Upon reduction, manganese changes to Mn²⁺ where it has an oxidation number of +2.

Therefore, the change in oxidation number for manganese is from +7 in MnO₄⁻ to +2 in Mn²⁺, which is a change of +7 to +2. The correct answer to the question is the oxidation number change for manganese, which is -5 (since it is a reduction, we subtract the higher oxidation number from the lower number, i.e., +2 - (+7) = -5).