How many electrons are needed for manganese in the balanced oxidation half-reaction? MnO4- (aq) + H+ (aq) → Mn2+ (aq)+H2O(l)?

5 electrons are needed.


Method 1 :
Balance the "substances" of the half equation :
MnO₄⁻(aq) + 8H⁺(aq) → Mn²⁺(aq) + 4H₂O(l)

Total charges on the left = (1-) + 8(1+) = 7+
Total charges on the right = 2+

To balance the charges on the both sides, 5e⁻ should be added on the left.


Method 2 :
Oxidation number of Mn in MnO₄⁻ = +7
Oxidation number of Mn in Mn²⁺ = +2

Change in oxidation number = (+2) - (+7) = -5
Hence, 5e⁻ should be added on the left.

Mn goes from a oxidation state of +7 (in MnO4^-1) to +2 (in Mn^+2) requiring 5 electrons.
Mn^+7 + 5 e --> Mn^+2
+7 +(-5) --> +2