In the MnO4− reduction half-reaction, how many H+ are involved?

In the MnO4− reduction half-reaction, how many H+ are involved?

• A. 4
• B. 8
• C. 10
• D. 12

Answer: (B)

Balancing a reduction half-reaction in acidic solution uses H+ to supply the hydrogens after fixing oxygen with water. For MnO4− to Mn2+, manganese drops from +7 to +2, so 5 electrons are involved. Balance manganese with one Mn on each side, then balance oxygen by putting 4 H2O on the product side. Those 4 H2O bring in 8 hydrogens, so add 8 H+ to the reactant side. Finally, add the 5 electrons to the reactant side to balance charge: MnO4− + 8 H+ + 5 e− → Mn2+ + 4 H2O. Thus eight H+ participate.

Balancing a reduction half-reaction in acidic solution uses H+ to supply the hydrogens after fixing oxygen with water. For MnO4− to Mn2+, manganese drops from +7 to +2, so 5 electrons are involved. Balance manganese with one Mn on each side, then balance oxygen by putting 4 H2O on the product side. Those 4 H2O bring in 8 hydrogens, so add 8 H+ to the reactant side. Finally, add the 5 electrons to the reactant side to balance charge: MnO4− + 8 H+ + 5 e− → Mn2+ + 4 H2O. Thus eight H+ participate.