Balance the following half reaction in acid solution: MnO₄^- $\to$ Mn²⁺ (aq)

Question

Quizplus · Accepted Answer

The oxidation state of S changes from -1 to 0, so this is the half-reaction for the reduction half. To balance the electrons, 5 electrons must be added to the left side:

S1U1P1-S1S1P0 + 5 eS1U1P1-S1S1P0 → S1U1P0

In the oxidation half, Mn changes from +7 to +2. However, the reaction takes place in an acidic solution, so H+ ions must be added to balance the equation. The final balanced half-reaction is:

MnO4- + 8 H+ + 5 e- → Mn2+ + 4 H2O

Multiplying the reduction half by 3 and adding it to the oxidation half yields the final balanced equation:

3 MnO4- + 24 H+ + 5 S1U1P1-S1S1P0 → 3 Mn2+ + 5 S1U1P0 + 12 H2O

Balance the Following Half Reaction in Acid Solution: MnO4- →\to→ Mn2+ (Aq) A)MnO4- → Mn2+ + 3 E- B)MnO4-

Balance the Following Half Reaction in Acid Solution: MnO₄^- $\to$ Mn²⁺ (Aq)
A)MnO₄^- → Mn²⁺ + 3 E^-
B)MnO₄^-