1. Balancing Redox reactions in an acid or a base

2. Example In an acidic solution MnO4- + H2O2 → Mn2+ + O2 Half reactions
MnO4- → Mn2+
H2O2 → O2

3. Top Equation MnO4- → Mn2+ MnO4- → Mn2+ + 4 H2O
MnO H+→ Mn H2O
MnO H++ 5 e-→ Mn H2O

4. Bottom Equation H2O2 → O2 H2O2 → O2 + 2 H+ H2O2 → O2 + 2 H+ + 2 e-
I need to equal 5 e- so…
That won’t work…
2MnO H++ 10 e-→ 2 Mn H2O
5 H2O2 → 5 O H e-

5. Add them together 2MnO4- + 16 H++ 10 e-→ 2 Mn2+ + 8 H2O
5 H2O2 → 5 O H e-
And you get
2 MnO H++ 5 H2O2
→ 2 Mn O2 + 8 H2O
Notice the H+ canceled out as well.

6. Balancing Redox Equations in a basic solution
Follow all rules for an acidic solution.
After you have completed the acidic reaction add OH- to each side to neutralize any H+.
Combine OH- and H+ to make H2O.
Cancel out any extra waters from both sides of the equation.

7. Example We will use the same equation as before In a basic solution
MnO4- + H2O2 → Mn2+ + O2
2 MnO H++ 5 H2O2
→ 2 Mn O2 + 8 H2O

8. Basic solution Since this is a basic solution we can’t have excess H+.
We will add OH- to each side to neutralize all H+
2 MnO H++ 5 H2O2 + 6OH-
→ 2 Mn O2 + 8 H2O + 6OH-
We added 6 OH- because there were 6H+

9. Cont.
H+ + OH- → H2O
Combine the hydroxide and hydrogen on the reactant side to make water
2 MnO H2O + 5 H2O2
→ 2 Mn O2 + 8 H2O + 6OH-
Cancel out waters on both sides
2 MnO H2O2
→ 2 Mn O2 + 2 H2O + 6OH-

10. Another example In a basic solution MnO4 − + SO32-→MnO4 2− + SO42-
Half reactions
MnO4 − → MnO4 2−
SO32-→ SO42-

11. Half reactions MnO4 − → MnO4 2− MnO4 - + e- → MnO4 2− SO32-→ SO42-
H2O + SO32-→ SO42-
H2O + SO32-→ SO H+
H2O + SO32-→ SO H+ +2e-
Double the top reaction

12. 2 MnO e- → 2 MnO4 2−
H2O + SO32-→ SO H+ +2e-
Combine them
2 MnO4 - + H2O + SO32-
→ 2 MnO4 2− +SO H+
Add OH-
2 MnO4 - + H2O + SO OH-
→ 2 MnO4 2− +SO H+ + 2 OH-

13. 2 MnO4 - + H2O + SO OH-
→ 2 MnO4 2− +SO H2O
finishing
2 MnO4 - + SO OH-
→ 2 MnO4 2− +SO42- + H2O