How to Balance Redox Reactions - Balancing Redox Reactions

2 I^- → I₂ + 2e^-

5 e^- + 8 H⁺ + MnO₄^- → Mn²⁺ + 4 H₂O

Now multiple the oxidations numbers so that the two half-reactions will have the same number of electrons and can cancel each other out:

5(2I^- → I₂ +2e^-)

2(5e^- + 8H⁺ + MnO₄^- → Mn²⁺ + 4H₂O)

• Balancing Redox Reactions - Half-Reaction Method
• Balancing Redox Reactions - Separate the Reactions
• Balancing Redox Reactions - Balance the Atoms
• Balancing Redox Reactions - Balance the Charge
• Balancing Redox Reactions - Add the Half-Reactions
• Balancing Redox Reactions - Check Your Work