How to Balance Redox Reactions - Balancing Redox Reactions

Balancing Redox 4 of 6 Reactions - Balance the Charge

Next, balance the charges in each half-reaction so that the reduction half-reaction consumes the same number of electrons as the oxidation half-reaction supplies. This is accomplished by adding electrons to the 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 5 of 6 Reactions - Add the Half-Reactions

Now add the two half-reactions:

10 I^- → 5 I₂ + 10 e^-

16 H⁺ + 2 MnO₄^- + 10 e^- → 2 Mn²⁺ + 8 H₂O

This yields the following final equation:

10 I^- + 10 e^- + 16 H⁺ + 2 MnO₄^- → 5 I₂ + 2 Mn²⁺ + 10 e^- + 8 H₂O

Get the overall equation by canceling out the electrons and H₂O, H⁺, and OH^- that may appear on both sides of the equation:

10 I^- + 16 H⁺ + 2 MnO₄^- → 5 I₂ + 2 Mn²⁺ + 8 H₂O

Balancing Redox 6 of 6 Reactions - Check Your Work

Check your numbers to make certain that the mass and charge are balanced. In this example, the atoms are now stoichiometrically balanced with a +4 net charge on each side of the reaction.