How to Balance Redox Reactions

To balance redox reactions, assign oxidation numbers to the reactants and products to determine how many moles of each species are needed to conserve mass and charge. First, separate the equation into two half-reactions, the oxidation portion and the reduction portion. This is called the half-reaction method of balancing redox reactions or the ion-electron method. Each half-reaction is balanced separately and then the equations are added together to give a balanced overall reaction. We want the net charge and number of ions to be equal on both sides of the final balanced equation.

For this example, let's consider a redox reaction between KMnO₄and HI in an acidic solution:

MnO₄^- + I^- → I₂ + Mn²⁺

 Separate the two half reactions:

I^- → I₂

MnO₄^- → Mn²⁺

 To balance the atoms of each half-reaction, first balance all of the atoms except H and O. For an acidic solution, next add H₂O to balance the O atoms and H⁺ to balance the H atoms. In a basic solution, we would use OH^- and H₂O to balance the O and H.

Balance the iodine atoms:

2 I^- → I₂

The Mn in the permanganate reaction is already balanced, so let's balance the oxygen:

MnO₄^- → Mn²⁺ + 4 H₂O

Add H⁺ to balance the 4 waters molecules:

MnO₄^- + 8 H⁺ → Mn²⁺ + 4 H₂O

The two half-reactions are now balanced for atoms:

MnO₄^- + 8 H⁺ → Mn²⁺ + 4 H₂O

 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)

 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

 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.