To balance redox reactions, you must assign oxidation numbers to the reactants and products to determine how many moles of each species are needed to conserve mass and charge.
The Half-Reaction Method
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 Reactions
Separate the two half-reactions:
I- → I2
MnO4- → Mn2+
Balance the Atoms
To balance the atoms of each half-reaction, first balance all of the atoms except H and O. For an acidic solution, next add 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 water molecules:
MnO4- + 8 H+ → Mn2+ + 4 H2O
The two half-reactions are now balanced for atoms:
MnO4- + 8 H+ → Mn2+ + 4 H2O
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
Next, multiply the oxidation numbers so that the two half-reactions have the same number of electrons and can cancel each other out:
5(2I- → I2 +2e-)
2(5e- + 8H+ + MnO4- → Mn2+ + 4H2O)
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 equation:
10 I- + 10 e- + 16 H+ + 2 MnO4- → 5 I2 + 2 Mn2+ + 10 e- + 8 H2O
Simplify 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 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.
In summary:
- Step 1: Break reaction into half-reactions by ions.
- Step 2: Balance the half-reactions stoichiometrically by adding water, hydrogen ions (H+) and hydroxyl ions (OH-) to the half-reactions.
- Step 3: Balance the half-reactions charges by adding electrons to the half-reactions.
- Step 4: Multiply each half-reaction by a constant so both reactions have the same number of electrons.
- Step 5: Add the two half-reactions together. The electrons should cancel out, leaving a balanced complete redox reaction.