# How to Balance Redox Reactions

## Keeping atoms and charges in balance This is a diagram that describes the half-reactions of a redox reaction or oxidation-reduction reaction. Cameron Garnham, Creative Commons License

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)

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