How to Balance Redox Reactions

When faced with a redox reaction, sadly, you can’t rely on the atoms and molecules alone to guide you in balancing. There will often be multiple solutions that appear to be right, yet don’t reflect chemical reality. To make sure that you get the balancing right, you have to look at the electrons as well. That makes sense, as a redox reaction is one where electrons are transferred as a part of the reaction.

The way we typically “look” at electrons in a redox reaction is by assigning oxidation numbers to all the atoms in the unbalanced chemical equation. (If you do not know how to assign oxidation numbers, see the article “Assigning Oxidation Numbers”.) If the oxidation number of an element changes from one side of the equation to the other, then we know it underwent oxidation or reduction.

Remember:
in oxidation an atom loses electrons, so it becomes more positive (or less negative)
in reduction an atom gains electrons, so it becomes more negative (or less positive)

Once the oxidized and reduced species are identified, you then break the reaction into two half-reactions, one representing the oxidation, one for the reduction. In each half-reaction, the electrons are included as either a reactant (reduction) or a product (oxidation). It may be necessary to add water, hydrogen or hydroxide ions to make the half reactions balance properly if the reaction occurs in an aqueous acid or base solution.

The key is to balance out the electrons, so find the least common multiple (LCM) for the numbers of electrons. Then multiply each half-reaction by whatever number is necessary to make the numbers of electrons equal. (Only coefficients in the equations are affected by the multiplication, chemical formulae are not.)

Once the electrons are balanced, the two half-reactions are then added together to make one, balanced equation. The electrons and any other identical items (such as water) on both sides of the equation cancel out, leaving the final answer.

EXAMPLES:

Example #1

Al + MnO4-1 MnO2 + Al(OH)4-1 [in basic solution]

Assign oxidation numbers. From left to right, Al = 0, Mn = +7, O = -2, Mn = +4, O = -2, Al = +3, O = -2, H = +1

Look for changes. Al went from 0 to +3 (oxidation), Mn from +7 to +4 (reduction).

Write half reactions. e = electrons

Al Al(OH)4-1 + 3e
MnO4-1 + 3e MnO2

Add in OH-1 and H2O to balance. (hydroxide, because the solution is basic)

Al + 4 OH-1 Al(OH)4-1 + 3e
MnO4-1 + 2 H2O + 3e MnO2 + 4 OH-1

Check whether the electrons are equal in the two reactions – they are.

Al + 4 OH-1 + MnO4-1 + 2 H2O + 3e Al(OH)4-1 + 3e + MnO2 + 4 OH-1

Cancel anything appearing on both sides (4 OH-1, 3e)

Al + MnO4-1 + 2 H2O Al(OH)4-1 + MnO2

Voila, the reaction is balanced.

Example #2

Cl2 Cl-1 + ClO-1 [in basic solution]

Oxidation numbers in order:
Cl = 0, Cl = -1, Cl = +1, O = -2

Half reactions:

Cl2 + e Cl-1
Cl2 ClO-1 + e

Balance half reactions:

Cl2 + 2e 2 Cl-1
Cl2 + 4 OH-1 2 ClO-1 + 2 H2O + 2e

Check electrons – they match.
Recombine equations.

2 Cl2 + 2e + 4 OH-1 2 Cl-1 + 2 ClO-1 + 2 H2O + 2e

Cancel electrons and reduce.

Cl2 + 2 OH-1 Cl-1 + ClO-1 + H2O

Example #3

As2O3 + NO3-1 H3AsO4 + NO [in acidic solution]

Oxidation Numbers (in order):
As = +3, O = -2, N = +5, O = -2, H = +1, As = +5, O = -2, N = +2, O = -2

As and N change.

As2O3 H3AsO4 + 2e
NO3-1 + 3e NO

Add H+1 and H2O to balance (acidic solution)

As2O3 + H2O + H+1 H3AsO4 + 2e
NO3-1 + 3e + 4 H+1 NO + 2 H2O

Multiply to get 6e in each equation.

[As2O3 + H2O + H+1 H3AsO4 + 2e] x3
[NO3-1 + 3e + 4 H+1 NO + 2 H2O] x2

3 As2O3 + 3 H2O + 3 H+1 3 H3AsO4 + 6e
2 NO3-1 + 6e + 8 H+1 2 NO + 4 H2O