Balancing Redox Reactions

By assigning oxidation numbers to the reactants and products, one can determine how many moles of each species are required for conservation of charge and mass, which is necessary to balance the equation. In general, to balance a redox reaction, both the net charge and the number of atoms must be equal on both sides of the equation. The most common method for balancing redox equations is the half-reaction method, also known as the ion-electron method, in which the equation is separated into two half-reactions—the oxidation part and the reduction part. Each half-reaction is balanced separately, and they are then added to give a balanced overall reaction, in which electrons do not appear explicitly by convention. Consider a redox reaction between KMnO₄ and HI in an acidic solution:

MnO₄⁻ + I⁻ → I₂ + Mn²⁺

1. Step 1: Separate the two half-reactions.
2. oxidation half-reaction: MnO₄⁻ → Mn²⁺
3. reduction half-reaction: I⁻ → I₂
4. Step 2: Balance the atoms of each half-reaction. First, balance all atoms except H and O. Next, in an acidic solution, add H₂O to balance the O atoms and then add H⁺ to balance the H atoms. (In a basic solution, use OH⁻ and H₂O to balance the O’s and H’s.)
5. To balance the iodine atoms, place a coefficient of two before the I⁻ ion.

2I⁻ → I₂

1. For the permanganate half-reaction, Mn is already balanced. Next, balance the oxygens by adding 4H₂O to the right side.

MnO₄⁻ → Mn²⁺ + 4H₂O

1. Finally, add H⁺ to the left side to balance the 4H₂Os. These two half-reactions are now balanced in mass (but not in charge).

MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

1. Step 3: Balance the charges of each half-reaction. The reduction half-reaction must consume the same number of electrons as are supplied by the oxidation half. For the oxidation reaction, add 2 electrons to the right side of the reaction:

2I⁻ → I₂ + 2e⁻

1. For the reduction reaction, a charge of +2 must exist on both sides. Add 5 electrons to the left side of the reaction to accomplish this:

5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O

1. Step 4: Both half-reactions must have the same number of electrons so that they will cancel. Multiply the oxidation half by 5 and the reduction half by 2 and add the two:

5(2I⁻ → I₂ + 2e⁻)

2(5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O)

1. The final equation is:

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

1. To get the overall equation, cancel out the electrons and any H₂Os, H⁺s, OH⁻s, or e⁻s that appear on both sides of the equation.

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

1. Step 5: Finally, confirm that mass and charge are balanced. There is a +4 net charge on each side of the reaction equation, and the atoms are stoichiometrically balanced.

As you may have noticed, balancing redox equations can be trickier and more involved than other types of reactions because often one needs to supply additional chemical species like water and protons to the equation, rather than simply playing with stoichiometric coefficients. The above scheme is the most general one that can be applied; it may be that many intermediate steps can be omitted if the equation is simpler. For example, for the equation:

Zn (s) + HCl (aq) → ZnCl₂ (aq) + H₂ (g)

one may balance it by just supplying coefficients.