In a reduction-oxidation reaction, one species gains electrons while the other loses them. We can divide up the overall redox reactions into two half-reactions, one for the oxidation and one for the reduction. Balancing these half reactions is a bit tricky, but necessary to properly balance redox reactions.

One good way to proceed is the following 5 step method. First, identify what's being oxidized or reduced: it's the thing that's changing it's oxidation number.

1. Balance the atom being oxidized or reduced.
2. Balance the oxidation number by adding electrons. Electrons have a -1 oxidation number.
3. Balance the charge by adding H⁺ if the reaction is done in an acidic solution or by adding OH^- ions if in a base.
4. Balance the hydrogens by adding water molecules
5. Check to see if the oxygen atoms are balanced: if they are, you correctly balanced the 1/2 reaction.

An example or two hopefully will make this clearer. Let's balance the following reaction in acid:

Cu⁺²(s) -> Cu(s)

1. Balance the atom being reduced: 1 copper on each side, already balanced: Cu⁺²(s) -> Cu(s)
2. Balance the oxidation number: Cu⁺² is +2, Cu(s) is 0. Since electrons are -1, adding two to the left gives both sides 0:Cu⁺²(s) + 2e^- -> Cu(s)
3. Balance charge: 0 on both sides, already balanced: Cu⁺²(s) + 2e^- -> Cu(s)
4. Balance hydrogens: 0 on both sides, already balanced: Cu⁺²(s) + 2e^- -> Cu(s)
5. Check oxygens: 0 on both sides, already balanced: Cu⁺²(s) + 2e^- -> Cu(s)

A more complex example should be helpful. Let's balance the following reaction in acid

MnO₄^-(aq) -> Mn⁺²(aq)

1. Balance the atom being reduced: 1 Mn on each side, already balanced: MnO₄^-(aq) -> Mn⁺²(aq)
2. Balance the oxidation number: the Mn in the MnO₄^- has an oxidation number of +7, the Mn⁺² has an oxidation number of +2. (Remember how to compute oxidation number?) Add 5 electrons to the left side: MnO₄^-(aq) + 5e^- -> Mn⁺²(aq)
3. Balance charge: Total charge on the left is -6, on the right is +2. We're in acid, so add 8 H⁺ ions to the left to get +2 on each side: MnO₄^-(aq) + 5e^- +8H⁺(aq) -> Mn⁺²
4. Balance the hydrogens by adding water: 8 H on the left, none on the right, so add four water to the right: MnO₄^-(aq) + 5e^- +8H⁺(aq) -> Mn⁺²(aq) + 4H₂O(l)
5. Check the number of oxygens: there are four on both sides, so we're done

Example: Balance the following 1/2 reaction in acid

NO₃^-(g) -> NO(g)

Solution. The nitrogen is being reduced.

1. >Balance the atom being reduced: 1 N on each side, already balanced: NO₃^-(g) -> NO(g)
2. Balance the oxidation number: The nitrogen in the NO₃^- has an oxidation number of +5 and the nitrogen in the NO has an oxidation number of +2. Add 3 electrons to the left: NO₃^-(g) + 3e^- -> NO(g)
3. Balance the charge: the charge is -4 on the left, 0 on the right. We're in acid, so add 4H⁺ to the left: NO₃^-(g) + 3e^- + 4H⁺(aq) -> NO(g)
4. Balance the hydrogens: 4 on the left, none on the right, add 2 water to the right: NO₃^-(g) + 3e^- + 4H⁺(aq) -> NO(g) + 2H₂O(l)
5. Check oxygens: three on each side, so we're done

Information provided by: http://learn.chem.vt.edu