Oxidation numbers method of balancing redox reaction equations

Before we will get to explanation very important disclaimer: oxidation numbers don't exist. They were invented to help in charge accounting needed when balancing redox reaction equations, but they don't refer to any real life chemical concept.

The general idea behind the oxidation numbers (ON) method for balancing chemical equations is that electrons are transferred between charged atoms. These charges - assigned to individual atoms - are called oxidation numbers, just to remind you that they don't reflect real structure of the reagents.

There are several simple rules used for assigning oxidation numbers to every atom present in any compound:

• First of all, charged mono atomic ion has oxidation number equal to its charge. Thus Na⁺ has oxidation number +1, Fe³⁺ has oxidation number +3, F^- has oxidation number of -1 and S^2- has oxidation number of -2.
• Second rule says that the oxidation number of a free element is always 0. Thus oxidation number of solid, metallic Cu is 0, oxidation number of O in O₂ is 0, the same holds for S in S₈ and so on.
• Oxygen in almost all compounds has oxidation number -2.
• Hydrogen in almost all compounds has oxidation number +1.
• Some elements usually have the same oxidation number in their compounds:
• alkali metals - Li, Na, K, Rb, Cs - oxidation numbers are +1
• alkaline earth metals - Be, Mg, Ca, Sr, Ba - oxidation numbers are +2
• halogens (except when they form compounds with oxygen or one another) - oxidation numbers are -1 (always true for fluorine)
• Last rule says that the charge of the ion or molecule equals sum of oxidation numbers of all atoms.

There are some exceptions to the rules 3 and 4 - for example oxygen in peroxides has oxidation number of -1, it is also not -2 in compounds with fluorine (where F is always -1), hydrogen in hydrides has oxidation number -1.

Before we will try to balance any equations let's use above rules to assign oxidation numbers to atoms in several substances.

For example - what is oxidation number of sulfur in SO₂? Particle is not charged, so oxidation number of sulfur must equal sum of oxidation numbers of oxygens, but with the opposite sign. Oxygen oxidation number is -2, there are two oxygens - that gives -4 together, so sulfur must have ON=+4.

What is oxidation number of atoms in CrO₄^2-? Oxygen is -2 and there are 4 oxygens - that gives overall of -8, ion has charge of -2, so central atom must have ON=+6.

How do we use oxidation numbers for balancing? First of all, we have to understand that oxidation means increase of oxidation number, while reduction means decrease of oxidation number. In both cases change of oxidation number is due to electrons lost (oxidation) or gained (reduction). We calculate oxidation numbers for all atoms present in the reaction equation (note that it is not that hard as it sounds, as for most atoms oxidation numbers will not change) and we look for a ratio that makes the number of electrons lost equal to the number of electrons gained. That gives us additional information needed for reaction balancing.

Let's try with following reaction:

KIO₃ + KI + H₂SO₄ → K₂SO₄ + H₂O + I₂

First of all - we don't need any spectators here, as they are only making things look more difficult then they are in reality. Quick glance tells us that the net ionic reaction is

IO₃^- + I^- + H⁺ → H₂O + I₂

Looks like IO₃^- is oxidizing agent here and I^- is reducting agent. I^- has oxidation number of -1, iodine in IO₃^- has oxidation number of +5. On the right side in I₂ both iodine atoms have oxidation number 0. It means that iodine in IO₃^- must gain 5 electrons. These electrons come from I^- - one for every I^- ion. Assuming (just like we do in the inspection method) that IO₃^- is the most complicated molecule and it's coefficient is 1 we will need five I^- for the redox process to complete:

1IO₃^- + 5I^- + H⁺ → H₂O + I₂

Now that the ratio between oxidizer and reducing agent is known we use simple techniques we know from the inspection method to balance remaining elements. There are six atoms of iodine on the left, so we need three I₂ molecules to balance iodine:

1IO₃^- + 5I^- + H⁺ → H₂O + 3I₂

And the final, trivial step is balancing oxygen, hydrogen and water:

IO₃^- + 5I^- + 6H⁺ → 3H₂O + 3I₂

Other case we can try is oxidation of Mn²⁺ with NaBiO₃ in acidic conditions:

Mn²⁺ + BiO₃^- + H⁺ → MnO₄^- + Bi³⁺ + H₂O

Using methods for oxidation numbers calculation we can easily check that manganese is oxidized from +2 to +7 (freeing five electrons) and bismuth is reduced from +5 to +3 (accepting two electrons). To balance electrons transferred we can put coefficients 2 and 5 on the left side of reaction equation:

2Mn²⁺ + 5BiO₃^- + H⁺ → MnO₄^- + Bi³⁺ + H₂O

Rest can be balanced by inspection and is not difficult to do, yielding:

2Mn²⁺ + 5BiO₃^- + 14H⁺ → 2MnO₄^- + 5Bi³⁺ + 7H₂O

Now the same equation can be also easily balanced as a full (non net-ionic) version:

4MnSO₄ + 10NaBiO₃ + 14H₂SO₄ → 4NaMnO₄ + 5Bi₂(SO₄)₃ + 14H₂O + 3Na₂SO₄