Oxidation Numbers

Oxidation numbers

By giving elements an oxidation number, it is possible to keep track of whether that element is losing or gaining electrons during a chemical reaction. The loss of electrons in one part of the reaction must be balanced by a gain of electrons in another part of the reaction.

Definition: Oxidation number

Oxidation number is the charge an atom would have if it was in a compound composed of ions.

There are a number of rules that you need to know about oxidation numbers, and these are listed below.

A molecule consisting of only one element always has an oxidation number of zero, since it is neutral.

For example the oxidation number of hydrogen in \(\text{H}_{2}\) is \(\text{0}\). The oxidation number of bromine in \(\text{Br}_{2}\) is also \(\text{0}\).
Monatomic ions (ions with only one element or type of atom) have an oxidation number that is equal to the charge on the ion.

For example, the chloride ion \(\text{Cl}^{-}\) has an oxidation number of \(-\text{1}\), and the magnesium ion \(\text{Mg}^{2+}\) has an oxidation number of \(\text{+2}\).
In a molecule or compound, the sum of the oxidation numbers for each element in the molecule or compound will be zero.

For example the sum of the oxidation numbers for the elements in water will be \(\text{0}\).
In a polyatomic ion the sum of the oxidation numbers is equal to the charge.

For example the sum of the oxidation numbers for the elements in the sulfate ion (\(\text{SO}_{4}^{2-}\)) will be \(-\text{2}\).
An oxygen atom usually has an oxidation number of \(-\text{2}\). One exception is in peroxides (e.g. hydrogen peroxide) when oxygen has an oxidation number of \(-\text{1}\).

For example oxygen in water will have an oxidation number of \(-\text{2}\) while in hydrogen peroxide (\(\text{H}_{2}\text{O}_{2}\)) it will have an oxidation number of \(-\text{1}\).
The oxidation number of hydrogen is often \(\text{+1}\). One exception is in the metal hydrides where the oxidation number is \(-\text{1}\).

For example the oxidation number of the hydrogen atom in water is \(\text{+1}\), while the oxidation number of hydrogen in lithium hydride (\(\text{LiH}\)) is \(-\text{1}\).
The oxidation number of fluorine is \(-\text{1}\).

Tip:

You will notice that some elements always have the same oxidation number while other elements can change oxidation numbers depending on the compound they are in.

Example 1: Oxidation Numbers

Question

Give the oxidation number of sulfur in a sulfate (\(\text{SO}_{4}^{2-}\)) ion

Step 1: Determine the oxidation number for each atom

Oxygen will have an oxidation number of \(-\text{2}\). (Rule \(\text{5}\), this is not a peroxide.) The oxidation number of sulfur at this stage is uncertain since sulfur does not have a set oxidation number.

Step 2: Determine the oxidation number of sulfur by using the fact that the oxidation numbers of the atoms must add up to the charge on the compound

In the polyatomic \(\text{SO}_{4}^{2-}\) ion, the sum of the oxidation numbers must be \(-\text{2}\) (rule \(\text{4}\)).

Let the oxidation number of sulfur be \(x\). We know that oxygen has an oxidation number of \(-\text{2}\) and since there are four oxygen atoms in the sulfate ion, then the sum of the oxidation numbers of these four oxygen atoms is \(-\text{8}\).

Putting this together gives: \begin{align*} x + (-8) & = -2 \\ x & = -2 + 8 \\ & = +6 \end{align*} So the oxidation number of sulfur is \(\text{+6}\).

Step 3: Write down the final answer

In the sulfate ion, the oxidation number of sulfur is \(\text{+6}\).

Example 2: Oxidation Numbers

Question

Give the oxidation number of both elements in ammonia (\(\text{NH}_{3}\)).

Step 1: Determine the oxidation number for each atom

Hydrogen will have an oxidation number of \(\text{+1}\) (rule 6, ammonia is not a metal hydride). At this stage we do not know the oxidation number for nitrogen.

Step 2: Determine the oxidation number of nitrogen by using the fact that the oxidation numbers of the atoms must add up to the charge on the compound

In the compound \(\text{NH}_{3}\), the sum of the oxidation numbers must be \(\text{0}\) (rule \(\text{3}\)).

Let the oxidation number of nitrogen be \(x\). We know that hydrogen has an oxidation number of \(\text{+1}\) and since there are three hydrogen atoms in the ammonia molecule, then the sum of the oxidation numbers of these three hydrogen atoms is \(\text{+3}\).

Putting this together gives: \begin{align*} x + (+3) & = 0 \\ & = -\text{3} \end{align*} So the oxidation number of nitrogen is \(-\text{3}\).

Step 3: Write the final answer

Hydrogen has an oxidation number of \(\text{+1}\) and nitrogen has an oxidation number of \(-\text{3}\).

Example 3: Oxidation Numbers

Question

Give the oxidation numbers for all the atoms in sodium chloride (\(\text{NaCl}\)).

Step 1: Determine the oxidation number for each atom in the compound

This is an ionic compound composed of \(\text{Na}^{+}\) and \(\text{Cl}^{-}\) ions. Using rule 2 the oxidation number for the sodium ion is \(\text{+1}\) and for the chlorine ion it is \(-\text{1}\).

This then gives us a sum of \(\text{0}\) for the compound.

Step 2: Write the final answer

The oxidation numbers for sodium is \(\text{+1}\) and for chlorine it is \(-\text{1}\).

Oxidation Numbers