In chemistry, an oxidation number is a label given to an atom in a chemical that shows its tendency to oxidize or reduce. A chemical with a high oxidation number is likely to be reduced, and a chemical with a low oxidation number is likely to be oxidized. The magnitude of a chemical's oxidation number is rarely higher than \(7\). In this article we look at how to find oxidation numbers of atoms within a chemical. This is helpful because in a redox reaction, we can find the oxidation number of the chemical being oxidized/reduced on both sides of the equation and determine if the chemical is getting oxidized or reduced.
Here are the first and easiest rules concerning oxidation numbers:
The oxidation number of any chemical in elemental form is \(0\). The oxidation number of a monoatomic ion has the same oxidation number as it does charge.
Here are three quick examples. Really, these are very easy to learn and understand if you understand how charges work.
Example 1: Argon atoms have no charge and an oxidation number of \(0\).
Example 2: The oxide ion, \(O^{2-}\), has an oxidation number of \(-2\).
Example 3: Find the oxidation number of all of the ions in the compound \(AlBr_3\).
Solution: In this compound, the aluminum ion has a charge of \(+3\), so its oxidation number is \(+3\). The bromide ions each have a charge of \(-1\), so each one has an oxidation number of \(-1\).
Here is the next rule, which was subtly hinted at in Example 3:
When an entire molecule or compound has a neutral charge, the sum of the oxidation numbers of all atoms in the chemical is \(0\).
This greatly increases the number of chemicals we can work with.
Example 4: Find the oxidation number of both types of ions in the chemical \(Os_3P_4\).
Solution: The phosphide ion present in this chemical has a charge of \(-3\), its oxidation number is \(-3\) as well. The sum of the oxidation numbers of the chemicals is \(0\) because this chemical has a neutral charge. Four phosphide ions have a total charge of \(-12\). Therefore the three osmium cations must have a total charge of \(+12\). Each one has the same charge, so each one has a charge of \(+4\), and thus an oxidation number of \(+4\).
Example 5: Find the oxidation number of chlorine in \(Cl_2\).
Solution 1: This molecule is chlorine’s elemental form, so each chlorine atom must have an oxidation number of \(0\).
Solution 2: The charge of the full molecule is \(0\). Since the orientation of the electrons linked to each chlorine atom is the same, there is no reason why their oxidation numbers would be different. In this case, each chlorine atom must have an oxidation number of \(0\)..
There is a strict procedure involving finding the oxidation numbers of nonmetals. Of course, noble gases rarely form bonds and thus their elemental forms all have oxidation numbers of \(0\). For other nonmetals, follow this list of rules. The first ones listed have the highest priority. If two rules conflict, follow the one that is listed first.
If hydrogen is present and all the atoms are nonmetals, hydrogen will have an oxidation number of \(+1\). If one or more metals are present, hydrogen instead has an oxidation number of \(-1\).
Alkaline metals have oxidation numbers of \(+1\).
Alkaline earth metals have oxidation numbers of \(+2\).
Fluorine always has an oxidation number of \(-1\).
Other halogens generally have oxidation numbers of \(-1\) as well, unless either oxygen or a halogen from an earlier period on the periodic table interferes with this.
Oxygen has an oxidation number of \(-2\), unless the compound also has fluorine (in which case the oxidation number could be different), or the oxygen appears in \(O_2^{2-}\), \(O_2^{-}\), or \(O_3^{-}\).
Example 6: Find the oxidation numbers of the hydrogen and the oxygen in \(H_2O\).
Solution: We start with the hydrogen. There are only nonmetals in this chemical, so each hydrogen atom has an oxidation number of \(-1\). In order for the entire molecule to have a neutral charge, the oxygen must have an oxidation number of \(+2\).
Example 7: The hydrocarbon ethane, \(C_2H_6\), contains two carbon atoms with single bonds to each other, and each one also has a single bond to three hydrogen atoms. What are the oxidation numbers of hydrogen and carbon in this molecule?
Solution: There are no metals in this molecule, so each hydrogen atom has an oxidation number of \(+1\). For the total charge of the molecule to be \(0\), each of the two carbon atoms must have an oxidation number of \(-3\).
Example 8: Find the oxidation number of each compound in \(HFO\).
Solution: Fluorine always has an oxidation number of \(-1\). Since there are nonmetals, hydrogen has an oxidation number of \(-1\) as well, so the oxygen must have an oxidation number of \(+2\).
For determining the oxidation numbers of atoms in polyatomic ions, all of the same rules apply as if the entire chemical had a neutral charge, except the sum of the oxidation numbers of each atom in the ion are equal to the whole charge of the ion rather than \(0\).
Example 9: Find the oxidation number of each atom in the \(NO_3^{-}\) ion.
Solution: Solution: In this molecule, we work with the oxygen first. According to the set of rules, each oxygen has an oxidation number of \(-2\). In order for the entire molecule to have a charge of \(-1\), the nitrogen atom must have an oxidation number of \(+5\).
Example 10: Find the oxidation number of each atom in the \(SCNO^{-}\) ion.
Solution: We have no halogens, so start with the oxygen; give it a \(-2\) oxidation number. The other three atoms must have a total charge of \(+1\) so that the entire ion has a total charge of \(-1\). From here, none of the other rules apply to the remaining atoms, so we use common oxidation numbers for the chemicals in other ions. With that in mind, let nitrogen have an oxidation number of \(-3\), and let carbon and sulfur each have oxidation numbers of \(+2\).
A half-reaction is an incomplete electron transfer. In oxidation, a chemical releases electrons, and in reduction, a chemical takes free electrons. Neither type of half-reaction will occur on its own, but if a chemical is oxidized, its oxidation number increases; if a chemical is reduced, its oxidation number decreases. The rules of oxidation numbers can be used to determine whether a chemical is being oxidized or reduced.
Example 11: Is this half-reaction oxidation or reduction?
$$HMnO_4 \rightarrow Mn^{4+}$$
Solution: The right side is just a manganese cation, so its oxidation number is \(+4\).
We must find the oxidation number of manganese on the left side. The compound contains a metal atom, so the oxidation number of hydrogen is \(+1\). In this case, as usual, each oxygen has an oxidation number of \(-2\), so for the chemical to have a neutral charge, the manganese on the left side must have an oxidation number of \(+7\). As a result, the manganese is being reduced.
Example 12: Is this an oxidation reaction or a reduction reaction?
$$ReO_2 \rightarrow H_2ReO_4$$
Solution: The left side has an ionic compound, so the oxide ions have a total charge of \(-4\), and a total oxidation number of \(-4\), so the oxidation number of the rhenium is \(+4\).
On the right side, rhenium is a metal, so each hydrogen has an oxidation number of \(+1\). As with most compounds containing oxygen, each one has an oxidation number of \(-2\), so in order for this compound to be neutral, rhenium must have an oxidation number of \(+6\) in this compound. As a result, rhenium is being oxidized, and this is an oxidation reaction.
This final section demonstrates the usefulness of oxidation numbers to electrochemistry. Once you understand and can apply the rules of oxidation numbers, the studies of electrochemistry become simpler; sometimes the balancing of redox reactions is made much easier with the use of the oxidation number techniques shown in this article.
Reference, courtesy of
"Chemical Principles: The Quest for Insight Second Edition" Atkins, Peter and Jones, Loretta