Oxidation States and Assigning Oxidation Numbers

It is important to know which atom is oxidized and which is reduced. Oxidation states or oxidation numbers are assigned to atoms in order to keep track of the redistribution of electrons during a redox reaction. In a redox reaction, the oxidation numbers of some atoms have to change to reflect the gain or loss of electrons. By keeping track of and comparing the oxidation numbers of the atoms on the reactant and the product side, it is possible to determine how many electrons are gained or lost by each atom. The oxidation number of an atom in a compound is assigned according to the following rules:

1. The oxidation number of free elements is zero. For example, the atoms in N₂, P₄, S₈, and He all have oxidation numbers of zero.
2. The oxidation number for a monatomic ion is equal to the charge of the ion. For example, the oxidation numbers for Na⁺, Cu²⁺, Fe³⁺, Cl⁻, and N³⁻ are +1, +2, +3, −1, and −3, respectively.
3. The oxidation number of each Group IA element in a compound is +1. The oxidation number of each Group IIA element in a compound is +2.
4. The oxidation number of each Group VIIA element (halogens) in a compound is −1, except when combined with an element of higher electronegativity. For example, in HCl, the oxidation number of Cl is −1; in HOCl, however, the oxidation number of Cl is +1 because of the oxygen (see rule 6 below).
5. The oxidation number of hydrogen is −1 in compounds with less electronegative elements than hydrogen (Groups IA and IIA). Examples include NaH and CaH₂. The more common oxidation number of hydrogen is +1.
6. In most compounds, the oxidation number of oxygen is −2. This is not the case, however, in molecules such as OF₂. Here, because F is more electronegative than O, the oxidation number of oxygen is +2. Also, in peroxides such as BaO₂, the oxidation number of O is −1 instead of −2 because of the structure of the peroxide ion, [O−O]²⁻. (Note that Ba, a group IIA element, cannot be a +4 cation.)
7. The sum of the oxidation numbers of all the atoms present in a neutral compound is zero. The sum of the oxidation numbers of the atoms present in a polyatomic ion is equal to the charge of the ion. Thus, for SO₄²⁻, the sum of the oxidation numbers must be −2.

Quick Quiz

Match the substance with its correct oxidation number below.

1. Na⁺
2. Cu²⁺
3. Cl⁻
4. Fe³⁺
   1. −1
   2. +3
   3. +2
   4. +1

Answers:

1. = (D)
2. = (C)
3. = (A)
4. = (B)

1. Example: Assign oxidation numbers to the atoms in the following reaction in order to determine the oxidized and reduced species and the oxidizing and reducing agents.

SnCl₂ + PbCl₄ → SnCl₄ + PbCl₂

1. Solution: All these species are neutral, so the oxidation numbers of all the atoms in each compound must add up to zero. In SnCl₂, since there are two chlorines present, and chlorine has an oxidation number of −1, Sn must have an oxidation number of +2. Similarly, the oxidation number of Sn in SnCl₄ is +4; the oxidation number of Pb is +4 in PbCl₄ and +2 in PbCl₂. Notice that the oxidation number of Sn goes from +2 to +4; i.e., it loses electrons and thus is oxidized, making it the reducing agent. Since the oxidation number of Pb has decreased from +4 to +2, it has gained electrons and been reduced. Pb is the oxidizing agent. The sum of the charges on both sides of the reaction is equal to zero, so charge has been conserved.

Note that even though we have been making the connection between oxidation states and charge distribution, the oxidation number is not in general the charge (nor even the formal charge) of the atom in a compound. It merely reflects a way of accounting for how electrons are transferred between species.