8.1 Oxidation numbers

To understand the type of reactant in a redox reaction, we must rationalize the direction of charge flow. For example, in the reaction below, zinc is the reducing agent (zinc undergoes oxidation) while copper is the oxidizing agent (copper undergoes reduction).

\[\mathrm{Zn}(s) + \mathrm{CuCl_2}(aq) \longrightarrow \mathrm{ZnCl_2}(aq) + \mathrm{Cu}(s)\]

We can assign oxidation numbers to each atom in a redox reaction with which we use to determine the type of agent each reactant is. We must follow a set of rules for assigning these numbers.

Oxidation number rules

Oxidation number = 0 for any atom in its pure, elemental state.
- Zn(s): Zn = 0
- H₂(g) = 0
Oxidation number = charge on a monatomic ion
- Na⁺: Na = 1
- Cl^–: Cl = –1
Sum of all oxidation numbers for the atoms in a polyatomic ion = charge on the polyatomic ion
- OH^–: O + H = = –2 + 1 = –1
Sum of all oxidation numbers = charge on the species
- H₂O: O + H + H = –2 + 1 + 1 = 0
- NaCl: Na + Cl = 1 + –1 = 0

Additionally, the rules below are prioritized in the following order:

Element	Oxidation Number	Exceptions
Fluorine	-1	none
Group 1A or 2A metal	+1 or +2, respectively	none
Hydrogen	1	If paired with Group 1A or 2A metal to form a metal hydride such as LiH and CaH₂
Oxygen	-2	If paired with any element listed above that would violate rules 1-4 such as H₂O₂ or KO₂
Group 7A (excluding Fluorine)	-1	If paired with any element listed above that would violate rules 1-4 such as ClF, BrO₄^–, and IO₃^–

Applying these rules to our reaction above gives the following oxidation numbers

\[\begin{alignat*}{3} & \mathrm{Zn}(s) ~+~ && \mathrm{CuCl_2}(aq) ~ \longrightarrow ~ && \mathrm{ZnCl_2}(aq) ~+~ & \mathrm{Cu}(s) \\ & ~(0) && (2)(-1) && (2)(-1) & ~(0) \end{alignat*}\]

Recognize that zinc begins with an oxidation state of zero and results in an oxidation state of 2+ whereas copper begins with a 2+ oxidation state and ends up with a 0 oxidation state. Therefore, zinc was oxidized (its oxidation number increased) whereas copper was reduced (its oxidation state decreased). We say that zinc lost electrons and copper gained electrons. The oxidation number for each chlorine did not change (it remained as –1). Chlorine is called a spectator ion in the reaction.

We can write these processes out in two half-reactions, one demonstrating the oxidation reaction and one demonstrating the reduction reaction. We explicitly include the electrons to keep both sides of the reaction arrow charge-balanced.

\[\begin{alignat*}{3} \mathrm{Oxidation~half~reaction:}~~ &&\mathrm{Zn}(s) &\longrightarrow \mathrm{Zn^{2+}}(aq) ~+~ 2e^-\\[2ex] \mathrm{Reduction~half~reaction:}~~ &&\mathrm{Cu^{2+}}(aq) ~+~ 2e^- &\longrightarrow \mathrm{Cu}(s) \end{alignat*}\]

The concept of oxidation comes from the reaction of a substance with elemental oxygen such as the oxidation of iron to form iron(III) oxide (i.e. rust).

\[\begin{alignat*}{3} & \mathrm{4Fe}(s) ~+~ && \mathrm{3O_2}(g) ~\longrightarrow~ && \mathrm{2Fe_2O}(s) \\ & ~(0) && ~(0) && (3)(-2) \end{alignat*}\]

Here, iron is oxidized (iron is the reducing agent). Oxygen is reduced (oxygen is the oxidizing agent).

Practice

Write out the oxidation numbers for each atom in the following reaction

\[\mathrm{Zn}(s) + \mathrm{Cl_2}(g) \longrightarrow \mathrm{ZnCl_2}(s)\]

and determine the following

Species oxidized
Species reduced
Oxidizing agent
Reducing agent

Solution

\[\begin{alignat*}{3} & \mathrm{Zn}(s) ~+~ && \mathrm{Cl_2}(g) ~ \longrightarrow ~ && \mathrm{ZnCl_2}(s) \\ & ~(0) && ~(0) && (2)(-1) \end{alignat*}\]

Species oxidized: Zn
Species reduced: Cl
Oxidizing agent: Cl₂
Reducing agent: Zn