Oxidation Numbers


Oxidation-reduction (redox) reactions involve the transfer of electrons from a reducing agent to an oxidizing agent. There are mnemonic devices to help us remember what oxidation and reduction is.

  1. OIL RIG - Oxidation is losing (electrons); Reduction is gaining (electrons)
  2. LEO the lion says GER - Loss (of) electrons (is) oxidation; Gain (of) electrons (is) reduction

Other terms to note are:

  1. Oxidizing agent - the species gaining electrons; the species undergoing reduction; the oxidizing agent oxidizes another species
  2. Reducing agent - the species losing electrons; the species undergoing oxidation; the reducing agent reduces another species

Think about a travel agent. A travel agent is someone you go to to set up travel plans. The travel agent is not doing the travelling… you are. The travel agent enables you to travel. An oxidizing agent allows something else to be oxidized. The oxidizing agent undergoes reduction to do this. Similar logic applies to the reducing agent.

Oxidation numbers allow us to determine what is being oxidized and what is being reduced in a chemical reaction. When the oxidation number of an atom increases (when going from reactants to products), the atom was oxidized (it lost electrons). When the oxidation number of an atom decreases (when going from reactants to products), the atom was reduced (it gained electrons).


Assigning Oxidation Numbers


A series of rules have been developed to assign oxidation numbers to atoms. There is not a rule for assigning an oxidation number to every element. Therefore, when assigning oxidation numbers, always start assigning numbers to elements that have rules.

Rules

1. The oxidation number equals 0 for an atom in its pure, elemental state

H2

H: 0

O2

O: 0

Xe

Xe: 0

C(s, graphite)

C: 0

Al(s)

Al: 0

2. The oxidation number equals the charge for a monatomic ion

Na+

Na: 1

Ca2+

Ca: 2

Al3+

Al: 3

3. The sum of all the oxidation numbers for atoms in a polyatomic ion equals the charge of the polyatomic ion

PO42–

P: +6

O: –2

The sum of the oxidation numbers equals the charge on the polyatomic ion.

+6 + (4 × –2) = –2

4. The sum of all the oxidation numbers of a species equals the overall charge of that species

H2O

H: +1

O: –2

The sum of the oxidation numbers equals the charge on the species.

(2 × 1) + (–2) = 0

5. Some elements can reliably be assigned certain oxidation numbers with some exceptions

Rules at the top of the list take priority over rules listed below.

Element Oxidation Number Exceptions
F –1 none
Group 1A/2A metals +1 and +2, respectively none
H +1 LiH - Li: +1 and H: –1
Oxygen –2 H2O2 - H: +1 and O: –1
Group 7A (except F) –1 CaF - Ca: +1 and F: –1


Example: Assign oxidation numbers to each atom

\[\mathrm{H_2SO_4}\]

Start with what you know: H and O

  • H: +1
  • O: –2

To assign the oxidation number to sulfur, take note of Rule 4. The overall charge of the molecule is zero. Sum over the oxidation numbers and solve for the missing one.

\[\begin{align*} (2 \times +1) + (x) + (4 \times -2) &= 0\\ x &= +6 \end{align*}\]

  • S: +6


Half-reactions


We can split a redox reaction into two half-reactions; the oxidation half-reaction and the reduction half-reaction. To do this, assign oxidation numbers to each atom in the reactants and the products. Determine what is oxidized and what is reduced then write out both half-reaction s.

Example: Write out the half-reactions for the following redox reaction

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

Reactants

  • Zn: 0
  • Cu: +2
  • Cl: –1

Products

  • Zn: +2
  • Cl: –1
  • Cu: 0

Here we see that Zn was oxidized (Zn is the reducing agent) and Cu was reduced (Cu is the oxidizing agent). The ion, Cl, is a spectator ion (its oxidation number did not change).

We can now write out the half-reactions.

Oxidation half-reaction \[\mathrm{Zn}(s) \longrightarrow \mathrm{ZnCl_2}(aq)\]

Reduction half-reaction \[\mathrm{CuCl_2}(aq) \longrightarrow \mathrm{Cu}(s)\]


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