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In a reduction-oxidation reaction, one species gains electrons while the other loses them. While it's easy to see what happens when you have a reaction like

Cu+2(aq) + 2e- -> Cu(s)
it's not always clear when you have a more complex reaction. Chemists use a bookkeeping device called the oxidation number to decide what is being oxidized and reduced.
  • If the oxidation number of a species in a reaction increases, the species was oxidized.
  • If the oxidation number of a species in a reaction decreases, the species was reduced.

To compute the oxidation number of an atom, ion or molecule is straightforward, but there are a few rules to remember.

  1. The oxidation number of an element in an elementary substance is 0.
  2. The oxidation number of an ion is equal to the charge on the ion.
  3. The sum of the oxidation number of the atoms in a neutral molecule is zero
  4. The sum of the oxidation number of the atoms in an ion is equal to the charge on the ion.
  5. Certain atoms almost always have the same oxidation number in all molecules
    • Group 1 metals (Li, Na...) are +1
    • Group 2 metals (Be, Mg...) are +2
    • Fluorine is -1
    • Oxygen is almost always -2. (Except for peroxides (O2-2) and superoxides.(O2-))
    • Hydrogen is usually +1, except in group 1 and 2 metal hydrides such as NaH and CaH2.

Some examples of the above rules are helpful.

  1. O2(g) is an elementary substance: oxygen has an oxidation number of 0 in the oxygen molecule.
  2. S-2 is a monoatomic ion; the oxidation number of sulfur is -2 in the sulfur ion.
  3. The overall oxidation number of Cr(OH)3 is 0
  4. The overall oxidation number of OH- is -1

The last three rules can be used to compute the oxidation number for all the atoms in a molecule. If you know the sum of the oxidation numbers and you can use the special cases to determine all but one atom, you can get the oxidation number. For example, what is the oxidation number on the manganese in MnO4-?

The overall oxidation number of the ion is -1. Oxygen is almost always -2, and there are four of them. Therefore,

Oxnum(Mn) + 4*Oxnum*(O) = -1
Oxnum(Mn) + 4*(-2) = -1
Oxnum(Mn) = +7

Example: What is the oxidation number on chromium in the chromate ion, CrO4-2?

Solution: The overall oxidation number for the ion is equal to the charge on the ion (Rule 2), so it's -2. The sum of the oxidation numbers is equal to the charge (Rule 4). There are four oxygens, each has -2 oxidation number. (Rule 5). Therefore,

Oxnum(Cr) + 4*Oxnum*(O) = -2
Oxnum(Cr) + 4*(-2) = -2
Oxnum(Cr) = +6


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