
home	test	exam	game	contact

Oxidation Numbers (Oxidation States)

Key Concepts

An oxidation number (oxidation state) is the charge an atom would carry if the molecule or ion were completely ionic.

Rules for Assigning Oxidation Numbers

Fluorine is assigned an oxidation number of -1 in compounds
HF oxidation number of fluorine = -1
Oxygen is assigned an oxidation number of -2 in compounds
H₂O oxidation number of oxygen is -2
Except
- Peroxides: oxidation number of oxygen is -1
  H₂O₂ oxidation number of oxygen is -1
- Superoxides: oxidation number of oxygen is -½
  KO₂ oxidation number of oxygen is ½
- Oxygen fluorides: OF₂ oxidation number of oxygen is +2
  O₂F₂ oxidation number of oxygen is +1
Hydrogen is assigned an oxidation of +1 in compounds
HCl oxidation number of hydrogen is +1
Group I elements (Alkali Metals) are assigned an oxidation number of +1 in compounds
NaNO₃ oxidation number of sodium is +1
Group II elements (Alkaline-earth metals) are assigned an oxidation number of +2 in compounds
MgBr₂ oxidation number of magnesium is +2
An atom of any element in the free state has an oxidation number of 0
S₈ oxidation number of each sulfur atom is 0
Any monatomic ion has an oxidation number equal to its charge
H^- oxidation number of hydrogen is -1
The sum of the oxidation numbers of all the atoms in a formula equals the electrical charge shown with the formula
- The sum of the oxidation numbers of all the atoms shown in a formula of a compound is 0
        CO₂: let x be the unknown oxidation number of carbon
        0 = x + (2 x -2)
        So x (oxidation number of C) = 0 + 4 = +4
- The sum of the oxidation numbers of all the atoms shown in the formula for a polyatomic ion or complex ion equals the electrical charge on the ion
        Cr₂O₇^2-: let y = unknown oxidation number of chromium
        -2 = 2y + (7 x -2)
        2y = -2 + 14
        2y = +12
        y (oxidation number of Cr) = 12 ÷ 2 = +6

Oxidation

An increase in oxidation number which corresponds to a loss of electrons
(or to an addition in oxygen or loss of hydrogen)

Example

Zn(s) -----> Zn²⁺ + 2e

Zinc has been oxidised since there has been a loss of electrons to form Zn²⁺.
The oxidation number of zinc has increased from 0 to 2+

Reduction

A reduction in oxidation number which corresponds to a gain of electrons
(or to an addition of hydrogen or loss of oxygen)

Examples

Ag⁺ + e -----> Ag(s)
The silver ion has been reduced since it has gained an electron to form Ag(s)
The oxidation number of silver has been reduced from +1 to 0

MnO₄^- + 8H⁺ + 5e -----> Mn²⁺ + 4H₂O
Oxidation number of manganese (x) in MnO₄^- is:
-1 = x + (4 x -2)
x = -1 + 8 = +7
oxidation number of manganese in MnO₄^- = +7
manganese has been reduced since there is a reduction in oxidation number from +7 to +2