Section 13.3 Oxidation Numbers
p. 583 - 595
Oxidation Numbers
An oxidation number is a "fictitious" number
assigned to an atom or ion to achieve
"electron book - keeping".
Rules:
1) The oxidation number ( O.N. ) of elemental
atoms is zero ( 0 ).
Examples: Cl2(g) has two atoms of Cl in one
molecule of Cl2(g). Each atom is assigned an
oxidation number of zero (0).
Fe(s) = 0
S8(s) = 0
2) The oxidation number ( O.N. ) of simple ions
equals the charge of the ion.
Examples: Cr 3+(aq) = + 3
S 2–(aq) = – 2
Fe 2+(aq) = + 2
3) The oxidation number ( O.N. ) of oxygen in
compounds is – 2 , except in peroxides, where
it is – 1.
Examples: SO2(g): oxygen's O.N. = – 2
H2O(l): oxygen's O.N. = – 2
Na2O(s): oxygen's O.N. = – 2
*H2O2(l): oxygen's O.N. = – 1
4) The oxidation number ( O.N. ) of hydrogen in
compounds is + 1 , except in metallic hydrides,
where it is – 1.
Examples: H2O(l): hydrogen's O.N. = + 1
NH3(g): hydrogen's O.N. = + 1
C4H10(g): hydrogen's O.N. = + 1
*NaH(s): hydrogen's O.N. = – 1
*CaH2(s): hydrogen's O.N. = – 1
5) The total of the ( O.N. ) of all the atoms in a molecule is zero ( 0 ).
Examples: N2O5(g)
Note: The sum of O.N. from 2 nitrogens plus 5 oxygens = total charge
of zero.
2 (x) + 5 (–2) = 0
2x – 10 = 0
2x = + 10
x=+5
Therefore the O.N. of each atom of nitrogen is + 5.
=0
P4O10(g): 4xx =++10(–2)
5
Note: The O.N. of halogens in compounds is usually –1.
S2Cl6(s):
NH3(g):
2x + 6(–1) = 0  x = + 3
x + 3(+1) = 0
 x=–3
6) The total of the ( O.N. ) of all the atoms in a complex
ion is equal to that complex ion's charge.
Examples:
 x = +7
MnO4–(aq):
x + 4(–2) = –1
Cr2O72–(aq):
2x + 7(–2) = –2
 x = +6
B4O72–(aq):
4x + 7(–2) = –2
 x = +3
C3H7Br(s):
3x + 7(+1) + (–1) = 0  x = – 2
Another example:
Find the oxidation number of chlorine in
NaClO3 (s)
Ions are Na+ and ClO3-
x + 3(-2) = -1
x = +5
Oxidation Numbers and Redox Reactions
• Oxidation number of an atom changes during a
chemical reaction.
• Based on oxidation numbers:
– Oxidation – increase in oxidation number
– Reduction – decrease in oxidation number
• Important: In a redox reaction, both oxidation and reduction
occur, therefore both, the increase and the decrease of the
oxidation numbers have to be present.
Example 1. (Sample problem 13.9, p. 587)
Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)
Example 2. (Sample problem 13.10, p. 587)
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Example 3. (Communication example 2, p. 588)
2Cr2O72-(aq) + 16H+(aq) + 3C2H5OH(aq) → 4Cr3+(aq) + 11H2O(l) + 3CH3COOH(aq)
Activity: Questions #1 – 5, p.585
Practice Questions # 6 – 9, p. 589
Balancing Redox Equations Using
Oxidation Numbers
Steps in Balancing Redox Equations:
1. Assign oxidation numbers and identify the atoms/ions whose
oxidation numbers change.
2. Using the change in oxidation numbers, write the number of
electrons transferred per atom.
3. Using the chemical formulas, determine the number of electrons
transferred per reactant. (Use the formula subscripts to do this.)
4. Calculate the simplest whole number coefficients for the reactants
that will balance the total number the total number of electrons
transferred. Balance the reactants and products.
5. Balance the O atoms using H2O(l), and then balance the H atoms
using H+(aq).
p. 590
p. 591
p. 592
Activity:
Practice Questions #11 – 15, p. 593
Section Review # 1 – 12, p. 595