Science 10
Part I:
Brakke
ECA – Chemistry Topic 04
CT04D01 – Oxidation Numbers and Redox Equations
Name……………………………………………………
Rules for Assigning Oxidation Number
1.
The oxidation number of an element (in its elementary state) is ZERO.
2.
The oxidation number of an element in a monatomic ion is equal to
the charge of that ion.
Examples: Na +1, Cl-1 in NaCl ; K+1, S –2 in K2S ; Al+3, O-2 in Al2O3.
3.
Certain elements have almost the same oxidation number in ALL or
almost all of their compounds.
Examples:
4.
Group 1 is always +1
Group 2 is always +2
Al is always +3
Oxygen is always –2 [except in peroxides where it is -1]
The sum of the oxidation numbers in a compound that is neutral is ZERO; the sum of the oxidation numbers in an ion is
equal to the charge on that ion.
In addition, “one cannot occur without the other”, and the “number of electrons lost by the substance being oxidized must be
gained by the substance being reduced”
Part II: Important terms:
a.
Oxidation
b.
Reduction
Part III: Using the terms and balancing oxidation-reduction reactions:
1. Balance the following oxidation reduction reaction using our normal technique.
2Na (s) + 2H2O (l)  2NaOH (aq) + H2 (g)
reduced
… …………………
oxidized
…… ……………….
N2 (g) + 3H2 (g)  2NH3 (g)
reduced
… …………………
oxidized
…… ……………….
2NH3 (g) + 2O2 (g)  N2O (g) + 3H2O (l)
reduced
… …………………
oxidized
…… ……………….
Science 10
2.
Brakke
ECA – Chemistry Topic 04
The following equations are NOT balanced. In the future (next year) we will learn how to balance these complex
equations based on their oxidation states and the transfer of electrons that can be determined via redox
determination. For now, your only task is to assign oxidation states and state which elements are being oxidized and
reduced (there may be more than one):
Fe S O4 + K Mn O4 + H2 S O4  Fe2 (S O4)3 + K H S O4 + Mn S O4 + H2 O
reduced
… …………………
oxidized
…… ……………….
H Cl + K Mn O4  Cl2 + H2 O + K Cl + Mn Cl2
reduced
… …………………
oxidized
…… ……………….
K O H + Co Cl2 + K Cl O3  Co2 O3 + K Cl + H2 O
reduced
… …………………
oxidized
…… ……………….
Br2 + K Cl O3 + H2 O + H2 S O4  H Br O3 + K2 S O4 + H Cl
reduced
… …………………
oxidized
…… ……………….
K2 Cr2 O7 + H Br  Cr Br2 + Br2 + H2 O + K Br
reduced
… …………………
oxidized
…… ……………….
Al + H N O3  Al (N O3)3 + N2 O + H2 O
reduced
… …………………
oxidized
…… ……………….
K Cl O3 + H2 S O4  K H S O4 + H Cl O4 + Cl O2 + H2 O
reduced
… …………………
oxidized
…… ……………….
Cr2 S3 + Mn (N O3)2 + Na2 C O3  Na2 Cr O4 + Na Mn O4 + Na2 S O4 + C O2 + N O
reduced
… …………………
oxidized
…… ……………….