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Lesson 4 – Oxidation numbers
An Oxidation number is the arbitrary assignment of a number to an element that reflects its gain or
loss of electrons. Assignment of oxidation numbers to elements is an easier way to see if an element
is oxidised or reduced without having to write half equations especially if it involves non-metal
elements.
For example – NO 3
NO
2
looking at this reaction it is difficult to know whether nitrogen is oxidised
or reduced. On the other hand Fe Fe
3+
is a lot easier to identify as oxidation as we can see
electrons have been lost.
The concept of an atom’s oxidation number or oxidation state is based on the following set of rules:
- An atom of a free element has an oxidation number of zero.
For example, Cl in Cl 2 has an oxidation state of 0, as does Na atoms in Na(s) and F atoms in F 2.
- A monatomic ion has an oxidation number equal to its charge.
For example, Na
has an oxidation number of +1, while Al
3+
has an oxidation number or oxidation
state of +3 and N
3 -
will have an oxidation number or state of - 3.
- Hydrogen has an oxidation number of +1,
The only exception is when it forms a compound with a metal (metal hydride) such as LiH where its
oxidation state is - 1.
- Group 1 metals have an oxidation state of +1 while group 2 metals have an oxidation state of +
- Oxygen has an oxidation number of - 2
The exception is in peroxides such as H 2
O
2
where oxygen has an oxidation number of - 1
and in F 2
O where it has an oxidation state of +1.
- The sum of the oxidation numbers for all atoms in a polyatomic compound is equal to the charge on
the compound. For example:
- MnO 4 - - the oxidation number of Mn + 4 X the oxidation number of O = - 1
- H 2 CO 3 - 2 X the oxidation number of hydrogen + the oxidation number of C + 3 X oxidation number of O = 0
- The oxidation number of fluorine is always – 1. Chlorine, bromine, and iodine usually have an
oxidation number of – 1, unless they’re in combination with an oxygen or fluorine.
*Note - oxidation numbers are written with the sign before the number eg - 2 or +1. This is different
for when we write the charges on an ion such as Cl
1 -
or Mn
5+
- Workout the oxidation number of the element that is underlined below.
a. ClO 3
Cl + 3 X - 2 = - 1 => Cl has an oxidation state of +
b. MnO 2
Mn + 2 X - 2 = 0 => Mn has an oxidation state of +
c. SO 4
2 -
S + 4 X - 2 = - 2 => S has an oxidation state of +
d. KMnO 4
1+ Mn + 4 X - 2 = 0 => Mn has an oxidation state of +
e. SO 3
S + 3 X - 2 = 0 => S has an oxidation state of +
f. Cr(H 2
O)
6
3+
Cr + 12 X 1 + 6 X - 2 = +3 => Cr has an oxidation state of +
g. CrCl 3
Cr + 3 X - 1 = 0 => Cr has an oxidation state of +
h. MnO 4
Mn + 4 X - 2 = - 1 => Mn has an oxidation state of +
i. Al
Al = 0
j. MoO 4
2 -
Mo + 4 X - 2 = - 2 => Mo has an oxidation state of +
k. Cl 2
Cl = 0
l. NaClO 4
1 + Cl + 4 X - 2 = 0 => Cl has an oxidation state of +
m. NaClO 2
1 + Cl + 2 X - 2 = 0 => Cl has an oxidation state of +
n. Ce 2
O
4
2 X Ce + 4 X - 2 = - 1 => Ce has an oxidation state of +3.
o. S 8
S= 0
p. O 3
O = 0
q. Al(s) Aluminium metal
Al = 0
r. IO 4
I + 4 X - 2 = 0 => I has an oxidation state of +
s. I 2
I = 0
t. IF 7
I + 7 X - 1 = 0 => I has an oxidation state of +
u. HIO 4
1 + I + 4 X - 2 = 0 => I has an oxidation state of +
c. 2Co
3+
(aq) + Ni(s) 2Co
2+
(aq) + Ni
2+
(aq)
This is a redox reaction.
oxidant – Co
3+
(aq) oxidation state +3 changes to CO
2+
(aq) with an oxidation
state of +2. It is reduced.
Reductant – Ni(s) oxidation state of 0 change to Zn
2+
(aq) with an oxidation
state of +2. It is oxidised.
d. Cu(s) + 2NO 3
(aq) + 4H
(aq) Cu
2+
(aq) + 2NO 2
(g) + 2H 2
O(l)
This is a redox reaction.
oxidant – NO 3
-
(aq) – nitrogen has an oxidation state of +5 changes to +4 in
NO
2
(g). It is reduced.
Reductant – Cu(s) oxidation state of 0 change to Cu
2+
(aq) with an oxidation
state of +2. It is oxidised.
e. 2MnO 4
(aq) + 6I
(aq) + 4H 2
O(l) 2MnO 2
(s) + 3I 2
(aq) + 8OH
(aq)
This is a redox reaction.
oxidant – MnO 4
-
(aq) - Mn has an oxidation state of +7 changes to +4 MnO 2
It is reduced.
Reductant – I
-
(aq) - oxidation state of - 1 changes to 0 in I 2
(aq). It is oxidised.
f. Ag
(aq) + Cl
(aq) AgCl(s)
This is not a redox reaction.
No change in oxidation numbers takes place.
Ag
+
remains Ag
+
in AgCl(s)
Cl
-
remains Cl
-
in AgCl(s)
g. 6CO 2
(g) + 6H 2
O(l) C 2
H
12
O
6
(aq) + 6O 2
(g)
This is a redox reaction.
Reductant – H 2
O(l) - O has an oxidation state of - 2 changes to 0 in O 2
. It is
oxidised.
Oxidant – CO 2
(g) - oxidation state of C in CO 2
changes to +4 to 0 in
C
6
H
12
O
6
(aq). It is reduced.
- Identify the conjugate pairs in the redox reactions below.
a. Zn(s) + 2Au
(aq) Au(s) + Zn
2+
(aq)
Zn(s)/Zn
2+
(aq) Au
(aq)/Au(s)
oxidant reductant
oxidant reductant
b. C 2
H
6
(g) + O 2
(g) CO 2
(g) + H 2
O(l)
C
2
H
6
(g) / CO 2
(g) O 2
(g) / H 2
O(l)
c. 2Co
3+
(aq) + Ni(s) 2Co
2+
(aq) + Ni
2+
(aq)
Co
3+
(aq) / Co
2+
(aq) Ni(s) / Ni
2+
(aq)
d. Cu(s) + 2NO 3
(aq) + 4H
(aq) Cu
2+
(aq) + 2NO 2
(g) + 2H 2
O(l)
Cu(s) / Cu
2+
(aq) NO 3
-
(aq) / NO 2
(g)
e. 2MnO 4
(aq) + 6I
(aq) + 4H 2
O(l) 2MnO 2
(s) + 3I 2
(aq) + 8OH
(aq)
MnO 4
-
(aq)/MnO 2
(s) I
-
(aq)/I 2
(aq)
f. Fe 2
O
3
(s) + 3CO(g) 2Fe(s) + CO 2
(g)
Fe 2
O
3
(s)/Fe(s) CO(g)/CO 2
(g)
g. CuO(s) + Mg(s) MgO(s) + Cu(s)
oxidant
reductant
oxidant reductant
oxidant
reductant
reductant oxidant
oxidant
reductant
oxidant reductant
oxidant
reductant
reductant
oxidant
oxidant
reductant
oxidant
reductant
oxidant
reductant
reductant oxidant
CuO(s) /Cu(s) Mg(s) / MgO(s)
