10 Reduction And Oxidation

1. 10 Reduction and Oxidation

2. Oxidation Numbers
•Group 1 ox no = +1
•Group 2 ox no = +2
•Group 3 ox no = +3
•H ox no = +1 except in an alkali metal
hydride e.g. NaH where it is -1
•O ox no = -2 except in a peroxide where it
is -1
•Halides usually -1 (except when bonded to
other halogens or oxygen)

3. Oxidation numbers add up to 0 for neutral
molecule
•e.g. NaCl Na = +1 CL = -1
MgCl
2
Mg = +2 Cl = -1
AlF
3
Al = +3 F = -1
For a charged ion ox no’s add up to the charge
MnO
4
-Mn = +7 O = -2

4. Find ox no of underlined elements
N
2
H
4
CrO
4
-2
ClO
3
4-
NH
4
+
VO
2
+
N = -2Cr = +6Cl = +2 N = -3V = +5

5. Oxidation is any process where there is an
increase in an oxidation state
Oxidation is loss of electrons
Reduction is any process where there is a
decrease in oxidation state
Reduction is gain of electrons

6. O oxidation
I is
L loss
R reduction
I is
G gain

7. increase in oxidation state
(loss of electrons)
-3 -2 -1 0 +1 +2 +3 +4 +5 +6 +7

decrease in oxidation state
(Gain of electrons)

8. Examples of oxidation
Fe
2+
- e  Fe
3+

Cu - 2e  Cu
2+
Example of reduction
Fe
3+
+ e  Fe
2+
MnO
4
-
+ 5e  Mn
2+
These are called half equations
In any reaction involving oxidation and reduction there must be
a species giving electrons and another species gaining
electrons. Hence the term redox reaction.

9. Deduce which of these reactions are redox reactions. Assign
oxidation no’s to each atom. In the case of redox
reactions identify the oxidising and reducing agents.
1) Ca
(s)
+

Cl
2(g)
 CaCl
2(s)
2) Cl
2(g)
+ H
2
S
(g)
 2HCl
(g)
+ S
(s)
3) Ba(NO
3
)
2(aq)
+ Na
2
SO
4(aq)
 BaSO
4(s)
+ 2NaNO
3(aq)
4) Ni
(s)
+ CuSO
4(s)
 NiSO
4(aq)
+ Cu
(s)
5) Cl
2
O
7(g)
+ H
2
O
(l)
 2HClO
4(aq)
6) 2CuCl
(aq)  CuCl
2(aq) + Cu
(s)

10. We can add 2 half equations together to make a full equation.
Oxidation and reduction must both occur together.
The species donating electrons is called the reducing agent
The species receiving electrons is called the oxidising agent

11. So we can add
Fe
2+
- e  Fe
3+
and
MnO
4
-
+ 5e  Mn
2+

to make a full equation
In this example Mn
7+
is being reduced and is the oxidising
agent. It is receiving electrons from the Fe
2+
Fe
2+
is being oxidised and is the reducing agent. It is giving
electrons to the Mn
7+
.

12. 1) Fe
2+
- e  Fe
3+
2) MnO
4
-
+ 5e  Mn
2+
In order to add the 2 half equations we

first have to balance the
electrons by multiplying 1) by 5
5Fe
2+
- 5e  5 Fe
3+
MnO
4
-
+ 5e  Mn
2+
5Fe
2+
+ MnO
4
-
 5Fe
3+
+ Mn
2+

13. Form the full equation for the reaction that occurs between
potassium manganate(VII) and hydrogen peroxide
The half equations are;
MNO
4
-
(aq)
+ 8H
+
(aq)
+ 5e  Mn
2+
(aq)

+ 4H
2
O
(l)
H
2
O
2(aq)
 O
2(g)
+ 2H
+
(aq)
+ 2e
2MnO
4
-
(aq)
+ 6H
+
(aq)
+ 5H
2
O
2(l)
 2Mn
2+
(aq)
+ 8H
2
O
(l)
+ 5O
2(g)

14. Write balance ion equations for the following and determine
whether oxidation or reduction occurs
b)chlorine to chloride ions
c)oxygen to oxide ions
d)aluminium to aluminium ions
e)magnesium to magnesium ions
f)tin(IV) ions to tin (II)ions
g)chromium(II) to chromium(III) ions
O
2
+ 4e  2O
2-
ox no 0 to -2 reduction
Al -3e  Al
3+
ox no 0 to +3 oxidation
Mg -2e  Mg
2+
ox no 0 to +2 oxidation
Sn
4+
+2e  Sn
2+
ox no

+4 to +2

reduction
Cl
2
+ 2e  2Cl
-
ox no 0 to -1 reduction

15. Construct full equations from the half equations in the data
sheets for the following reactions.
H
2O
2 reacting with either Fe
2+
or Fe
3+
S
2
O
3
2-
reacting with either I
2
or I
-
MnO
4
-
with either Sn
2+
or Sn
4+