Year 12 Chemistry: Chapter 5 Analysing Oxidants and Reductants
5.1 What is a redox reaction?
 One of the reactants loses electrons – this process is called oxidation.
 One of the reactants gains these electrons – this is called reduction.
 Oxidation and Reduction occur simultaneously.
Oxidation:
Reduction:
Mg  Mg2+ + 2eO2 + 4e-  2O2-
Overall:
O
I
L
R
I
G
Substances that cause oxidation to occur are called oxidising agents or oxidants. Note
that the oxidant causes oxidation but itself is always reduced. Oxygen is the oxidant.
Substances that cause reduction to occur are called reducing agents or reductants.
Reductants cause reduction but are always themselves oxidised. Magnesium is the
reductant.
5.2 Oxidation numbers: classifying redox reactions
To determine whether a reaction is a redox reaction, oxidation numbers are assigned to
elements involved in the reaction. Oxidation numbers have no physical meaning.
Oxidation numbers are also known as oxidation states.
Oxidation number Rules
 Free element is zero
o Na, Cl, Cl2, P, H2, Mg etc
 A simple ion is equal to the charge on the ion
o Na+ = +1
o Cl= -1
2+
o Mg
= +2
2o S
= -2
3+
o Al
= +3
 In compounds, elements have fixed numbers except for a few exceptional
circumstances.
o Main group metals have an oxidation number equal to their valency.
o Fluorine has an oxidation number of -1.
o Hydrogen has an oxidation number of +1 when it forms compounds with
non-metals and -1 in metal hydrides such as NaH and CaH2.
o Oxygen almost always has an oxidation number of -2. In F2O it has an
oxidation number of +2 (Fluorine is more electronegative) and in peroxides
such as H2O2 and BaO2, its oxidation number is -1.
 The sum of the oxidation number in a neutral compound is zero
MgCl2
Na2S
 The sum of the oxidation number in a polyatomic ion is equal to the charge on the
ion.
AlCl4 The most electronegative element in a compound has the negative oxidation
number.
Examples:
1. Hydrogen = +1 and Oxygen = -2
 HNO3
2. Ionic charge 2- and oxygen = -2
 CO323. Fluorine = -1 and Oxygen = -2
 F 2O
Variable oxidation number
The oxidation numbers of the transition metals vary depending upon the compound. The
different oxidation states of the transition metals often have characteristic colours.
*Copy figure 5.5
Changing oxidation numbers
 An increased oxidation number means the element has been oxidised.
 A decreased oxidation number means the element has been reduced.
Example:
 C(s) + O2(g)  CO2(g)
 2C(s) + O2(g)  CO(g)
 2CO(g) + O2(g)  CO(g)
5.3 Writing Half Equations
16.4 Half equations for complex redox reactions
K O H E S
Any school laboratory could have a series of strong oxidants (potassium permanganate,
potassium dichromate and potassium chromate) that can be employed when a substance
needs to be oxidised. For safety reasons, these are often stored together and well away
from flammable materials.
Complex redox reaction:
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
Iron (II) ions to iron (III) ion:
MnO4- ion to Mn2+ ion
(K) Step 1: Balance all atoms in the half equation except Oxygen or Hydrogen
(O) Step 2: Balance the oxygen atoms by adding WATER.
(H) Step 3: Balance the hydrogen atoms by adding HYDROGEN (H+)
(E) Step 4: Balance the charges by adding ELECTRONS.
(S) Step 5: Add STATES
*Write the overall equation, show the two half equations.
Example: Potassium dichromate (K2Cr2O7) reacts with potassium iodide (KI) in acidified
solution, The dichromate ion (Cr2O7) is reduced to Cr3+, and the iodide ion (I-) is oxidised
to I2.
Write:
a) oxidation equation of the I- to I2.
b) Reduction equation of the Cr2O7 to Cr3+.
K
O
H
E
S
c) Overall equation
QUESTIONS: 1, 3, 6, 16, 7, 10, 11, 19, 12, 13, 20, 24, 25, 29, 33, 34
Year 12 Chemistry: Chapter 5 Redox reaction
5.4 Volumetric Analysis
Redox titration involves reaction of an oxidant and reductant. For some reactions, for
example those involving the permanganate ion (MnO4-), the equivalence point will be
indicated by a colour change in the reacting solutions, whereas for others an indicator
must be added in order to detect the equivalence point.
Volumetric analysis involving redox reactions can be used to determine the composition
of many substances.
Item
Wine
Ingredient for analysis
Ethanol
Wine
Fruit Juice
Household bleach
Sulfur dioxide
Vitamin C (ascorbic acid)
Hypochlorite ion
Hair bleach
Hydrogen peroxide
Titrate with
React with potassium dichromate
then titre with iron (II) solution
Iodine solution
Iodine solution
React with acidified potassium iodide
solution and titrate with sodium
thiosulfate solution
Potassium permanganate solution
Example: A 10.0 mL sample of white wine was placed in a volumetric flask and water was
added to make 100.0 mL of solution. Then 20.0 mL aliquots of the diluted wine were
titrated against 0.100M acidified potassium dichromate (K2Cr2O7) solution. The mean
titre was 24.61 mL. Calculate the concentration of ethanol in the sample of white wine.
2Cr2O72- (aq) + 3CH3CH2OH (aq) + 16H+ (aq)  4Cr3+ (aq) + CH3COOH (aq) + 11H2O (l)
Step 2: Find the amount of the known
Step 3: mole ratio
Step 4: Find the amount of the unknown by using the dilution factor
Step 5: Find the concentration of the unknown
Questions