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CHAPTER 7
OXIDATION AND
REDUCTION REACTIONS
REDOX REACTIONS

Redox reactions:
- oxidation and reduction reactions that
occurs simultaneously.

Oxidation:
1) addition of oxygen
2) loss of hydrogen
3) loss of electron
4) increase in the oxidation state / oxidation
number

Reduction:
1) loss of oxygen
2) addition of hydrogen
3) gain of electrons
4) decrease in the oxidation state /
oxidation number
Electron Transfer Reactions
• Oxidation reactions: half-reaction that involves loss
of electrons
• Reduction reactions: half-reaction that involves gain
of electrons
• Oxidizing agents: substance that accepts electrons
(substances that are reduced)
• Reducing agents: substances that donates electrons
(substances that are oxidized)
Electron Transfer Reactions
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
2Mg
O2 + 4e-
2O2-
Reduction half-reaction (gain e-)
2Mg + O2 + 4e-
2Mg + O2
2Mg2+ + 2O2- + 4e-
2MgO
Examples:
1)
Zn (s) + CuSO4 (aq)
Zn
ZnSO4 (aq) + Cu (s)
Zn2+ + 2e- Zn is oxidized
Cu2+ + 2e-
Cu Cu2+ is reduced
Zn is the reducing agent
Cu2+ is the oxidizing agent
2) Copper wire reacts with silver nitrate to form silver metal.
What is the oxidizing agent in the reaction?
Cu (s) + 2AgNO3 (aq)
Cu
Cu(NO3)2 (aq) + 2Ag (s)
Cu2+ + 2e-
Ag+ + 1e-
Ag
Ag+ is reduced Ag+ is the oxidizing agent
Oxidation number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an oxidation
number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal
to the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually –2. In
H2O2 and O22- it is –1.
4. The oxidation number of hydrogen is +1 except
when it is bonded to metals in binary compounds.
In these cases, its oxidation number is –1.
5. Group IA metals are +1, IIA metals are +2 and
fluorine is always –1.
6. The sum of the oxidation numbers of all the
atoms in a molecule or ion is equal to the charge
on the molecule or ion.
7. Oxidation numbers do not have to be integers.
Oxidation number of oxygen in the superoxide
ion, O2-, is –½.
The Oxidation Numbers of Elements in their Compounds
Examples:
What are the oxidation numbers of all the elements in
this folowing compounds or ion?
i) HCO3-
O = –2
H = +1
3x(–2) + 1 + ? = –1
C = +4
ii) NaIO3
Na = +1
O = -2
3x(-2) + 1 + ? = 0
I = +5
iii) IF7
iv) K2Cr2O7
v) Li2O
vi) HNO3
Titrations in Redox Reactions
5Fe2+ + MnO4- + 8H+
Mn2+ + 5Fe3+ + 4H2O
Dark purple (MnO4-) to light pink (Mn2+)
16.42 mL of 0.1327 M KMnO4 solution is needed to oxidize
25.00 mL of an acidic FeSO4 solution. What is the molarity of
the iron solution?
5Fe2+ + MnO4- + 8H+
volume red
M
red
moles red
16.42 mL = 0.01642 L
0.01642 L x
0.1327 mol KMnO4
1L
Mn2+ + 5Fe3+ + 4H2O
x
rxn
coef.
moles oxid
V
oxid
M oxid
25.00 mL = 0.02500 L
5 mol Fe2+
1 mol KMnO4
x
1
0.02500 L Fe2+
= 0.4358 M
Balancing Redox Equations

Ion electron method
Balancing Redox Equations for
Reaction Takes Place in Acidic
Solution
Following is the guideline for balancing
an oxidation-reduction equation in acidic
solution by the ion-electron method.
 Step 1:
- Write the two half-reactions that contain
the elements being oxidised and reduced
using the entire formula of the ion or
molecule.

Step 2:
- Balance all the elements except
oxygen and hydrogen.
 Step 3:
- Balance oxygen using H2O and
hydrogen using H+.
 Step 4:
- Balance the charge using electrons.

If necessary, multiplying one or both balanced
half-reaction by integers to equalize the
number of electrons transferred in the two half
reactions.

Step 5:
- Add the half-reactions and cancel
identical species.
- Verifying – check to be sure that the
elements and charges are balanced.

Example 1:
Balance the following equation by the ionelectron method.
Cr2O72- + ClCr3+ + Cl2

Solution:
Determine the oxidation number of the
species involved.
2Oxn(Cr) + 7Oxn(O) = -2
2Oxn(Cr) + 7(-2) = -2
Oxn(Cr) = +6
The oxidation number of Cr is +6
The oxidation number of Cl- is -1 whereas
the oxidation number of Cl2 is zero.
Step 1:The ionic equation is divided in 2
distinct half-reactions. One
representing oxidation and the other
representing reduction.
Cr2O72- (aq)
Cr3+ (aq) ..(1)(reduction)
Cl- (aq)
Cl2
..(2)(oxidation)
Step 2 : Balance the atoms in each halfequation separately (except H and O)
Cr2O72- (aq)
2Cl- (aq)
2Cr3+ (aq) ……(3)
Cl2
……(4)
Step 3 : For reaction in acid medium, add
H2O to balance the O atoms and H+
to balance the H atoms.
14H+ + Cr2O72- (aq)
2Cl- (aq)
2Cr3+ (aq) + 7H2O
…(5)
Cl2
…(6)
Step 4: Balance the charges of each halfreaction by adding electron(s) to
either the left or right side of the
equation.
If the electron(s) appears on the
right, the process is an oxidation
process.
If electron(s) appears on the left, it is
a reduction process.
-
-
The net charge on the left is +12. The net charge
of Cr on the right is +6.
Therefore, we add 6 electrons (-6) to the left so
that the net charges on both sides are equal +6
6 e- + 14H+ + Cr2O72- (aq)
-
-
2Cr3+ (aq) + 7H2O
…(7)
For the 2nd half-equation, the net charges on the
left is -2 while the net charge on the right is 0.
Therefore, we add 2 electrons to the right so that
the net charges on both sides are equal -2.
2Cl- (aq)
Cl2 + 2 e…(8)
-
-
Equalise the number of electrons in both
half-equations by multiplying one or two
half-equations with appropriate
coefficients.
By looking at the 2 half equations, the 2nd
half-equation is multiplied by 3 to balance
the electrons on both half-reaction.
Cr2O72- (aq) + 14H+ + 6 e2Cl- (aq)
(equation (8)  3) 6Cl- (aq)
2Cr3+ (aq) + 7H2O
Cl2 + 2 e3Cl2 + 6 e-
…(7)
…(8)
…(9)
Step 5: Then, add the 2 half-reaction
equations together and balance the
final equation by inspection.
The electrons on both sides should
be cancelled.
Cr2O72- (aq) + 14H+ + 6 e6Cl- (aq)
(7) + (9) Cr2O72-
2Cr3+ (aq) + 7H2O …(7)
3Cl2 + 6 e…(9)
(aq) + 14H+ + 6Cl- (aq)
2Cr3+ (aq) + 7H2O
+ 3Cl2