Download Redox Reactions

Chapter 19
Redox Equilibrium I:
Redox Reactions
19.1 Redox Reactions
19.2 Balancing Redox Equations
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New Way Chemistry for Hong Kong A-Level Book 2
19.1 Redox Reactions (SB p.172)
Redox Reactions
Oxidation
Reduction
Addition of oxygen
Removal of oxygen
Removal of hydrogen
Addition of hydrogen
Loss of electron
Gain of electron
Increase in O.N.
Decrease in O.N.
(An 'imaginary charge')
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New Way Chemistry for Hong Kong A-Level Book 2
19.1 Redox Reactions (SB p.175)
Rules for Assigning Oxidation Number
1. The oxidation number of an element is 0.
2. For a simple ionic compound, the oxidation
number of a constituent element is the same as
the charge on the ion.
3. The oxidation number of some elements in their
compounds are always the same.
4. In an uncharged compound, the sum of
oxidation numbers of all constituent atoms is 0.
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New Way Chemistry for Hong Kong A-Level Book 2
19.1 Redox Reactions (SB p.176)
5. In polyatomic ion, the sum of oxidation numbers
of all constituent atoms is equal to the charge of
the ion.
6. The oxidation number of a constituent element in a
compound can be obtained by arithmetic
calculation. This is done by first assigning
reasonable oxidation numbers to the other
elements.
7. The oxidation number of an element can be
different in different compounds.
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New Way Chemistry for Hong Kong A-Level Book 2
19.1 Redox Reactions (SB p.172)
Definition of oxidizing and reducing agents
Oxidizing agent
1. Undergoes reduction.
Reducing agent
Undergoes oxidation.
2. Oxidizes a reducing agent. Reduces an oxidizing agent.
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3. Gain of electron(s).
Loss of electron(s).
4. Decrease in oxidation
number.
Increase in oxidation
number.
New Way Chemistry for Hong Kong A-Level Book 2
19.2 Balancing Redox Equations (SB p.180)
Half Equation Method
Step Reduction
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MnO4-(aq)
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Balance the no. of O atoms by adding H2O
Balance the no. of I atoms:
molecules:
2I-(aq)
I2(aq)
2+
MnO4 (aq)
Mn (aq) + 4H2O(l)
Balance the no. of H atoms by adding H+(aq)
ions:
MnO4-(aq) + 8H+(aq) Mn2+(aq) + 4H2O(l)
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Balance the no. of charges:
Balance the no. of charges:
(with a charge of +2 on both sides)
(with a charge of –2 on both
sides)
MnO4-(aq) + 8H+(aq) + e2I-(aq)
I2(aq) + 2e-…(2)
Mn2+(aq) + 4H2O(l)…(1)
Combine the two half equations and eliminate electrons:
For (1) x 2 =(3) and (2) x 5 =(4)
(3) + (4) : 2MnO4-(aq) + 10I-(aq) + 16H+(aq)
2Mn2+(aq) + 5I2(aq) + 8H2O(l)
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Oxidation
Mn2+(aq)
I -(aq)
New Way Chemistry for Hong Kong A-Level Book 2
I2(aq)
19.2 Balancing Redox Equations (SB p.182)
Oxidation Number Method
Example
1. BrO3-(aq) + I-(aq)
+5
Br-(aq) + I2(aq)
-1
-1
0
O.N. increased by 1
2. BrO3-(aq) + I-(aq)
+5
-1
Br-(aq) + I2(aq)
-1
0
O.N. decreased by 6
3. BrO3-(aq) +6 I-(aq)
Br-(aq) + 3 I2(aq)
4. BrO3-(aq) +6 I-(aq)
Br-(aq) + 3 I2(aq) + 3H2O(l)
5. BrO3-(aq) +6 I-(aq) + 6H+
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Br-(aq) + 3 I2(aq) + 3H2O(l)
New Way Chemistry for Hong Kong A-Level Book 2
The END
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New Way Chemistry for Hong Kong A-Level Book 2