1. Chapter 5: Introduction to Reactions in Aqueous Solutions
Chemistry 140 Fall 2002
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 5: Introduction to Reactions in Aqueous Solutions
Philip Dutton
University of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
(modified 2003 by Dr. Paul Root and 2005 by Dr. David Tramontozzi)
General Chemistry: Chapter 5
Prentice-Hall © 2002

2. General Chemistry: Chapter 5
Contents
5-1 The Nature of Aqueous Solutions
5-2 Precipitation Reactions
5-3 Acid-Base Reactions
5-4 Oxidation-Reduction: Some General Principles
5-5 Balancing Oxidation-Reduction Equations
5-6 Oxidizing and Reducing Agents
5-7 Stoichiometry of Reactions in Aqueous Solutions: Titrations
Focus on Water Treatment
General Chemistry: Chapter 5
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3. Example 5-6
Balancing the Equation for a Redox Reaction in Acidic Solution.
The reaction described below is used to determine the sulfite ion concentration present in wastewater from a papermaking plant. Write the balanced equation for this reaction in acidic solution..
SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq)
Determine the oxidation states: 4+ 7+ 6+ 2+
SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq)
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4. Example 5-6 Write the half-reactions: SO32-(aq) → SO42-(aq)
MnO4-(aq) → Mn2+(aq)
Balance atoms other than H and O:
Already balanced for elements.
SO32-(aq) → SO42-(aq)
MnO4-(aq) → Mn2+(aq)
General Chemistry: Chapter 5
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5. Example 5-6 Balance O by adding H2O: H2O(l) + SO32-(aq) → SO42-(aq)
MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
Balance hydrogen by adding H+:
H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq)
8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
Check that the charges are balanced: Add e- if necessary.
H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq) + 2e-
5e- + 8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
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6. Example 5-6 Multiply the half-reactions to balance all e-:
[ H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq) + 2e- ] x 5
[ 5e- + 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l) ] x 2
5 H2O(l) + 5 SO32-(aq) → 5 SO42-(aq) + 10 e H+(aq) 6 3
16 H+(aq) + 10 e- + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l)
Add both equations and simplify (eliminate things common on both sides):
5 SO32-(aq) + 2 MnO4-(aq) + 6H+(aq) →
5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l)
Check the balance (atoms and charges)!
General Chemistry: Chapter 5
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7. General Chemistry: Chapter 5
Balancing in Acid
Write the equations for the half-reactions.
Balance all atoms except H and O.
Balance oxygen using H2O.
Balance hydrogen using H+.
Balance charge using e-.
Equalize the number of electrons.
Add the half reactions.
Check the balance.
General Chemistry: Chapter 5
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8. Balancing in Basic Solution
OH- appears instead of H+.
Treat the equation as if it were in acid.
Then add OH- to each side to neutralize H+.
Remove H2O appearing on both sides of equation.
Check the balance.
General Chemistry: Chapter 5
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9. Example 5-7
Balancing the Equation for a Redox Reaction in Basic Solution.
Balance the equation for the reaction in which permanganate ion oxidizes cyanide ion to cyanate ion in basic solution and is itself reduced to MnO2.
MnO4-(aq) + CN-(aq) → MnO2(s) + OCN-(aq)
The first steps are the same as for balancing in acidic solution. Write half-reactions, balance elements, balance O, balance H, balance charge and combine to remove electrons.
2 MnO4-(aq) + 3 CN-(aq) + 2 H+(aq) →
2 MnO2(s) + 3 OCN-(aq) + H2O (l)
General Chemistry: Chapter 5
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10. General Chemistry: Chapter 5
Example 5-7
Knowing that H+ and OH- combine to form water we now add enough OH- to remove all of the H+
2 H2O(l)
2 MnO4-(aq) + 3 CN-(aq) + 2 H+(aq) + 2 OH-(aq) →
2 MnO2(s) + 3 OCN-(aq) + H2O(l) + 2 OH-(aq)
Eliminate water which is common to both sides to get the final balanced equation in basic media
2 MnO4-(aq) + 3 CN-(aq) + 2 H2O(l) →
2 MnO2(s) + 3 OCN-(aq) + H2O(l) + 2 OH-(aq)
2 MnO4-(aq) + 3 CN-(aq) + H2O(l) →
2 MnO2(s) + 3 OCN-(aq) + 2 OH-(aq)
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11. 5-6 Oxidizing and Reducing Agents
An oxidizing agent (oxidant):
Contains an element whose oxidation state decreases (is reduced) in a redox reaction
A reducing agent (reductant):
Contains an element whose oxidation state increases (is oxidized) in a redox reaction
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12. Redox Strong reducing agent N2O4(l) + 2 N2H4(l)  3 N2(g) + 4 H2O(g)
Strong oxidizing agent
‘GREEN’ Products
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13. General Chemistry: Chapter 5
Example 5-8
Identifying Oxidizing and Reducing Agents.
Hydrogen peroxide, H2O2, is a versatile chemical. Its uses include bleaching wood pulp and fabrics and substituting for chlorine in water purification. One reason for its versatility is that it can be either an oxidizing or a reducing agent. For the following reactions, identify whether hydrogen peroxide is an oxidizing or reducing agent.
General Chemistry: Chapter 5
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14. General Chemistry: Chapter 5
Chemistry 140 Fall 2002
Example 5-8
H2O2(aq) + 2 Fe2+(aq) + 2 H+(aq) → 2 H2O(l) + 2 Fe3+(aq)
Iron is oxidized and peroxide is reduced.
5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) →
8 H2O(l) + 2 Mn2+(aq) + 5 O2(g)
Fe(2) is oxidized to Fe(3). Therefore peroxide is an oxidizing agent. Peroxide is reduced to water.
Mn(7) is reduced to Mn(2). Therefore peroxide is a reducing agent. Peroxide is oxidized to molecular oxygenl.
Manganese is reduced and peroxide is oxidized.
General Chemistry: Chapter 5
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15. 5-7 Stoichiometry of Reactions in Aqueous Solutions: Titrations.
Carefully controlled addition of one solution to another.
Equivalence Point
Both reactants have reacted completely.
Indicators
Substances which change color near an equivalence point.
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16. General Chemistry: Chapter 5
Indicators
General Chemistry: Chapter 5
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17. General Chemistry: Chapter 5
Example 5-10
Standardizing a Solution for Use in Redox Titrations.
A piece of iron wire weighing g is converted to Fe2+(aq) and requires mL of a KMnO4(aq) solution for its titration. What is the molarity of the KMnO4(aq)?
5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) →
4 H2O(l) + 5 Fe3+(aq) + Mn2+(aq)
General Chemistry: Chapter 5
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18. General Chemistry: Chapter 5
Example 5-10
5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) →
4 H2O(l) + 5 Fe3+(aq) + Mn2+(aq)
Determine KMnO4 consumed in the reaction: 4
Determine the concentration:
General Chemistry: Chapter 5
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19. General Chemistry: Chapter 5
Chemistry 140 Fall 2002
Chapter 5 Questions
1, 3, 5, 6, 8, 14, 16, 17, 19, 24, 25, 27, 33, 37, 41, 43, 51, 53, 59, 68, 71, 82, 96.
General Chemistry: Chapter 5
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