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

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 Prentice-Hall © 2002

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) General Chemistry: Chapter 5 Prentice-Hall © 2002

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 Prentice-Hall © 2002

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) General Chemistry: Chapter 5 Prentice-Hall © 2002

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- + 10 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 Prentice-Hall © 2002

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 Prentice-Hall © 2002

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 Prentice-Hall © 2002

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 Prentice-Hall © 2002

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) General Chemistry: Chapter 5 Prentice-Hall © 2002

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 General Chemistry: Chapter 5 Prentice-Hall © 2002

Redox Strong reducing agent N2O4(l) + 2 N2H4(l)  3 N2(g) + 4 H2O(g) Strong oxidizing agent ‘GREEN’ Products General Chemistry: Chapter 5 Prentice-Hall © 2002

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 Prentice-Hall © 2002

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 Prentice-Hall © 2002

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. General Chemistry: Chapter 5 Prentice-Hall © 2002

General Chemistry: Chapter 5 Indicators General Chemistry: Chapter 5 Prentice-Hall © 2002

General Chemistry: Chapter 5 Example 5-10 Standardizing a Solution for Use in Redox Titrations. A piece of iron wire weighing 0.1568 g is converted to Fe2+(aq) and requires 26.42 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 Prentice-Hall © 2002

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 Prentice-Hall © 2002

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 Prentice-Hall © 2002