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Chemistry: Atoms First

16.1 Balancing Oxidation-Reduction Reactions

Chemistry: Atoms First16.1 Balancing Oxidation-Reduction Reactions
  1. Preface
  2. 1 Essential Ideas
    1. Introduction
    2. 1.1 Chemistry in Context
    3. 1.2 Phases and Classification of Matter
    4. 1.3 Physical and Chemical Properties
    5. 1.4 Measurements
    6. 1.5 Measurement Uncertainty, Accuracy, and Precision
    7. 1.6 Mathematical Treatment of Measurement Results
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  3. 2 Atoms, Molecules, and Ions
    1. Introduction
    2. 2.1 Early Ideas in Atomic Theory
    3. 2.2 Evolution of Atomic Theory
    4. 2.3 Atomic Structure and Symbolism
    5. 2.4 Chemical Formulas
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  4. 3 Electronic Structure and Periodic Properties of Elements
    1. Introduction
    2. 3.1 Electromagnetic Energy
    3. 3.2 The Bohr Model
    4. 3.3 Development of Quantum Theory
    5. 3.4 Electronic Structure of Atoms (Electron Configurations)
    6. 3.5 Periodic Variations in Element Properties
    7. 3.6 The Periodic Table
    8. 3.7 Molecular and Ionic Compounds
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  5. 4 Chemical Bonding and Molecular Geometry
    1. Introduction
    2. 4.1 Ionic Bonding
    3. 4.2 Covalent Bonding
    4. 4.3 Chemical Nomenclature
    5. 4.4 Lewis Symbols and Structures
    6. 4.5 Formal Charges and Resonance
    7. 4.6 Molecular Structure and Polarity
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  6. 5 Advanced Theories of Bonding
    1. Introduction
    2. 5.1 Valence Bond Theory
    3. 5.2 Hybrid Atomic Orbitals
    4. 5.3 Multiple Bonds
    5. 5.4 Molecular Orbital Theory
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  7. 6 Composition of Substances and Solutions
    1. Introduction
    2. 6.1 Formula Mass and the Mole Concept
    3. 6.2 Determining Empirical and Molecular Formulas
    4. 6.3 Molarity
    5. 6.4 Other Units for Solution Concentrations
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  8. 7 Stoichiometry of Chemical Reactions
    1. Introduction
    2. 7.1 Writing and Balancing Chemical Equations
    3. 7.2 Classifying Chemical Reactions
    4. 7.3 Reaction Stoichiometry
    5. 7.4 Reaction Yields
    6. 7.5 Quantitative Chemical Analysis
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  9. 8 Gases
    1. Introduction
    2. 8.1 Gas Pressure
    3. 8.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
    4. 8.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions
    5. 8.4 Effusion and Diffusion of Gases
    6. 8.5 The Kinetic-Molecular Theory
    7. 8.6 Non-Ideal Gas Behavior
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  10. 9 Thermochemistry
    1. Introduction
    2. 9.1 Energy Basics
    3. 9.2 Calorimetry
    4. 9.3 Enthalpy
    5. 9.4 Strengths of Ionic and Covalent Bonds
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  11. 10 Liquids and Solids
    1. Introduction
    2. 10.1 Intermolecular Forces
    3. 10.2 Properties of Liquids
    4. 10.3 Phase Transitions
    5. 10.4 Phase Diagrams
    6. 10.5 The Solid State of Matter
    7. 10.6 Lattice Structures in Crystalline Solids
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  12. 11 Solutions and Colloids
    1. Introduction
    2. 11.1 The Dissolution Process
    3. 11.2 Electrolytes
    4. 11.3 Solubility
    5. 11.4 Colligative Properties
    6. 11.5 Colloids
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  13. 12 Thermodynamics
    1. Introduction
    2. 12.1 Spontaneity
    3. 12.2 Entropy
    4. 12.3 The Second and Third Laws of Thermodynamics
    5. 12.4 Free Energy
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  14. 13 Fundamental Equilibrium Concepts
    1. Introduction
    2. 13.1 Chemical Equilibria
    3. 13.2 Equilibrium Constants
    4. 13.3 Shifting Equilibria: Le Châtelier’s Principle
    5. 13.4 Equilibrium Calculations
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  15. 14 Acid-Base Equilibria
    1. Introduction
    2. 14.1 Brønsted-Lowry Acids and Bases
    3. 14.2 pH and pOH
    4. 14.3 Relative Strengths of Acids and Bases
    5. 14.4 Hydrolysis of Salt Solutions
    6. 14.5 Polyprotic Acids
    7. 14.6 Buffers
    8. 14.7 Acid-Base Titrations
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  16. 15 Equilibria of Other Reaction Classes
    1. Introduction
    2. 15.1 Precipitation and Dissolution
    3. 15.2 Lewis Acids and Bases
    4. 15.3 Multiple Equilibria
    5. Key Terms
    6. Key Equations
    7. Summary
    8. Exercises
  17. 16 Electrochemistry
    1. Introduction
    2. 16.1 Balancing Oxidation-Reduction Reactions
    3. 16.2 Galvanic Cells
    4. 16.3 Standard Reduction Potentials
    5. 16.4 The Nernst Equation
    6. 16.5 Batteries and Fuel Cells
    7. 16.6 Corrosion
    8. 16.7 Electrolysis
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  18. 17 Kinetics
    1. Introduction
    2. 17.1 Chemical Reaction Rates
    3. 17.2 Factors Affecting Reaction Rates
    4. 17.3 Rate Laws
    5. 17.4 Integrated Rate Laws
    6. 17.5 Collision Theory
    7. 17.6 Reaction Mechanisms
    8. 17.7 Catalysis
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  19. 18 Representative Metals, Metalloids, and Nonmetals
    1. Introduction
    2. 18.1 Periodicity
    3. 18.2 Occurrence and Preparation of the Representative Metals
    4. 18.3 Structure and General Properties of the Metalloids
    5. 18.4 Structure and General Properties of the Nonmetals
    6. 18.5 Occurrence, Preparation, and Compounds of Hydrogen
    7. 18.6 Occurrence, Preparation, and Properties of Carbonates
    8. 18.7 Occurrence, Preparation, and Properties of Nitrogen
    9. 18.8 Occurrence, Preparation, and Properties of Phosphorus
    10. 18.9 Occurrence, Preparation, and Compounds of Oxygen
    11. 18.10 Occurrence, Preparation, and Properties of Sulfur
    12. 18.11 Occurrence, Preparation, and Properties of Halogens
    13. 18.12 Occurrence, Preparation, and Properties of the Noble Gases
    14. Key Terms
    15. Summary
    16. Exercises
  20. 19 Transition Metals and Coordination Chemistry
    1. Introduction
    2. 19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
    3. 19.2 Coordination Chemistry of Transition Metals
    4. 19.3 Spectroscopic and Magnetic Properties of Coordination Compounds
    5. Key Terms
    6. Summary
    7. Exercises
  21. 20 Nuclear Chemistry
    1. Introduction
    2. 20.1 Nuclear Structure and Stability
    3. 20.2 Nuclear Equations
    4. 20.3 Radioactive Decay
    5. 20.4 Transmutation and Nuclear Energy
    6. 20.5 Uses of Radioisotopes
    7. 20.6 Biological Effects of Radiation
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  22. 21 Organic Chemistry
    1. Introduction
    2. 21.1 Hydrocarbons
    3. 21.2 Alcohols and Ethers
    4. 21.3 Aldehydes, Ketones, Carboxylic Acids, and Esters
    5. 21.4 Amines and Amides
    6. Key Terms
    7. Summary
    8. Exercises
  23. A | The Periodic Table
  24. B | Essential Mathematics
  25. C | Units and Conversion Factors
  26. D | Fundamental Physical Constants
  27. E | Water Properties
  28. F | Composition of Commercial Acids and Bases
  29. G | Standard Thermodynamic Properties for Selected Substances
  30. H | Ionization Constants of Weak Acids
  31. I | Ionization Constants of Weak Bases
  32. J | Solubility Products
  33. K | Formation Constants for Complex Ions
  34. L | Standard Electrode (Half-Cell) Potentials
  35. M | Half-Lives for Several Radioactive Isotopes
  36. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
    18. Chapter 18
    19. Chapter 19
    20. Chapter 20
    21. Chapter 21
  37. Index

Learning Objectives

By the end of this section, you will be able to:
  • Define electrochemistry and a number of important associated terms
  • Split oxidation-reduction reactions into their oxidation half-reactions and reduction half-reactions
  • Produce balanced oxidation-reduction equations for reactions in acidic or basic solution
  • Identify oxidizing agents and reducing agents

Electricity refers to a number of phenomena associated with the presence and flow of electric charge. Electricity includes such diverse things as lightning, static electricity, the current generated by a battery as it discharges, and many other influences on our daily lives. The flow or movement of charge is an electric current (Figure 16.2). Electrons or ions may carry the charge. The elementary unit of charge is the charge of a proton, which is equal in magnitude to the charge of an electron. The SI unit of charge is the coulomb (C) and the charge of a proton is 1.602 ×× 10−19 C. The presence of an electric charge generates an electric field. Electric current is the rate of flow of charge. The SI unit for electrical current is the SI base unit called the ampere (A), which is a flow rate of 1 coulomb of charge per second (1 A = 1 C/s). An electric current flows in a path, called an electric circuit. In most chemical systems, it is necessary to maintain a closed path for current to flow. The flow of charge is generated by an electrical potential difference, or potential, between two points in the circuit. Electrical potential is the ability of the electric field to do work on the charge. The SI unit of electrical potential is the volt (V). When 1 coulomb of charge moves through a potential difference of 1 volt, it gains or loses 1 joule (J) of energy. Table 16.1 summarizes some of this information about electricity.

Common Electrical Terms
Quantity Definition Measure or Unit
Electric charge Charge on a proton 1.602 ×× 10−19 C
Electric current The movement of charge ampere = A = 1 C/s
Electric potential The force trying to move the charge volt = V = J/C
Electric field The force acting upon other charges in the vicinity
Table 16.1
Three photographs are shown in this figure. The first shows lightning against a dark evening sky. The second shows a child at the open base of a green plastic playground tube slide. The child’s hair is sticking up and the child’s shadow on the base of the slide shows the child’s hair sticking up and out in all directions. The final picture shows a 9 volt battery from which red and blue coated wire that is twisted together extend from the battery terminals to the lower region of a yellow platform or board. Above this region are six resistors in a horizontal row, evenly spaced horizontally across the span of the board. Green, blue, and white wires are also visible on the board. 6 orange L E D light bulbs extend from the upper edge of the platform in a horizontal line parallel to the pegs.
Figure 16.2 Electricity-related phenomena include lightning, accumulation of static electricity, and current produced by a battery. (credit left: modification of work by Thomas Bresson; credit middle: modification of work by Chris Darling; credit right: modification of work by Windell Oskay)

Electrochemistry studies oxidation-reduction reactions, which were first discussed in an earlier chapter, where we learned that oxidation was the loss of electrons and reduction was the gain of electrons. The reactions discussed tended to be rather simple, and conservation of mass (atom counting by type) and deriving a correctly balanced chemical equation were relatively simple. In this section, we will concentrate on the half-reaction method for balancing oxidation-reduction reactions. The use of half-reactions is important partly for balancing more complicated reactions and partly because many aspects of electrochemistry are easier to discuss in terms of half-reactions. There are alternate methods of balancing these reactions; however, there are no good alternatives to half-reactions for discussing what is occurring in many systems. The half-reaction method splits oxidation-reduction reactions into their oxidation “half” and reduction “half” to make finding the overall equation easier.

Electrochemical reactions frequently occur in solutions, which could be acidic, basic, or neutral. When balancing oxidation-reduction reactions, the nature of the solution may be important. It helps to see this in an actual problem. Consider the following unbalanced oxidation-reduction reaction in acidic solution:

MnO4(aq)+Fe2+(aq)Mn2+(aq)+Fe3+(aq)MnO4(aq)+Fe2+(aq)Mn2+(aq)+Fe3+(aq)

We can start by collecting the species we have so far into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction. Each of these half-reactions contain the same element in two different oxidation states. The Fe2+ has lost an electron to become Fe3+; therefore, the iron underwent oxidation. The reduction is not as obvious; however, the manganese gained five electrons to change from Mn7+ to Mn2+.

oxidation (unbalanced):Fe2+(aq)Fe3+(aq)reduction (unbalanced):MnO4(aq)Mn2+(aq)oxidation (unbalanced):Fe2+(aq)Fe3+(aq)reduction (unbalanced):MnO4(aq)Mn2+(aq)

In acidic solution, there are hydrogen ions present, which are often useful in balancing half-reactions. It may be necessary to use the hydrogen ions directly or as a reactant that may react with oxygen to generate water. Hydrogen ions are very important in acidic solutions where the reactants or products contain hydrogen and/or oxygen. In this example, the oxidation half-reaction involves neither hydrogen nor oxygen, so hydrogen ions are not necessary to the balancing. However, the reduction half-reaction does involve oxygen. It is necessary to use hydrogen ions to convert this oxygen to water.

charge not balanced: MnO4(aq)+8H+(aq)Mn2+(aq)+4H2O(l)charge not balanced: MnO4(aq)+8H+(aq)Mn2+(aq)+4H2O(l)

The situation is different in basic solution because the hydrogen ion concentration is lower and the hydroxide ion concentration is higher. After finishing this example, we will examine how basic solutions differ from acidic solutions. A neutral solution may be treated as acidic or basic, though treating it as acidic is usually easier.

The iron atoms in the oxidation half-reaction are balanced (mass balance); however, the charge is unbalanced, since the charges on the ions are not equal. It is necessary to use electrons to balance the charge. The way to balance the charge is by adding electrons to one side of the equation. Adding a single electron on the right side gives a balanced oxidation half-reaction:

oxidation (balanced): Fe2+(aq)Fe3+(aq)+eoxidation (balanced): Fe2+(aq)Fe3+(aq)+e

You should check the half-reaction for the number of each atom type and the total charge on each side of the equation. The charges include the actual charges of the ions times the number of ions and the charge on an electron times the number of electrons.

Fe: Does(1×1)=(1×1)?Yes.Charge: Does[1×(+2)]=[1×(+3)+1×(−1)]?Yes.Fe: Does(1×1)=(1×1)?Yes.Charge: Does[1×(+2)]=[1×(+3)+1×(−1)]?Yes.

If the atoms and charges balance, the half-reaction is balanced. In oxidation half-reactions, electrons appear as products (on the right). As discussed in the earlier chapter, since iron underwent oxidation, iron is the reducing agent.

Now return to the reduction half-reaction equation:

reduction (unbalanced): MnO4(aq)+8H+(aq)Mn2+(aq)+4H2O(l)reduction (unbalanced): MnO4(aq)+8H+(aq)Mn2+(aq)+4H2O(l)

The atoms are balanced (mass balance), so it is now necessary to check for charge balance. The total charge on the left of the reaction arrow is [(−1) ×× (1) + (8) ×× (+1)], or +7, while the total charge on the right side is [(1) ×× (+2) + (4) ×× (0)], or +2. The difference between +7 and +2 is five; therefore, it is necessary to add five electrons to the left side to achieve charge balance.

Reduction (balanced): MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)Reduction (balanced): MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)

You should check this half-reaction for each atom type and for the charge, as well:

Mn: Does(1×1)=(1×1)?Yes.H: Does(8×1) =(4×2)?Yes.O: Does(1×4) =(4×1)?Yes.Charge: Does[1×(−1)+8×(+1)+5×(−1)]=[1×(+2)]?Yes.Mn: Does(1×1)=(1×1)?Yes.H: Does(8×1) =(4×2)?Yes.O: Does(1×4) =(4×1)?Yes.Charge: Does[1×(−1)+8×(+1)+5×(−1)]=[1×(+2)]?Yes.

Now that this half-reaction is balanced, it is easy to see it involves reduction because electrons were gained when MnO4MnO4 was reduced to Mn2+. In all reduction half-reactions, electrons appear as reactants (on the left side). As discussed in the earlier chapter, the species that was reduced, MnO4MnO4 in this case, is also called the oxidizing agent. We now have two balanced half-reactions.

oxidation:Fe2+(aq)Fe3+(aq)+ereduction:MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)oxidation:Fe2+(aq)Fe3+(aq)+ereduction:MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)

It is now necessary to combine the two halves to produce a whole reaction. The key to combining the half-reactions is the electrons. The electrons lost during oxidation must go somewhere. These electrons go on to cause reduction. The number of electrons transferred from the oxidation half-reaction to the reduction half-reaction must be equal. There can be no missing or excess electrons. In this example, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five; therefore, it is necessary to multiply every term in the oxidation half-reaction by five and every term in the reduction half-reaction by one. (In this case, the multiplication of the reduction half-reaction generates no change; however, this will not always be the case.) The multiplication of the two half-reactions by the appropriate factor followed by addition of the two halves gives

oxidation:5×(Fe2+(aq)Fe3+(aq)+e)reduction:MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)¯overall:5Fe2+(aq)+MnO4(aq)+8H+(aq)5Fe3+(aq)+Mn2+(aq)+4H2O(l)oxidation:5×(Fe2+(aq)Fe3+(aq)+e)reduction:MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)¯overall:5Fe2+(aq)+MnO4(aq)+8H+(aq)5Fe3+(aq)+Mn2+(aq)+4H2O(l)

The electrons do not appear in the final answer because the oxidation electrons are the same electrons as the reduction electrons and they “cancel.” Carefully check each side of the overall equation to verify everything was combined correctly:

Fe: Does(5×1)=(5×1)?Yes.Mn: Does(1×1)=(1×1)?Yes.H: Does(8×1)=(4×2)?Yes.O: Does(1×4)=(4×1)?Yes.Charge: Does[5×(+2)+1×(−1)+8×(+1)]=[5×(+3)+1×(+2)]?Yes.Fe: Does(5×1)=(5×1)?Yes.Mn: Does(1×1)=(1×1)?Yes.H: Does(8×1)=(4×2)?Yes.O: Does(1×4)=(4×1)?Yes.Charge: Does[5×(+2)+1×(−1)+8×(+1)]=[5×(+3)+1×(+2)]?Yes.

Everything checks, so this is the overall equation in acidic solution. If something does not check, the most common error occurs during the multiplication of the individual half-reactions.

Now suppose we wanted the solution to be basic. Recall that basic solutions have excess hydroxide ions. Some of these hydroxide ions will react with hydrogen ions to produce water. The simplest way to generate the balanced overall equation in basic solution is to start with the balanced equation in acidic solution, then “convert” it to the equation for basic solution. However, it is necessary to exercise caution when doing this, as many reactants behave differently under basic conditions and many metal ions will precipitate as the metal hydroxide. We just produced the following reaction, which we want to change to a basic reaction:

5Fe2+(aq)+MnO4(aq)+8H+(aq)5Fe3+(aq)+Mn2+(aq)+4H2O(l)5Fe2+(aq)+MnO4(aq)+8H+(aq)5Fe3+(aq)+Mn2+(aq)+4H2O(l)

However, under basic conditions, MnO4MnO4 normally reduces to MnO2 and iron will be present as either Fe(OH)2 or Fe(OH)3. For these reasons, under basic conditions, this reaction will be

3Fe(OH)2(s)+MnO4(aq)+2H2O(l)3Fe(OH)3(s)+MnO2(s)+OH(aq)3Fe(OH)2(s)+MnO4(aq)+2H2O(l)3Fe(OH)3(s)+MnO2(s)+OH(aq)

(Under very basic conditions MnO4MnO4 will reduce to MnO42−,MnO42−, instead of MnO2.)

It is still possible to balance any oxidation-reduction reaction as an acidic reaction and then, when necessary, convert the equation to a basic reaction. There are very few examples in which the acidic and basic reactions will involve the same reactants and products. However, balancing a basic reaction as acidic and then converting to basic will work. To convert to a basic reaction, it is necessary to add the same number of hydroxide ions to each side of the equation so that all the hydrogen ions (H+) are removed and mass balance is maintained. Hydrogen ion combines with hydroxide ion (OH) to produce water.

Let us now try a basic equation. We will start with the following basic reaction:

Cl(aq)+MnO4(aq)ClO3(aq)+MnO2(s)Cl(aq)+MnO4(aq)ClO3(aq)+MnO2(s)

Balancing this as acid gives

Cl(aq)+2MnO4(aq)+2H+(aq)ClO3(aq)+2MnO2(s)+H2O(l)Cl(aq)+2MnO4(aq)+2H+(aq)ClO3(aq)+2MnO2(s)+H2O(l)

In this case, it is necessary to add two hydroxide ions to each side of the equation to convert the two hydrogen ions on the left into water:

Cl(aq)+2MnO4(aq)+(2H++2OH)(aq)ClO3(aq)+2MnO2(s)+H2O(l)+2OH(aq)Cl(aq)+2MnO4(aq)+(2H++2OH)(aq)ClO3(aq)+2MnO2(s)+H2O(l)+2OH(aq)
Cl(aq)+2MnO4(aq)+(2H2O)(l)ClO3(aq)+2MnO2(s)+H2O(l)+2OH(aq)Cl(aq)+2MnO4(aq)+(2H2O)(l)ClO3(aq)+2MnO2(s)+H2O(l)+2OH(aq)

Note that both sides of the equation show water. Simplifying should be done when possible. In this case, it is necessary to remove one H2O from each side of the reaction arrow.

Cl(aq)+2MnO4(aq)+H2O(l)ClO3(aq)+2MnO2(s)+2OH(aq)Cl(aq)+2MnO4(aq)+H2O(l)ClO3(aq)+2MnO2(s)+2OH(aq)

Again, check each side of the overall equation to make sure there are no errors:

Cl: Does(1×1)=(1×1)?Yes.Mn: Does(2×1)=(2×1)?Yes.H: Does(1×2)=(2×1)?Yes.O: Does(2×4+1×1)=(3×1+2×2+2×1)?Yes.Charge: Does[1×(−1)+2×(−1)]=[1×(−1)+2×(−1)]?Yes.Cl: Does(1×1)=(1×1)?Yes.Mn: Does(2×1)=(2×1)?Yes.H: Does(1×2)=(2×1)?Yes.O: Does(2×4+1×1)=(3×1+2×2+2×1)?Yes.Charge: Does[1×(−1)+2×(−1)]=[1×(−1)+2×(−1)]?Yes.

Everything checks, so this is the overall equation in basic solution.

Example 16.1

Balancing Acidic Oxidation-Reduction Reactions

Balance the following reaction equation in acidic solution:
MnO4(aq)+Cr3+(aq)Mn2+(aq)+Cr2O72−(aq)MnO4(aq)+Cr3+(aq)Mn2+(aq)+Cr2O72−(aq)

Solution

This is an oxidation-reduction reaction, so start by collecting the species given into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction.
oxidation (unbalanced):Cr3+(aq)Cr2O72−(aq)reduction (unbalanced):MnO4(aq)Mn2+(aq)oxidation (unbalanced):Cr3+(aq)Cr2O72−(aq)reduction (unbalanced):MnO4(aq)Mn2+(aq)

Starting with the oxidation half-reaction, we can balance the chromium

oxidation (unbalanced): 2Cr3+(aq)Cr2O72−(aq)oxidation (unbalanced): 2Cr3+(aq)Cr2O72−(aq)

In acidic solution, we can use or generate hydrogen ions (H+). Adding seven water molecules to the left side provides the necessary oxygen; the “left over” hydrogen appears as 14 H+ on the right:

oxidation (unbalanced): 2Cr3+(aq)+7H2O(l)Cr2O72−(aq)+14H+(aq)oxidation (unbalanced): 2Cr3+(aq)+7H2O(l)Cr2O72−(aq)+14H+(aq)

The left side of the equation has a total charge of [2 ×× (+3) = +6], and the right side a total charge of [−2 + 14 ×× (+1) = +12]. The difference is six; adding six electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):

oxidation (balanced): 2Cr3+(aq)+7H2O(l)Cr2O72−(aq)+14H+(aq)+6eoxidation (balanced): 2Cr3+(aq)+7H2O(l)Cr2O72−(aq)+14H+(aq)+6e

Checking the half-reaction:

Cr: Does(2×1)=(1×2)?Yes.H: Does(7×2)=(14×1)?Yes.O: Does(7×1) =(1×7)?Yes.Charge: Does[2×(+3)]=[1×(−2)+14×(+1)+6×(−1)]?Yes.Cr: Does(2×1)=(1×2)?Yes.H: Does(7×2)=(14×1)?Yes.O: Does(7×1) =(1×7)?Yes.Charge: Does[2×(+3)]=[1×(−2)+14×(+1)+6×(−1)]?Yes.

Now work on the reduction. It is necessary to convert the four oxygen atoms in the permanganate into four water molecules. To do this, add eight H+ to convert the oxygen into four water molecules:

reduction (unbalanced): MnO4(aq)+8H+(aq)Mn2+(aq)+4H2O(l)reduction (unbalanced): MnO4(aq)+8H+(aq)Mn2+(aq)+4H2O(l)

Then add five electrons to the left side to balance the charge:

reduction (balanced): MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)reduction (balanced): MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)

Make sure to check the half-reaction:

Mn: Does(1×1)=(1×1)?Yes.H: Does(8×1)=(4×2)?Yes.O: Does(1×4)=(4×1)?Yes.Charge: Does[1×(−1)+8×(+1)+5×(−1)]=[1×(+2)]?Yes.Mn: Does(1×1)=(1×1)?Yes.H: Does(8×1)=(4×2)?Yes.O: Does(1×4)=(4×1)?Yes.Charge: Does[1×(−1)+8×(+1)+5×(−1)]=[1×(+2)]?Yes.

Collecting what we have so far:

oxidation:2Cr3+(aq)+7H2O(l)Cr2O72−(aq)+14H+(aq)+6ereduction:MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)oxidation:2Cr3+(aq)+7H2O(l)Cr2O72−(aq)+14H+(aq)+6ereduction:MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)

The least common multiple for the electrons is 30, so multiply the oxidation half-reaction by five, the reduction half-reaction by six, combine, and simplify:

10Cr3+(aq)+35H2O(l)+6MnO4(aq)+48H+(aq)5Cr2O72−(aq)+70H+(aq)+6Mn2+(aq)+24H2O(l)10Cr3+(aq)+35H2O(l)+6MnO4(aq)+48H+(aq)5Cr2O72−(aq)+70H+(aq)+6Mn2+(aq)+24H2O(l)
10Cr3+(aq)+11H2O(l)+6MnO4(aq)5Cr2O72−(aq)+22H+(aq)+6Mn2+(aq)10Cr3+(aq)+11H2O(l)+6MnO4(aq)5Cr2O72−(aq)+22H+(aq)+6Mn2+(aq)

Checking each side of the equation:

Mn: Does(6×1)=(6×1)?Yes.Cr: Does(10×1)=(5×2)?Yes.H: Does(11×2)=(22×1)?Yes.O: Does(11×1+6×4)=(5×7)?Yes.Charge: Does[10×(+3)+6×(−1)]=[5×(−2)+22×(+1)+6×(+2)]?Yes.Mn: Does(6×1)=(6×1)?Yes.Cr: Does(10×1)=(5×2)?Yes.H: Does(11×2)=(22×1)?Yes.O: Does(11×1+6×4)=(5×7)?Yes.Charge: Does[10×(+3)+6×(−1)]=[5×(−2)+22×(+1)+6×(+2)]?Yes.

This is the balanced equation in acidic solution.

Check your learning

Balance the following equation in acidic solution:
Hg22++AgHg+Ag+Hg22++AgHg+Ag+

Answer:

Hg22+(aq)+2Ag(s)2Hg(l)+2Ag+(aq)Hg22+(aq)+2Ag(s)2Hg(l)+2Ag+(aq)

Example 16.2

Balancing Basic Oxidation-Reduction Reactions

Balance the following reaction equation in basic solution:
MnO4(aq)+Cr(OH)3(s)MnO2(s)+CrO42−(aq)MnO4(aq)+Cr(OH)3(s)MnO2(s)+CrO42−(aq)

Solution

This is an oxidation-reduction reaction, so start by collecting the species given into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction
oxidation (unbalanced):Cr(OH)3(s)CrO42−(aq)reduction (unbalanced):MnO4(aq)MnO2(s)oxidation (unbalanced):Cr(OH)3(s)CrO42−(aq)reduction (unbalanced):MnO4(aq)MnO2(s)

Starting with the oxidation half-reaction, we can balance the chromium

oxidation (unbalanced): Cr(OH)3(s)CrO42−(aq)oxidation (unbalanced): Cr(OH)3(s)CrO42−(aq)

In acidic solution, we can use or generate hydrogen ions (H+). Adding one water molecule to the left side provides the necessary oxygen; the “left over” hydrogen appears as five H+ on the right side:

oxidation (unbalanced): Cr(OH)3(s)+H2O(l)CrO42−(aq)+5H+(aq)oxidation (unbalanced): Cr(OH)3(s)+H2O(l)CrO42−(aq)+5H+(aq)

The left side of the equation has a total charge of [0], and the right side a total charge of [−2 + 5 ×× (+1) = +3]. The difference is three, adding three electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):

oxidation (balanced): Cr(OH)3(s)+H2O(l)CrO42−(aq)+5H+(aq)+3eoxidation (balanced): Cr(OH)3(s)+H2O(l)CrO42−(aq)+5H+(aq)+3e

Checking the half-reaction:

Cr: Does(1×1)=(1×1)?Yes.H: Does(1×3+1×2)=(5×1)?Yes.O: Does(1×3+1×1)=(4×1)?Yes.Charge: Does [0=[1×(−2)+5×(+1)+3×(−1)]?Yes.Cr: Does(1×1)=(1×1)?Yes.H: Does(1×3+1×2)=(5×1)?Yes.O: Does(1×3+1×1)=(4×1)?Yes.Charge: Does [0=[1×(−2)+5×(+1)+3×(−1)]?Yes.

Now work on the reduction. It is necessary to convert the four O atoms in the MnO4 minus the two O atoms in MnO2 into two water molecules. To do this, add four H+ to convert the oxygen into two water molecules:

reduction (unbalanced): MnO4(aq)+4H+(aq)MnO2(s)+2H2O(l)reduction (unbalanced): MnO4(aq)+4H+(aq)MnO2(s)+2H2O(l)

Then add three electrons to the left side to balance the charge:

reduction (balanced): MnO4(aq)+4H+(aq)+3eMnO2(s)+2H2O(l)reduction (balanced): MnO4(aq)+4H+(aq)+3eMnO2(s)+2H2O(l)

Make sure to check the half-reaction:

Mn: Does(1×1)=(1×1)?Yes.H: Does(4×1)=(2×2)?Yes.O: Does(1×4)=(1×2+2×1)?Yes.Charge: Does[1×(−1)+4×(+1)+3×(−1)]=[0]?Yes.Mn: Does(1×1)=(1×1)?Yes.H: Does(4×1)=(2×2)?Yes.O: Does(1×4)=(1×2+2×1)?Yes.Charge: Does[1×(−1)+4×(+1)+3×(−1)]=[0]?Yes.

Collecting what we have so far:

oxidation:Cr(OH)3(s)+H2O(l)CrO42−(aq)+5H+(aq)+3ereduction:MnO4(aq)+4H+(aq)+3eMnO2(s)+2H2O(l)oxidation:Cr(OH)3(s)+H2O(l)CrO42−(aq)+5H+(aq)+3ereduction:MnO4(aq)+4H+(aq)+3eMnO2(s)+2H2O(l)

In this case, both half reactions involve the same number of electrons; therefore, simply add the two half-reactions together.

MnO4(aq)+4H+(aq)+Cr(OH)3(s)+H2O(l)CrO42−(aq)+MnO2(s)+2H2O(l)+5H+(aq)MnO4(aq)+4H+(aq)+Cr(OH)3(s)+H2O(l)CrO42−(aq)+MnO2(s)+2H2O(l)+5H+(aq)
MnO4(aq)+Cr(OH)3(s)CrO42−(aq)+MnO2(s)+H2O(l)+H+(aq)MnO4(aq)+Cr(OH)3(s)CrO42−(aq)+MnO2(s)+H2O(l)+H+(aq)

Checking each side of the equation:

Mn: Does(1×1)=(1×1)?Yes.Cr: Does(1×1)=(1×1)?Yes.H: Does(1×3)=(2×1+1×1)?Yes.O: Does(1×4+1×3)=(1×4+1×2+1×1)?Yes.Charge: Does[1×(−1)]=[1×(−2)+1×(+1)]?Yes.Mn: Does(1×1)=(1×1)?Yes.Cr: Does(1×1)=(1×1)?Yes.H: Does(1×3)=(2×1+1×1)?Yes.O: Does(1×4+1×3)=(1×4+1×2+1×1)?Yes.Charge: Does[1×(−1)]=[1×(−2)+1×(+1)]?Yes.

This is the balanced equation in acidic solution. For a basic solution, add one hydroxide ion to each side and simplify:

OH(aq)+MnO4(aq)+Cr(OH)3(s)CrO42−(aq)+MnO2(s)+H2O(l)+(H++OH)(aq)OH(aq)+MnO4(aq)+Cr(OH)3(s)CrO42−(aq)+MnO2(s)+H2O(l)+(H++OH)(aq)
OH(aq)+MnO4(aq)+Cr(OH)3(s)CrO42−(aq)+MnO2(s)+2H2O(l)OH(aq)+MnO4(aq)+Cr(OH)3(s)CrO42−(aq)+MnO2(s)+2H2O(l)

Checking each side of the equation:

Mn: Does(1×1)=(1×1)? Yes.Cr: Does(1×1)=(1×1)?Yes.H: Does(1×1+1×3)=(2×2)?Yes.O: Does(1×1+1×4+1×3)=(1×4+1×2+2×1)?Yes.Charge: Does[1×(−1)+1×(−1)]=[1×(−2)]?Yes.Mn: Does(1×1)=(1×1)? Yes.Cr: Does(1×1)=(1×1)?Yes.H: Does(1×1+1×3)=(2×2)?Yes.O: Does(1×1+1×4+1×3)=(1×4+1×2+2×1)?Yes.Charge: Does[1×(−1)+1×(−1)]=[1×(−2)]?Yes.

This is the balanced equation in basic solution.

Check Your Learning

Balance the following in the type of solution indicated.

(a) H2+Cu2+Cu(acidic solution)H2+Cu2+Cu(acidic solution)

(b) H2+Cu(OH)2Cu(basic solution)H2+Cu(OH)2Cu(basic solution)

(c) Fe+Ag+Fe2++AgFe+Ag+Fe2++Ag

(d) Identify the oxidizing agents in reactions (a), (b), and (c).

(e) Identify the reducing agents in reactions (a), (b), and (c).

Answer:

(a) H2(g)+Cu2+(aq)2H+(aq)+Cu(s);H2(g)+Cu2+(aq)2H+(aq)+Cu(s); (b) H2(g)+Cu(OH)2(s)2H2O(l)+Cu(s);H2(g)+Cu(OH)2(s)2H2O(l)+Cu(s); (c) Fe(s)+2Ag+(aq)Fe2+(aq)+2Ag(s);Fe(s)+2Ag+(aq)Fe2+(aq)+2Ag(s); (d) oxidizing agent = species reduced: Cu2+, Cu(OH)2, Ag+ (e) reducing agent = species oxidized: H2, H2, Fe.

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