textbook redox notes

CHAPTER 12

Oxidizing and Reducing in a

Redox Reaction

BLM 12.1.1

OVERHEAD

Reduction

is the gain of electrons.

Oxidation

is the loss of electrons.

In a redox reaction, the reducing agent is oxidized and the oxidizing agent

is reduced.

CHAPTER 12

Balancing an Equation for a

Reaction that Occurs in an Acidic

Solution

BLM 12.2.1

OVERHEAD

gains 2e

Zn(s) + Cu

_{(aq) → Cu(s) + Zn}

_(aq)

loses 2e

Zn(s) + Cu

_{(aq) → Cu(s) + Zn}

_(aq)



oxidizing agent

accepts electrons

undergoes reduction

reducing agent

Sample Problem

Sulfur is oxidized by nitric acid in an aqueous solution, producing sulfur dioxide, nitrogen

monoxide, and water, as shown by the unbalanced equation. Use the half-reaction method to balance the equation:

S(s) + HNO3(aq) → SO2(g) + NO(g) + H2O(ℓ)

(Note: To reduce the complexity, the subscript, 8, has been omitted from the sulfur. After the final step, you could add the subscript to the sulfur and multiply all other coefficients by 8.)

Solution

First, write the total ionic equation: S(s) + H+_{(aq) + NO}

3−(aq) → SO2(g) + NO(g) + H2O(ℓ)

Step 1Write the unbalanced half-reactions. Write only those compounds that contain the atom that is oxidized or the atom that is reduced. For this step, ignore the fact that one side of the equation might have oxygen atoms and the other side has none.

Oxidation half-reaction: S(s) → SO2(g)

Reduction half-reaction: NO3−(aq) → NO(g)

Step 2Balance atoms other than oxygen and hydrogen. The sulfur and nitrogen atoms are already balanced.

Step 3Balance oxygen atoms by adding water molecules.

Oxidation half-reaction: There are two oxygen atoms on the right side of the equation, so you must add two water molecules to the left side:

S(s) + 2H2O(ℓ) → SO2(g)

Reduction half-reaction: There are three oxygen atoms on the left side of the equation and one on the right. Add two water molecules to the right side:

NO3−(aq) → NO(g) + 2H2O(ℓ)

Step 4Balance hydrogen atoms by adding hydrogen ions.

Oxidation half-reaction: There are four hydrogen atoms in the water on the left side of the equation, so add four hydrogen ions to the right side:

S(s) + 2H2O(ℓ) → SO2(g) + 4H+(aq)

Reduction half-reaction: There are four hydrogen atoms in the water molecules on the right side of the equation, so add four hydrogen ions to the left side:

NO3−(aq) + 4H+(aq) → NO(g) + 2H2O(ℓ)

The nitric acid makes this an acid solution, so skip to Step 8 of the steps for balancing half-reactions.

Step 8Balance the charges by adding electrons.

Oxidation half-reaction: There are zero net charges on the left side of the equation and four positive charges on the right. Therefore, give the right side a net charge of zero by adding four electrons to the right side:

S(s) + 2H2O(ℓ) → SO2(g) + 4H+(aq) + 4e−

Reduction half-reaction: There are four positive charges and one negative charge on the left side of the equation, giving it a net positive charge of three. The net charge on the right side of the equation is zero. Add three electrons to the left side to give it a net zero charge:

NO3−(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(ℓ)

Balancing an Equation for a

Reaction that Occurs in an Acidic

Solution

OVERHEAD

The half-reactions are balanced. Now balance the entire equation by following the steps in the method for balancing reactions using half-reactions.

Step 1Determine the lowest common multiple of the numbers of electrons in the oxidation and reduction half-reactions. There are four electrons in the oxidation half-reaction and three electrons in the reduction half-reaction. The lowest common multiple of 4 and 3 is 12.

Step 2Multiply one or both half-reactions by the number that will bring the number of electrons to the lowest common multiple.

Oxidation half-reaction:

3[S(s) + 2H2O(ℓ) → SO2(g) + 4H+(aq) + 4e−]

3S(s) + 6H2O(ℓ) → 3SO2(g) + 12H+(aq) + 12e−

Reduction half-reaction:

4[NO3−(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(ℓ)]

4NO3−(aq) + 16H+(aq) + 12e− → 4NO(g) + 8H2O(ℓ)

Step 3Add the balanced half-reactions:

3S(s) + 6H2O(ℓ) + 4NO3−(aq) + 16H+(aq) + 12e− → 3SO2(g) + 12H+(aq) + 12e− + 4NO(g) + 8H2O(ℓ)

Step 4Cancel the electrons and any other identical molecules or ions present on both sides of the equation:

3S(s) + 6H2O(ℓ) + 4NO3−(aq) + 16H+(aq) + 12e– → 3SO2(g) + 12H+(aq) + 12e– + 4NO(g) + 8H2O(ℓ)

Subtract 6H2O(ℓ) and 12 H+(aq) from both sides of the equation, as well as the 12 electrons:

3S(s) + 4NO3−(aq) + 4H+(aq) →3SO2(g) + 4NO(g) + 2H2O(ℓ)

3S8(s) + 32NO3−(aq) + 32H+(aq) →24SO2(g) + 32NO(g) + 16H2O(ℓ)

CHAPTER 12

The Half-Reaction Method of

Balancing Equations

BLM 12.2.2

Acidic conditions Basic conditions

Neutral conditions

Balance atoms other than O or H

Balance O atoms by adding H2O

Balance H atoms by adding H+ ions

Remove electrons, ions, or molecules present on both sides of the equation

Write unbalanced net ionic equation

Write two half-reactions and balance independently

Balance atoms

Balance charges by adding electrons

Write half-reactions with LCM number of electrons

Add half-reactions

Balance assuming acidic conditions

Add OH– ions to “neutralize” H+ ions

Combine H+ and OH– ions into H2O

Remove H2O molecules present on both sides

Balance the equations below using the half-reaction method. Refer to the above concept organizer for the steps of the half-reaction method.

1. Balance the following equations, assuming acidic conditions:

(a) Br–(aq) + MnO4–(aq) →Mn2+(aq) + Br2(ℓ)

CHAPTER 12

The Half-Reaction Method of

(b) HBiO3(aq) + Mn2+(aq) →MnO4–(aq) + Bi3+(aq)

CHAPTER 12

The Half-Reaction Method of

Balancing Equations Answer Key

BLM 12.2.2A

ANSWER KEY

1. (a) 10Br–(aq) + 2MnO4–(aq) + 16H+(aq) → 5Br 2(ℓ) + 2Mn2+(aq) + 8H2O(ℓ)

(b)2Mn2+_{(aq) + 13HBiO}

3(aq) + 33H+(aq) → 2MnO4–(aq) + 13Bi3+(aq) + 23H2O(ℓ)

Write unbalanced net ionic equationAssign oxidation numbers to identify redox reaction

Identify element oxidized and element reduced

Find increase and decrease in oxidation numbers

Find smallest whole-number ratio of oxidized and reduced elements

Balance atoms of oxidized and reduced elements

Balance other elements by inspection

Acidic conditions

Include H2O and H+, as necessary

Neutral conditions

Basic conditions

Balance assuming acidic conditions

Add OH– to “neutralize” H+ ions present

Simplify resulting equation, if possible

The Oxidation Number Method of

Balancing Equations

ASSESSMENT

Balance the following equations using the oxidation number method. Assume neutral conditions. Refer to the above concept organizer for the steps of the oxidation number method.

1.NaI(aq) + HClO(aq) → NaIO3(aq) + HCl(aq)

The Oxidation Number Method of

Balancing Equations

ASSESSMENT

3.C3H8O(ℓ) + CrO3(aq) + H2SO4(aq) → Cr2(SO4)3(aq) + C3H6O(ℓ) + H2O(ℓ)

4.KCl(aq) + MnO2(s) + H2SO4(aq) → K2SO4(aq) + MnSO4(aq) + Cl2(g) + H2O(ℓ)

5.Cu(s) + H2SO4(aq) → CuSO4(aq) + H2O(ℓ) + SO2(g)

6. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2S(g) + H2O(ℓ)

1.NaI(aq) + 3HClO(aq) → NaIO3(aq) + 3HCl(aq)

2.I2(s) + 10HNO3(aq) → 2HIO3(aq) + 10NO2(g) + 4H2O(ℓ)

3.3C3H8O(ℓ) + 2CrO3(aq) + 3H2SO4(aq) → Cr2(SO4)3(aq) + 3C3H6O(ℓ) + 6H2O(ℓ)

4.2KCl(aq) + MnO2(s) + 2H2SO4(aq) → K2SO4(aq)+ MnSO4(aq) + Cl2(g) + 2H2O(ℓ)

5.Cu(s) + 2H2SO4(aq) → CuSO4(aq) + 2H2O(ℓ) + SO2(g)

6.4Zn(s) + 5H2SO4(aq) → 4ZnSO4(aq) + H2S(g) + 4H2O(ℓ)

Investigation 12.B: Redox

Reactions and Balanced

Equations

HANDOUT

You have practised balancing equations for redox reactions, but can you predict the products of a redox reaction? Can you determine whether a reaction has occurred and, if so, whether it was a redox reaction? In this investigation, you will develop these skills.

Question

How can you tell whether a redox reaction occurs when reactants are mixed? Can you observe the transfer of electrons in the mixture?

Predictions

• Predict which metals of magnesium, zinc, copper, and aluminium can be oxidized by aqueous hydrogen ions (acidic solution). Explain your reasoning.

• Predict whether metals that cannot be oxidized by hydrogen ions can be dissolved in acids. Explain your reasoning.

• Predict whether the combustion of a hydrocarbon is a redox reaction. What assumptions have you made about the products?

Safety Precautions

• The acid solutions are corrosive. Handle them with care.

Investigation 12.B: Redox

Reactions and Balanced

Equations

HANDOUT

Materials

• small pieces of each of the following metals: – magnesium

– zinc – copper – aluminium

• hydrochloric acid (1 mol/L) • sulfuric acid (1 mol/L) • well plate

Procedure

Part 1 Reactions of Acids with Metals

1. Place one small piece of each metal into a well on the well plate. Add a few drops of hydrochloric acid to each metal. Record your observations. If you are unsure of your observations, repeat the procedure on a larger scale in a small test tube.

2. Place another small piece of each metal into clean sections of the well plate. Add a few drops of sulfuric acid to each metal. Record your observations. If you are unsure of your observations, repeat the procedure on a larger scale in a small test tube.

3. Dispose of the mixtures into the beaker supplied by your teacher.

Part 2 Combustion of Hydrocarbons

4. Light a Bunsen burner and observe the combustion of natural gas. Adjust the colour of the flame by varying the quantity of oxygen admitted to the burner. Describe how the colour depends on the quantity of oxygen.

CHAPTER 12

Investigation 12.B: Redox

Reactions and Balanced

Equations

BLM 12.2.4

HANDOUT

Analysis

Part 1 Reactions of Acids with Metals

1. Write a complete balanced equation for each of the reactions of an acid with a metal.

2. Write each equation from Question 1 in net ionic form.

CHAPTER 12

Investigation 12.B: Redox

Reactions and Balanced

Equations

BLM 12.2.4

HANDOUT

3. Determine which of the reactions from Question 1 are redox reactions.

5. Describe and explain any trends that you see in your answers to Question 4.

6. In the reactions you observed, are the hydrogen ions acting as an oxidizing agent, a reducing agent, or neither?

CHAPTER 12

Investigation 12.B: Redox

Reactions and Balanced

Equations

BLM 12.2.4

HANDOUT

8. Your teacher might demonstrate the reaction of copper with concentrated nitric acid to produce copper(II) ions and brown, toxic nitrogen dioxide gas. Write a balanced net ionic equation for this reaction. Do the hydrogen ions behave in the same way as you described in Question 7? Identify the oxidizing agent and the reducing agent in this reaction.

9. From your observations of the reaction of copper with hydrochloric acid in your investigation and the reaction of copper with nitric acid in the demonstration described above, can you tell whether hydrogen ions or nitrate ions are the better oxidizing agent? Explain.

Part 2 Combustion of Hydrocarbons

10.The main component of natural gas is methane, CH4(g). The products of the combustion of this

gas in a Bunsen burner depend on how the burner is adjusted. A blue flame indicates complete combustion. What are the products of complete combustion? Write a balanced chemical equation for this reaction.

CHAPTER 12

Investigation 12.B: Redox

Reactions and Balanced

Equations

BLM 12.2.4

HANDOUT

11.A yellow or orange flame from a Bunsen burner indicates incomplete combustion and the presence of carbon in the flame. Write a balanced chemical equation for this reaction.

13.Assume that the fuel in the burning candle is paraffin. Although paraffin is a mixture of

hydrocarbons, you can represent it by the formula C25H52(s). Write a balanced chemical equation

for the complete combustion of paraffin.

14.Write two different balanced equations that both represent a form of incomplete combustion of paraffin.

15.How do you know that at least one of the incomplete combustion reactions is taking place when a candle burns?

16.Are combustion reactions also redox reactions? Does your answer depend on whether the combustion is complete or incomplete? Explain.

CHAPTER 12

Investigation 12.B: Redox

Reactions and Balanced

Equations

BLM 12.2.4

HANDOUT

Conclusion

Applications

18.Gold is very unreactive and does not dissolve in most acids. However, it does dissolve in aqua regia (Latin for “royal water”), which is a mixture of concentrated hydrochloric acid and nitric acid. The unbalanced ionic equation for the reaction is as follows:

Au(s) + NO3−(aq) + Cl−(aq) → AuCl4−(s) + NO2(g)

Balance the equation, and identify the oxidizing agent and reducing agent.

19.Natural gas is burned in gas furnaces. Give at least three reasons why this combustion reaction should be as complete as possible. How would you try to ensure complete combustion?

CHAPTER 12

Investigation 12.B: Redox

Reactions and Balanced

Equations Answer Key

BLM 12.2.4A

ANSWER KEY

Answers to Prediction Questions

 Based on your experience with the activity series, you will likely predict that all metals mentioned, except copper, are oxidized by aqueous hydrogen ions.

 Metals that cannot be oxidized by hydrogen ions would not dissolve in acid. To dissolve, metals need to form ions—they must be oxidized.

Answers to Analysis Questions

1. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

Cu(s) + HCl(aq) → NR Cu(s) + H2SO4(aq) → NR

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)

2. Note that for each metal, the net ionic equations are the same for both acids. Therefore, only one equation is shown for each metal.

Mg(s) + 2H+_{(aq) → Mg}2+_{(aq) + H} 2(g)

Zn(s) + 2H+_{(aq) → Zn}2+_{(aq) + H} 2(g)

Cu(s) + H+_{(aq) → NR}

2Al(s) + 6H+_{(aq) → 2Al}3+_{(aq) + 3H} 2(g)

3. The reactions of aluminium, zinc, and magnesium with acid are redox reactions.

4. Note that for each metal, the half-reactions are the same for both acids. Therefore, only one reduction and oxidation is shown for each metal. The half-reactions are as follows:

Oxidation

Al(s) → Al3+_{(aq) + 3e}–

Mg(s) → Mg2+_{(aq) + 2e}–

Zn(s) → Zn2+_{(aq) + 2e}–

Reduction

2H+ _{+ 2e}– _{→ H} 2(g)

2H+ _{+ 2e}– _{→ H} 2(g)

5. You may notice that copper did not react with the hydrogen ions. You may recall that copper ions are fairly good oxidizing agents; you could suggest that copper ions are better oxidizing agents than hydrogen ions.

6. The hydrogen ions are reduced. Therefore, they act as the oxidizing agent.

7. A neutralization reaction is a double displacement reaction. Therefore, it is not a redox reaction. The hydrogen ions have an oxidation number of –1 on both the reactant and product sides of the equation. In other words, they do not act as oxidizing agents.

CHAPTER 12

Investigation 12.B: Redox

Reactions and Balanced

Equations Answer Key

BLM 12.2.4A

ANSWER KEY

8. When nitric acid is added to copper the reaction mixture turns green (copper ions) and a yellow-brown gas (NO2) is produced.

The balanced chemical equation is:

Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(ℓ)

The balanced net ionic equation is:

Cu(s) + 2NO3– (aq) + 4H+(aq) → Cu2+(aq) + 2NO2(g) + 2H2O(ℓ)

In this reaction, copper metal is the reducing agent. Its oxidation number increases from 0 to –2. Nitrate ion is the oxidizing agent. Its oxidation number of nitrogen decreases from –5 to –4.

9. The reaction of nitrate ions and copper atoms generates a positive cell potential while the reaction of copper atoms with hydrogen ions does not. Therefore, nitrate ions are the better oxidizing agents.

10.The products for the complete combustion of a hydrocarbon are water and carbon dioxide. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

11.Assuming that the incomplete combustion does not involve the production of carbon monoxide, the balanced chemical equation is:

CH4(g) + O2(g) → C(s) + 2H2O(g)

12.Assuming that the incomplete combustion does not involve the production of carbon, the balanced chemical equation is:

2CH4(g) + 3O2(g) → 2CO(g) + 4H2O(g)

13.The balanced chemical equation for the complete combustion of paraffin wax, C25H52(g) is:

C25H52(g) + 38O2(g) → 25CO2(g) + 26H2O(g)

14.Two balanced equations for the incomplete combustion of paraffin wax are: C25H52(g) + 13O2(g) → 25C(s) + 26H2O(g)

2C25H52(g) + 51O2(g) → 50CO(g) + 52H2O(g)

15.It is usually possible to observe the production of soot (carbon) when a candle burns.

16.All combustion reactions involve the reduction of oxygen. In any combustion reaction, oxygen with an oxidation number of 0 forms compounds with hydrogen and possibly carbon.

accompanied by a corresponding oxidation, and since both complete and incomplete combustion involve the reduction of oxygen, all combustion reactions must be redox reactions.

Answer to Conclusion Question

17.If it is clear that a neutral element (such as a metal or hydrogen) is produced, the reaction must be a redox reaction. The oxidation number of an atom of an element is zero, while the oxidation number of an atom of an element in a compound is not. Electron transfer cannot be observed directly.

CHAPTER 12

Investigation 12.B: Redox

Reactions and Balanced

Equations Answer Key

BLM 12.2.4A

ANSWER KEY

Answers to Applications Questions

18.Au(s) + 3NO3–(aq) + 6H+(aq) + 4Cl–(aq) → AuCl4–(aq) + 3NO2(g) + 3H2O(ℓ)

The oxidizing agent is the nitrate ion. The reducing agent is gold.

19.Complete combustion maximizes the energy release from the fuel used, prevents the formation of deadly carbon monoxide gas, and prevents the production of soot. To ensure complete

CHAPTER 12

Balancing a Disproportionation

Reaction

BLM 12.2.5

OVERHEAD

Problem

Balance the following unbalanced equation for the disproportionation, in an acidic solution, of nitrous acid, HNO2(aq), forming nitric acid, HNO3(aq); nitrogen monoxide, NO(g); and water:

HNO2(aq) → HNO3(aq) + NO(g) + H2O(ℓ)

Solution

First, write the ionic and then the net ionic equations. H+_{(aq) + NO}

2−(aq) → H+(aq) + NO3−(aq) + NO(g)+ H2O(ℓ)

NO2−(aq) → NO3−(aq) + NO(g) + H2O(ℓ)

Step 1Write the unbalanced half-reactions that show the formulas of the given reactant(s) and product(s).

Oxidation half-reaction: NO2−(aq) → NO3−(aq)

Reduction half-reaction: NO2−(aq) → NO(g)

Step 2Balance any atoms other than oxygen and hydrogen first. Oxidation half-reaction: The nitrogen atoms are already balanced. Reduction half-reaction: The nitrogen atoms are already balanced.

Step 3Balance any oxygen atoms by adding water molecules.

Oxidation half-reaction: There are two oxygen atoms in the nitrite ion on the left side of the equation and three oxygen atoms in the nitrate ion on the right side. Therefore, add one water molecule to the left side:

NO2−(aq) + H2O(ℓ) → NO3−(aq)

Reduction half-reaction: There are two oxygen atoms in the nitrite ion on the left side of the equation and one oxygen atom in the nitrogen monoxide on the right side. Therefore, add one water molecule to the right side:

NO2−(aq) → NO(g) + H2O(ℓ)

Step 4Balance any hydrogen atoms by adding hydrogen ions.

Oxidation half-reaction: There are two hydrogen atoms in the water molecule on the left side of the equation and zero on the right. Therefore, add two hydrogen ions to the right side:

NO2−(aq) + H2O(ℓ) → NO3−(aq) + 2H+(aq)

Reduction half-reaction: There are two hydrogen atoms in the water molecule on the right side of the equation and zero on the left. Therefore, add two hydrogen ions to the left side:

NO2−(aq) + 2H+(aq) → NO(g) + H2O(ℓ)

The solution is acidic, so skip to Step 8.

Step 8Balance the charges by adding electrons.

Oxidation half-reaction: There is one negative charge on the left side of the equation and a net charge of plus one on the right. To balance the charges, add two electrons to the right side: NO2−(aq) + H2O(ℓ) → NO3−(aq) + 2H+(aq) + 2e−

Reduction half-reaction: There is a net charge of plus one on the left side of the equation and zero on the right. To balance the charges, add one electron to the left side:

CHAPTER 12

Balancing a Disproportionation

Reaction

BLM 12.2.5

OVERHEAD

The half-reactions are balanced. Now balance the entire equation.

Step 1Determine the lowest common multiple of the numbers of electrons in the oxidation and reduction half-reactions. There were two electrons lost in the oxidation half-reaction and one electron was gained in the reduction half-reaction. The lowest common multiple of 1 and 2 is 2.

Step 2Multiply one or both half-reactions by the number that will bring the number of electrons to the lowest common multiple.

Oxidation half-reaction:

NO2−(aq) + H2O(ℓ) → NO3−(aq) + 2H+(aq) + 2e−

Reduction half-reaction:

2[NO2−(aq) + 2H+(aq) + e− → NO(g) + H2O(ℓ)]

2NO2−(aq) + 4H+(aq) + 2e− → 2NO(g) + 2H2O(ℓ)

Step 3Add the balanced half-reactions:

NO2−(aq) + H2O(ℓ) + 2NO2−(aq) + 4H+(aq) + 2e− → NO3−(aq) + 2H+(aq) + 2e− + 2NO(g) + 2H2O(ℓ)

Step 4Remove the electrons and any other identical molecules or ions that are present on both sides of the equation:

NO2−(aq) + H2O(ℓ) + 2NO2−(aq) + 4H+(aq) + 2e− →NO3−(aq) + 2H+(aq) + 2e− + 2NO(g) + 2H2O(ℓ)

NO2−(aq) + 2NO2−(aq) + 2H+(aq) → NO3−(aq) + 2NO(g) + H2O(ℓ)

3NO2−(aq) + 2H+(aq) → NO3−(aq) + 2NO(g) + H2O(ℓ)

Step 5If spectator ions were removed when forming half-reactions, add them back to the equation. One hydrogen ion was removed when forming half-reactions:

3NO2−(aq) + 3H+(aq) →H+(aq) + NO3−(aq) + 2NO(g) + H2O(ℓ)

CHAPTER 12

The Smelting Process

OVERHEAD

Smelting involves the chemical reduction of iron ions to iron atoms with

carbon as the reducing agent. The process takes place in several steps inside

a large blast furnace. Note the size of the person relative to the blast

furnace.

Writing Balanced Ionic, Net

Ionic, and Half-Reaction

Equations

ASSESSMENT

Examine the following situations. If appropriate, write balanced ionic and net ionic equations to describe the situations. If the situations are redox reactions, write oxidation and reduction half-reaction equations. Refer to the activity series for metals and solubility guidelines if necessary.

1. A solution of lead(II) nitrate is mixed with a solution of potassium iodide. balanced ionic equation:

net ionic equation:

oxidation half-reaction:

reduction half-reaction:

2. Liquid sodium chloride is electrolyzed. balanced ionic equation:

net ionic equation:

oxidation half-reaction:

Writing Balanced Ionic, Net

Ionic, and Half-Reaction

Equations

ASSESSMENT

3. Aqueous gold(III) chloride is stirred with a cadmium rod. balanced ionic equation:

net ionic equation:

oxidation half-reaction:

reduction half-reaction:

4. Zinc nitrate solution is placed in a solid silver cup. balanced ionic equation:

net ionic equation:

oxidation half-reaction:

reduction half-reaction:

Writing Balanced Ionic, Net Ionic,

and Half-Reaction Equations

Answer Key

ANSWER KEY

1. balanced ionic equation:

Pb2+_{(aq) + 2NO}3–_{(aq) + 2K}+_{(aq) + 2I}–_{(aq) → PbI}

2(s) + 2K+ (aq) + 2NO3–(aq)

net ionic equation: Pb2+_{(aq) + 2I}–_{(aq) → PbI}

2(s)

oxidation half-reaction equation: none reduction half-reaction equation: none

2. balanced ionic equation:

2Na+_{(aq) + 2Cl}–_{(aq) → 2Na(s) + Cl} 2(g)

net ionic equation:

2Na+_{(aq) + 2Cl}–_{(aq) → 2Na(s) + Cl} 2(g)

oxidation half-reaction equation: 2Cl–_{(ℓ) → Cl}

2(g) + 2e–

reduction half-reaction equation: 2Na+_{(aq) + 2e}– _{→ 2Na(s)}

3. balanced ionic equation:

2Au3+_{(aq) + 6Cl}–_{(aq) + 3Cd(s) → 2Au(s) + 3Cd}2+_{(aq) + 6Cl}–_(aq)

net ionic equation:

2Au3+_{(aq) + 3Cd(s) → 2Au(s) + 3Cd}2+_(aq)

oxidation half-reaction equation: Cd(s) → Cd2+_{(aq) + 2e}–

reduction half-reaction equation: Au3+_{(aq) + 3e}– _{→ Au(s)}

CHAPTER 12

Assigning Oxidation Numbers

OVERHEAD

Sample Problem

Assign an oxidation number to each atom in the following compounds:

(a) SiBr4(ℓ) (b) HClO4(aq) (c) Cr2O72−(aq)

Solution

(a) • Because the compound SiBr4(ℓ) does not contain hydrogen or oxygen, Rule 5 from Table 12.2

applies.

• Because SiBr4(ℓ) is a neutral compound, Rule 6 also applies.

• Silicon has an electronegativity of 1.9. Bromine has an electronegativity of 3.0. From Rule 5, therefore, you can assign bromine an oxidation number of −1.

• The oxidation number of silicon is unknown, so call it x. You know from Rule 6 that the sum of the oxidation numbers is 0. Therefore, the oxidation numbers of

1 Si atom + 4 Br atoms = 0 x + 4(–1) = 0

x – 4 = 0 x = +4

( b) • Because the compound HClO4(aq) contains hydrogen and oxygen, Rules 3 and 4 apply.

• Because HClO4(aq) is a neutral compound, Rule 6 also applies.

• Hydrogen has its usual oxidation number of +1.

• Oxygen has its usual oxidation number of −2.

• The oxidation number of chlorine is unknown, so call it x.

• You know from Rule 6 that the sum of the oxidation numbers is 0. Therefore, the oxidation numbers of

1 H atom + 1 Cl atom + 4 O atoms = 0 (+1) + x + 4 (–2) = 0

1 – 8 + x = 0 x –7 = 0 x = +7

• The oxidation number of hydrogen is +1.

• The oxidation number of chlorine is +7.

• The oxidation number of oxygen is −2.

(c) • Because the dichromate ion, Cr2O72−(aq), contains oxygen, Rule 4 applies.

• Because it is a polyatomic ion, Rule 7 also applies.

• Oxygen has its usual oxidation number of −2.

• The oxidation number of chromium is unknown, so call it x.

• You know from Rule 7 that the sum of the oxidation numbers is −2. Therefore, the oxidation numbers of

2 Cr atoms + 7 O atoms = −2 2x + 7(–2) = −2

2x − 14 = −2 2x = 14 −2 2x = 12 x = +6

CHAPTER 12

Balancing a Redox Equation in

Basic Solution

BLM 12.3.3

OVERHEAD

Sample Problem

Write a balanced net ionic equation to show the formation of iodine by bubbling oxygen gas through a basic solution that contains iodide ions.

Solution

Step 1Write an unbalanced equation from the given information: O2(g) + I−(aq) → I2(s)

Step 2Assign an oxidation number to each atom in the equation to determine whether it is a redox reaction:

O2(g) + I−(aq) → I2(s)

0 –1 0

Because iodide is oxidized to iodine, the reaction is a redox reaction. The product that contains oxygen is unknown at this stage. However, oxygen must be reduced.

Step 3Because the reaction is a redox reaction, identify the atom or atoms that undergo an increase in oxidation number and the atom or atoms that undergo a decrease in oxidation number. Iodine is the element that undergoes an increase in oxidation number. Oxygen is the element that undergoes a decrease in oxidation number.

Step 4Find the numerical values of the increase and the decrease in oxidation numbers. Iodine undergoes an increase in its oxidation number from −1 to 0, an increase of 1. Assume that the

oxidation number of oxygen after reduction is its normal value, which is −2. Thus, oxygen undergoes a decrease in its oxidation number from 0 to −2, a decrease of 2.

Step 5Determine the lowest common multiple of the increase in oxidation number and the decrease in oxidation number.

Step 6Apply the numbers found in Step 5 to balance the number of atoms oxidized (iodine) and the number of atoms reduced (oxygen). Make sure there are two iodine atoms for every oxygen atom:

O2(g) + 4I−(aq) → 2I2(s)

Step 7Balance the other elements by inspection, if possible. No other reactants or products can be balanced by inspection.

Step 8For reactions that occur in acidic or basic solutions, include water molecules, hydrogen ions, or hydroxide ions as needed to balance the equation. The reaction occurs in a basic solution. As you learned in Section 12.2, for basic conditions, start by assuming that the conditions are acidic. Add water molecules and hydrogen ions as necessary to balance the atoms:

O2(g) + 4I−(aq) → 2I2(s) + 2H2O(ℓ)

O2(g) + 4I−(aq) + 4H+(aq) → 2I2(s) + 2H2O(ℓ)

Add hydroxide ions to adjust for basic conditions. Simplify the resulting equation:

O2(g) + 4I−(aq) + 4H+(aq) + 4OH−(aq) → 2I2(s) + 2H2O(ℓ) + 4OH−(aq)

O2(g) + 4I−(aq) + 4H2O(ℓ) → 2I2(s) + 2H2O(ℓ) + 4OH−(aq)

CHAPTER 12

Redox Titrations

OVERHEAD

Sample Problem

You are using a 0.011 43 mol/L KMnO4(aq) solution to determine the percentage by mass of an

aqueous solution of H2O2(aq). You know that the peroxide solution is about 3% H2O2(aq) by mass.

You prepare the sample by adding 1.423 g of the hydrogen peroxide solution to an Erlenmeyer flask. (Although the hydrogen peroxide is in aqueous solution, you can determine the mass by placing an empty flask on a balance and adding peroxide with a pipette.) You add about 75 mL of water to dilute the solution. You also add some dilute sulfuric acid to acidify the solution. You reach the light-purple-coloured endpoint of the titration when you have added 40.22 mL of the KMnO4(aq)

solution. What is the percent, by mass, of the peroxide solution?

(Percent by mass is defined as: × 100%)

What Is Required?

You need to determine the mass of H2O2(aq) in the sample. You need to express your result as a

mass percent.

What Is Given?

Concentration of KMnO4(aq) = 0.011 43 mol/L

Volume of KMnO4(aq) = 40.22 mL

Mass of H2O2(aq) solution = 1.423 g

Plan Your Strategy

Step 1Write the balanced chemical equation for the reaction.

Step 2Calculate the amount (in mol) of permanganate ion added, based on the volume and concentration of the potassium permanganate solution.

Step 3Determine the amount (in mol) of hydrogen peroxide needed to reduce the permanganate ions.

CHAPTER 12

Redox Titrations

BLM 12.4.1

OVERHEAD

Act on Your Strategy

Step 1The redox equation was provided on the previous page. It is already balanced: 5H2O2(aq) + 2MnO4−(aq) + 6H+(aq) → 5O2(g) + 2Mn2+(aq) + 8H2O(ℓ)

Step 2The concentration of MnO4−(aq) is the same as the concentration of KMnO4(aq):

n = cV

n = ( 0.011 43 ) (0.040 22 ) n = 4.597 × 10-4 _mol

Step 3Permanganate ions react with hydrogen peroxide in a 2:5 ratio, as shown by the coefficients in the balanced equation:

Amount (in moles) H2O2(aq):

Step 4

Check Your Solution

The units are correct. The value for the mass of pure H2O2(aq) that you obtained is less than the mass

of the H2O2(aq) sample solution, as you would expect. The mass percent you obtained for the

solution is close to the expected value. It makes sense that the value is somewhat less than 3%, since H2O2(aq) decomposes in solution, forming water and oxygen.

CHAPTER 13

The Motion of Ions and

Electrons

BLM 13.1.1

OVERHEAD

CHAPTER 13

The Hydrogen Half-Cell with

Zinc

BLM 13.1.2

OVERHEAD

CHAPTER 13

Thought Lab: Assigning

Reference Values

BLM 13.1.3

HANDOUT

Many scales of measurement have zero values that are arbitrary. For example, on Earth, average sea level is often assigned as the zero of altitude. In this Thought Lab, you will investigate what happens to calculated cell potentials when the reference half-cell is changed.

Procedure

1. Choose the half reaction for Al3+ _{and Al as your reference point and assign a value of 0 V for this}

half-reaction. To make the standard cell potential for the Al3+_{/Al half-reaction equal to zero, you}

would have to add 1.66 V to the accepted standard reduction potential. To adjust all the reduction potentials to the new reference, you add 1.66 V to each value.

Reduction half-reaction Accepted E° (V) Adjusted E°(V) [+ 1.66 (V)]

F2(g) + 2e− → 2F−(aq) +2.87

Fe3+_{(aq) + e}− _{→ Fe}2+_{(aq) +0.77}

2H+_{(aq) + 2e}− _{→ H}

2(g) 0.00

Al3+_{(aq) + 3e}− _{→ Al(s) −1.66}

Li+_{(aq) + e}− _{→ Li(s) −3.04}

2. Use the given standard reduction potentials to calculate the standard cell potentials for the following redox reactions:

(a) 2Li(s) + 2H+_{(aq) → 2Li}+_{(aq) + H} 2(g)

(b) 2Al(s) + 3F2(g) → 2Al3+(aq) + 6F–(aq)

(c) 2FeCl3(aq) + H2(g) → 2FeCl2(aq) + 2HCl(aq)

(d) Al(NO3)3(aq) + 3Li(s) → 3LiNO3(aq) + Al(s)

Thought Lab: Assigning

Reference Values

HANDOUT

Analysis

1. Compare your calculations from Procedure Steps 2 and 3. What effect does changing the zero on the scale of reduction potentials have on:

(a) reduction potentials?

(b) cell potentials?

2. Find the difference between the temperatures at which water boils and freezes on the following scales (assume that a difference is positive, rather than negative):

(a) the Celsius temperature scale

(b) the Kelvin temperature scale

3. What do your answers to Question 2 tell you about these two temperature scales?

4. How are temperature scales and reduction potentials similar?

CHAPTER 13

Thought Lab: Assigning

Reference Values Answer Key

BLM 13.1.3A

ANSWER KEY

1.

Reduction half-reaction Accepted E° (V) Adjusted E°(V) [+ 1.66 (V)]

F2(g) + 2e− → 2F−(aq) +2.87 +4.53

Fe3+_{(aq) + e}− _{→ Fe}2+_{(aq) +0.77 +2.43}

2H+_{(aq) + 2e}− _{→ H}

2(g) 0.00 +1.66

Al3+_{(aq) + 3e}− _{→ Al(s) −1.66 0.00}

Li+_{(aq) + e}− _{→ Li(s) −3.04 −1.39}

2. (a) Eºcell = Eºcathode – Eºanode = 0.00 V – (–3.04 V) = +3.04 V

(b) Eºcell = Eºcathode – Eºanode = 2.87 V – (–1.66 V) = +4.53 V

(c) Eºcell = Eºcathode – Eºanode = 0.77 V – 0.000 V = +0.77 V

(d) Eºcell = Eºcathode – Eºanode = –1.66 V – (–3.04 V) = +1.38 V

3. (a) Eºcell = Eºcathode – Eºanode = 1.66 V – (–1.38 V) = +3.04 V

(b) Eºcell = Eºcathode – Eºanode = 4.53 V – 0.00 V = +4.53 V

(c) Eºcell = Eºcathode – Eºanode = 2.43 V – 1.66 V = +0.77 V

(d) Eºcell = Eºcathode – Eºanode = 0.00 V – (–1.38 V) = +1.38 V

Answers to Analysis Questions

1. (a) Shifting the zero value by +1.66 V increased each reduction potential by +1.66 V.

(b) Changing the zero on the scale of reduction potentials has no effect on the calculated cell potentials.

2. (a) On the Celsius temperature scale the difference between the temperature at which water boils and the temperature at which water freezes is 100.0 °C – 0.0 °C = 100 °C.

(b) On the Kelvin temperature scale the difference between the temperature at which water boils and the temperature at which water freezes is 373.15 K – 273.15 K = 100 K.

3. The Celsius and Kelvin temperature scales have two different zero values, but a change of 1 degree is the same on both scales. Zero on the Celsius scale is 273.15 K on the Kelvin scale. Zero on the Kelvin scale is –273.15 °C. The zero value of the Celsius scale is defined based on the freezing point of water under standard conditions. For the Kelvin scale, the zero is shifted to occur at absolute zero – the temperature at which particles of matter have zero energy under standard conditions.

4. The Celsius scale and reduction potentials have a zero value that is arbitrarily set.

Voltaic Cell Notation

ASSESSMENT

Answer the following questions in the space provided.

1. (a) Label each component of the following shorthand representation of a galvanic cell.

(b)What does the location (left or right of the double vertical line) of each substance in the shorthand representation tell you about the cell?

(c) If you were to draw a diagram of the cell, would it matter whether the cathode was shown to the right or to the left of the anode? Explain your answer.

2. Use galvanic cell notation to represent a galvanic cell that has a copper electrode in copper(II) sulfate solution and an iron electrode in iron(II) nitrate solution.

Voltaic Cell Notation

3. Consider the following half-reactions: Oxidation: 2I–_{(aq) → I}

2(s) + 2e–

Reduction: MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(ℓ)

(a) Write the overall reaction.

(b)A galvanic cell based on this reaction uses inert electrodes, such as graphite electrodes. Explain why.

(c) Write the shorthand representation of the cell using galvanic cell notation.

4. Use the following shorthand representation to sketch a possible design for the following cell:

Voltaic Cell Notation Answer

Key

ANSWER KEY

1. (a)

(b) The anode is on the left. The cathode is on the right.

(c)No.

2. Fe | Fe2+ _{|| Cu}2+ _{| Cu}

3. (a)2MnO4–(aq) + 16H+(aq) + 10I–(aq) → 5I2(s) + 2Mn2+(aq) + 8H2O(ℓ)

(b) None of the reactants or products would make suitable electrodes.

(c)graphite | I2(s) | I–(aq) || H+(aq), MnO4–(aq), Mn2+(aq) | graphite

CHAPTER 13

Fuel Cell Technology

OVERHEAD

Hydrogen gas from tanks and oxygen in air provide a continuous supply of

fuel for the fuel cell. The electrolyte is a solid polymer that allows

CHAPTER 13

Electrolytic Cells

OVERHEAD

Adding an external voltage to reverse the electron flow converts a Daniell

cell from a voltaic cell into an electrolytic cell.

Voltaic cell

Electrolytic cell

Oxidation half-reaction:

Oxidation half-reaction:

Reduction half-reaction:

Reduction half-reaction:

CHAPTER 13

Electrolytic Cells Answer Key

ANSWER KEY

Voltaic cell Electrolytic cell Oxidation half-reaction:

Zn(s) → Zn2+_{(aq) +2e}–

Oxidation half-reaction:

Cu(s) → Cu2+_{(aq) + 2e}–

Reduction half-reaction:

Cu2+_{(aq) + 2e}– _{→ Cu(s)}

Reduction half-reaction:

Zn2+_{(aq) + 2e}– _{→ Zn(s)}

Cell reaction:

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

Cell reaction:

Cu(s) + ZnSO4(aq) → CuSO4(aq) + Zn(s)

CHAPTER 13

The Chlor-alkali Cell

OVERHEAD

CHAPTER 13

Investigation 13.B: Electrolysis of

Aqueous Potassium Iodide

BLM 13.3.3

HANDOUT

When an aqueous solution is electrolyzed, the electrolyte or water can undergo electrolysis. In this investigation, you will build an electrolytic cell, carry out the electrolysis of an aqueous solution, and identify the products.

Question

What are the products from the electrolysis of a 1 mol/L aqueous solution of potassium iodide? Are the observed products the ones predicted using reduction potentials?

Predictions

Use the relevant standard reduction potentials from the table in Appendix G, and the non-standard reduction potentials you used previously for water, to predict the electrolysis products. Predict which product(s) are formed at the anode and which product(s) are formed at the cathode.

CHAPTER 13

Investigation 13.B: Electrolysis of

Aqueous Potassium Iodide

Answer Key

BLM 13.3.3A

ANSWER KEY

Answers to Prediction Question

1. The K+_{(aq) and I}–_{(aq) concentrations are 1 mol/L, so the standard reduction potentials for the}

half-reactions that involve these ions should be used. The non-standard values for water should be used.

I2(s) + 2e– → 2I–(aq) Eº = 0.54 V

O2(g) + 4H+(aq) + 4e– → 2H2O(ℓ) Eº = 0.81 V

2H2O(ℓ) + 2e– → H2(g) + 2OH–(aq) Eº = –0.41 V

K+_{(aq) + e}– _{→ K(s) Eº = –3.04 V}

The four possible overall reactions are: 2K+_{(aq) + 2I}–_{(aq) → 2K(s) + 2I}

2(s); Eºcell = –3.58 V

2H2O(ℓ) → 2H2(g) + O2(g); Eº cell = –1.23 V

4K+_{(aq) + 2H}

2O(ℓ) → 4K(s) + O2(g) + 4H+ (aq); Eº cell = –3.85 V

The cell reaction that produces hydrogen gas, hydroxide ions, and iodine requires the lowest external voltage. Therefore, you should predict the formation of these products. The hydrogen gas and hydroxide ions should form at the cathode, and the iodine should form at the anode.

Answers to Analysis Questions

1. 1.

2. Adding the starch solution to the electrolyte at the anode results in a purple solution, indicating the presence of molecular iodine. At the cathode, the clear, colourless phenolphthalein solution will turn pink, indicating that the solution is basic and suggesting the presence of hydroxide ions. (Hydrogen gas is also formed at the cathode, but students do not test for the gas.)

3. 2I–_{(aq) → I}

2(s) + 2e–

4. 2H2O(ℓ) + 2e– → H2(g) + 2OH–(aq)

5. 2I–_{(aq) + 2H}

2O(ℓ) → I2(aq) + H2(g) + 2OH–(aq)

CHAPTER 13

Investigation 13.B: Electrolysis of

Aqueous Potassium Iodide Answer

Key

BLM 13.3.3A

ANSWER KEY

6. Eºcell = Eºcathode – Eºanode = –0.83 V – (+0.54 V) = –1.37 V

The calculated external voltage needed is 1.36 V. The actual voltage needed to carry out this electrolytic process is greater than this. As you learned on p. 505 of the textbook, this excess is called overpotential. This is the reason that the external voltage used in this investigation was significantly higher than the calculated value.

Answer to Conclusion Question

Answers to Application Questions

8. The observations would probably not change. Consider the four possible reactions. The electrolysis of aqueous sodium iodide requires greater external voltage than the electrolysis of water. The reaction producing sodium metal and hydrogen ions also requires greater external voltage than the electrolysis of water. However, the reaction producing iodine, hydroxyl ions, and hydrogen gas requires less external voltage than the electrolysis of water. These are the same products that were observed in the investigation.

CHAPTER 13

Calculating the Mass of

Electrolysis Product

BLM 13.4.1

Problem

Calculate the mass of aluminium produced by the electrolysis of molten aluminium chloride, if a current of 500 mA passes between the half-cells for 1.50 h.

What Is Required?

You need to calculate the mass of aluminium produced.

What Is Given?

You know the identity of the electrolyte, the current, and the time. electrolyte: AlCl3(ℓ)

current: 500 mA time: 1.50 h

You know the charge on one mole of electrons: 9.65 × 104 _C/mol.

Plan Your Strategy

Use the current and the time to find the quantity of electric charge that passed from the anode to the cathode. From the charge, find the amount of electrons that passed through the circuit. Use the stoichiometry of the relevant half-reaction to relate the amount of electrons to the amount of aluminium produced. Use the molar mass of aluminium to convert the amount of aluminium to a mass of aluminium.

CHAPTER 13

Calculating the Mass of Electrolysis

Product

BLM 13.4.1

Act on Your Strategy

To calculate the quantity of electrical charge in coulombs, convert the data to SI units: 1000 mA = 1 A

q = IΔt

q = (0.500 A)(5400 s) = 2.700 × 103 _C

Find the amount of electrons. One mole of electrons has a charge of 9.65 × 104 _C:

Amount of electrons

The half-reaction for the reduction of aluminium ions to aluminium atoms is: Al3+ _{+ 3e}– _{→ Al}

Convert the amount of aluminium to a mass: mAl = nMAl

= 0.252 g

Check Your Solution

The answer is expressed in units of mass. To check your answer, use estimation. If the current were 1 A, then 1 mol of electrons would pass in 9.65 × 104 _{s. In this example, the current is less than 1 A,}

and the time is much less than 9.65 × 104 _{s. Therefore, much less than 1 mol of electrons would be}

used, and much less than 1 mol (27 g) of aluminium would be formed.

CHAPTER 13

Investigation 13.C: Electroplating

HANDOUT

You have learned that electroplating is a process in which a metal is deposited, or plated, onto the cathode of an electrolytic cell. In this investigation, you will build an electrolytic cell and electrolyze a copper(II) sulfate solution to plate copper onto the cathode. You will use Faraday’s law to relate the mass of metal deposited to the amount of electric charged used.

Question

Does the measured mass of copper plated onto the cathode of an electrolytic cell agree with the mass calculated according to Faraday’s law?

Prediction

Predict whether the measured mass of copper plated onto the cathode of an electrolytic cell will be greater than, equal to, or less than the mass calculated using Faraday’s law.

Safety Precautions

• Nitric acid is corrosive. Note that the CuSO4 solution contains sulfuric acid and hydrochloric acid.

Wash any spills on your skin with plenty of cold water. Inform your teacher immediately. • Avoid touching the parts of the electrodes that have been washed with nitric acid.

• Acetone is flammable. Use acetone in the fume hood.

• Make sure your hands and your lab bench are dry before handling any electrical equipment.

Materials

• 3 cm × 12 cm × 1 mm Cu strip

• 150 mL 1.0 mol/L HNO3 in a 250 mL beaker

• deionized water in a wash bottle • 50 cm 16-gauge bare solid copper wire • 120 mL acidified 0.50 mol/L CuSO4 solution

(with 5 mL of 6 mol/L H2SO4 and 3 mL of

0.1 mol/L HCl added) • fine sandpaper

• 250 mL beaker

• 2 electrical leads with alligator clips

• adjustable D.C. power supply with ammeter • drying oven

Procedure

1. Clean any tarnish off the copper strip by sanding it gently. Working in a fume hood, dip the bottom of the copper strip in the nitric acid for a few seconds, and then rinse the strip carefully with de-ionized water. Avoid touching the section that has been cleaned by the acid.

2. Place the copper strip in the beaker, with the clean part of the strip at the bottom. Bend the top of the strip over the rim of the beaker so that the copper strip is secured in a vertical position, as shown in the diagram on the next page. This copper strip will serve as the anode.

CHAPTER 13

Investigation 13.C: Electroplating

(continued)

BLM 13.4.2

HANDOUT

3. Wrap the copper wire around a pencil to make a closely spaced coil. Leave 10 cm of the wire unwrapped. Measure and record the mass of the wire. Working in a fume hood, dip the coil in the nitric acid, and rinse the coil with de-ionized water.

4. Use the 10 cm of uncoiled wire to secure the coil on the opposite side of the beaker from the anode, as shown in the diagram. This copper wire will serve as the cathode.

5. Pour 120 mL of the acidified CuSO4(aq) solution into the beaker.

Attach the lead from the negative

terminal of the power supply to the cathode. Attach the positive terminal to the anode.

6. Turn on the power supply and set the current to 1.0 A. Maintain this current for 20 min by adjusting the variable current knob as needed.

7. After 20 min, turn off the power. Remove the cathode and rinse it very gently with deionized water. Place the cathode in a drying oven for 20 min.

8. Measure and record the new mass of the cathode.

2. Use the measured current and the time for which the current passed to calculate the amount of charge used.

CHAPTER 13

Investigation 13.C: Electroplating

(continued)

BLM 13.4.2

HANDOUT

3. Use your answers to Questions 1 and 2 to calculate the mass of copper plated onto the cathode.

4. Compare the calculated mass from Question 3 with the measured increase in mass of the cathode. Give possible reasons for any difference between the two values.

Conclusion

5. How did the mass of copper electroplated onto the cathode of the electrolytic cell compare with the mass calculated using Faraday’s law? Compare your answer with your prediction from the beginning of this investigation.

Applications

CHAPTER 13

Investigation 13.C: Electroplating

(continued)

BLM 13.4.2

HANDOUT

7. Suppose you repeated the investigation with the copper(II) sulfate solution, but you passed the current for only half as long as before. How would the masses of copper plated onto the cathode compare in the two investigations? Explain your answer.

CHAPTER 13

Investigation 13.C: Electroplating

Answer Key

BLM 13.4.2A

ANSWER KEY

Answers to Prediction Questions

A prediction made using Faraday’s law involves a theoretical calculation, which does not take into account the experimental conditions, variations in materials used, and expertise of the experimenter. Therefore, you should have predicted that the mass of the copper plated onto the cathode would be less than predicted using Faraday’s law.

Answers to Analysis Questions

1. The half-reaction occurring at the cathode (reduction of copper) is: Cu2+_{(aq) + 2e}– _{→ Cu(s)}

2. You should use your measured results to calculate quantity of electricity, as done in the Sample Problem on p. 515 of the textbook. Assuming a current of 1.00 A and 1200 s:

Quantity of electricity = 1.00 A × 1200 s = 1.2 × 103 _C

3.

The half-reaction is: Cu2+_{(aq) + 2e}– _{→ Cu(s)}

4. Assuming no trivial errors, the differences between the calculated mass and the actual mass will likely be the result of improper drying or differences in the actual current flow (quantity of electrons) moving through the circuit.

Answers to Conclusion Question

5. Typically the mass of the copper electroplated onto the cathode will be less than predicted, assuming you have completely dried the anode. The answer for the second part of this question will be based on your predictions at the start of the investigation.

Answers to Applications Questions

6. Assuming that the iron(II) ions will plate out in the reaction, compare the mass of iron that should theoretically plate out to the mass of the copper plated out. In this case, since the mass of 1 mol of iron is less than that of 1 mol of copper, and since the number of moles of electrons required per mole of iron(II) ions is the same as required for copper(II) ions, the total mass of iron plated out should be less than the mass of copper plated out.

7. With the current running for half the time, you would expect the mass of copper plated to be half the original amount. If the current runs for half the time, then only half the number of electrons are provided, and only half of the reduction reactions can take place.

Investigation 13.C: Electroplating

Answer Key

ANSWER KEY