CHAPTER Chapter 17 Oxidation Numbers ... 7 2−. 2 Assign known oxidation numbers. According to Rule g, the oxidation number of the O atoms is −2, so this

602

C H A P T E R

Copyright © by Holt, Rinehart and Winston. All rights reserved.

If you run out of gasoline in a car, you might have to walk to the nearest gasstation for more fuel. But running out of energy when you are millions of

kilometers from Earth is a different story! A robot designed to collect data onother bodies in our solar system needs a reliable, portable source of energy thatcan work in the absence of an atmosphere to carry out its mission. Many of thepower sources for these explorer robots are batteries. In this chapter, you willlearn about the processes of oxidation and reduction and how they are used inbatteries to provide energy. You will also learn how these processes are used topurify metals and protect objects from corrosion.

Pre-Reading QuestionsWhat type of charge results from losing electrons? from gaining electrons?

Name a device that converts chemical energy into electrical energy.

Why are batteries marked with positive and negative terminals?

SAFETY PRECAUTIONS

1

2

3

CONTENTSSECTION 1

Oxidation-ReductionReactions

SECTION 2

Introduction toElectrochemistry

SECTION 3

Galvanic Cells

SECTION 4

Electrolytic Cells

17START-UPACTIVITY

603

Lights OnPROCEDURE

1. Assemble batteries, a light-emitting diode, and wires so that thediode lights. Make a diagram of your construction.

2. Remove the batteries, and reconnect them in the opposite direction.Record the results.

ANALYSIS

1. A light-emitting diode allows electrons to move through it in onlyone direction—into the short leg and out of the longer leg. Basedon your results, from which end of the battery must electrons leave?

2. How are the atoms changing in the end of the battery from whichelectrons leave?

3. What must happen to the electrons as they enter into the other end of the battery?

4. Why does a battery eventually run down?


604 Chapter 17

KEY TERMS

• oxidation

• reduction

• oxidation-reductionreaction

• oxidation number

• half-reaction

• oxidizing agent

• reducing agent

Topic LinkTopic LinkRefer to the “Ions and IonicCompounds” and “Covalent

Compounds” chapters formore information about

chemical bonding.

OBJECTIVES

Identify atoms that are oxidized or reduced through electron transfer.

Assign oxidation numbers to atoms in compounds and ions.

Identify redox reactions by analyzing changes in oxidation numbersfor different atoms in the reaction.

Balance equations for oxidation-reduction reactions through the half-reaction method.

Electron Transfer and Chemical ReactionsYou already know that atoms with very different electronegativities bondby an electron transfer. For example, sodium chloride is formed by thetransfer of electrons from sodium atoms to chlorine atoms in the reactionshown in Figure 1 and described by the following equation:

2Na(s) + Cl2(g) ⎯→ 2NaCl(s)

Though NaCl is the way the formula of sodium chloride is usually written,the compound is made up of ions. Therefore, it might be helpful to thinkof sodium chloride as if its formula were written Na+Cl− so that youremember the ions.

When the electronegativity difference between the atoms is smaller, apolar covalent bond can form when the atoms join, as shown below.

2C(s) + O2(g) ⎯→ 2CO(g)

The C–O bond has some ionic character because there is an unequalsharing of electrons between the carbon atom and the oxygen atom. Theoxygen atom attracts the shared electrons more strongly than the carbonatom does.

Oxidation Involves a Loss of ElectronsIn the examples above, electrons were transferred at least in part fromone atom to another. The sodium atom lost an electron to the chlorineatom. The carbon atom lost some of its control over its electrons to theoxygen atom. The loss, wholly or in part, of one or more electrons iscalled

Thus, in making NaCl, the sodium atom is oxidized from Na to Na+.Likewise, the carbon atom is oxidized when CO forms, even though thecarbon atom does not become an ion.

oxidation.

1

2

3

4

Oxidation-Reduction Reactions1S E C T I O N

oxidation

a reaction that removes one ormore electrons from a substancesuch that the substance’s valenceor oxidation state increases


Reduction Involves a Gain of ElectronsIn making NaCl, the electrons lost by the sodium atoms do not justdisappear. They are gained by the chlorine atoms. The gain of electrons isdescribed as The chlorine atoms are reduced as they changefrom Cl2 to 2Cl−.

When joining with carbon atoms to make CO, oxygen atoms do not gainelectrons but gain only a partial negative charge. But because the electronsin the C—O bonds spend more time near the oxygen atoms, the change isstill a reduction.

More than one electron may be gained in a reduction. In the formationof Li3N, described by the equation below, three electrons are gained byeach nitrogen atom.

6Li(s) + N2(g) ⎯→ 2Li3N(s)

Oxidation and Reduction Occur TogetherWhen oxidation happens there must also be reduction taking place. Youwill learn later in this chapter that oxidation and reduction can happen atdifferent places. In most situations, however, oxidation and reduction hap-pen in a single place. Consider HgO being broken down into its elements,as described by the following equation:

2HgO(s) ⎯→ 2Hg(l) + O2(g)

In this reaction, mercury atoms are reduced, while oxygen atoms are oxi-dized. A single reaction in which an oxidation and a reduction happen iscalled an or redox reaction.oxidation-reduction reaction

reduction.

Oxidation, Reduction, and Electrochemistry 605

Figure 1Sodium metal and chlorinegas react violently to formsodium chloride. Oxidationand reduction happentogether in this reaction.

reduction

a chemical change in which electrons are gained, either bythe removal of oxygen, the addition of hydrogen, or the addition of electrons

oxidation-reductionreaction

any chemical change in whichone species is oxidized (loseselectrons) and another species is reduced (gains electrons); alsocalled redox reaction

www.scilinks.orgTopic: Redox ReactionsSciLinks code: HW4110

2Na(s) + Cl2(g) ⎯→ 2NaCl(s)

Sodiumatom, Na

Chlorinemolecule, Cl2

Chloride ion, Cl−

Sodium ion,Na+


oxidation numbers of +1, +2, and+3, respectively.

f. The oxidation number of eachhydrogen atom in a compound is+1, unless it is combined with ametal atom; then it is −1.

g. The oxidation number of eachoxygen atom in compounds isusually −2. When combined withfluorine atoms, oxygen becomes+2. In peroxides, such as H2O2,an oxygen atom has an oxidationnumber of −1.

3. Calculate remaining oxidationnumbers, and verify the results.• Use the total oxidation number of

each element’s atoms (the oxidationnumber for an atom of the elementmultiplied by the subscript for theelement) and the following rules tocalculate missing oxidation numbers.h. The sum of the oxidation numbers

for all the atoms in a molecule iszero.

i. The sum of the oxidation numbersfor all atoms in a polyatomic ion isequal to the charge on that ion.

Assigning Oxidation Numbers

1. Identify the formula.• If no formula is provided, write the

formula of the molecule or ion.

2. Assign known oxidation numbers.• Place an oxidation number above

each element’s symbol according tothe following rules.a. The oxidation number of an atom

of any free (uncombined) elementin atomic or molecular form iszero.

b. The oxidation number of amonatomic ion is equal to thecharge on the ion.

c. The oxidation number of an atomof fluorine in a compound isalways −1 because it is the mostelectronegative element.

d. An atom of the more electronega-tive element in a binary compoundis assigned the number equal tothe charge it would have if it werean ion.

e. In compounds, atoms of the elements of Group 1, Group 2,and aluminum have positive

SKILLSSKILLS

606 Chapter 17

Oxidation NumbersTo identify whether atoms are oxidized or reduced, chemists use a modelof which can help them identify differences in anatom of an element in different compounds. By following the set of rulesdescribed in Skills Toolkit 1 below, you can assign an oxidation number toeach atom in a molecule or in an ion. Sample Problem A shows how to usethe rules. You can see three different oxidation numbers for atoms ofmanganese in Figure 2.

By tracking oxidation numbers, you can tell whether an atom is oxi-dized or reduced. If the oxidation number of an atom increases during areaction, the atom is oxidized. If the oxidation number decreases, theatom is reduced. Like other models, oxidation numbers have limits. Youshould consider them a bookkeeping tool to help keep track of electrons.In some cases, additional rules are needed to find values that make sense.

oxidation numbers,oxidation number

the number of electrons thatmust be added to or removedfrom an atom in a combinedstate to convert the atom into the elemental form

11


SAM PLE PROBLE M ASAM PLE PROBLE M A

Determining Oxidation NumbersAssign oxidation numbers to the sulfur and oxygen atoms in the pyrosulfate ion, S2O7

2−.

1 Identify the formula.The pyrosulfate ion has the formula S2O7

2−.

2 Assign known oxidation numbers.According to Rule g, the oxidation number of the O atoms is −2, so thisnumber is written above the O symbol in the formula. Because the oxi-dation number of the sulfur atoms is unknown, x is written above the Ssymbol. Thus the formula is as follows:

S2

xO−2

72−

3 Calculate remaining oxidation numbers, and verify the results.• Multiplying the oxidation numbers by the subscripts, we see that the

S atoms contribute 2x and the O atoms contribute 7(−2) = −14 to thetotal oxidation number. To come up with the correct total charge, (Rulei), 2x + (–14) = −2. Solve this equation to find x = +6.

• In S2O72−, the oxidation number of the S atoms is +6, and the oxidation

number of the O atoms is −2. The sum of the total oxidation numbersfor each element is 2(+6) + 7(−2) = −2, which is the charge on the ion.

In this book, the oxidation number for asingle atom is writtenabove its chemicalsymbol. However, besure to use the totalnumber of atoms foreach element when finding the sum of the oxidation numbersfor all atoms in the molecule or ion.

PRACTIC E

PRACTICE HINT

Determine the oxidation number for each atom in each of the following.

1 a. NH4+

b. Al

c. H2O

d. Pb2+

e. H2

f. PbSO4

g. KClO3

h. BF3

i. Ca(OH)2

j. Fe2(CO3)3

k. H2PO4−

l. NH4NO3

PRACTICE HINT

PROBLEM

SOLVING

SKILLPROBLEM

SOLVING

SKILL


Figure 2An atom of manganese in its elemental form has anoxidation number of 0. InMnO2, the oxidation numberof the manganese atom is +4.In the permanganate ion,MnO4

−, the oxidation numberof the manganese atom is +7.

Mn0 K

+1+Mn+4

O−2

2

Mn+7

O−2−

4


608 Chapter 17

Identifying Redox ReactionsFigure 3 shows the reaction of Zn with HCl. Is this a redox reaction?Hydrochloric acid is a solution in water of Cl−, which plays no part in thereaction, and H3O+. The net change in this reaction is

2H3O+(aq) + Zn(s) ⎯→ H2(g) + 2H2O(l) + Zn2+(aq)

Using rules a, b, f, g, h, and i from Skills Toolkit 1, you can give oxidationnumbers to all atoms as follows:

2H+1

3O−2+(aq) + Zn

0(s) ⎯→ H

02(g) + 2H

+12O

−2(l) + Zn

+2 2+(aq)

Comparing oxidation numbers, you see that the zinc atom changes from0 to +2 and that two hydrogen atoms change from +1 to 0. So, this is aredox reaction. In a redox reaction, the oxidation numbers of atoms thatare oxidized increase, and those of atoms that are reduced decrease.

Half-ReactionsIn the reaction shown in Figure 3, each zinc atom loses two electrons andis oxidized. One way to show only this half of the overall redox reactionis by writing a for the change.

Zn(s) ⎯→ Zn2+(aq) + 2e−

Note that electrons are a product. Of course, there is also a half-reactionfor reduction in which electrons are a reactant.

2e− + 2H3O+(aq) ⎯→ H2(g) + 2H2O(l)

By adding the two half-reactions together, you get the overall redox reac-tion shown earlier. Notice that enough electrons are in each half-reaction tokeep the charges balanced. Keep in mind that free electrons do not actuallyleave the zinc atoms and float around before being picked up by the hydro-nium ions. Instead, they are “handed off” directly from one to the other.

half-reaction

Figure 3Zinc metal reacts withhydrochloric acid, makingbubbles of hydrogen gas.

half-reaction

the part of a reaction that involvesonly oxidation or reduction

Zinc atom,Zn

Water molecule,H2O

Chlorideion, Cl−

Water molecule,H2O

Hydrogenmolecule, H2

Zinc ion,Zn2+

Hydroniumion, H3O+

Chlorideion, Cl−


SKILLSSKILLS

Balancing Redox Equations Using theHalf-Reaction Method

1. Identify reactants and products.• Write the unbalanced equation in ionic form, excluding any

spectator ions.• Assign oxidation numbers, and identify the atoms that change

their oxidation numbers. Ignore all species whose atoms do notchange their oxidation number.

2. Write and balance the half-reactions.• Separate the equation into its two half-reactions.• For each half-reaction, do the following:

a. Balance atoms other than hydrogen and oxygen.b. Balance oxygen atoms by adding water molecules as needed.c. Balance hydrogen atoms by adding one hydronium ion for

each hydrogen atom needed and then by adding the samenumber of water molecules to the other side of the equation.

d. Balance the overall charge by adding electrons as needed.

3. Make the electrons equal, and combine half-reactions.• Multiply each half-reaction by an appropriate number so that

both half-reactions have the same number of electrons. Now theelectrons lost equal the electrons gained, so charge is conserved.

• Combine the half-reactions, and cancel anything that is com-mon to both sides of the equation.

4. Verify your results.• Double-check that all atoms and charge are balanced.

22


Balancing Oxidation-Reduction EquationsEquations for redox reactions are sometimes difficult to balance. Use thesteps in Skills Toolkit 2 below to balance redox equations for reactions inacidic aqueous solution. An important step is to identify the key ions ormolecules that contain atoms whose oxidation numbers change. Theseatoms are the starting points of the unbalanced half-reactions. For thereaction of zinc and hydrochloric acid, the unbalanced oxidation andreduction half-reactions would be as follows:

Zn(s) ⎯→ Zn2+(aq) and H3O+(aq) ⎯→ H2(g)

These reactions are then separately balanced. Finally, the balanced equa-tions of the two half-reactions are added together to cancel the electrons.


610 Chapter 17

To help avoid confusionbetween charges and

oxidation numbers, thesign of a charge is

written last and the sign of an oxidation

number is written first.Thus the Fe2+ ion has a

2+ charge and a +2 oxidation number.

SAM PLE PROBLE M BSAM PLE PROBLE M B

PROBLEM

SOLVING

SKILLPROBLEM

SOLVING

SKILL

PRACTICE HINTPRACTICE HINT

PRACTIC EUse the half-reaction method to write a balanced equation for each of the following reactions in acidic, aqueous solution.

1 The reactants are Fe(s) and O2(aq), and the products are Fe3+(aq) and H2O(l).

2 Al(s) is placed in the acidic solution and forms H2(g) and Al3+(aq).

3 The reactants are sodium bromide and hydrogen peroxide, and theproducts are bromine and water.

4 The reactants are manganese dioxide and a soluble copper(I) salt, andthe products are soluble manganese(II) and copper(II) salts.

The Half-Reaction MethodWrite and balance the equation for the reaction when an acidic solutionof MnO4

− reacts with a solution of Fe2+ to form a solution containingMn2+ and Fe3+ ions.

1 Identify reactants and products.• The unbalanced equation in ionic form is as follows:

H3O+(aq) + MnO4−(aq) + Fe2+(aq) ⎯→ Mn2+(aq) + Fe3+(aq)

• Oxidation numbers for the atoms are as follows:

H+1

3O−2+ + Mn

+7O−2

4− + Fe

+2 2+ ⎯→ Mn+2 2+ + Fe

+3 3+

• Atoms of Mn and Fe change oxidation numbers.

2 Write and balance the half-reactions.Unbalanced: Fe2+ ⎯→.Fe3+ MnO4

− ⎯→.Mn2+

Balance O: Fe2+ ⎯→.Fe3+ MnO4− ⎯→.Mn2+ + 4H2O

Balance H: Fe2+ ⎯→.Fe3+ 8H3O+ + MnO4− ⎯→.Mn2+ + 12H2O

Balance e–: Fe2+ ⎯→.Fe3+ + e− 5e−+ 8H3O+ + MnO4− ⎯→.Mn2+ + 12H2O

3 Make the electrons equal, and combine half-reactions.Multiply the half-reaction that involves iron by 5 to make the numbersof electrons the same in each half-reaction. Add the half-reactions, andcancel the electrons to get the final balanced equation:

8H3O+(aq) + MnO4−(aq) + 5Fe2+(aq) ⎯→ Mn2+(aq) + 5Fe3+(aq) + 12H2O(l)

4 Verify your results.Note that there are equal numbers of all atoms and a net charge of +17 on each side of the equation.


Identifying Agents in Redox ReactionsIn Sample Problem B, the permanganate ion caused the oxidation of theiron(II) ion. Substances that cause the oxidation of other substances arecalled They accept electrons easily and so are reduced.Common oxidizing agents are oxygen, hydrogen peroxide, and halogens.

cause reduction to happen and are themselves oxi-dized. The iron(II) ion caused the reduction of the permanganate ion andwas the reducing agent. Common ones are metals, hydrogen, and carbon.

Reducing agents

oxidizing agents.


UNDERSTANDING KEY IDEAS

1. Explain oxidation and reduction in terms ofelectron transfer.

2. How can you identify a reaction as a redoxreaction?

3. Describe how an oxidation-reduction reaction may be broken down into two half-reactions, and explain why the latterare useful in balancing redox equations.

4. Compare the number of electrons lost in an oxidation half-reaction with the numberof electrons gained in the correspondingreduction half-reaction.

5. Describe what an oxidizing agent and areducing agent are.

PRACTICE PROBLEMS

6. Assign oxidation numbers to the atoms in each of the following:

a. H2SO3 c. SF6

b. Cl2 d. NO3−

7. Assign oxidation numbers to the atoms in each of the following:

a. CH4 c. NaHCO3

b. HSO3− d. NaBiO3

8. Identify the oxidation number of a Cr atom in each of the following: CrO3, CrO, Cr(s),CrO2, Cr2O3, Cr2O7

2−, and CrO42−.

9. Which of the following equations representredox reactions? For each redox reaction,determine which atom is oxidized and whichis reduced, and identify the oxidizing agentand the reducing agent.

a. MgO(s) + H2CO3(aq) ⎯→MgCO3(s) + H2O(l)

b. 2KNO3(s) ⎯→ 2KNO2(s) + O2(g)

c. H2(g) + CuO(s) ⎯→ Cu(s) + H2O(l)

d. NaOH(aq) + HCl(aq) ⎯→NaCl(aq) + H2O(l)

e. H2(g) + Cl2(g) ⎯→ 2HCl(g)

f. SO3(g) + H2O(l) ⎯→ H2SO4(aq)

10. Use the half-reaction method in acidic,aqueous solution to balance each of thefollowing redox reactions:

a. Cl−(aq) + Cr2O72−(aq) ⎯→

Cl2(g) + Cr3+(aq)

b. Cu(s) + Ag+(aq) ⎯→ Cu2+(aq) + Ag(s)

c. Br2(l) + I−(aq) ⎯→ I2(s) + Br−(aq)

d. I−(aq) + NO2−(aq) ⎯→ NO(g) + I2(s)

11. Determine which atom is oxidized andwhich is reduced, and identify the oxidizingagent and the reducing agent for each reac-tion in item 10.

CRITICAL THINKING

12. How is it possible for hydrogen peroxide tobe both an oxidizing agent and a reducingagent?

Section Review1

oxidizing agent

the substance that gains electronsin an oxidation-reduction reactionand is reduced

reducing agent

a substance that has the potentialto reduce another substance


++

__

612 Chapter 17

KEY TERMS

• electrochemistry

• voltage

• electrode

• electrochemical cell

• cathode

• anode

Figure 4When the switch of a flashlight is closed, electrons“pushed” by the battery areforced through the thintungsten filament of thebulb. This flow of electronsmakes the filament hotenough to emit light.

OBJECTIVES

Describe the relationship between voltage and themovement of electrons.

Identify the parts of an electrochemical cell and their functions.

Write electrode reactions for cathodes and anodes.

Chemistry Meets ElectricityThe science of deals with the connections betweenchemistry and electricity. It is an important subject because it is involvedin many of the things you use every day. Electrochemical devices changeelectrical energy into chemical energy and vice versa. A simple flashlightis an everyday example of something that converts chemical energy intoelectrical energy, which is then converted into light energy.

Figure 4 shows the components of a typical flashlight. The powersource is two batteries or cells. Each cell has a metal terminal at each end.If you examine a battery, you will find a � or + symbol near the top iden-tifying the positive terminal. The bottom is the negative terminal, and ispossibly unmarked or marked with a � or − symbol. By closing the switch,you turn the flashlight on. Electrons move from the negative terminal ofthe lower battery, through a metal circuit that includes the light bulb, andcontinue to the positive terminal of the upper battery. The circuit is com-pleted by electrons and ions moving charge through each of the batteriesand electrons moving charge from the upper battery to the lower one.When you turn off the flashlight, you break the pathway, and the move-ment of electrons and ions stops.

electrochemistry

1

2

3

Introduction to Electrochemistry2S E C T I O N

electrochemistry

the branch of chemistry that isthe study of the relationshipbetween electric forces andchemical reactions

www.scilinks.orgTopic: Electrical EnergySciLinks code: HW4045


“Electrical Pressure” Is Expressed in VoltsElectrochemical reactions in a battery cause a greater electron densityin the negative terminal than in the positive terminal. Electrons repeleach other, so there is a higher “pressure” on the electrons in the nega-tive terminal, which drives the electrons out of the battery and throughthe flashlight.

Electrical “pressure,” often called electric potential, or isexpressed in units of volts. The voltage of an ordinary flashlight cell is 1.5volts or 1.5 V. When two cells are placed end-to-end, as in Figure 4, thevoltages add together, so the overall voltage driving electrons is 3.0 V.Themovement of electrons or other charged particles is described as electriccurrent and is expressed in units of amperes.

Components of Electrochemical CellsA flashlight battery is an electrochemical cell. An consists of two electrodes separated by an electrolyte. An is aconductor that connects with a nonmetallic part of a circuit. You havelearned about two kinds of conductors. One kind includes metals, whichconduct electric current through moving electrons. The second kindincludes electrolyte solutions, which conduct through moving ions.

The Cathode Is Where Reduction OccursElectrode reactions happen on the surfaces of electrodes. Figure 5 showshalf of an electrochemical cell. The copper strip is an electrode because itis a conductor in contact with the electrolyte solution.The reaction on thiselectrode is the reduction described by the following equation:

Cu2+(aq) + 2e− ⎯→ Cu(s)

The copper electrode is a because reduction happens on it.cathode

electrodeelectrochemical cell

voltage, voltage

the potential difference or electromotive force, measured in volts; it represents the amountof work that moving an electriccharge between two pointswould take

electrode

a conductor used to establishelectrical contact with a non-metallic part of a circuit, such asan electrolyte

electrochemical cell

a system that contains two electrodes separated by an electrolyte phase

cathode

the electrode on whose surfacereduction takes place

Copper(II) sulfate,CuSO4, solution

Copper metal

Cathodic reaction: Cu2+(aq) + 2e− ⎯→ Cu(s)

Figure 5Copper(II) ions, Cu2+(aq),are reduced to atoms as they gain electrons on thecathode.Water molecule,

H2O

Copper atom, Cu

Copper(II) ion, Cu2+

Sulfate ion, SO2−4



614 Chapter 17

The Anode Is Where Oxidation OccursThe electrons that cause reduction at the cathode are pushed there froma reaction at the second electrode of a cell.The is the electrode onwhich oxidation occurs. Figure 6 shows the second half of the electro-chemical cell. The zinc strip is an electrode because it is a conductor incontact with the solution. The reaction on this electrode is the oxidationdescribed by the following equation:

Zn(s) ⎯→ Zn2+(aq) + 2e−

The zinc electrode is an anode because oxidation happens on it.The electrode reactions described here would have been called half-

reactions in the last section. The difference is that a half-reaction is ahelpful model, but the electrons shown are never really free. An elec-trode reaction describes reality because the electrons actually move fromone electrode to another to continue the reaction.

Pathways for Moving ChargesThe electrode reactions cannot happen unless the electrodes are part of acomplete circuit. So, a cell must have pathways to move charges.Wires areoften used to connect the electrodes through a meter or a light bulb.Electrons carry charges in the wires and electrodes.

Ions in solution carry charges between the electrolytes to complete thecircuit. A porous barrier, as shown in Figure 7, or a salt bridge keeps thesolutions from mixing, but lets the ions move. In the cell shown, charge iscarried through the barrier by a combination of Zn2+(aq) ions moving tothe right and SO4

2−(aq) ions moving to the left. Understand that positivecharge may be carried from left to right through this cell either by nega-tive particles (electrons or anions) moving from right to left or by cationsmoving from left to right.

anodeanode

the electrode on whose surfaceoxidation takes place; anionsmigrate toward the anode, andelectrons leave the system fromthe anode

Zinc sulfate,ZnSO4, solution

Zinc metal

Anodic reaction: Zn(s) ⎯→ Zn2+(aq) + 2e−

Electrodes Porous barrier

Figure 7The light bulb is powered bythe reaction in this cell.

Figure 6Zinc atoms are oxidized to zinc ions,Zn2+(aq), as they lose electrons at the anode. The zinc strip dissolves as the reaction continues.

Water molecule,H2O

Zinc ion,Zn2+

Sulfate ion, SO2−4

Zinc atom,Zn



1. What is voltage?

2. How does voltage relate to the movementof electrons?

3. List the components of an electrochemicalcell, and describe the function of each.

4. What are the names of the electrodes in anelectrochemical cell? What type of reactionhappens on each?

5. Describe the difference in how charge flowsin wires and in electrolyte solutions.

6. A 12 V car battery has six cells connectedend-to-end. What is the voltage of a singlecell?

7. Will the reaction below happen at an anodeor a cathode? Explain.

Fe(CN)63−(aq) + e− ⎯→ Fe(CN)6

4−(aq)

8. Describe the changes in oxidation numberthat happen in an anode reaction and in acathode reaction.

CRITICAL THINKING

9. What would happen if you put one of thebatteries in backward in a two-cell flashlight?

10. What would happen if you put both batter-ies in backward in a two-cell flashlight?

11. If an electrode reaction has dissolved oxy-gen, O2(aq), as a reactant, is the electrodean anode or a cathode? Explain.

12. Write an electrode reaction in which youchange Br−(aq) to Br2(aq). Would this reac-tion happen at an anode or a cathode?

13. Write an electrode reaction in whichSn4+(aq) is changed to Sn2+(aq). Would thisreaction happen at an anode or a cathode?

14. If you wanted to use an electrochemical cellto deposit a thin layer of silver metal onto abracelet, which electrode would you makethe bracelet? Explain using the equation forthe electrode reaction that would occur.

15. What would happen at each electrode ifbatteries were connected to the cell inFigure 7 so that electrons flowed in the opposite direction of the direction describedin the text?

16. Compare the equations for electrode reac-tions with the equations for half-reactions.

17. Write the electrode reactions for a cell thatinvolves only Cu(s) and Cu2+(aq) in whichthe anode reaction is the reverse of thecathode reaction. What is the net result ofoperating this cell?

18. Is it correct to say that the net chemicalresult of an electrochemical cell is a redoxreaction? Explain.


Section Review2

www.scilinks.orgTopic: Electrochemical CellsSciLinks code: HW4046

The Complete CellFigure 7 shows a complete cell composed of the electrodes shown sepa-rately in earlier figures. The overall process when the anode reaction isadded to the cathode reaction for this cell is the same as the followingredox reaction:

Zn(s) + Cu2+(aq) ⎯→ Zn2+(aq) + Cu(s)

Although the two electrode reactions occur at the same time, they occurat different places in the cell. This is an important distinction from theredox reactions discussed in the last section.


616 Chapter 17

Key Terms• corrosion

• standard electrodepotential

Anode CathodeZinc sulfate,

ZnSO4, solution

Zinc metal

Copper(II) sulfate,CuSO4, solution

Copper metal

e–

e–

Porous barrier

Figure 8As a result of the reaction in a galvanic cell, the bulblights up as electrons movein the wires from the anodeto the cathode.

OBJECTIVES

Describe the operation of galvanic cells, including dry cells, lead-acidbatteries, and fuel cells.

Identify conditions that lead to corrosion and ways to prevent it.

Calculate cell voltage from a table of standard electrode potentials.

Types of Galvanic CellsA battery is one kind of galvanic cell, a device that can change chemicalenergy into electrical energy. In these cells, a spontaneous reaction hap-pens that causes electrons to move. Figure 8 shows a kind of galvanic cellknown as a Daniell cell. Daniell cells were used as energy sources in theearly days of electrical research. Of course, Daniell cells would beimpractical to use in a radio or portable computer today. There are manyother kinds of galvanic cells. These include dry cells, lead-acid batteries,and fuel cells.

1

2

3

Galvanic Cells3S E C T I O N

e–

e–

Water molecule,H2O

Water molecule,H2O

Water molecule,H2O

Zinc ion,Zn2+

Zinc atom,Zn

Sulfate ion,SO2−

4

Zinc ion,Zn2+

Sulfate ion,SO2−

4

Copper(II) ion,Cu2+

Copper atom,Cu

Copper(II) ion,Cu2+

Sulfate ion,SO2−

4


Porousseparator

Moist paste ofC, MnO2,and NH4Cl

Carbon rod(cathode)

Zinc shell(anode) Porous

separator

KOH electrolyte

Brass current collector

Graphite-MnO2cathode mix

Zn-KOHanode paste

Steel jacket

Figure 9Two familiar kinds of dry cells use different electrolytes. The electrolytesmake the cell on the leftacidic and the cell on theright alkaline.

Dry CellsAlthough the Daniell cell was useful for supplying energy in the lab, itwasn’t very portable because it held solutions. The energy source that youknow as a battery and use in radios and remote controls is a dry cell. In adry cell, moist electrolyte pastes are used instead of solutions. This cellwas invented over a century ago by Georges Leclanché, a French chemist.His original design was close to that shown on the left in Figure 9. A car-bon rod, the battery’s positive terminal, connects with a wet paste of car-bon; ammonium chloride, NH4Cl; manganese(IV) oxide, MnO2; starch;and water. When the cell is used, the carbon rod is the cathode, and thefollowing electrode reaction happens:

Cathode: 2MnO2(s) + 2NH4+(aq) + 2e− ⎯→ Mn2O3(s) + 2NH3(aq) + H2O(l)

The zinc case serves as the negative terminal of the battery. When the cellis used, zinc dissolves in the following electrode reaction:

Anode: Zn(s) + 4NH3(aq) ⎯→ 2e− + Zn(NH3)42+(aq)

The white powder that you see on old corroded batteries is the chloridesalt of this Zn(NH3)4

2+(aq) complex ion.The NH4

+(aq) ion is a weak acid, and for this reason the Leclanché cell is called the acidic version of the dry cell. The alkaline cell, shown inFigure 9, is a newer, better version. The ingredients of the alkaline cell aresimilar to the acidic version, but the carbon cathode is replaced by a pieceof brass, and ammonium chloride is replaced by potassium hydroxide.The presence of this strong base gives the alkaline cell its name. The elec-trode reactions that occur when the cell is used are described by the equa-tions below.

Cathode: 2MnO2(s) + H2O(l) + 2e− ⎯→ Mn2O3(s) + 2OH−(aq)

Anode: Zn(s) + 2OH−(aq) ⎯→ 2e− + Zn(OH)2(s)

A sturdy steel shell is needed to prevent the caustic contents from leakingout of the battery. Because of this extra packaging, alkaline cells are moreexpensive than cells of the older, acidic version.



618 Chapter 17

SAFETY PRECAUTIONS

Lead-Acid BatteriesThe batteries just discussed are called dry cells because the water is notfree, but absorbed in pastes. In contrast, the Daniell cell and the lead-acidbattery use aqueous solutions of electrolytes, so they should be used in anupright position.

Most car batteries are lead-acid storage batteries. Usually they havesix cells mounted side-by-side in a single case, as shown in Figure 10.Though many attempts have been made to replace the heavy lead-acidbattery by lighter alternatives, no other material has been found that canreliably and economically give the large surges of electrical energyneeded to start a cold engine.

A fully charged lead-acid cell is made up of a stack of alternating leadand lead(IV) oxide plates isolated from each other by thin porous sepa-rators. All these components are in a concentrated solution of sulfuricacid. The positive terminal of one cell is linked to the negative terminal ofthe next cell in the same way that the batteries in the flashlight were con-nected. This arrangement of cells causes the outermost terminals of thelead-acid battery to have a voltage of 12.0 V. When the cell is used, it actsas a galvanic cell with the following reactions:

Cathode: PbO2(s) + HSO4−(aq) + 3H3O+(aq) + 2e− ⎯→ PbSO4(s) + 5H2O(l)

Anode: Pb(s) + HSO4−(aq) + H2O(l) ⎯→ 2e− + PbSO4(s) + H3O+(aq)

Notice that PbSO4 is produced at both electrodes.Unlike the Daniell and Leclanché cells, the lead-acid cell is recharge-

able. So, when the battery runs down, you do not need to replace it. Instead,an electric current is applied in a direction opposite to that discussed above.As a result of the input of energy, the reactions are reversed. The cell iseventually restored to its charged state. During recharge, the cell functionsas an electrolytic cell, which you will learn about in the next section.

LABListen Up

Quick

PROCEDURE1. Press a zinc strip and a copper

strip into a raw potato. Thestrips should be about 0.5 cmapart but should not touchone another.

2. While listening to the ear-phone, touch one wire fromthe earphone to one of the

metal strips and the otherwire from the earphone tothe second metal strip usingalligator clips. Record yourobservations.

3. While listening to the earphone, touch both wires to a single metal strip.Record your observations.

ANALYSIS1. Compare your results from

step 2 with your results fromstep 3. Suggest an explanationfor any similarities or differences.

2. Suggest an explanation forthe sound.


Intercell connectors

H2SO4(aq)Intercell dividers

PbO2 plates

Pb plates

Fuel CellsIn a fuel cell, the oxidizing and reducing agents are brought in, often asgases, from outside of the cell, rather than being part of it. Unlike a dry cell,a fuel cell can work forever, in principle, changing chemical energy intoelectrical energy. Figure 11 models a fuel cell that uses the reactions below.

Cathode: O2(g) + 2H2O(l) + 4e− ⎯→ 4OH−(aq)

Anode: 2H2(g) + 4OH−(aq) ⎯→ 4e− + 4H2O(l)

Because fuel cells directly change chemical energy into electricalenergy, they are very efficient and are cleaner than the burning of fuels inpower plants to generate electrical energy. Research into fuel cells con-tinues, and fuel cells are used in a few experimental power plants.


Figure 10The lead-acid battery is used to store energy inalmost all vehicles. Althoughthe cutaway view shows asingle PbO2 plate and a single Pb plate in each cell,there are actually several of each.

Excess H2(g) + H2O(g) Excess O2(g) + H2O(g)

H2(g) O2(g)

Porous graphiteanode

Porous graphitecathode

Electrolyte solution

e–e–

K+

OH_

Figure 11The reactions in this fuel celltake place at carbon electrodesthat contain metal catalysts.The water formed is removed as a gas.


620 Chapter 17

Corrosion CellsOxygen is so reactive that many metals spontaneously oxidize in air.Fortunately, many of these reactions are slow. The disintegration of metalsis called Usually O2(g) is the oxidizing agent, but the directreaction of O2 with the metal is not usually how corrosion happens.

Water is usually involved in corrosion. Consider the corrosion of iron.Hydrated iron(III) oxide, or rust, forms by the following overall reaction:

4Fe(s) + 3O2(aq) + 4H2O(l) ⎯→ 2Fe2O3•2H2O(s)

However, the reaction mechanism is more complicated than thisequation suggests. An even simpler version of what happens is shown inFigure 12. The iron dissolves by the oxidation half-reaction

Fe(s) ⎯→ Fe3+(aq) + 3e−

Any oxidation must be accompanied by a reduction taking place at thesame time, but not necessarily at the same location. In fact, the electronsproduced by the oxidation are consumed by the reduction of oxygen at acathodic site elsewhere on the iron’s surface.The reduction half-reaction is

O2(aq) + 2H2O(l) + 4e− ⎯→ 4OH−(aq)

Electrons move in the metal and ions move in the water layerbetween the two reaction sites, as in an electrochemical cell. In fact, a cor-rosion cell is an unwanted galvanic cell. Chemical energy is converted intoelectrical energy, which heats the metal.

The three ingredients—oxygen, water, and ions—needed for the cor-rosion of metals are present almost everywhere on Earth. Even purerainwater contains a few H3O+ and HCO3

− ions from dissolved carbondioxide. Higher ion concentrations—from airborne salt near the ocean,from acidic air pollutants, or from salts spread on icy roads—make corro-sion worse in certain areas.

corrosion.

Iron, Fe Paint

Water layer Ion conduction

Electron conduction

e−

Fe(s) ⎯→ 3e− + Fe3+(aq) O2(aq) + 2H2O(l) + 4e− ⎯→ 4OH−(aq)

Rust

Figure 12The cathodic reaction happens where the O2concentration is high. Theanodic reaction happens in a region where the O2concentration is low, such as in a pit in the metal.

corrosion

the gradual destruction of ametal or alloy as a result ofchemical processes such as oxidation or the action of achemical agent

Water molecule, H2O

Oxygen molecule, O2

Iron(III) ion, Fe3+

Hydroxide ion, OH−


Methods to Prevent CorrosionCorrosion is a major economic problem. About 20% of all the iron andsteel produced is used to repair or replace corroded structures. That iswhy the prevention of corrosion is a major focus of research in materialsscience and electrochemistry.An obvious response to corrosion is to paintthe metal or coat it with some other material that does not corrode.However, once a crack or scrape occurs in the coating, corrosion canbegin and often spread even faster than on an uncoated surface.

Some metals corrode more easily than others do. Electronegativity isone factor. Gold, the metal with the highest electronegativity, is the mostcorrosion resistant. The alkali metals, with the lowest electronegativities,easily corrode.The properties of the metal oxides that form are also impor-tant. Despite having low electronegativities, aluminum, chromium, andtitanium are corrosion-resistant metals. This is because the oxides of thesemetals form layers that cover the underlying metal, stopping oxidation. Incontrast, rust is a porous powder that flakes off, so it does not protect theiron surface.

Surprisingly, it is better to coat steel with another metal that does cor-rode. Trash cans, for example, are made of zinc-coated steel. This coatingdoes not stop corrosion. But the zinc corrodes first, making the steelunderneath last much longer than it would without the zinc layer.

Whenever two metals are in electrical contact, a corrosion cell is likelyto form. In fact, the metal that is the anode in the cell corrodes faster thanit would if it were not connected to another metal. This idea explains theuse of sacrificial anodes on ships and pipelines, as shown in Figure 13. Asthe anode corrodes, it gives electrons to the cathode. The corrosion of theanode slows or stops the corrosion of the important structural metal in aprocess called cathodic protection.


Figure 13The Alaskan oil pipeline iscathodically protected by aparallel zinc cable.

www.scilinks.orgTopic: CorrosionSciLinks code: HW4147


622 Chapter 17

Determining the Voltage of a CellA voltmeter is an electronic instrument that measures the voltagebetween its two leads. The student in Figure 14 is using the meter to meas-ure the difference in potentials between the electrodes in a Daniell cell.With such a meter, measuring the voltage of a cell is easy. However, thevoltage of a cell depends on such factors as temperature and concentra-tion. And because there are so many combinations of electrode reactions,it would be very difficult to measure the voltage for each combination.

Standard Electrode PotentialsPicking one electrode as a standard and determining electrode potentials inreference to that standard is much easier than measuring the potentialbetween every combination of electrodes is. The electrode that has beenchosen as a standard is the standard hydrogen electrode (SHE). It consistsof a platinum electrode in a 1.00 M H3O+ solution in the presence of H2 gasat 1 atm pressure and 25°C.The SHE is assigned a potential of 0.0000 V andits reaction is

2H3O+(aq) + 2e− ⎯→←⎯ H2(g) + 2H2O(l)

When measuring potentials, a salt bridge, a narrow tube filled with a con-centrated solution of a salt, must be used to link the compartments. Whena SHE is joined to another electrode by a salt bridge, a voltmeter can beused to determine the E°, for the electrode.Some standard electrode potentials are shown in Table 1, on the last pageof this section.

The standard electrode potential is sometimes called the standardreduction potential because it is listed by the reduction half-reactions.However, a voltmeter allows no current in the cell during the measure-ment. Therefore, the conditions are neither galvanic nor electrolytic—thecell is at equilibrium. As a result, the half-reactions listed in the table areshown as reversible. If the reaction occurs in the opposite direction, as anoxidation half-reaction, E° will have the opposite sign.

standard electrode potential,standard electrode potential

the potential developed by ametal or other material immersedin an electrolyte solution relativeto the potential of the hydrogenelectrode, which is set at zero

Figure 14It would be difficult tomeasure every possible com-bination of electrodes usinga technique like the oneshown here. This is why theSHE is used as a referencefor all other electrodes.


SAM PLE PROBLE M CSAM PLE PROBLE M C

Calculating Cell VoltageCalculate the voltage of a cell for the naturally occurring reactionbetween a liquid mercury electrode in a solution of mercury(I) nitrateand a cadmium metal electrode in a solution of cadmium nitrate.

1 Gather information.From Table 1 the standard electrode potentials are

Hg22+(aq) + 2e− ⎯→ 2Hg(l) E° = +0.7973 V

Cd2+(aq) + 2e− ⎯→ Cd(s) E° = –0.4030 V

2 Plan your work.The mercury electrode has the more positive E°, so it is the cathode.

3 Calculate.E°cell = E°Hg − E°Cd = (+0.7973 V) − (−0.4030 V) = +1.2003 V

4 Verify your results.The positive value of E°cell shows that the reaction is spontaneous andwill occur naturally with the mercury electrode as the cathode.

When asked to calculatethe voltage of a cell fora particular chemicalequation, you mustdetermine which atom isoxidized and which isreduced based on thechange of oxidationnumbers.

PRACTICE HINTPRACTICE HINT


Calculating the Voltage of a CellThink of E° as a measure of the ability of an electrode to gain electrons.A more positive value means the electrode is more likely to be a cathode.The standard cell voltage—the voltage of a cell under standard condi-tions—can be found by subtracting the standard potentials of the twoelectrodes, as follows:

E°cell = E°cathode − E°anode

To determine the reaction that will happen naturally, use the electrode withthe most positive E° value as the cathode, as shown in Sample Problem C.Otherwise, a given reaction happens naturally if E°cell is positive. If E°cell isnegative, the reaction could be made to happen if energy is added.

PRACTIC E

PROBLEM

SOLVING

SKILLPROBLEM

SOLVING

SKILL

Use data from Table 1 to answer the following:

1 Calculate the voltage of a cell if the reactions are as follows:Fe(s) ⎯→ Fe3+(aq) + 3e–

O2(g) + 2H2O(l) + 4e– ⎯→ 4OH−(aq)

2 Calculate the voltage of a cell for the naturally occurring reactionbetween a copper electrode in a copper(II) solution and a zinc elec-trode in a solution containing zinc ions.

3 Calculate the voltage of a cell for the naturally occurring reactionbetween a silver electrode in a solution containing silver ions and a copper electrode in a copper(II) solution.


624 Chapter 17


1. How does a fuel cell differ from a battery?

2. How do an acidic dry cell and an alkalinedry cell differ?

3. Of the metals Zn, Fe, and Ag, which willcorrode the easiest? Explain.

4. Describe how a standard electrode potentialis measured.

PRACTICE PROBLEMS

5. Calculate the voltage and identify the anodefor a cell in which the following electrodereactions take place:

Ag(s) + Cl−(aq) ⎯→ AgCl(s) + e−

Cl2(g) + 2e− ⎯→ 2Cl−(aq)

6. Calculate the voltage and identify the cathode for a cell in which the natural reaction between the following electrodeshappens:

Zn2+(aq) + 2e− ⎯→←⎯ Zn(s)

2H2O(l) + 2e− ⎯→←⎯ H2(g) + 2OH−(aq)

CRITICAL THINKING

7. Write the overall equation for the reactionoccurring in a lead-acid cell during dis-charge. What happens to the sulfuric acidconcentration during this process? Why is itpossible to use a hydrometer, which meas-ures the density of a liquid, to determine if a lead-acid cell is fully charged?

8. A sacrificial anode is allowed to corrode.Why is use of a sacrificial anode consideredto be a way to prevent corrosion?

Section Review3

Table 1 Standard Electrode Potentials

Electrode reaction E°(V) Electrode reaction E°(V)

Li+(aq) + e− ⎯→←⎯ Li(s) −3.0401 Cu2+(aq) + 2e− ⎯→←⎯ Cu(s) +0.3419

K+(aq) + e− ⎯→←⎯ K(s) −2.931 O2(g) + 2H2O(l) + 4e− ⎯→←⎯ 4OH−(aq) +0.401

Na+(aq) + e− ⎯→←⎯ Na(s) −2.71 I2(s) + 2e− ⎯→←⎯ 2I−(aq) +0.5355

2H2O(l) + 2e− ⎯→←⎯ H2(g) + 2OH−(aq) −0.828 Fe3+(aq) + e− ⎯→←⎯ Fe2+(aq) +0.771

Zn2+(aq) + 2e− ⎯→←⎯ Zn(s) −0.7618 Hg22+(aq) + 2e− ⎯→←⎯ 2Hg(l) +0.7973

Fe2+(aq) + 2e− ⎯→←⎯ Fe(s) −0.447 Ag+(aq) + e− ⎯→←⎯ Ag(s) +0.7996

PbSO4(s) + H3O+(aq) + 2e− ⎯→←⎯ −0.42 Br2(l) + 2e− ⎯→←⎯ 2Br−(aq) +1.066Pb(s) + HSO4

−(aq) + H2O(l)

Cd2+(aq) + 2e− ⎯→←⎯ Cd(s) −0.4030 Cl2(g) + 2e− ⎯→←⎯ 2Cl−(aq) +1.358

Pb2+(aq) + 2e− ⎯→←⎯ Pb(s) −0.1262 PbO2(s) + 4H3O+(aq) + 2e− ⎯→←⎯ +1.455Pb2+(aq) + 6H2O(l)

Fe3+(aq) + 3e− ⎯→←⎯ Fe(s) −0.037 PbO2(s) + HSO4−(aq) + 3H3O+(aq) + +1.691

2e− ⎯→←⎯ PbSO4(s) + 5H2O(l)

2H3O+(aq) + 2e− ⎯→←⎯ H2(g) + 2H2O(l) 0.0000 Ce4+(aq) + e− ⎯→←⎯ Ce3+(aq) +1.72

AgCl(s) + e− ⎯→←⎯ Ag(s) + Cl−(aq) +0.222 F2(g) + 2e− ⎯→←⎯ 2F−(aq) +2.866

Refer to Appendix A for additional standard electrode potentials.



Fuel CellsHistorical PerspectiveIn 1839, Sir William Robert Grove, aBritish lawyer and physicist, built thefirst fuel cell. More than 100 yearslater, fuel cells finally found a practicalapplication—in space exploration.During short space missions, batteriescan provide enough energy to keep theastronauts warm and to run electricalsystems. But longer missions needenergy for much longer periods oftime, and fuel cells are better suited forthis than batteries are. Today, fuel cells

are critical to the space shuttle missions and to future missions onthe international space station.

Blood Alcohol TestingA more down-to-Earth use of fuel cells is found in traffic-lawenforcement. Police officers need quick and simple ways todetermine a person’s blood alcohol level in the field. In the timeit takes to bring a person to the station or to a hospital for ablood or urine test, the person’s blood alcohol content (BAC)might change. Fuel cells, such as the one in the device shownabove, provide a quick and accurate way to measure BAC from a breath sample. The alcohol ethanol from the person’s breath isoxidized to acetic acid at the anode. At the cathode, gaseous oxygen is reduced and combined with hydronium ions (releasedfrom the anode) to form water. The reactions generate an electric current. The size of this current is related to the BAC.

Questions1. Research at least two more uses for fuel cells. Identify the

reactants and products for the cell in each use.2. Research careers open to chemical engineers. Determine

the level of education, the approximate salaries, and theareas of study needed to become a chemical engineer.

C A R E E R A P P L I C AT I O N

Chemical EngineerChemical engineers do much of theongoing fuel-cell research. There aremany careers open to chemical engi-neers. They can work to find alternative,renewable fuel sources, to design newrecyclable materials, and to devise newrecycling methods. These scientistscombine knowledge of chemistry, physics,and mathematics to link laboratorychemistry with its industrial applications.As with any scientist, they also must begood problem solvers.

Chemical engineers use lab tech-niques that you may know, includingdistillation, separation, and mixing. Thechemical engineer in the photo is part of a team that is developing a compound tohelp locate explosives, such as those usedin land mines. The compound detectssmall amounts of nitrogen-containingcompounds in the air that are often foundwith explosives. The bright fluorescence ofthe compound dims when it contactsthese compounds.

SCIENCE AND TECHNOLOGY

A small pump sends a precisesample size into the fuel cellinside this device.

www.scilinks.orgTopic: Fuel Cells SciLinks code: HW4061


626 Chapter 17

KEY TERMS

• electrolytic cell

• electrolysis

• electroplating

Power source

Impure copperanode

Anode sludge Copper(II) sulfate/sulfuric acid solution

Pure coppercathode

e–

e–

Figure 15The experiment shown here mirrors the industrialrefining of copper.

OBJECTIVES

Describe how electrolytic cells work.

Describe the process of electrolysis in the decomposition of waterand in the production of metals.

Describe the process of electroplating.

Cells Requiring EnergyGalvanic cells generate electrical energy, but another kind of cell con-sumes electrical energy. This energy is used to drive a chemical reaction.The cell shown in Figure 15 is a laboratory-scale version of the industrialprocess used to refine copper.The anode is impure copper, which includessuch metals as zinc, silver, and gold. The oxidation reaction at the anodechanges Cu atoms in the impure sample to Cu2+(aq). The opposite reac-tion happens at the cathode. The Cu2+(aq) ions are reduced to Cu atoms.Pure copper is formed, adding to the pure copper cathode.

Impurities such as Zn and other active metals also dissolve as cations.But they are not reduced at the cathode. Inactive metals, such as Au andAg, fall to the bottom as an anode sludge. This sludge is a valuable sourceof these more-expensive metals in the industrial process.

1

2

3

Electrolytic Cells4S E C T I O N

Water molecule,H2O

Sulfate ion,SO2−

4

Copper atom, Cu

Hydroniumion, H3O+

Hydrogen sulfate ion,

HSO−4

Copper(II)ion, Cu2+

Water molecule,H2O

Sulfate ion,SO2−

4

Hydroniumion, H3O+

Hydrogen sulfate ion,

HSO−4

Copper(II) ion, Cu2+

Zinc ion,Zn2+

Zinc ion,Zn2+

Ag

Cu

Au

Zn


ElectrolysisIn an chemical changes are brought about by drivingelectrical energy through an electrochemical cell. In fact, the words elec-trolysis and electrolytic mean “splitting by electricity” and does refer to the decomposition of a compound, usually into its elements.However, many changes other than decomposition can happen when youuse an electrolytic cell. For example, adiponitrile, one of the raw materi-als used to make nylon, is synthesized in an electrolytic cell. What makesan electrolytic cell useful is that the overall reaction is a nonspontaneousprocess that is forced to happen by an input of energy.

Electrolysis of WaterThe electrolysis of water, shown in Figure 16, leads to the overall reactionin which H2O is broken down into its elements, H2 and O2. Pure waterdoes not have enough ions in it and is not conductive enough for elec-trolysis. An electrolyte, such as sodium sulfate, must be added. TheNa+(aq) and SO4

2−(aq) ions play no part in the electrode reactions, whichare as follows:

Anode: 6H2O(l) ⎯→ 4e− + O2(g) + 4H3O+(aq)

Cathode: 4H2O(l) + 4e− ⎯→ 2H2(g) + 4OH−(aq)

As always, oxidation happens on the anode, while reduction happens on thecathode. Note that hydronium ions form at the anode. Thus, the solutionnear the anode becomes acidic. But hydroxide ions form at the cathode. So,the solution near the cathode becomes basic.

electrolysis

electrolytic cell,


electrolytic cell

an electrochemical device inwhich electrolysis takes placewhen an electric current is in the device

Figure 16Electrical energyfrom the battery isused to break downwater. Hydrogenforms at the cathode,and oxygen forms atthe anode.

electrolysis

the process in which an electriccurrent is used to produce achemical reaction, such as thedecomposition of water

Topic LinkTopic LinkRefer to the “Causes of

Change” chapter formore information

about spontaneous and nonspontaneous reactions.


628 Chapter 17

Sodium Production by ElectrolysisSodium is such a reactive metal that preparing it through a chemicalprocess can be dangerous. Sir Humphry Davy first isolated it in 1807 by theelectrolysis of molten sodium hydroxide.Today, sodium is made by the elec-trolysis of molten sodium chloride in a Downs cell, as shown in Figure 17.

Pure sodium chloride melts at 801°C. The addition of calcium chlo-ride, CaCl2, to the NaCl lowers the melting point.The Downs cell can thenwork at 590°C, and less energy is needed to run the cell. The equationsbelow describe the major reactions that occur.

Anode: 2Cl−(l) ⎯→ 2e− + Cl2(g)

Cathode: 2Na+(l) + 2e− ⎯→ 2Na(s)

Because chlorine reacts with most metals, the anode is made ofgraphite. The cathode is steel. Because sodium melts at 98°C and is lessdense than the molten salts, it floats to the top and can be removed.

In addition to sodium, a small amount of calcium also forms at thecathode by the reaction:

Cathode: Ca2+(l) + 2e− ⎯→ Ca(s)

The calcium is more dense than the molten salts, so it falls to the bottomof the cell, where it slowly changes back to Ca2+ by the following reaction:

Ca(l) + 2Na+(l) ⎯→ 2Na(l) + Ca2+(l)

Power source

Anode

Cathode

Molten NaCl

Cl2 outlet

Inlet forNaCl

Liquid Na metal

Na outlet

7 Ve– e–

Figure 17In a Downs cell the electrolysis ofmolten NaCl forms the elementssodium and chlorine.


Aluminum Production by ElectrolysisAluminum is the most abundant metal in Earth’s crust.Aluminum is light,weather resistant, and easily worked. You have seen aluminum in use indrink cans, in food packaging, and even in airplanes. However, aluminumis never found in nature as a pure metal. Instead, it is isolated from its orethrough electrolysis.

The process used to get aluminum from its ore, bauxite, is the electro-chemical Hall-Héroult process. This process is the largest single user ofelectrical energy in the United States—nearly 5% of the national total.This need for energy makes the manufacture of aluminum expensive.Recycling aluminum saves almost 95% of the cost! Aluminum recycling isone of the most economically worthwhile recycling programs that hasbeen developed.

The bauxite is processed to extract and purify hydrated alumina,Al2O3. The alumina is fed into huge carbon-lined tanks, like the one inFigure 18. There the alumina dissolves in molten cryolite, Na3AlF6, at970°C. Liquid aluminum forms at the cathode. Being more dense than themolten cryolite, aluminum sinks to the floor of the tank. As reduction con-tinues, the level of aluminum rises. As needed, the liquid aluminum isdrained and allowed to cool.

Carbon rods serve as the anode. The carbon is oxidized during theanodic reaction, forming CO2. The rods are eaten away by this oxidationand must be replaced from time to time.

The Hall-Héroult process has been in use for more than a century. Butscientists do not completely understand how alumina dissolves and whatexactly the species are that participate in the electrode reactions.Although scientists still debate how the process works, they agree that theoverall reaction is

2Al2O3(l) + 3C(s) ⎯→ 4Al(l) + 3CO2(g)


Powersource

Carbon anode

Solution ofalumina, Al2O3, incryolite, Na3AlF6

Molten aluminum, Al Carbon cathode

6.2 V

e–

e–

Figure 18The Hall-Héroult process isused to make aluminum bythe electrolysis of dissolvedalumina, Al2O3.


630 Chapter 17

ElectroplatingMany of the metal things that you use every day—forks and spoons, cansfor food and drinks, plumbing fixtures, jewelry and decorative ornaments,automobile and appliance parts, nails, nuts, and bolts—have been treatedto change their surfaces. Often this involves putting a layer of anothermetal on top of the main metal. is one way of applyingthese finishes. Forks, spoons, and jewelry are often electroplated to givethe objects the appearance of silver or gold while still keeping the cost ofthe objects low. The chrome parts on automobiles have been electro-plated to improve the parts’ appearance and protect them from corrosion.Electroplating is also used for many electronic and computer parts to givethem a certain physical property or to make them last longer by protect-ing them against corrosion.

To electroplate a bracelet with silver, as shown in the simplified modelin Figure 19, the bracelet is made the cathode of an electrolytic cell. Theanode is a strip of pure silver metal. Both electrodes are placed in a solu-tion of silver ions. The net reactions that happen are very simple. Theanode slowly dissolves by the following oxidation reaction:

Anode: Ag(s) ⎯→ e− + Ag+(aq)

The cathode reaction on the surface of the bracelet is the reverse of theanode reaction.

Cathode: Ag+(aq) + e− ⎯→ Ag(s)

The result is that a thin coating of silver forms on the bracelet. The longerthe plating is continued, the thicker the silver layer becomes.

Electroplatingelectroplating

the electrolytic process of platingor coating an object with a metal

www.scilinks.orgTopic: ElectroplatingSciLinks code: HW4049

Silver strip, Ag

AnodeCathode

Ag+

AgCNsolution

Ag+Ag+

Power source

e–

e–

CN–

Figure 19The bracelet in this cell is beingcoated with a thin layer of silver.Silver ions are replaced in the solutionas the pure silver anode dissolves.


Benefits and Concerns About ElectroplatingA major benefit of electroplating a metal is that it becomes more resist-ant to corrosion. Chrome-plated car parts and zinc-plated food cans aretwo uses of electroplating to reduce corrosion. Another benefit of electro-plating is that it improves the appearance of an object. An item that isgold-plated or silver-plated is much cheaper than the same item in solidgold or silver but looks the same.

However, there are also drawbacks with the electroplating process. Inpractice, electroplating is not as simple as the description above suggests.It is difficult to get metal to deposit in depressions, so the metal layeroften is not uniform. Sometimes, deposits are loose and powdery.Additives are used and conditions such as temperature and pH are care-fully controlled to overcome these problems. Over time, impurities buildup in the solutions used for electroplating. Eventually, the spent solutionsmust be discarded. They can contain high concentrations of such toxicmetals as cadmium or chromium and require careful disposal to protectthe environment.



1. How does an electrolytic cell differ from agalvanic cell?

2. What chemical process occurs at the anodeof an electrolytic cell?

3. What chemical process occurs at the cath-ode of an electrolytic cell?

4. What form of energy is used to drive anelectrolytic cell?

5. List three commercial products made byusing electrolytic cells.

6. What does electrolysis mean?

7. Write the equation for the cathodic reactionthat occurs in the Downs cell used to makesodium.

8. How is aluminum manufactured?

9. Describe the electroplating process.

CRITICAL THINKING

10. In the copper refining process, why doeszinc not also deposit on the cathode?

11. Elemental aluminum was first prepared in1827 by the reaction of aluminum chloridewith potassium.

a. Write the balanced equation for this reaction.

b. Determine if the reaction is a redox reaction.

c. Would this reaction need to happen in anelectrolytic cell? Explain.

12. Cryolite, Na3AlF6, is an ionic mineral used inthe preparation of aluminum. Explain whythe sodium ions are not reduced during theelectrolytic process that produces aluminum.

13. The following reaction happens naturally:

Zn(s) + Cu2+(aq) ⎯→ Cu(s) + Zn2+(aq)

Explain why it is necessary to use an elec-tric current to deposit a layer of zinc on acopper bracelet.

14. Explain why a galvanic cell is often used inan electrolytic cell. What function does thegalvanic cell serve?

15. Why is it so important to recycle rather thandiscard aluminum products?

Section Review4


632 Chapter 17

CHAPTER HIGHLIGHTS17

Assigning Oxidation NumbersSkills Toolkit 1 p. 606Sample Problem A p. 607

Balancing Redox Equations Using the Half-Reaction MethodSkills Toolkit 2 p. 609Sample Problem B p. 610

Calculating Cell VoltageSample Problem C p. 623

K E Y S K I L L S

SECTION ONE Oxidation-Reduction Reactions• The loss or gain of electrons in a chemical reaction is called oxidation or

reduction, respectively.

• In a redox reaction, oxidation and reduction occur at the same time.

• An oxidation number may be assigned to each atom in a molecule or ion.

• Half-reactions, in which only the oxidation or the reduction is described, areuseful in balancing redox equations.

• Reducing agents readily donate electrons; oxidizing agents readily accept electrons.

SECTION TWO Introduction to Electrochemistry• An electrochemical cell is made up of two electrodes linked by one or more

ionic conductors.

• When electric current is in a cell, electrode reactions take place.

• Oxidation happens at the anode. Reduction happens at the cathode.

SECTION THREE Galvanic Cells• Many examples of galvanic cells are power sources that generate electrical

energy from chemical energy.

• Fuel cells differ from batteries in that their oxidizing and reducing agents aregases introduced to the cell from outside.

• The corrosion of metals generally happens in a galvanic cell.

• Whether there is electric current or not, the electrodes of a cell have differentpotentials. The difference between these is the voltage of the cell.

SECTION FOUR Electrolytic Cells• Electrical energy is used to power an electrolytic cell.

• In electrolysis, a compound is decomposed, usually to its elements; H2, O2, Al,Na, and Cl2 are among the elements that can be isolated by electrolysis.

• A layer of a second metal is deposited cathodically in electroplating.

oxidationreductionoxidation-reduction

reactionoxidation numberhalf-reactionoxidizing agentreducing agent

electrochemistryvoltageelectrodeelectrochemical cellcathodeanode

K E Y T E R M SK E Y I D E A S

corrosionstandard electrode

potential

electrolytic cellelectrolysiselectroplating


PART AFor each item, write on a separate piece of paper the number of the word, expression, orstatement that best answers the item.

1. A redox reaction always involves(1) a change in oxidation number.(2) a change in state.(3) a transfer of protons.(4) formation of ions.

2. In any redox reaction, the number of electrons gained is (1) equal to the number of electrons lost.(2) less than the number of electrons lost.(3) greater than the number of electrons lost.(4) unrelated to the number of electrons lost.

3. The reaction CuO + CO ⎯→ CO2 + Cu is anexample of (1) reduction, only.(2) oxidation, only.(3) both reduction and oxidation.(4) neither reduction nor oxidation.

4. A sulfur ion, S2−, is oxidized to a sulfur atom by (1) losing electrons.(2) gaining electrons.(3) losing protons.(4) gaining protons.

5. When a Fe3+ ion is changed to a Fe2+ ion,the Fe3+ ion is (1) reduced and loses one electron.(2) reduced and gains one electron.(3) oxidized and loses one electron.(4) oxidized and gains one electron.

6. What is the oxidation number of hydrogenin NaH? (1) +1 (3) −1 (2) +2 (4) −2

7. The oxidation number of all elements in thecompound K2Cr2O7 must add up to (1) 0 (3) +2(2) −2 (4) +12

8. What is the oxidation number of oxygen inOF2?(1) 0 (3) −2(2) −1 (4) −2

9. In which substance does oxygen have anoxidation number of −2?(1) CO2 (3) O2

(2) OF2 (4) H2O2

10. The half reaction Zn(s) ⎯→ Zn+2(aq) + 2e−

represents the(1) reduction of zinc atoms.(2) oxidation of zinc atoms.(3) reduction of zinc ions.(4) oxidation of zinc ions.

11. When a substance is oxidized, it (1) acts as a reducing agent.(2) acts as an oxidizing agent.(3) gains protons.(4) loses protons.

Sometimes, only one part of a graph or table isneeded to answer a question. In such cases,focus only on that information to answer thequestion.


17Physical Setting/ChemistryREGENTS EXAM PRACTICE


18. What is the voltage for an electrochemicalcell that has reached equilibrium?(1) 0(2) 1(3) less than 1(4) greater than 1

19. In order for a redox reaction to be sponta-neous, the potential (E0) for the overallreaction must be(1) greater than zero.(2) zero.(3) between zero and −1.(4) less than −1.

20. An electrolytic cell differs from and electro-chemical cell in that the electrolytic cell (1) involves a redox reaction.(2) is spontaneous.(3) generates electrical energy.(4) requires an external electric current.

21. The decomposition of water, 2H2O ⎯→2H2 + O2, is forced to occur by using anexternally applied electric current. Thisprocess is called(1) hydrolysis.(2) electrolysis.(3) neutralization.(4) cathodic reduction.

22. Flatware can be plated with silver by attach-ing the object to be plated to an externalsource of electric current. Positive silverions will migrate toward the object if theobject has (1) a negative charge.(2) fewer electrons than protons.(3) no charge.(4) a positive charge.

12. What happens to reducing agents in redoxreactions?(1) They are reduced.(2) They are oxidized.(3) They gain protons.(4) They gain electrons.

13. The type of reaction that occurs sponta-neously in an electrochemical cell is bestdescribed as a(n)(1) reduction reaction.(2) oxidation reaction.(3) redox reaction.(4) transfer of protons between electrodes.

14. In an electrochemical cell, electrons flowfrom the(1) cathode, where reduction occurs.(2) cathode, where oxidation occurs.(3) anode, where reduction occurs.(4) anode, where oxidation occurs.

15. The electrode in an electrochemical cellwhere reduction always occurs is the(1) cathode, where a substance loses

electrons.(2) anode, where a substance loses electrons.(3) cathode, where a substance gains

electrons.(4) anode, where a substance gains electrons.

16. In a galvanic cell when chemical energy ischanged into electrical energy, the anode (1) can be either positive or negative.(2) is positive.(3) is negative.(4) is not charged.

17. When an electrochemical device is beingrecharged, the cell acts as a(n) (1) fuel cell.(2) electrolytic cell.(3) galvanic cell.(4) dry cell.

634 Chapter 17


PART B–1 For each item, write on a separate piece of paper the number of the word, expression, orstatement that best answers the item.

29. Which half reaction shows the conservationof both mass and charge?(1) Cl2 ⎯→ 2Cl− + 2e−

(2) Cl2 ⎯→ 2Cl− + e−

(3) Cl2 + 2e− ⎯→ 2Cl−

(4) Cl− + 2e− ⎯→ Cl2

30. What happens to the zinc atom in the reac-tion below?

Zn(s) + CuSO4(aq) ⎯→ Cu(s) + ZnSO4(aq)

(1) Zinc gains two electrons and is reduced.(2) Zinc loses two electrons and is reduced.(3) Zinc gains two electrons and is oxidized.(4) Zinc loses two electrons and is oxidized.

31. Examine the following redox reaction.

2H2S + 3O2 ⎯→ 2SO2 + 2H2O

The oxidizing agent is (1) oxygen because it loses electrons.(2) oxygen because it gains electrons.(3) sulfur because it loses electrons.(4) sulfur because it gains electrons.

32. Examine the following unbalanced equation.

__Mg(s) + __H+(aq) ⎯→ __Mg2+(aq) + __H2(g)

When this equation is balanced, the coefficient of H+ is(1) 1. (3) 3.(2) 2. (4) 4.

33. A cell contains two electrodes. The firstelectrode is a strip of zinc metal in a solu-tion containing zinc ions. The second elec-trode is a strip of copper metal in a solutionof copper ions. When this cell operates as agalvanic cell,(1) Cu is oxidized and Zn2+ is reduced.(2) Cu is reduced and Zn2+ is oxidized.(3) Cu2+ is oxidized and Zn is reduced.(4) Cu2+ is reduced and Zn is oxidized.

23. In the following reaction, which species isreduced?

2K + Br2 ⎯→ 2K+ + 2Br−

(1) K (3) Br2

(2) K+ (4) Br−

24. Compared to the mass and total charge ofthe reactants in a redox reaction, the massand total charge of the products is (1) the same. (3) greater.(2) less. (4) equal to zero.

25. Which equation represents a redox reaction?(1) NaCl(aq) + AgNO3(aq) ⎯→

NaNO3(aq) + AgCl(s)(2) H+(aq) + OH−(aq) ⎯→ H2O(l)(3) 4Na(s) + O2(g) ⎯→ 2Na2O(s)(4) NaOH(aq) + HCl(aq) ⎯→

NaCl(aq) + H2O(l)

26. What is the oxidation number of chromiumin the compound K2Cr2O7?(1) 0 (3) +6(2) +2 (4) +12

27. In the reaction H2O + Cl2 ⎯→ HCl + HClO,the oxidation number of chlorine(1) increases, only.(2) decreases, only.(3) remains unchanged.(4) both increases and decreases.

28. The reaction 2KClO3 ⎯→ 2KCl + 3O2 involves (1) oxidation, only.(2) reduction, only.(3) both oxidation and reduction.(4) neither oxidation nor reduction.



636 Chapter 17

PART B–2Answer the following items.

34. Assign an oxidation number to the N atom in each of the following substances.a. N2 e. N2O3

b. NO2 r. N2O4

c. NO g. N2O5

d. N2O h. NH4NO3

35. Examine the following redox reaction.

H2S + H+ + NO3− ⎯→ HSO4

− + NO2 + H2O

a. Assign oxidation numbers to all species.b. Write the reduction half-reaction.c. Write the oxidation half-reaction.d. Identify the reducing agent.e. Identify the oxidizing agent.f. Balance the reaction using the half-reaction method.

36. Write the balanced half-reaction for the conversion of Fe(s) to Fe2+(aq).

37. Write the balanced half-reaction for the conversion of Cl2(g) to Cl−(aq).

38. Using half reactions, balance the redox equation for the reaction of solidzinc and aqueous iron(III) ions to form aqueous zinc ions and aqueousiron(II) ions.

39. Using a table of standard electrode potentials, calculate the voltage for aredox reaction involving the reduction of chlorine gas to chloride ions andthe oxidation of copper metal to copper(II) ions.

Questions 40 to 47 refer to the diagram below.

40. What observations provide evidence that a chemical reaction has taken place?

41. Write a balanced equation for the reaction shown. Explain why this is classified as a redox reaction.

Zn

CuCl–

Cu2+ Zn2+



42. Identify what substance is oxidized and what substance is reduced in the reaction.

43. Identify the reducing agent and the oxidizing agent.

44. Describe how the mass of zinc metal will change as the reaction takes place.

45. Describe how the mass of copper metal will change as the reaction takes place.

46. Describe how the concentration of copper and zinc ions will change as the reaction takes place.

47. What role, if any, do the chloride ions play in this reaction?

48. When plating an object with silver, the object is placed at the negativeelectrode of a source of electrical energy.a. Explain the significance of the electrode’s negative charge.b. What half-reaction occurs at this electrode?

49. A weak solution of hydrogen peroxide, H2O2, is used as an antiseptic forsmall cuts. This solution is kept in a dark bottle because exposure to lightcauses the hydrogen peroxide to decompose into water and oxygen.a. Write the balanced equation for this reaction.b. Assign oxidation numbers to each atom in this reaction.c. Are the oxygen atoms in H2O2 oxidized or reduced? Explain your answer.

PART CAnswer the following items.

Questions 50–51 are based on the following passage.

Skunk-Spray Remedy

So that pretty black cat with the white stripe down its back wasn’t a cat afterall? Well, hold off on dumping Fido into a tomato-juice bath. Chemistry has amuch better way of conquering skunk spray.

Paul Krebaum, the inventor of a new de-skunking formula, says thatwhile working as a materials engineer, he constantly had to deal with the less-than-pleasant smell of the hydrogen sulfide gas that was released from one ofhis experiments. Mr. Krebaum was losing popularity with his neighbors, andventing off the gas only partially solved the problem. A better solution, hedecided, would be to find a way to eliminate the smell entirely.

Mr. Krebaum rifled through his old chemistry books and found thathydrogen peroxide could oxidize these sulfur-containing compounds to muchless smelly compounds. He immediately whipped up a hydrogen peroxide mix-ture, and it worked like a charm. As an additional benefit, the volatility of thesulfur-containing compounds was decreased, preventing the odorous gasesfrom reaching our noses quite so easily.


The reaction by which the hydrogen sulfide was destroyed as shownbelow, produced sulfate compounds that do not have the unpleasant odor.

2NaOH + 4H2O2 + H2S ⎯→ Na2SO4 + 6H2O

“The receptors that are in your nose sensitive to sulfur in its low oxidationstate,” says Mr. Krebaum. “However, they are not sensitive to sulfur in its highoxidation state.” Some time later, a friend of Mr. Krebaum’s complained tohim that a skunk had sprayed his pet. Because the odor in a skunk’s spray alsocomes from compounds containing sulfur in a low oxidation state, Mr.Krebaum thought his solution might also work on this age-old problem. Hemixed up a milder version to try out on the pet: 1 quart 3% hydrogen peroxidesolution, 1/4 cup baking soda, and 1 teaspoon liquid soap. His friend tried itout, and the result was one wet and unhappy–but much less smelly–pet.

Mr. Krebaum says that the hydrogen peroxide in the remedy actually oxi-dizes the compounds, while the baking soda reduces the acidity of the mixtureand the soap helps to wash out the greasy skunk spray. The pet should bewashed thoroughly with the mixture, taking care to avoid its eyes. If the mix-ture is left on for a few minutes–long enough for the reaction to occur–andthen rinsed away with tap water, the smell will disappear. He does have onewarning: mix the formula just before using it because the mixture breaks downquickly. The reaction releases oxygen, so the formula should not be put into asealed container; it will build up pressure and could eventually blow the top.For this reason, bottles of “Krebaum’s Skunkinator” will not be appearing ondrug-store shelves any time soon.

50. Assign the oxidation numbers to sulfur in H2S and in Na2SO4. Write thehalf reaction for the oxidation of sulfur.

51. What element in hydrogen peroxide acts as the oxidizing agent? Write thehalf-reaction for its reduction.

638 Chapter 17



TECHNOLOGY AND LEARNING

56. Graphing Calculator

Calculate the Equilibrium Constant, usingthe Standard Cell Voltage

The graphing calculator can run a programthat calculates the equilibrium constant foran electrochemical cell using an equationcalled the Nernst equation, given the stand-ard potential and the number of electronstransferred. Given that the standard poten-tial is 2.041 V and that two electrons aretransferred, you will calculate the equilib-rium constant. The program will be used tomake the calculations.

Go to Appendix C. If you are using a TI-83Plus, you can download the programNERNST and data and run the application asdirected. If you are using another calculator,your teacher will provide you with key-strokes and data sets to use. After you have run the program, answer the followingquestions.a. What is the equilibrium constant when

the standard potential is 0.099?b. What is the equilibrium constant when

the standard potential is 1.125?c. What is the equilibrium constant when

the standard potential is 2.500?

FOCUS ON GRAPHING

Study the graph below, and answer the questions that follow.For help in interpreting graphs, see Appendix B, “Study Skills for Chemistry.”

52. What is the rate of production, in g/h, foreach metal?

53. What is the rate of production, in mol/h, foreach metal?

54. How much time is needed for 1.00 mol ofelectrons to be absorbed?

55. Write the half-reaction for the productionof silver metal from silver ions. Write thehalf-reaction for the production of goldmetal from gold(III) ions. Using these half-reactions, provide an explanation for thedifference in the rates of production of the metals.

Time (h)

Mass of Metal Deposited by Electroplating Using a 5 A Current

Mas

s (g

)

0

50

100

150

200

250

300

350

400

450

500

0 5 10 15 20

Ag

Au


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CHAPTER Chapter 17 Oxidation Numbers ... 7 2−. 2 Assign known oxidation numbers. According to Rule g, the oxidation number of the O atoms is −2, so this