CHAPTER 18: OXIDATION-REDUCTIONREACTIONS AND ELECTROCHEMISTRY

INTRODUCTION

There are many important oxidation-reduction reactions. These reactions are characterized byelectron transfer. One of the most annoying and costly of the oxidation-reduction reactions is therusting of automobile bodies. In this chapter you will learn what actually happens during anoxidation-reduction reaction, and what means are available to keep your car from rusting.

GOALS FOR THIS CHAPTER

1. Define oxidation and reduction, and identify oxidation-reduction reactions between a metaland a nonmetal. (Section 18.1)

2. Be able to assign oxidation states to all atoms in molecules and in ions. (Section 18.2)3. Be able to determine which elements are oxidized and which are reduced, and which

species is the oxidizing agent and which is the reducing agent (Section 18.3)4. Be able to-balance a redox reaction by the half-reaction method. (Section 18.4)5. Know how a redox reaction can be separated into half reactions for the purpose of

generating electrical current. (Section 18.5)6. Understand how the lead storage battery works, and how common dry cell batteries work.

(Section 18.6)7. Understand how corrosion occurs, and what methods are available to combat corrosion.

(Section 18.7)8. Understand how electrolysis is used to produce a chemical reaction which does not occur

naturally. (Section 18.8)

QUICK DEFINITIONS

Oxidation-reductionreactions

Oxidation

Reduction

Oxidation states

Also called redox reactions, they are reactions where onespecies loses electrons, and another species gains electrons.(Section 18.1)

The loss of electrons. Oxidation is loss (OIL). Also defined asan increase in oxidation state. (Section 18.1)

The gain of electrons. Reduction is Kain (RIG). Also defined asa decrease in oxidation state. (Section 18.1)

An imaginary assignment of electrons in a molecule or ion tothe most electronegative atom to help determine where electronsin a redox reaction are lost or gained. (Section 18.2)

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Oxidizing agent

Reducing agent

Half-reaction

Electrochemistry

Galvanic cell

Anode

Cathode

Electrolysis

The species in a redox reaction which is reduced, the electronacceptor. (Section 18.3)

The species in a redox reaction which is oxidized, the electrondonor. (Section 18.3)

An individual reduction or oxidation taken from a completeequation. Electrons are shown as either reactants or products.(Section 18.4)

The study of the methods for converting chemical energy toelectrical energy. (Section 18.5)

Also called an electrochemical battery. A device whichproduces electrical current by separating the oxidation and thereduction half-reactions and forcing the electrons to move

H_. . __~e~'Y~~I!lh.e~ t\V?l>.Y.~~ire. (S~ctiO!!}J·_~L ~_____ _ ~ _._

The electrode in a galvanic cell where oxidation (loss ofelectrons) occurs. (Section 18.5)

The electrode in a galvanic cell where reduction (gain ofelectrons) occurs. (Section 18.5)

The process of adding electrical energy to a cell to cause achemical reaction which by itself would not occur. (Sections18.5 and 18.8)

Lead storage battery

Potential

Dry cell batteries

Corrosion

The automobile battery which depends upon the oxidation of Pband the reduction of PbOz to generate energy. (Section 18.6)

The "pressure" on electrons to flow from the anode to thecathode. (Section 18.6)

Small efficient batteries which do not use a liquid electrolyte.(Section 18.6)

The oxidation of a metal. This term is usually associated withthe process which converts metals from a useful form, such assolid iron, to a non-useful form, such as rust (Section 1ST?)

Quick Definitions 379

Cathodic protection

PRETEST

A method of protecting metals from corrosion. A metal moreeasily oxidized than the metal you are protecting is attachedwith an insulated wire to the protected metal. The more easilyoxidized metal is oxidized, leaving the protected metal intact.(Section 18.7)

1. When a molecule of fluorine reacts to form two fluoride ions, is this reaction an exampleof oxidation or reduction?

2. For the reaction below, which element is oxidized and which is reduced?

Zn(s) + 2HCI(aq) --Iloo- ZnCI2(aq) + H2(g)

3. What is the oxidation state for each atom in the molecules below?

a. Ni02b. XeOF4

4. What is the oxidation state for each element in the reaction below?

5. For the reaction below, which atom is oxidized and which is reduced, and identify theoxidizing and reducing agents.

6. For the reaction below, write the reduction and the oxidation half-reactions.

7. Balance the reaction below by the half-reaction method. The reaction occurs in acidicsolution.

380 18 Oxidation-Reduction

8. Based on the direction of electron flow, in which container does oxidation occur?

e- e-

9. What major differences are there between an acid dry cell battery and an alkaline dry cell?

10. List two methods for preventing the corrosion of metals.

PRETEST ANSWERS

1. When a molecule of fluorine reacts to form two fluoride ions, fluorine gains electrons.This is an example of reduction.

(18.1)

2. In the reaction between zinc metal and hydrochloric acid, zinc loses electrons. It isoxidized. Hydrogen ions gain electrons to form hydrogen gas. Hydrogen ions arereduced.

Zn(s) -... Zn2+(aq) + 2e-2H+(aq)+ 2e- -... Hz(g) (18.1)

3. The rules for assigning oxidation states are found in section 18.2 of your text book.

a. Oxygen is usually 2-. In NiOz there are two oxygens for a total of 4-. Nickel mustbe 4+ so that the sum of charge is O.

NiOz/,4+ 2- each

b. In XeOF4, oxygen is usually 2-. Each fluorine is 1- for a total of 4-. Xenon must be

6+ so that the sum of charges is O.

XeOF4

It'6+ 2- 1-each (18.2)Pretest Answers 381

4.1205 (s)

/\5+ each 2- each

+ 5CO(g)-... 12(s) + 5C02 (g)

t \ t t \2+ 2- 0 4+ 2- each

(18.2)

5. To identify which atoms are oxidized and which are reduced we need to first assignoxidation states to each atom in the reaction.

PbS(s) + 4H202 (1) -... PbS04(s) + 4H20(1)

/\ t\ /t\ t\2+ 2- 1+ 1- 2+ 2+ 2- 1+ 2-

each each each each

The oxidation state of oxygen in H202, hydrogen peroxide, is 1-. Oxygen in H202 has an

oxidation state of 1-, while oxygen in Hz0 has a oxidation state of 2-. The oxidation state

decreases, so oxygen is reduced. H20Z is the oxidizing agent. Sulfur in PbS has an

oxidation state of 2-, and sulfur in PbS04 has an oxidation state of 6+. The oxidation state

of sulfur increases, so sulfur is oxidized. PbS is the reducing agent. (18.3)

6. For the reaction between aluminum metal and fluorine gas to produce solid aluminumfluoride, the half-reactions are

2AI(s) Ae+caq)

3Fz(g) 6F(aq)oxidation

reduction (18.4)

7. Step 1: Write the equations for the oxidation and reduction half-reactions.

Mnz+ (aq) + Bi03 (s) ......Mn04- (aq) + Bi3+ (aq)

+ +, +' +2+ 6+ 2- each 7+ 2-each 3+

The Mnz+ ion loses five electrons to become the permanganate ion. Manganese isoxidized. The oxidation half-reaction is

Mn2+ ...... MnO·4 oxidation half-reaction

Bismuth in Bi03 gains three electrons to become the Bi3+ ion. Bismuth is reduced.

The reduction half-reaction is

BiO ...... Bi3+3

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reduction half-reaction

Step 2a: For both the oxidation and reduction half-reactions, all elements other thanoxygen are balanced.

Step 2b: The oxidation half-reaction has four oxygen atoms on the right and noneon the left, so add four molecules of water to the left

The reduction half-reaction has three oxygen atoms on the left and none on the right,so add three water molecules to the right.

Step 2c: The oxidation half-reaction has eight hydrogens on the left, so add 8H+ tothe right.

The reduction half-reaction has six hydrogens on the right, so add 6H+ to the left.

Step 2d: The oxidation half-reaction has an excess of 5+ charge on the right side, soadd Se to the right side to balance the charge.

charge

4~O + Mn2+ -flio- Mn04- + 8H+2+ -flio- 7+

The reduction half-reaction has an excess of 3+ charge on the left side, so add 3e- tothe left side to balance the charge.

charge

6H+ + Bi03 --.... Bi3+ +3~O

6+ --.... 3+

Step 3: In the oxidation half-reaction five electrons are transferred, and in thereduction half-reaction three electrons are transferred. Multiply the oxidationhalf-reaction by three and the reduction half-reaction by five to balance the numberof electrons transferred.

Pretest Answers 383

3(4Hz0 + Mnz+ Mn04- + 8H+ + Se')

12HzO+ 3Mnz+ 3Mn04- + 24H+ + 15e-

5(3e- + 6H+ + Bi03 Bi3+ + 3HzO)

15e- + 30H+ + 5Bi03 5Bi3+ + 15HzO

Step 4: Now add the two half-reactions together.

12Hz0 + 3Mnz+ ....... 3Mn04- + 24H+ + 15e"

15e- + 30H+ + 5Bi03 ....... 5Be+ + 15Hz0

The fifteen electrons on each side cancel, and the number of hydrogen ions andwater molecules can be reduced. There are twenty-four hydrogen ions on the rightand thirty on the left side. Cancel the twenty-four on each side and we are left withsix hydrogen ions on the left. Cancel twelve water molecules on each side and weare left with three water molecules on the right. The equation becomes

Step 5: Let's check the elements and the charges on each side. There are sixhydrogen atoms, five bismuth atoms, fifteen oxygen atoms and three manganeseatoms on each side. There is a 12+ charge on each side so the equation is balanced.

6H+(aq) + 5Bi03(s) + 3Mnz+(aq) 5Bi3+(aq) + 3Mn04-(aq) + 3Hz0(l)

elements 6 H 5 Bi 150 3 Mn 6 H 5 Bi 150 3 Mncharge 12+ 12+ (18.4)

8. During oxidation, electrons are lost, so oxidation occurs in the left container and reductionin the right container. (18.5)

9. Both acid and alkaline dry cells contain zinc, which is oxidized to Znz+ with a loss of twoelectrons. However, in the alkaline cell, hydroxide ions are present instead of ammonium

chloride.

Zn ....... Znz+ + 2e- acid dry cell oxidation

Zn + 20H- ....... ZnO(s) + Hz0 + 2e- alkaline dry cell oxidation

384 18 Oxidation-Reduction

In both the acid and alkaline dry cell versions, Mn02 is reduced to Mn203, but the alkalineversion contains OH- ions.

2e- + 2Mn02 +~O~ Mn203 + 20H- alkaline dry cell reduction (18.6)

10. The most common method for preventing corrosion is to coat the metal to be protectedwith paint or with a metal plating. The metal plating keeps oxygen necessary for corrosionaway from the metal underneath. Cathodic protection is used to prevent corrosion ofburied fuel tanks and pipelines. A metal such as magnesium, which is very easilyoxidized, is connected by a wire to the metal which is to be protected. The magnesium ispreferentially oxidized and the metal tank is left unharmed. The magnesium must bereplaced periodically as it is oxidized to Mg2+ ions. Iron is often alloyed with metals suchas chromium and nickel, both ofwhich form protective oxide coats. (18.7)

CHAPTER REVIEW

18.1 OXIDATION-REDUCTION REACTIONS

How Does an Oxidation-Reduction Reaction OccurBetweena Metal and a Nonmetal?

Reactions between metals and nonmetals involve the transfer of electrons. When the metal Lireacts with Br2, the result is the ionic compound LiBr.

2Li + Br2~ 2LiBr

Both Li and Br, began as neutral species, but after the reaction, both were ions. Li became theLi+ cation, and Br2 became the Br anion. Electrons must have been transferred between lithiumand bromine for the reaction to have occurred. Reactions of this type are called oxidation-reduction reactions, or often redox reactions. In this reaction, Li has lost an electron tobecome the Li+ cation. This process is called oxidation. Each Br atom has gained an electron toform Br'. This process is called reduction. In every oxidation-reduction reaction, one species isoxidized, and another is reduced.

Chapter Review 385

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18.2 OXIDATION STATES

How Are Oxidation States Useful, and How Are They Determined?

In some redox reactions it is not easy to tell which species has been oxidized and which has beenreduced. The assignment of oxidation states or oxidation numbers can help you determinewhich species is oxidized and which reduced, and whether a reaction is really a redox reaction ornot. An oxidation state is an imaginary number assigned to each element in a chemical reaction.The number comes from the charge that element would have if it were an ion. How the electronsare assigned to the atoms is governed by a set of rules, but is basically determined by the electro-

negativity of the atom. Some species are easy to assign oxidation states to. All of the metalswhich form cations with a 1+ charge have oxidation states of 1+. So the oxidation state of K+would be 1+. Many atoms in chemical reactions are not ions, but are covalently bonded to otheratoms; that is, the electrons are shared between two atoms. Assign the electrons as though the

atoms were ions. The most electronegative atom is assigned both the shared electrons. Becausemolecules are electrically neutral, the sum of the oxidation states of all the atoms must be zero.

Example:The oxidation states of N and H in ammonia,~, can be determined by assigning each of

the pairs of shared electrons to the most electronegative atom, N. One of the rules says

that hydrogen almost always has a 1+ oxidation state, so each hydrogen in the ammoniamolecule is 1+. Because the molecule must be electrically neutral, nitrogen has a 3-

oxidation state to balance the 3+ from the three hydrogen atoms. All the rules fordetermining oxidation state are presented in detail in Section 18.2, but are summarized

below.

Rules for Assigning Oxidation Numbers1. Uncombined elements have oxidation states of O.2. All monatomic ions have the same oxidation state as their charge.3. Oxygen has an oxidation state of 2- except when combined as a peroxide.4. Hydrogen usually has an oxidation state of 1+, except when combined with a metal.5. In compounds between two elements, assign the shared electrons to the most

electronegative atom. The charge on the electronegative atom will be equal to the chargeon the ion, when that atom forms ions.

6. The sum of all oxidation states in a molecule must be zero.7. The sum of all oxidation states in an ion must be equal to the charge on the ion.

386 18 Oxidation-Reduction

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18.3 OXIDATION-REDUCTION REACTIONS BETWEEN NONMETALS

How Can We Tell Which Atoms Are Oxidized orReduced in Covalent Compounds?

In reactions between metals and nonmetals, it is often easy to determine which atoms areoxidized and which are reduced, because the reactants are often pure elements which haveoxidation states of O. The oxidation states of the ions formed in the reaction are the same as thecharges on the ions. But in the case of reactions between nonmetals, the compounds are oftencovalently bonded and the oxidation states of the reactants are often not zero. It is not alwayspossible to tell by inspection what is oxidized and what is reduced. By assigning oxidation statesto each atom in the reaction, and comparing the oxidation states on each side of the equation, it ispossible to decide which element is oxidized and which is reduced. When the oxidation stateincreases, an element has been oxidized, and when the oxidation state decreases, the element hasbeen reduced.

Example:The net ionic equation for a reaction between N~S208 and NaI is

We can determine which element is oxidized and which is reduced by assigning oxidationnumbers to each of the atoms. On the left side of the equation the T ion has an oxidationstate of 1-, the same as the charge on the ion. In the S208-2 ion, the peroxydisulfate ion,

oxygen has an oxidation state of 2-. There are 8 oxygens, so the total number for oxygensis 16-. There are two sulfur atoms in the ion; each one of them must have an oxidationstate of 7+ because two times the oxidation state of sulfur plus eight times the oxidationstate of oxygen equals the charge on the peroxydisulfate ion, which is 2-. There is a net 2-charge on the ion, so the sum of all oxidation numbers must equal 2-. On the right side ofthe equation, 12 has an oxidation number of O. Oxygen in the sulfate ion is equal to 2-.

There are four oxygens, for a total number of 8-. The oxidation state of sulfur musttherefore be 6+ because four times the oxidation state of oxygen plus the oxidation state ofsulfur equals the charge on the sulfate ion, which is 2-.

S2082- + 2r-......... h + zso>/ \ t + +\

7+ each 2- each 1-each 0 each 6+ 2- each

In this reaction, iodine increases in oxidation state, from 1- to O. It is oxidized. Sulfurdecreases in oxidation state, from 7+ to 6+. It is reduced. We say that the species which isoxidized is the reducing agent, because it causes the reduction of another species. T is

oxidized so I' is the reducing agent. The species which is reduced is called the oxidizing

agent because it causes the oxidation of another species. S20t is the oxidizing agent.

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hrhsCalloutLEO the lion goes GERor OIL RIG

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18.4 BALANCING OXIDATION-REDUCTION REACTIONSBY THE HALF-REACTION METHOD

How Can Redox Reactions Be Balanced?

In Chapter 6 you learned how to balance simple chemical reactions by inspection. Balancingredox reactions is more difficult, and can rarely be done by inspection. Another method isneeded for balancing these reactions. One method for balancing redox reactions is called thehalf-reaction method. A half-reaction is part of a complete chemical equation. In a redoxreaction, there is always a reduction half-reaction, and an oxidation half-reaction. We can writeeach of them separately, and balance the difference in oxidation states on each side of theequation by adding electrons to either the right or the left side. Section 18.4 in your textbookgives some general steps to be used when balancing redox reactions, and five specific steps touse when balancing redox reactions which take place in acidic solution. We will use the fivespecific steps to balance the reaction below.

Example:Balance the redox reaction below which takes place in acidic solution.

Step 1: Write the equations for the oxidation and reduction half-reactions.First, decide which element is oxidized and which is reduced by assigning oxidation statesto each element. In this reaction, Au is oxidized. It increases in oxidation state, from 0 to3+. cr participates in the reaction, but does not change in oxidation state. Nitrogen isreduced. It decreases in oxidation state, from 5+ to 4+.

Au + cr + N03- -liIIo- AuCI4- + NOzt t tt t t tto 1- 5+ 2- 3+ 1- 4+ 2-

We can now write the half-reactions.

Au + cr -liIIo- AuCl4- oxidation half-reaction

N03- -liIIo- NOz reduction half-reaction

Step 2a: Balance all the elements except hydrogen and oxygen.There are four chlorine atoms on the right, so we must add four on the left. Theunbalanced equation shows that the chlorine atoms are present as chloride ions.

The reduction half-reaction has all atoms except hydrogen and oxygen balanced.

388 18 Oxidation-Reduction

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Step 2b: Balance oxygen using H20.The oxidation half-reaction contains no oxygen atoms. Add a water molecule to the rightside of the reduction half-reaction to balance the three oxygen atoms on the left.

Step 2c: Balance hydrogen using H+.The oxidation half-reaction contains no hydrogen atoms. Add two hydrogen ions to theleft side of the reduction half-reaction to balance the two hydrogens on the right.

Step 2d: Balance the charge using electrons.In the oxidation half-reaction the total charge on the left is 4- and on the right is 1-, so addthree electrons to the right.

charge

- --- --- -Au+ 4Cf-- AuC14-

4- 1-

In the reduction half-reaction the total charge on the left is 1+ and on the right is 0, so addan electron to the left side.

charge

2H++ N03- N02 + H201+ 0

Step 3: Equalize the number of electrons transferred in the oxidation and reductionhalf-reactions.The oxidation half-reaction transfers three electrons, but the reduction half-reactiontransfers only one. Multiply the reduction half-reaction by three to equalize the number ofelectrons transferred.

3(e- + 2H+ + NO- N02 + Hz0)

3e- + 6H+ + 3N03- 3N02 + 3Hz0

Step 4. Add the half-reactions and cancel identical species which appear on both sides.

Chapter Review 389

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Au + 4Cr-.... AuCI4- + 3e-

3e- + 6H++ 3N03--.... 3N02 +3~O

We can cancel the three electrons which appear on both sides, so the equation is

6H+Caq) + 3N03-(aq) + Au(s) + 4Cr(aq)-"" 3N02(g) + 3~O(l) + AuCI4-(aq)

Step 5. Check to be sure the elements and the charges balance.Each side has three nitrogens, nine oxygens, one gold, and four chlorine atoms. Thecharge on each side is the same, 1-, so the equation is balanced.

elementscharge

3N03- + 6H++ Au + 4Cr-.... 3N02 +3~O + AuCI4-3N 90 6H lAu 4Cl -.... 3N 90 6H 1Au 4Cl

1- 1-

18.5 ELECTROCHEMISTRY--AN INTRODUCTION

Chemical and electrical energy can be interchanged. The study of the interchange of these twoforms of energy is called electrochemistry. The interchange of energy forms can occur in eitherdirection. You can convert chemical energy to electrical energy,or you can use an electricalcurrent to produce a chemical reaction. Let's first consider how a chemical reaction can be usedto produce electrical energy.

How Can a Chemical ReactionProducean Electrical Current?

During a redox reaction, electrons are transferred whenever the reactants collide in solution. It isnot possible to use the electron transfer to generate electrical energy under these circumstances.In order to harness the energy of a redox reaction, it is necessary to physically separate the twohalf-reactions in two separate containers which are connected by a wire. The electrons which aretransferred between the oxidation half-reaction and the reduction half-reaction travel along thewire, producing an electrical current.

cathode

reductionhalf-reaction

390 18 Oxidation-Reduction

1saltbridge

wire

anode

oxidationhalf-reaction

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A reaction between two half-cells connected only by a wire will not happen unless there isanother connection between the two containers which allows ions to flow freely back and forth.As electrons leave one container, and travel along the wire to the other container, differences incharges in the two containers would occur. The container with the oxidation half-reaction wouldbuild up a positive charge from the loss of electrons. The container with the reductionhalf-reaction would build up a negative charge from the gain of electrons. The extra connectionwhich contains ions allows negative ions to travel to the container losing electrons, and positiveions to travel to the container gaining electrons, so the net charge in each container is zero. Thisconnection is called a salt bridge. The current which is produced in a cell such as this can beused to do useful work, and is the principle upon which batteries are made. Cells powered bytwo separate half-reactions connected by a wire and by some type of connection to allow ionexchange are called galvanic cells. The electrode where electrons are lost is called the anode,and the electrode where electrons are gained is called the cathode.

18.6 BATTERIES

What Is a Battery?

A battery is a galvanic-cell:---Bifferentbatteries--use-different chemical reactions to generate acurrent.

How Does the Lead Storage Battery Work?

The lead storage battery is responsible for generating the current which starts automobiles. Thehalf-reactions and the overall reaction of the lead storage battery are shown below.

Pb(s) + PbOis) + 2H2SOiaq)-eo-- 2PbS04(s) + 2H20(l) overall

The solid Pb is cast into a metal grid, for easy contact with the liquid HzS04' and the solid Pb02is coated onto a Pb grid, also for contact with HzS04' Notice that HzS04 participates in both theoxidation and the reduction half-reactions. Solid Pb, Pb02 and H2S04 are all used up in thereaction to generate a current.

What Are Dry Cell Batteries?

A dry cell battery is one which contains no liquid electrolytes. The components are either solids,or moist pastes. Dry cells are produced in both an acidic and a basic (alkaline) version. Thehalf-reactions in both these cells are complex, but are presented below.

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Acidic

oxidation half-reaction

2NH4+ + 2Mn0z+ 2e---'" MnZ03 + 2NH3 + HzO reduction half-reaction

Basic

Zn + 20H---'" ZnO(s) + HzO + 2e- oxidation half-reaction

2MnOz+ HzO+ 2e---'" Mnz03 + 20H- reduction half-reaction

18.7 CORROSION

What Is Corrosion and How Can It Be Prevented?

Corrosion is the oxidation of metals. The most common example is the rusting of iron in thepresence of Oz. Oxidation of metals often results in weakening of the structuralproperties of themetals. Sometimes-the metafoxide forms a coating on the outside of the metal which can protectthe metal against further oxidation. This is true in the case of aluminum. The metal oxidecoating of iron flakes off, exposing fresh metal to the rusting process. So rusted iron is notprotected against further corrosion.

Corrosion of iron can be prevented by producing stainless steel, an alloy which containschromium and nickel. The added metals form tough metal oxide coats which do not flake off.Metals can also be painted with protective paint, and surfaces can be plated with another metalwhich forms a stable oxide coat.

Another method of protecting iron from corrosion is cathodic protection. Cathodic protectioncan be achieved by attaching a piece of metal, such as magnesium, which oxidizes more readilythan iron does, to the piece of iron. The magnesium is oxidized, leaving the iron intact.

18.8 ELECTROLYSIS

What Is Electrolysis?

Automobile lead storage batteries last several years. The reason we can reuse the battery to startour cars many times is that the battery is recharged while we drive. When we apply an electriccurrent through the alternater of the car, we force the reaction to produce Pb, Pb02 and ~S04'This is just the opposite of the reaction which produces a current to start our cars. This is anexample of electrolysis, the use of electrical energy to drive a chemical reaction that-would nototherwise occur. By charging the battery, we force the reaction toward the left, and we keep theamount of Pb, PbOz and HzS04 high enough so that the car will start when we want it to.

392 18 Oxidation-Reduction

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How Can Pure Aluminum Be Produced By Electrolysis?

Aluminum is found in nature as aluminum oxide. Production of aluminumfrom its ore proved tobe difficult and expensive, because the oxide is a very stable molecule. For this reason,aluminum utensils were very expensive until a better method for producing pure aluminum wasdiscovered. Pure aluminum can be produced from aluminum oxide by electrolysis. In theprocess, Ae+ gains three electrons to become AI.

LEARNING REVIEW

1. For each of the partial reactions, decide whether oxidation or reduction is occurring.

a.b.c.

Li --lIiIo- Li+ + e"Br2 + 2e- --- 2Br"S2- S + Ze'

2. For each reaci1onbeIow~idenfifywlilcneIement is oxidizedandwhichis reduced.

a. Ca(s) + I2(g) --- CaIz(s)b. 2K(s) + S(s) --- K2S(s)c. 6Na(s) + Nzeg) --lIiIo- 2N~N(s)

3. Determine the oxidation states for each element in the substances below.

a. CH4b. SO 2-4c. NaHC03d. N20Se. HI04

4. Determine the oxidation state for each element in the reactions below.

a. 4Fe(s) + 302(g) + 12HCl(aq) ....... 4FeC13(aq) + 6H20(l)b. Zn(s) + 2AgN03(aq) --lIiIo- Zn(N03}z(aq) + 2Ag(s)c. MgC12(l) ---liRo- Mg(s) + C12(g)

5. During a redox reaction, does the reactant which is the reducing agent contain an elementwhich is oxidized or reduced?

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6. For each of the reactions below identify which atom is oxidized and which is reduced, andidentify the oxidizing and reducing agents.

a. 2C2H6(g) + 702(g) -.... 4C02(g) + 6H20(g)b. 2KN03(l) -.... 2KN02(l) + ozeg)c. 3CuO(s) + 2NH3(aq) -.... 3Cu(s) + Nzeg) + 3H20(l)d. ~Cr207(aq) + 14HI(aq) -.... 2CrI3(s) + 2KI(aq) + 312(s) +7~O(l)

7. Balance each of the reactions below by the half-reaction method.

a. Zn(s) + Cu2+(aq) -.... Zn2+(aq) + Cu(s)b. ReS\aq) + Sb3+(aq) -.... Re4+(aq) + Sbs+(aq)

8. Each of the reactions below occurs in acidic solution. Balance each one by thehalf-reaction method.

a. H2S(aq) + N03-(aq) -.... S(s) + NO(g)b. HsI0 6(aq) --I'(aq) -.... ~(s)c. Cr20/(aq) + Sn

2+(aq) -.... Sn4+(aq) + Cr3+(aq)

d. 12(s) + N03-(aq) -.... I03-(aq) + N02(g)

9. Normally, when a redox reaction occurs, no useful work is produced. How can a redoxreaction be made to perform useful work?

10. Label what is needed to complete the electrical circuit and allow the redox reaction toproceed.

..dbreductionoxidation

11. Briefly explain how the lead storage battery works.

12. Aluminum metal easily loses electrons to form A1203• How can aluminum metal be

produced from its oxide?

394 18 Oxidation-Reduction