-
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)
378
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
-
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
382 18 Oxidation-Reduction
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
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
-
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
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
-
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.
Chapter Review 387
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsCalloutLEO the lion goes GERor OIL RIG
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
-
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
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
-
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
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
-
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
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
-
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.
Chapter Review 391
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
-
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
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
hrhsHighlight
-
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?
Learning Review 393
hrhsHighlight
hrhsHighlight
-
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