Chapter 14:
OxidationReduction
Reactions
14-1
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Questions for Consideration
1.
2.
3.
4.
5.
6.
7.
What occurs in an oxidation-reduction reaction?
How do we account for the loss and gain of electrons in
an oxidation-reduction reaction?
How do chemical reactions provide electricity in
batteries?
How are simple oxidation-reduction reactions balanced?
How are complex oxidation-reduction reactions
balanced?
How do oxidation-reduction reactions generate
electricity in spontaneous reactions? How is electricity
used to drive oxidation-reduction reactions that do not
normally occur?
How can corrosion be prevented?
14-2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Chapter 14 Topics:
1.
2.
3.
4.
5.
6.
7.
What Is an Oxidation-Reduction Reaction?
Oxidation Numbers
Batteries
Balancing Simple Oxidation-Reduction
Reactions
Balancing Complex Oxidation-Reduction
Reactions
Electrochemistry
Corrosion Prevention
14-3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
14.1 What is an Oxidation-Reduction
Reaction?



A reaction in which electrons are transferred
Also called redox reactions
An example:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
In this reaction, electrons are transferred from zinc
to copper.
Figure 14.5
14-4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Oxidation-Reduction Reaction
Figure 14.6
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
14-5
Oxidation and Reduction

Oxidation


Reduction


The process of losing one or more electrons
The process of gaining one or more electrons
Oxidation and reduction are coupled
reactions.

Oxidation does not occur without reduction.
14-6
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Oxidation and Reduction
Oxidation
0
2+ 1
2+ 1
0
Zn(s) + CuCl2(aq) → ZnCl2(aq) + Cu(s)
Reduction

Only the reactants that change charge are involved in
the redox reaction.



Zinc is oxidized because it changed charge from no charge
(0) to +2, and thus lost two electrons.
Copper is reduced because it changed charge from a +2 to
no charge (0), and thus gained two electrons.
Chlorine is not involved in the reaction because it did not
change its charge.
14-7
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Oxidizing and Reducing Agents

Oxidizing agent
 A reactant that gains electrons and is reduced is
the oxidizing agent because it accepts the electrons
that are lost by the reactant that is oxidized.

Reducing agent
 A reactant that loses electrons and is oxidized is
the reducing agent because it provides electrons to
the reactant that gets reduced.
14-8
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Oxidizing and Reducing Agents
Oxidized & reducing agent
0
2+ 1
2+ 1
0
Zn(s) + CuCl2(aq) → ZnCl2(aq) + Cu(s)
Reduced & oxidizing agent


Zinc is oxidized and loses two electrons to copper,
making zinc the reducing agent.
Copper is reduced and gains two electrons from zinc,
making copper the oxidizing agent.
14-9
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Oxidation-Reduction Reactions

As shown in the figure, when iron metal is placed in a
solution of copper(II) nitrate, Cu(NO3)2, a singledisplacement reaction occurs in which copper deposits on
the surface of the iron and aqueous iron(III) nitrate
forms.
a) Write a balanced chemical equation for this reaction.
b) Identify the element oxidized and the element reduced.
c) Identify the oxidizing and reducing agents.
d) On a molecular level, what’s happening at the surface
of the iron metal?
Figure from p. 576
14-10
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: OxidationReduction Reactions
Write a balanced chemical equation for this reaction.
2Fe(s) + 3Cu(NO3)2(aq) → 2Fe(NO3)3(aq) + 3Cu(s)
b) Identify the element oxidized and the element reduced.
a)
Oxidation – loses electrons, increases charge
0
2+ 5+ 2
3+ 5+ 2-
0
2Fe(s) + 3Cu(NO3)2(aq) → 2Fe(NO3)3(aq) + 3Cu(s)
Reduction – gains electrons, decreases
charge
Iron is oxidized
Copper is reduced
14-11
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: OxidationReduction Reactions
c)
Identify the oxidizing and reducing agents.
Oxidized and reducing agent
0
2+ 5+ 2
3+ 5+ 2
0
2Fe(s) + 3Cu(NO3)2(aq) → 2Fe(NO3)3(aq) + 3Cu(s)
Reduced and oxidizing agent
Iron is oxidized, and therefore acts as the reducing
agent.
Copper is reduced, and therefore acts as the
oxidizing agent.
14-12
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: OxidationReduction Reactions
d)
On a molecular level, what’s happening at the
surface of the iron metal?
On the surface of the iron metal, iron atoms
act according to the half-reaction:
Fe(s) → Fe3+ + 3e
and move from the nail as a solid into
solution as ions.
The copper atoms act according to the half-reaction:
Cu2+ + 2e → Cu(s)
and move from the solution as ions onto the
nail as a solid.
14-13
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Oxidation Numbers


A charge assigned to the atoms in any compound
Also called oxidation states



For ionic compounds, the oxidation number is the same as
the ion’s charge.
We treat covalent compounds as if they’re ionic
compounds, assigning oxidation numbers to elements
using oxidation number rules.
We use rules to assign oxidation numbers (Table
14.1).

The rules are a hierarchy. The first rule that applies takes
precedence over any subsequent rules that may apply.
14-14
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Rules to Assign Oxidation Numbers
14-15
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Some Notes on the Rules for Oxidation
Numbers

For an isolated atom or a molecule that contains
only one element, only rule 1 need be applied. An
uncombined element, whether occurring as an atom
such as He or a molecule such as H2, has an
oxidation number of 0. Thus, the oxidation number
of sulfur in S, S2, and S8 is 0.

A monoatomic ion has an oxidation number equal to
its ionic charge, according to rule 2.
14-16
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Some Notes on the Rules for Oxidation
Numbers



Rule 2 can be applied whenever oxidation numbers
have been assigned to all but one element.
Rule 5 uses position in the periodic table to cover
situations that are not handled by one of the other
rules.
To assign oxidation numbers to binary compounds,
handle one of the elements with an appropriate rule
and the other with rule 2.
Total positive
oxidation numbers
+
Total negative
= net charge
oxidation numbers
14-17
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Assigning Oxidation
Numbers

Assign oxidation numbers to each element in the
following formulas:
1. Mg
2. Mn2O3
3. Na2SO4
4. Cr2O72
14-18
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Assigning
Oxidation Numbers
1.
2.
Mg – uncombined element, oxidation number is 0
Mn2O3 – the charge on the oxygen is 2 (according to
Rule 6); thus, the oxidation number on Mn must be 3+
2
Mn2O3
According to Rule 2, the sum of the oxidation
numbers must equal the total charge, in this case 0.
Using y to designate the unknown charge on the Mn:
2y + (3 × 2) = 0
y = +3
14-19
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Assigning
Oxidation Numbers
3.
Na2SO4 - the charge on the oxygen is 2 (according
to Rule 6) and the charge on the Na is 1+
(according to Rule 5); thus, the oxidation number
on S must be 6+
1+ y 2
Na2SO4
According to Rule 2, the sum of the oxidation
numbers must equal the total charge, in this case 0.
Using y to designate the unknown charge on S:
(2 × +1) + y + (4 × –2) = 0
y = +6
14-20
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Assigning
Oxidation Numbers
4.
Cr2O72 - the charge on the oxygen is 2- (according
to Rule 6) and the total charge for the polyatomic
ion is 2-; thus, the oxidation number on Cr must
be 6+
y
2
Cr2O72
According to Rule 2, the sum of the oxidation
numbers must equal the total charge, in this case 2
(the charge on the polyatomic ion). Using y to
designate the unknown charge on the Cr:
2y + (7 × 2) = 2
y = +6
14-21
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Identifying Oxidation-Reduction
Reactions

An oxidation-reduction reaction occurs if one or
more elements changes its oxidation number.



When the oxidation number increases, electrons have been
lost and oxidation has occurred.
When the oxidation number decreases, electrons have been
gained and reduction has occurred.
If a reaction has an uncombined element on one side
that is combined on the opposite side, the reaction is
an oxidation-reduction reaction.
14-22
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Identifying OxidationReduction Reactions


Determine which of the following represent
oxidation-reduction reactions.
For reactions that involve oxidation-reduction,
identify the oxidizing and reducing agents.
1.
2.
3.
4.
2H2O(l) → 2H2(g) + O2(g)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
4Al(s) + 3O2(g) → 2Al2O3(s)
Mg(s) + Zn(NO3)2(aq) → Mg(NO3)2(aq) + Zn(s)
14-23
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Identifying
Oxidation-Reduction Reactions
1.
2H2O(l) → 2H2(g) + O2(g)
This is an oxidation-reduction reaction because
the oxidation numbers change in the reaction.
The oxidation numbers for the reaction are:
Oxidation and Reducing agent
1
2
0
0
2H2O(l) → 2H2(g) + O2(g)
Reduction and
Oxidizing agent
14-24
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Identifying
Oxidation-Reduction Reactions
1+ 1
2.
1+ 2 1+
1+ 1
1+ 2
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
This is not an oxidation-reduction reaction. It is
an acid-base neutralization. Notice how none of
the elements change charge in the reaction.
14-25
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Identifying
Oxidation-Reduction Reactions
3.
4 Al(s) + 3O2(g) → 2Al2O3(s)
This is an oxidation-reduction reaction because
the oxidation numbers change in the reaction.
The oxidation numbers for the reaction are:
Oxidation and Reducing agent
0
0
3+
2
4Al(s) + 3O2(g) → 2Al2O3(s)
Reduction and
Oxidizing agent
14-26
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Identifying
Oxidation-Reduction Reactions
4.
Mg(s) + Zn(NO3)2(aq) → Mg(NO3)2(aq) + Zn(s)
This is an oxidation-reduction reaction
because the oxidation numbers change. The
oxidation numbers for the reaction are:
Oxidation and Reducing agent
0
2+ 5+ 2–
2+ 5+ 2–
0
Mg(s) + Zn(NO3)2(aq) → Mg(NO3)2(aq) + Zn(s)
Reduction and Oxidizing agent
14-27
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Oxidation and Reduction


Half reactions represent either the oxidation or the
reduction that occurs in separate compartments of a
voltaic cell.
Each compartment is called a half-cell.

Oxidation half-reaction
Zn(s) → Zn2+(aq) + 2e–

Reduction half-reaction
Cu2+(aq) + 2e– → Cu(s)


Notice that because the electrons exist on either side of the
arrow in the two reactions, they cancel out.
The separation of the oxidation and reduction
processes into half-cells harnesses the energy released
from a chemical reaction to generate electricity.
14-28
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Oxidation and Reduction in
Batteries

The reaction that occurs in most camera batteries is:
Zn(s) + Ag2O(s) → ZnO(s) + 2Ag(s)

What is the oxidizing agent?
What is the reducing agent?

14-29
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Oxidation and
Reduction in Batteries

The reaction that occurs in most camera batteries is:
0
1+
2–
2+ 2–
0
Zn(s) + Ag2O(s) → ZnO(s) + 2Ag(s)

Zinc is oxidized and thus, the reducing agent.

Silver is reduced and thus, the oxidizing agent.
14-30
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
14.4 Balancing Simple OxidationReduction Equations

The reaction between silver nitrate and copper can
be written simply as:
Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s)

Does this equation appear balanced to you? Do the
atoms balance? Does the charge balance?
14-31
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
14.4 Balancing Simple OxidationReduction Equations

The reaction between silver nitrate and copper can
be written simply as:
Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s)

Does this equation seem balanced to you? Do the
atoms balance? Does the charge balance?
You probably noticed that the total of the charges of
the reactants is 1+, and the total of the charges of the
products is 2+. Electrons lost do not equal electrons
gained.
The equation is NOT balanced.


14-32
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Balancing Simple Equations
Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s)


Separating the reaction into half-reactions:
Oxidation: Cu(s) → Cu2+(aq) + 2e–
Reduction: Ag+(aq) + e– → Ag(s)
The electrons lost must equal the electrons gained, so
we need to multiply the reduction half-reaction by 2:
2[Ag+(aq) + e– → Ag(s)]
2Ag+(aq) + 2e– → 2Ag(s)
14-33
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Balancing Simple Equations

Now we can add the two half-reactions:
Cu(s) → Cu2+(aq) + 2e–
2 Ag+(aq) + 2e– → 2Ag(s)
Cu(s) + 2Ag+(aq) + 2e– → Cu2+(aq) + 2e– + 2Ag(s)

Notice that 2 electrons appear as a reactant and a
product. They cancel, and the overall reaction is:
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
14-34
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Balancing Simple Equations
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Double-checking the numbers of atoms and the
overall charge:
Reactants
Products
Cu
1
1
Ag
2
2
Charge
2+
2+
14-35
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Balancing Simple Equations
Figure 14.17
14-36
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Balancing Simple Equations

Some questions to guide you through the balancing process:
1.
Which substance is oxidized and which is reduced? To
answer this question, assign oxidation numbers to all the
elements in the reaction.
2.
What are the half-reactions for the oxidation and
reduction processes? For each half-reaction,
a) Which element changes in oxidation number?
b) Can any spectator ions be ignored until the final
balancing step?
c) How many electrons must be added to the appropriate
side of the equation to account for the change in
oxidation number?
14-37
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Balancing Simple Equations

Some questions to guide you through the balancing
process:
3.
4.


What factor can be used to multiply each coefficient
in the balanced half-reactions to equalize the number
of electrons gained or lost?
When adding the two half-reactions, can any
substances that are present in equal amounts on both
sides of the equation be canceled out?
After answering these questions, you should have a
balanced overall equation.
Make sure the sum of the atoms and the sum of the
charges equal on either side of the equation.
14-38
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Balancing Simple Equations
Magnesium metal reacts with aqueous chromium(III) nitrate to
produce magnesium nitrate, Mg(NO3)2, and solid chromium.
Write a balanced equation for this reaction.
14-39
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Balancing Simple
Equations
Magnesium metal reacts with aqueous chromium(III)
nitrate to produce magnesium nitrate, Mg(NO3)2, and solid
chromium. Write a balanced equation for this reaction.
We begin by writing formulas for reactant and products
and writing a skeletal equation:
Mg(s) + Cr(NO3)3(aq) → Mg(NO3)2(aq) + Cr(s)
To determine which substances are oxidized and
reduced, we write oxidation numbers for the reactants
and products:
0
3+ 5+ 2–
2+ 5+ 2–
0
Mg(s) + Cr(NO3)3(aq) → Mg(NO3)2(aq) + Cr(s)
14-40
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Balancing Simple
Equations
Since the nitrate, NO3–, is not involved in the
oxidation-reduction reaction, we can write a skeletal
ionic equation:
Mg(s) + Cr3+(aq) → Mg2+(aq) + Cr(s)
Next, we can write half-reactions for the oxidation
and reduction reactions:
Oxidation: Mg(s) → Mg2+(aq) + 2e–
Reduction: Cr3+(aq) + 3e– → Cr(s)
Notice: The electrons lost do not equal the electrons
gained.
14-41
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Balancing Simple
Equations
The magnesium atom loses 2 electrons, but the
chromium atom gains 3 electrons. We must multiply
the oxidation reaction by 3 and the reduction
reaction by 2:
3[Mg(s) → Mg2+(aq) + 2e–]
2[Cr3+(aq) + 3e– → Cr(s)]
3Mg(s) → 3Mg2+(aq) + 6e–
2Cr3+(aq) + 6e– → 2Cr(s)
Now electrons lost equal electrons gained.
14-42
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Balancing Simple
Equations
Add the two reactions together:
3Mg(s) → 3Mg2+(aq) + 6e–
2Cr3+(aq) + 6e– → 2Cr(s)
3 Mg(s) + 2Cr3+(aq) + 6e– → 2Cr(s) + 3Mg2+(aq) + 6e–
The balanced ionic equation is:
3Mg(s) + 2Cr3+(aq) → 2Cr(s) + 3Mg2+(aq)
Returning the nitrates to the equation:
3Mg(s) + 2Cr(NO3)3(aq) → 2Cr(s) + 3Mg(NO3)2(aq)
14-43
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Balancing Simple
Equations
3Mg(s) + 2Cr(NO3)3(aq) → 2Cr(s) + 3Mg(NO3)2(aq)
Double-check the number of atoms and
charges:
Reactants
Products
Mg
3
3
Cr
2
2
N
6
6
O
18
18
Charge
0
0
14-44
Copyright © McGraw-Hill Education. Permission required for reproduction or display.