Transcript Chapter 17

INTRODUCTORY CHEMISTRY
Concepts and Critical Thinking
Sixth Edition by Charles H. Corwin
Chapter 17
Oxidation and
Reduction
by Christopher Hamaker
© 2011 Pearson Education, Inc.
Chapter 17
1
Oxidation–Reduction Reactions
• Oxidation–reduction reactions are reactions
involving the transfer of electrons from one
substance to another.
• We have seen several oxidation–reduction
reactions so far.
• Whenever a metal and a nonmetal react, electrons
are transferred.
2 Na(s) + Cl2(g) → 2 NaCl(s)
• Combustion reactions are also examples of
oxidation–reduction reactions.
© 2011 Pearson Education, Inc.
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2
Example of Oxidation–Reduction
• The rusting of iron is also an example of an
oxidation–reduction reaction.
• Iron metal reacts with oxygen in air to produce the
ionic compound iron(III) oxide, which is
composed of Fe3+ and O2- ions.
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
• Iron loses electrons and is oxidized.
Fe → Fe3+ + 3 e-
• Oxygen gains electrons and is reduced.
O2 + 4 e- → 2 O2© 2011 Pearson Education, Inc.
Chapter 17
3
Oxidation Numbers
•
The oxidation number describes how many
electrons have been lost or gained by an atom.
•
Oxidation numbers are assigned according to
seven rules:
1. A metal or a nonmetal in the free state has an oxidation
number of 0.
2. A monoatomic ion has an oxidation number equal to its
ionic charge.
3. A hydrogen atom is usually assigned an oxidation number of
+1.
© 2011 Pearson Education, Inc.
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4
Rules for Oxidation Numbers
4. An oxygen atom is usually assigned an oxidation number
of –2.
5. For a molecular compound, the more electronegative
element is assigned a negative oxidation number equal to
its charge as an anion.
6. For an ionic compound, the sum of the oxidation numbers
for each of the atoms in the compound is equal to 0.
7. For a polyatomic ion, the sum of the oxidation numbers for
each of the atoms in the compound is equal to the ionic
charge on the polyatomic ion.
© 2011 Pearson Education, Inc.
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5
Assigning Oxidation Numbers
• What is the oxidation number for magnesium
metal, Mg?
– Mg = 0 according to Rule #1.
• What is the oxidation number for sulfur in the
sulfide ion, S2-?
– S = –2 (Rule #2).
• What is the oxidation number for barium and
chloride in BaCl2?
– Ba is present as Ba2+, so Ba = +2 (Rules #2 and #6).
– Cl is present as Cl-, so Cl = –1 (Rules #2 and #6).
© 2011 Pearson Education, Inc.
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6
Oxidation Numbers in Compounds
• What are the oxidation numbers for each element
in oxalic acid, H2C2O4?
– H = +1 (Rule #3).
– O = 2 (Rule #4).
• To find the oxidation number for carbon, recall
that the sum of the oxidation numbers equals zero.
2(+1) + 2(ox no C) + 2(–2) = 0
2 + 2(ox no C) – 8 = 0
2(ox no C) = +6
C = +3
© 2011 Pearson Education, Inc.
Chapter 17
7
Oxidation Numbers in Compounds,
Continued
• What are the oxidation numbers for each element
in carbon tetrachloride, CCl4?
– Cl = –1 (Rule #5).
• To find the oxidation number for carbon, recall
that the sum of the oxidation numbers equals zero.
(ox no C) + 4(–1) = 0
(ox no C) – 4 = 0
(ox no C) = +4
C = +4
© 2011 Pearson Education, Inc.
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8
Oxidation Numbers in Polyatomic Ions
• What are the oxidation numbers for chlorine and
oxygen in the perchlorate ion, ClO4-?
– O = –2 (Rule #4).
• To find the oxidation number for carbon, recall
that the sum of the oxidation numbers equals the
charge on the ion (Rule #7).
(ox no Cl) + 4(–2) = –1
(ox no Cl) – 8 = –1
(ox no Cl) = +7
Cl = +7
© 2011 Pearson Education, Inc.
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9
Redox Reactions
• Recall that a chemical reaction involving the
transfer of electrons is an oxidation–reduction
reaction, or a redox reaction.
• For example, iron metal is heated with sulfur to
produce iron(II) sulfide.
Fe(s) + S(s) → FeS(s)
• The sulfur changes from 0 to –2 and the iron
changes from 0 to +2.
© 2011 Pearson Education, Inc.
Chapter 17
10
Oxidation and Reduction
• The iron loses electrons and is oxidized.
Fe → Fe2+ + 2 e-
• The sulfur gains electrons and is reduced.
S + 2 e- → S2-
© 2011 Pearson Education, Inc.
Chapter 17
11
Oxidizing and Reducing Agents
• Oxidation is the loss of electrons, and reduction
is the gain of electrons.
• An oxidizing agent is a substance that causes
oxidation by accepting electrons. The oxidizing
agent is reduced.
• A reducing agent
is a substance that
causes reduction by
donating electrons.
The reducing agent
is oxidized.
© 2011 Pearson Education, Inc.
Chapter 17
12
Redox Reactions
• In a redox reaction,
one substance must be
oxidized and one
substance must be
reduced.
• The total number of
electrons lost is equal
to the total electrons
gained.
© 2011 Pearson Education, Inc.
Chapter 17
13
Redox Reactions, Continued
• Identify the reducing agent, the oxidizing agent,
and the oxidation and reduction in the following
reaction:
CuS(s) + H2(g) → Cu(s) + H2S(g)
• Cu is reduced from +2 to 0.
• H is oxidized from 0 to +1.
© 2011 Pearson Education, Inc.
Chapter 17
14
Ionic Equations
• Redox reactions in aqueous solutions are most
often shown in the ionic form.
• Ionic equations readily show us the change in
oxidation number.
5 Fe2+(aq)+MnO4-(aq)+8 H+(aq)→5 Fe3+(aq)+Mn2+(aq)+4 H2O(l)
• We can easily tell that the oxidation number of
iron changes from +2 to +3; iron is oxidized.
• Manganese is reduced from +7 in MnO4- to +2 in
Mn2+; manganese is reduced.
© 2011 Pearson Education, Inc.
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Ionic Equations, Continued
• We can map the reaction to show the oxidation
and reduction processes and to determine the
oxidizing and reducing agents:
© 2011 Pearson Education, Inc.
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16
Balancing Redox Reactions
•
When we balance redox reactions, the number of
electrons lost must equal the number of electrons
gained.
•
We will balance redox reactions using the
oxidation number method, which has three
steps:
1. Inspect the reaction and the substances undergoing a
change in oxidation number.
a. Write the oxidation number above each element.
b. Diagram the number of electrons lost by the oxidized substance
and the number of electrons gained by the reduced substance.
© 2011 Pearson Education, Inc.
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Oxidation Number Method
2. Balance each element in the equation using a coefficient.
Remember that the electrons lost must equal the electrons
gained. If they are not the same, balance the electrons as
follows:
a. In front of the oxidized substance, place a coefficient equal to
the number of electrons gained by the reduced substance.
b. In front of the reduced substance, place a coefficient equal to
the number of electrons lost by the oxidized substance.
© 2011 Pearson Education, Inc.
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Oxidation Number Method, Continued
3. After balancing the equation, verify that the coefficients are
correct.
a. Place a checkmark above the symbol for each element to verify
that the number of atoms is the same on both sides.
b. For ionic equations, verify that the total charge on the left side
of the equation is the same as the total charge on the right side
of the equation.
© 2011 Pearson Education, Inc.
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Balancing a Redox Reaction
• Balance the following redox reaction using the
oxidation number method:
Fe2O3(l) + CO(g) → Fe(l) + CO2(g)
• Since the total electrons gained and lost must be
equal, we must find the lowest common multiple.
For this reaction, it is 6.
© 2011 Pearson Education, Inc.
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20
Balancing a Redox Reaction,
Continued
• Each iron gains three electrons, so place a 2 in
front of the Fe. There are two iron atoms in Fe2O3,
so no coefficient is necessary.
• Each carbon loses two electrons, so place a 3 in
front of CO and CO2.
√
√
√ √
√
√ √
Fe2O3(l) + 3 CO(g) → 2 Fe(l) + 3 CO2(g)
• Check to see that the number of each type of atom
is the same on both sides:
– There are 2 Fe atoms, 6 O atoms, and 3 C atoms on
each side.
© 2011 Pearson Education, Inc.
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21
Balancing Redox Equations
•
An alternative method for balancing redox
reactions is the half-reaction method.
•
A half-reaction shows the oxidation or reduction
process of a redox reaction separately.
•
The steps are as follows:
1. Write the half-reaction for both the oxidation and reduction
processes.
© 2011 Pearson Education, Inc.
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22
Half-Reaction Method
2. Balance the atoms in each half-reaction using coefficients.
a. Balance all elements except oxygen and hydrogen.
b. Balance oxygen using H2O.
c. Balance hydrogen using H+.
d. For reactions in basic solution, add one OH- to each side for each
H+ and combine H+ and OH- to H2O.
e. Balance the ionic charges using electrons.
© 2011 Pearson Education, Inc.
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Half-Reaction Method, Continued
3. Multiply each half-reaction by a whole number so that the
total number of electrons in each is the same.
4. Add the two half-reactions together and cancel the identical
species, including electrons.
5. After balancing, verify that the coefficients are correct by
making sure there are the same number of each atom on
each side of the reaction and that the overall charge is the
same on both sides.
© 2011 Pearson Education, Inc.
Chapter 17
24
Balancing a Redox Equation
• Balance the following redox reaction using the
half-reaction method:
Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
• The two unbalanced half-reactions are:
Fe2+ → Fe3+
MnO4- → Mn2+
• We balance the two half-reactions as follows:
Fe2+ → Fe3+ + e-
5 e- + 8 H+ + MnO4- → Mn2+ + 4 H2O
© 2011 Pearson Education, Inc.
Chapter 17
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Balancing a Redox Equation,
Continued
• Since Fe2+ loses one electron and MnO4- gains five
electrons, we have to multiply the iron halfreaction by 5:
5 Fe2+ → 5 Fe3+ + 5 e5 e- + 8 H+ + MnO4- → Mn2+ + 4 H2O
• We add the two half-reactions together and cancel
out the five electrons on each side to get the
following balanced equation:
5 Fe2+ + 8 H+ + MnO4- → Fe2+ + Mn2+ + 4 H2O
© 2011 Pearson Education, Inc.
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Spontaneous Redox Reactions
• Chemical reactions that occur without any input of
energy are spontaneous.
• The reaction of zinc metal with aqueous copper
sulfate is spontaneous:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
• Cu2+ has a greater tendency to gain electrons than
Zn2+.
• We can compare metals and arrange them in a
series based on their ability to gain electrons.
© 2011 Pearson Education, Inc.
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Reduction Potentials
• The tendency for a
substance to gain
electrons is its
reduction potential.
• The strongest
reducing agent is
the most easily
oxidized.
• At right is a table of
reduction potentials
for several metals.
© 2011 Pearson Education, Inc.
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Spontaneous Reactions
• A species can be
reduced by any
reducing agent
lower in the table.
• Any metal below H2
can react with acid
and be oxidized.
• A species can be
oxidized by any
oxidizing agent
above it on the table.
© 2011 Pearson Education, Inc.
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29
Predicting Spontaneous Reactions
• A reaction will be spontaneous when the stronger
oxidizing and reducing agents are the reactants,
and the weaker oxidizing and reducing agents are
the products.
• Predict whether the following reaction will be
spontaneous:
Ni2+(aq) + Sn(s) →
weaker
oxidizing agent
weaker
reducing agent
Ni(s) + Sn2+(aq)
stronger
reducing agent
stronger
oxidizing agent
• The reaction is nonspontaneous as written.
© 2011 Pearson Education, Inc.
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30
Voltaic Cells
• The conversion of chemical energy to electrical
energy in a redox reaction is electrochemistry.
• If we can physically separate the oxidation and
reduction half-reactions, we can use the electrons
from the redox reaction to do work. This is called
an electrochemical cell.
• Let’s look at the following reaction of zinc metal
with copper(II) sulfate:
Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)
© 2011 Pearson Education, Inc.
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31
Voltaic Cells, Continued
• We place a zinc electrode in aqueous ZnSO4 and a
copper electrode in aqueous CuSO4. The
electrodes are connected by a wire to allow the
flow of electrons.
• A salt bridge is used to complete the circuit.
• Zinc metal is
oxidized and copper
ions are reduced in
each half cell.
© 2011 Pearson Education, Inc.
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32
Voltaic Cells, Continued
• Oxidation occurs at the anode of an
electrochemical cell.
• Reduction occurs at the cathode of an
electrochemical cell.
• Electrons flow through the wire from the anode to
the cathode in a voltaic cell.
• Negatively charged ions travel through the salt
bridge away from the cathode and toward the
anode in a voltaic cell.
© 2011 Pearson Education, Inc.
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Batteries
• A battery is any electrochemical cell that
spontaneously produces electrical energy.
• A battery has stored chemical energy that can be
converted to electrical energy by a chemical
reaction.
• A battery is made up of one or
more voltaic cells.
• An automobile battery is made
up of six cells.
• A battery without an electrolyte
solution is a dry cell.
© 2011 Pearson Education, Inc.
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34
Electrolytic Cells
• Electrolytic cells are electrochemical cells that do
not operate spontaneously. The process is referred
to as electrolysis.
• A source of electricity is required to drive an
electrolytic cell.
• An example of an electrolysis reaction is the
recharging of the battery in a cell phone.
© 2011 Pearson Education, Inc.
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35
Chemistry Connection: Hybrid Vehicles
• Hybrid vehicles are designed to operate on both
gasoline and lightweight voltaic cells.
• Hybrids generate energy while braking or
coasting, and store it in batteries.
• The General Motors Volt is estimated to get better
than 100 miles per gallon.
© 2011 Pearson Education, Inc.
Chapter 17
36
Chapter Summary
• A redox reaction is a reaction involving the
transfer of electrons from one substance to
another.
• The oxidation number describes how many
electrons have been lost or gained by an atom.
• Oxidation is the loss of electrons.
• Reduction is the gain of electrons.
© 2011 Pearson Education, Inc.
Chapter 17
37
Chapter Summary, Continued
• An oxidizing agent is a substance that causes
oxidation by accepting electrons. The oxidizing
agent is reduced.
• A reducing agent is a substance that causes
reduction by donating electrons. The reducing
agent is oxidized.
• In redox reactions, the number of electrons lost
must equal the number of electrons gained.
© 2011 Pearson Education, Inc.
Chapter 17
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Chapter Summary, Continued
•
There are two methods to balance redox
reactions:
1. The oxidation number method
2. The half-reaction method
•
•
•
The tendency for a substance to gain electrons is
its reduction potential.
The conversion of chemical energy to electrical
energy in a redox reaction is electrochemistry.
We can physically separate oxidation and
reduction half-reactions and use the electrons
from the redox reaction to do work in an
electrochemical cell.
© 2011 Pearson Education, Inc.
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Chapter Summary, Continued
© 2011 Pearson Education, Inc.
Chapter 17
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