Transcript Chapter 18 Oxidation–Reduction Reactions and Electrochemistry Chapter 18 Table of Contents 18.1 18.2 18.3 18.4 18.5 18.6 18.7 18.8 Oxidation–Reduction Reactions Oxidation States Oxidation–Reduction Reactions Between Nonmetals Balancing Oxidation–Reduction Reactions by the Half-Reaction Method Electrochemistry: An Introduction Batteries Corrosion Electrolysis.

Chapter 18
Oxidation–Reduction
Reactions and
Electrochemistry
Chapter 18
Table of Contents
18.1
18.2
18.3
18.4
18.5
18.6
18.7
18.8
Oxidation–Reduction Reactions
Oxidation States
Oxidation–Reduction Reactions Between Nonmetals
Balancing Oxidation–Reduction Reactions by the
Half-Reaction Method
Electrochemistry: An Introduction
Batteries
Corrosion
Electrolysis
2
Section 18.1
Oxidation–Reduction Equations
•
Oxidation–reduction reaction (redox
reaction) – a chemical reaction
involving the transfer of electrons.
 Oxidation – loss of electrons
 Reduction – gain of electrons
http://www.youtube.com/watch?v=Ftw7a5ccubs
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Section 18.1
Oxidation–Reduction Equations
Exercise
In the reaction below Sn(II) _____________.
Sn2+ + 2Fe3+ → Sn4+ + 2Fe2+
a)
b)
c)
d)
gains electrons
is reduced
is oxidized
is neither oxidized nor reduced
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Section 18.2
Oxidation States
• Allow us to keep track of electrons in oxidation–
reduction reactions by assigning charges to the
various atoms in a compound.
Oxidation States for the Transition Metals
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5
Section 18.2
Oxidation States
Rules for Assigning Oxidation States
1. Oxidation state of an atom in an elemental state = 0
2. Oxidation state of monatomic ion = charge of the ion
3. Oxygen = 2 in covalent compounds (except in
peroxides where it = 1)
4. Hydrogen = +1 in covalent compounds
5. Fluorine = 1 in compounds
6. Sum of oxidation states = 0 in compounds
7. Sum of oxidation states = charge of the ion in ions
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Section 18.2
Oxidation States
Exercise
Find the oxidation states for each of the
elements in each of the following
compounds:
•
•
•
•
•
K2Cr2O7
CO32MnO2
PCl5
SF4
K = +1; Cr = +6; O = –2
C = +4; O = –2
Mn = +4; O = –2
P = +5; Cl = –1
S = +4; F = –1
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Section 18.2
Oxidation States
What are the Oxidation Numbers for each element in the
following?
H2O
N2
KMnO4
CO2
CH4
CHCl3
He
Cu
Na2Cr2O7
+1 for H, -2 for O
Zero for N, elemental state
+1 for K, -2 for O, +7 for Mn
-2 for O, +4 for C
+1 for H, -4 for C
+1 for H, -1 for Cl, +2 for C
Zero for He, elemental state
Zero for Cu, elemental state
+1 for Na, -2 for O, +6 for Cr
1(+1 K)=+1
4(-2 O)= -8
-7
1(+1 H)=+1
3(-1 Cl)= -3
-2
2(+1 Na)=+2
7(-2 O)= -14
-12
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Section 18.2
Oxidation States
More
Practice!
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9
Section 18.3
Oxidation–Reduction Reactions Between Nonmetals
• 2Na(s) + Cl2(g)  2NaCl(s)
• Na  oxidized
 Na is also called the reducing agent (electron
donor).
• Cl2  reduced
 Cl2 is also called the oxidizing agent
(electron acceptor).
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10
Section 18.3
Oxidation–Reduction Reactions Between Nonmetals
• CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
• C  oxidized
 CH4 is the reducing agent.
• O2  reduced
 O2 is the oxidizing agent.
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11
Section 18.3
Oxidation–Reduction Reactions Between Nonmetals
Redox Characteristics
•
•
•
•
Transfer of electrons
Transfer may occur to form ions
Oxidation – increase in oxidation state
(loss of electrons); reducing agent
Reduction – decrease in oxidation state
(gain of electrons); oxidizing agent
Oxidation
0
2+ 1Zn(s) + CuCl2(aq)
•
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2+ 10
ZnCl2(aq) + Cu(s)
Reduction
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Section 18.3
Oxidation–Reduction Reactions Between Nonmetals
Concept Check
Which of the following are oxidation–reduction
reactions? Identify the oxidizing agent and the
reducing agent.
a)Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
b)Cr2O72-(aq) + 2OH-(aq)  2CrO42-(aq) + H2O(l)
c)2CuCl(aq)  CuCl2(aq) + Cu(s)
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Section 18.4
Balancing Oxidation–Reduction Reactions by the Half-Reaction Method
Half–Reactions
•
•
The overall reaction is split into two half–reactions, one
involving oxidation and one reduction.
Has electrons as reactants or products
8H+ + MnO4– + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
Reduction: 8H+ + MnO4– + 5e– → Mn2+ + 4H2O
Oxidation: 5Fe2+ → 5Fe3+ + 5e–
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Section 18.4
Balancing Oxidation–Reduction Reactions by the Half-Reaction Method
The Half–Reaction Method for Balancing Equations for Oxidation–
Reduction Reactions Occurring in Acidic Solution
1. Identify and write the equations for the
oxidation and reduction half–reactions.
2. For each half–reaction:
A.
B.
C.
D.
Balance all the elements except H and O.
Balance O using H2O.
Balance H using H+.
Balance the charge using electrons.
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Section 18.4
Balancing Oxidation–Reduction Reactions by the Half-Reaction Method
The Half–Reaction Method for Balancing Equations for Oxidation–
Reduction Reactions Occurring in Acidic Solution
3. If necessary, multiply one or both
balanced half–reactions by an integer to
equalize the number of electrons
transferred in the two half–reactions.
4. Add the half–reactions, and cancel
identical species.
5. Check that the elements and charges are
balanced.
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Section 18.4
Balancing Oxidation–Reduction Reactions by the Half-Reaction Method
Cr2O72-(aq) + SO32-(aq)  Cr3+(aq) + SO42-(aq)
• How can we balance this equation?
• First Steps:
 Separate into half-reactions.
 Balance elements except H and O.
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Section 18.4
Balancing Oxidation–Reduction Reactions by the Half-Reaction Method
Method of Half Reactions
• Cr2O72-(aq)  2Cr3+(aq)
• SO32-(aq)  SO42-(aq)
• Balance O’s with H2O and H’s with H+
• 14H+(aq) + Cr2O72-(aq)  2Cr3+(aq) + 7H2O(aq)
• H2O(l) + SO32-(aq)  SO42-(aq) + 2H+(aq)
• How many electrons are involved in each half
reaction? Balance the charges.
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Section 18.4
Balancing Oxidation–Reduction Reactions by the Half-Reaction Method
Method of Half Reactions (continued)
6 e- + 14H+(aq) + Cr2O72-(aq)  2Cr3+(aq) + 7H2O(aq)
H2O(l) + SO32-(aq)  SO42-(aq) + 2H+(aq) + 2eMultiply whole reactions by a whole number to make
the number of electrons gained equal the number of
electrons lost.
6 e- + 14H+(aq) + Cr2O72-(aq)  2Cr3+(aq) + 7H2O(aq)
3(H2O(l) + SO32-(aq)  SO42-(aq) + 2H+(aq) + 2e-)
Combine half reactions cancelling out those reactants
and products that are the same on both sides,
especially the electrons.
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Section 18.4
Balancing Oxidation–Reduction Reactions by the Half-Reaction Method
Method of Half Reactions (continued)
8
4
6e- +
+ Cr2O7

+ 7H2O(aq)
3H2O(l) + 3SO32-(aq)  3SO42-(aq) + 6H+(aq) + 6e14H+(aq)
2-(aq)
2Cr3+(aq)
• Final Balanced Equation:
Cr2O72- + 3SO32- + 8H+  2Cr3+ + 3SO42- + 4H2O
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Section 18.4
Balancing Oxidation–Reduction Reactions by the Half-Reaction Method
Exercise
When the reaction Ce2+ + Co2+ → Ce3+ + Co
is balanced, the coefficient in front of Ce2+ is
a)
b)
c)
d)
0
1
2
3
Ce2+ → Ce3+ +1e2e- + Co2+ → Co
2Ce2+ + Co2+ → 2Ce3+ + Co
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Section 18.4
Balancing Oxidation–Reduction Reactions by the Half-Reaction Method
Exercise
Balance the following oxidation–reduction
reaction that occurs in acidic solution.
Br–(aq) + MnO4–(aq)  Br2(l)+ Mn2+(aq)
10Br–(aq) + 16H+(aq) + 2MnO4–(aq)  5Br2(l)+ 2Mn2+(aq) + 8H2O(l)
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Section 18.5
Electrochemistry: An Introduction
Electrochemistry
• The study of the interchange of chemical and
electrical energy.
• Two types of processes:
 Production of an electric current from a
chemical reaction.
 The use of electric current to produce a
chemical change.
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Section 18.5
Electrochemistry: An Introduction
Making an Electrochemical Cell
8H+ + MnO4– + 5e– → Mn2+ + 4H2O
Fe2+ → Fe3+ + e–
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Section 18.5
Electrochemistry: An Introduction
Making an Electrochemical Cell
• If electrons flow through
the wire charge builds up.
• Solutions must be
connected to permit ions
to flow to balance the
charge.
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Section 18.5
Electrochemistry: An Introduction
Making an Electrochemical Cell
• A salt bridge or porous disk connects the half
cells and allows ions to flow, completing the
circuit.
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Section 18.5
Electrochemistry: An Introduction
Electrochemical Battery (Galvanic Cell)
• Device powered by an oxidation–reduction reaction
where chemical energy is converted to electrical energy.
• Anode – electrode where oxidation occurs
• Cathode – electrode where reduction occurs
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Section 18.5
Electrochemistry: An Introduction
Electrolysis
• Process where electrical energy is used to
produce a chemical change.
 Nonspontaneous
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Section 18.6
Batteries
Lead Storage Battery
• Anode reaction – oxidation
Pb + H2SO4  PbSO4 + 2H+ + 2e
• Cathode reaction – reduction
PbO2 + H2SO4 + 2e + 2H+  PbSO4 + 2H2O
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29
Section 18.6
Batteries
Lead Storage Battery – Overall Reaction
Pb(s) + PbO2(s) + 2H2SO4(aq)  2PbSO4(s) + 2H2O(l)
Hydrometer to
measure
H2SO4
concentration.
As the battery discharges the sulfate of the acid
precipitates with the lead taking it out of solution and
reducing the acid concentration. As the battery is
recharged the current goes into dissolving the lead
sulfate restoring the acid concentration.
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Section 18.6
Batteries
Electric Potential
• The “pressure” on electrons to flow from anode
to cathode in a battery, like water flow.
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Section 18.6
Batteries
Dry Cell Batteries
• Do not contain a liquid
electrolyte.
• Acid version
• Anode reaction – oxidation
Zn  Zn2+ + 2e
•
Cathode reaction – reduction
2NH4+ + 2MnO2 + 2e  Mn2O3 + 2NH3 + 2H2O
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Section 18.6
Batteries
Dry Cell Batteries
• Do not contain a liquid electrolyte.
 Alkaline version
– Anode reaction – oxidation
Zn + 2OH  ZnO + H2O + 2e
–
Cathode reaction – reduction
2MnO2 + H2O + 2e  Mn2O3 + 2OH
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Section 18.6
Batteries
Dry Cell Batteries
• Do not contain a liquid electrolyte.

Other Types
•
•
Silver cell – Zn anode, Ag2O cathode
Mercury cell – Zn anode, HgO cathode
•
Nickel-cadmium – rechargeable
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Section 18.7
Corrosion
•
The oxidation of metals to form mainly oxides and
sulfides.
 Some metals, such as aluminum, protect themselves
with their oxide coating.
 Corrosion of iron can be
prevented by coatings, by
alloying and cathodic
protection.
 Cathodic protection of an
underground pipe.
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Section 18.8
Electrolysis
• Forcing a current through a cell to produce a
chemical change that would not otherwise
occur.
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