Transcript Chapter Electrochemistry Chemistry 4th Edition McMurry/Fay Dr. Paul Charlesworth Michigan Technological University Redox Reactions Prentice Hall ©2004 Chapter 18 Slide 2

Chapter
18
Electrochemistry
Chemistry 4th Edition
McMurry/Fay
Dr. Paul Charlesworth
Michigan Technological University
Redox Reactions
Prentice Hall ©2004
Chapter 18
01
Slide 2
Redox Reactions
01
Redox reaction are those involving the oxidation
and reduction of species.
OIL – Oxidation Is Loss of electrons.
RIG – Reduction Is Gain of electrons.
Oxidation and reduction must occur together.
They cannot exist alone.
Prentice Hall ©2004
Chapter 18
Slide 3
Redox Reactions
Prentice Hall ©2004
Chapter 18
02
Slide 4
Redox Reactions
02
Oxidation Half-Reaction: Zn(s)  Zn2+(aq) + 2 e–.
• The Zn loses two electrons to form Zn2+.
•
Prentice Hall ©2004
Chapter 18
Slide 5
Redox Reactions
03
Reduction Half-Reaction: Cu2+(aq) + 2 e–  Cu(s)
• The Cu2+ gains two electrons to form copper.
•
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Chapter 18
Slide 6
Redox Reactions
•
04
Overall: Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
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Chapter 18
Slide 7
Electrochemical Cells
01
•
Electrodes: are usually metal strips/wires
connected by an electrically conducting wire.
•
Salt Bridge: is a U-shaped tube that contains a gel
permeated with a solution of an inert electrolyte.
•
Anode: is the electrode where oxidation takes
place.
•
Cathode: is the electrode where reduction takes
place.
Prentice Hall ©2004
Chapter 18
Slide 8
Electrochemical Cells
•
02
Convention for expressing the cell:
Anode Half-Cell || Cathode Half-Cell
Electrode | Anode Soln || Cathode Soln | Electrode
Zn(s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu(s)
Pt(s) | H2 (1 atm) | H+ (1 M) || Fe3+(aq), Fe2+(aq) | Pt(s)
•
Electrons flow from anode to cathode. Anode is
placed on left by convention.
Prentice Hall ©2004
Chapter 18
Slide 9
Electrochemical Cells
•
03
The standard potential of any galvanic cell is the
sum of the standard half-cell potentials for the
oxidation and reduction half-cells.
E°cell = E°oxidation + E°reduction
•
Standard half-cell potentials are always quoted as
a reduction process. The sign must be changed for
the oxidation process.
Prentice Hall ©2004
Chapter 18
Slide 10
Electrochemical Cells
04
•
The standard half-cell potentials are determined
from the difference between two electrodes.
•
The reference point is called the standard hydrogen
electrode (S.H.E.) and consists of a platinum
electrode in contact with H2 gas (1 atm) and
aqueous H+ ions (1 M).
•
The standard hydrogen electrode is assigned an
arbitrary value of exactly 0.00 V.
Prentice Hall ©2004
Chapter 18
Slide 11
Electrochemical Cells
Prentice Hall ©2004
Chapter 18
05
Slide 12
Electrochemical Cells
Prentice Hall ©2004
Chapter 18
06
Slide 13
Electrochemical Cells
•
When selecting two half-cell reactions the more
negative value will form the oxidation half-cell.
•
Consider the reaction between zinc and silver:
Ag+(aq) + e–  Ag(s)
Zn2+(aq) + 2 e–  Zn(s)
•
07
E° = 0.80 V
E° = – 0.76 V
Therefore, zinc forms the oxidation half-cell:
Zn(s)  Zn2+(aq) + 2 e–
Prentice Hall ©2004
Chapter 18
E° = – (–0.76 V)
Slide 14
Electrochemical Cells
08
•
What is the standard emf of an electrochemical cell
made of a Cd electrode in a 1.0 M Cd(NO3)2
solution and a Cr electrode in a 1.0 M Cr(NO3)2
solution?
•
What is the standard emf of an electrochemical cell
made of a Mg electrode in a 1.0 M Mg(NO3)2
solution and a Ag electrode in a 1.0 M AgNO3
solution?
Prentice Hall ©2004
Chapter 18
Slide 15
Spontaneity of a Reaction
01
•
The value of E˚cell is related to the thermodynamic
quantities of ∆G˚ and K.
•
The value of E˚cell is related to ∆G˚ by:
∆G˚ = –nFE˚cell
•
The value of K is related to ∆G˚ by:
∆G˚ = –RT ln K
Prentice Hall ©2004
Chapter 18
Slide 16
Spontaneity of a Reaction
•
02
These are combined:
–nFE˚cell = –RT ln K
•
and rearranged at 25°C to give:
or
•
E˚cell = (0.0257n) ln K
E˚cell = (0.0592n) log K
(Units = volts, n = moles, F = 96,500 coulombs)
Prentice Hall ©2004
Chapter 18
Slide 17
Spontaneity of a Reaction
•
03
Calculate the standard free energy change (∆G˚)
and the equilibrium constant (K) for the following
reactions at 25°C:
1. Sn(s) + 2 Cu2+(aq) æ Sn2+(aq) + 2 Cu+(aq)
2. Fe2+(aq) + 2 Ag(s) æ Fe(s) + 2 Ag+(aq)
3. 4 Fe2+(aq) + O2(g) + 4 H+(aq)  4 Fe3+(aq) + 2 H2O(l)
Prentice Hall ©2004
Chapter 18
Slide 18
The Nernst Equation
•
01
Cell potentials can be modified by temperature and
composition changes according to the equation:
∆G = ∆G° + RT lnQ

•
Giving:
Prentice Hall ©2004
–nFE = –nFE° + RT lnQ
RT
E E 
lnQ
nF
o
Chapter 18
Slide 19
The Nernst Equation
03
•
Consider the reaction of metallic zinc with
hydrochloric acid. Calculate the cell potential at
25°C when [H+] = 1.0 M, [Zn2+] = 0.0010 M, and
PH2 = 0.10 atm.
•
Consider the reaction of metallic copper with
iron(III) to give copper(II) and iron(II). What is the
potential of a cell when [Fe3+] = 0.0001 M, [Cu2+] =
0.25 M, and [Fe2+] = 0.20 M?
Prentice Hall ©2004
Chapter 18
Slide 20
The Nernst Equation
•
04
A particularly important use of the Nernst equation
is in the electrochemical determination of pH.
Pt | H2 (1 atm) | H+ (? M) || Reference Cathode
Ecell = EH2  H+ + Eref
•
The Nernst equation can be applied to the
half-reaction:
H2(g)  2 H+(aq) + 2 e–
Prentice Hall ©2004
Chapter 18
Slide 21
The Nernst Equation
•
For the half-reaction: H2(g)  2 H+(aq) + 2 e–

E H 2  H+
•
05
=
o
E H2  H +
2
+
H
0.0592
V
–
log
n
PH
2
E° = 0 V for this reaction (standard hydrogen
electrode). PH2 is 1 atm.

EH 2  H+
Prentice Hall ©2004
V log H + 2
= 0 V – 0.0592
n
Chapter 18
Slide 22
The Nernst Equation
•
07
The overall potential is given by:
Ecell  0.0592 V pH  Eref
•
Which rearranges to give an equation for the
determination of pH:
Ecell  Eref
 pH
0.0592 V
Prentice Hall ©2004
Chapter 18
Slide 23
The Nernst Equation
08
•
A higher cell potential indicates a higher pH,
therefore we can measure pH by measuring Ecell.
•
A glass electrode (Ag/AgCl wire in dilute HCl) with a
calomel reference is the most common
arrangement.
Glass: Ag(s) + Cl–(aq)  AgCl(s) + e–
E° = –0.22 V
Calomel: Hg2Cl2(s) + 2 e–  2 Hg(l) + 2 Cl–(aq)
E° = 0.28 V
Prentice Hall ©2004
Chapter 18
Slide 24
The Nernst Equation
09
• The glass pH probe is constructed as follows:
Ag(s) | AgCl(s) | HCl(aq) | glass | H+(aq) || reference
•The difference in [H+] from one side of the glass
membrane to the other causes a potential to develop,
which adds to the measured Ecell.
Ecell  Eref
 pH
0.0592 V
Prentice Hall ©2004
Chapter 18
Slide 25
The Nernst Equation
•
10
The following cell has a potential of 0.55 V at 25°C:
Pt(s) | H2 (1 atm) | H+ (? M) || Cl– (1 M) | Hg2Cl2(s) | Hg(l)
•
What is the pH of the solution at the anode?
•
The following cell has a potential of 0.28 V at 25°C:
Pt(s) | H2 (1 atm) | H+ (? M) || Pb2+ (1 M) | Pb(s)
•
What is the pH of the solution at the anode?
Prentice Hall ©2004
Chapter 18
Slide 26
Batteries
01
•
Batteries are the most important practical
application of galvanic cells.
•
Single-cell batteries consist of one galvanic cell.
•
Multicell batteries consist of several galvanic cells
linked in series to obtain the desired voltage.
Prentice Hall ©2004
Chapter 18
Slide 27
Batteries
•
02
Lead Storage Battery: A typical 12 volt battery
consists of six individual cells connected in series.
•
Anode: Lead grid packed with spongy lead.
Pb(s) + HSO4–(aq)  PbSO4(s) + H+(aq) + 2 e–
Cathode: Lead grid packed with lead oxide.
PbO2(s) + 3 H+(aq) + HSO4–(aq) + 2 e–  PbSO4(s) + 2 H2O(l)
Electrolyte: 38% by mass sulfuric acid.
Cell Potential: 1.924 V
Prentice Hall ©2004
Chapter 18
Slide 28
Batteries
•
03
Zinc Dry-Cell: Also called a Leclanché cell, uses a
viscous paste rather than a liquid solution.
Anode: Zinc metal can on outside of cell.
Zn(s)  Zn2+(aq) + 2 e–
•
Cathode: MnO2 and carbon black paste on graphite.
•2
MnO2(s) + 2 NH4+(aq) + 2 e–  Mn2O3(s) + 2 NH3(aq) +
2 H2O(l)
•
Electrolyte: NH4Cl and ZnCl2 paste.
•
Cell Potential: 1.5 V but deteriorates to 0.8 V with use.
Prentice Hall ©2004
Chapter 18
Slide 29
Batteries
•
04
Alkaline Dry-Cell: Modified Leclanché cell which
replaces NH4Cl with NaOH or KOH.
Anode: Zinc metal can on outside of cell.
Zn(s) + 2 OH–(aq)  ZnO(s) + H2O(l) + 2 e–
Cathode: MnO2 and carbon black paste on graphite.
2 MnO2(s) + H2O(l) + 2 e–  Mn2O3(s) + 2 OH–(aq)
Electrolyte: NaOH or KOH, and Zn(OH)2 paste.
Cell Potential: 1.5 V but longer lasting, higher power,
and more stable current and voltage.
Prentice Hall ©2004
Chapter 18
Slide 30
Batteries
•
05
Mercury Dry-Cell: Modified Leclanché cell which
replaces MnO2 with HgO and uses a steel cathode.
Anode: Zinc metal can on outside of cell.
Zn(s) + 2 OH–(aq)  ZnO(s) + H2O(l) + 2 e–
Cathode: HgO in contact with steel.
2 HgO (s) + H2O(l) + 2 e–  Hg(l) + 2 OH–(aq)
Electrolyte: KOH, and Zn(OH)2 paste.
Cell Potential: 1.3 V with small size, longer lasting,
and stable current and voltage.
Prentice Hall ©2004
Chapter 18
Slide 31
Batteries
•
06
Nickel–Cadmium Battery: Modified Leclanché cell
which is rechargeable.
Anode: Cadmium metal.
Cd(s) + 2 OH–(aq)  Cd(OH)2(s) + 2 e–
Cathode: Nickel(III) compound on nickel metal.
NiO(OH) (s) + H2O(l) + e–  Ni(OH)2(s) + OH–(aq)
Electrolyte: Nickel oxyhydroxide, NiO(OH).
Cell Potential: 1.30 V
Prentice Hall ©2004
Chapter 18
Slide 32
Batteries
•
•
07
Nickel–Metal–Hydride
(NiMH):
Replaces toxic Cd anode with
a hydrogen atom impregnated
ZrNi2 metal alloy.
•
During oxidation at the anode,
hydrogen atoms are released
as H2O.
•
Recharging reverses this
reaction.
Prentice Hall ©2004
Chapter 18
Slide 33
Batteries
•
08
Lithium Ion (Li–ion): The newest rechargeable
battery is based on the migration of Li+ ions.
•Anode:
Li metal, or Li atom impregnated graphite.
Li(s)  Li+ + e–
Cathode: Metal oxide or sulfide that can accept Li+.
MnO2(s) + Li+(aq) + e–  LiMnO2(s)
Electrolyte: Lithium-containing salt such as LiClO4, in
organic solvent. Solid state polymers can also be used.
Cell Potential: 3.0 V
Prentice Hall ©2004
Chapter 18
Slide 34
Batteries
•
09
Fuel Cell: Uses externally fed CH4 or H2, which
react to form water. Most common is H2.
•
Anode: Porous carbon containing metallic catalysts.
2 H2(s) + 4 OH–(aq)  4 H2O(l) + 4 e–
Cathode: Porous carbon containing metallic catalysts.
O2(s) + 2 H2O(l) + 4 e–  4 OH–(aq)
Electrolyte: Hot aqueous KOH solution.
Cell Potential: 1.23 V, but is only 40% of cell capacity.
Prentice Hall ©2004
Chapter 18
Slide 35
Batteries
10
•
Fuel cells are not batteries
because they are not selfcontained.
•
Fuel cells typically have
about 40% conversion to
electricity; the remainder is
lost as heat.
•
Excess heat can be used to
drive turbine generators.
Prentice Hall ©2004
Chapter 18
Slide 36
Corrosion
01
•
Corrosion is the oxidative deterioration of metal.
•
25% of steel produced in USA goes to replace steel
structures and products destroyed by corrosion.
•
Rusting of iron requires the presence of BOTH
oxygen and water.
•
Rusting results from tiny galvanic cells formed by
water droplets.
Prentice Hall ©2004
Chapter 18
Slide 37
Corrosion
02
Oxidation: Fe(s)  Fe2+(aq) + 2 e–
Reduction: O2(g) + 4 H+(aq) + 4 e–  2 H2O(l)
Overall:
Prentice Hall ©2004
2 Fe(s) + O2(g) + 4 H+(aq)  2 Fe2+(aq) + 2 H2O(l)
Chapter 18
Slide 38
Corrosion
03
•
Galvanizing: is the coating of iron with zinc. Zinc is
more easily oxidized than iron, which protects and
reverses oxidation of the iron.
•
Cathodic Protection: is the protection of a metal
from corrosion by connecting it to a metal (a
sacrificial anode) that is more easily oxidized.
•
All that is required is an electrical connection to the
sacrificial anode (usually magnesium or zinc).
Prentice Hall ©2004
Chapter 18
Slide 39
Corrosion
•
04
Cathodic Protection with Zinc Layer
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Chapter 18
Slide 40
Corrosion
•
05
Cathodic Protection with Magnesium Anode
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Chapter 18
Slide 41
Electrolysis
Prentice Hall ©2004
01
Chapter 18
Slide 42
Electrolysis
01
•
Electrolysis: is the process in which electrical
energy is used to drive a nonspontaneous chemical
reaction.
•
An electrolytic cell is an apparatus for carrying
out electrolysis.
•
Processes in an electrolytic cell are the reverse of
those in a galvanic cell.
Prentice Hall ©2004
Chapter 18
Slide 43
Electrolysis
05
•
Electrolysis of Water: Requires an electrolyte
species, that is less easily oxidized and reduced
than water, to carry the current.
•
Anode: Water is oxidized to oxygen gas.
2 H2O(l)  O2(g) + 4 H+(aq) + 4 e–
•
Cathode: Water is reduced to hydrogen gas.
4 H2O(l) + 4 e–  2 H2(g) + 4 OH–(aq)
Prentice Hall ©2004
Chapter 18
Slide 44
Electrolysis
02
•Electrolysis
ElectrolysisofofAqueous
Molten Sodium
SodiumChloride:
Chloride: uses
different processes to molten sodium chloride.
Based on cell potentials, water (–0.83 V) would be
preferentially reduced over sodium ions (–2.71 V).
Based on cell potentials, water (+1.23 V) should be
preferentially oxidized over chloride ions (+1.36 V).
The observed product at the anode is Cl2, not O2,
because of a phenomenon called overvoltage.
Prentice Hall ©2004
Chapter 18
Slide 45
Electrolysis
04
•
Overvoltage: Additional voltage needed to
maintain the rate of electron transfer at the
electrode–solution interface.
•
Overvoltage is required when a half-reaction has a
significant activation energy, and so a slow rate.
•
Overvoltage for formation of O2 or H2 is much
greater than for formation of Cl2.
Prentice Hall ©2004
Chapter 18
Slide 46
Electrolysis
07
•
Quantitative Electrolysis: The amount of
substance produced at an electrode by
electrolysis depends on the quantity of charge
passed through the cell.
•
Reduction of 1 mol of sodium ions requires 1 mol
of electrons to pass through the system.
•
The charge on 1 mol of electrons is 96,500
coulombs.
Prentice Hall ©2004
Chapter 18
Slide 47
Electrolysis
•
08
To determine the moles of electrons passed, we
measure the current and time that the current
flows:
Charge (C) = Current (A) x Time (s)
•
Because the charge on 1 mol of e– is 96,500 C, the
– passed through the cell is:
number
of
moles
of
e

–
1
mole
e
–
Moles of e = Charge (C) 
96,500 C
Prentice Hall ©2004
Chapter 18
Slide 48
Electrolysis
09
•
A constant current of 30.0 A is passed through an
aqueous solution of NaCl for a time of 1.00 hour.
How many grams of Na and how many liters of Cl2
gas at STP are produced?
•
How many amperes must be passed through a
Downs cell to produce sodium metal at a rate of
30.0 kg/hour?
Prentice Hall ©2004
Chapter 18
Slide 49
Electrolysis Applications
•
01
Manufacture of Sodium (Downs Cell):
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Chapter 18
Slide 50
Electrolysis Applications
•
02
Manufacture of Cl2 and NaOH (Chlor–Alkali):
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Chapter 18
Slide 51
Electrolysis Applications
•
03
Manufacture of Aluminum (Hall–Heroult):
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Chapter 18
Slide 52
Electrolysis Applications
•
04
Electrorefining and Electroplating:
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Chapter 18
Slide 53