Download Chapter 20 Electrochemistry

Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 20
Electrochemistry
John D. Bookstaver
St. Charles Community College
Cottleville, MO
Electrochemistry
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Electrochemical Reactions
In electrochemical reactions, electrons
are transferred from one species to
another.
Electrochemistry
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Oxidation Numbers
In order to keep
track of what loses
electrons and what
gains them, we
assign oxidation
numbers.
Electrochemistry
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Oxidation and Reduction
• A species is oxidized when it loses electrons.
– Here, zinc loses two electrons to go from neutral
zinc metal to the Zn2+ ion.
Electrochemistry
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Oxidation and Reduction
• A species is reduced when it gains electrons.
– Here, each of the H+ gains an electron, and they
combine to form H2.
Electrochemistry
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Oxidation and Reduction
• What is reduced is the oxidizing agent.
– H+ oxidizes Zn by taking electrons from it.
• What is oxidized is the reducing agent.
– Zn reduces H+ by giving it electrons.
Electrochemistry
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Oxidizing and Reducing Agents
• Examples of good oxidizing agents:
H2O2, MnO4-, Cr2O72-, Ce4+, O3, halogens
• Examples of good reducing agents:
alkali and alkaline earth metals
Electrochemistry
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Assigning Oxidation Numbers
1. Elements in their elemental form have
an oxidation number of 0.
2. The oxidation number of a monatomic
ion is the same as its charge.
Electrochemistry
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Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
– Oxygen has an oxidation number of −2,
except in the peroxide ion, which has an
oxidation number of −1.
– Hydrogen is −1 when bonded to a metal
and +1 when bonded to a nonmetal.
Electrochemistry
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Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
– Fluorine always has an oxidation number
of −1.
– The other halogens have an oxidation
number of −1 when they are negative;
they can have positive oxidation
numbers, however, most notably in
oxyanions.
Electrochemistry
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Assigning Oxidation Numbers
4. The sum of the oxidation numbers in a
neutral compound is 0.
5. The sum of the oxidation numbers in a
polyatomic ion is the charge on the
ion.
Electrochemistry
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Sample Exercise 20.1 (p. 845)
The nickel-cadmium (nicad) battery, a
rechargeable “dry cell” used in batteryoperated devices, uses the following redox
reaction to generate electricity:
Cd(s) + NiO2(s) + 2 H2O(l)  Cd(OH)2(s) + Ni(OH)2(s)
Identify the substances that are oxidized and
reduced, and indicate which is the oxidizing
agent and which is the reducing agent.
Electrochemistry
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Practice Exercise 20.1
Identify the oxidizing and reducing
agents in the following oxidationreduction equation:
2 H2O(l) + Al(s) + MnO4-(aq)  Al(OH)4-(aq) + MnO2(s)
Electrochemistry
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Balancing Oxidation-Reduction
Equations
Perhaps the easiest way to balance the
equation of an oxidation-reduction
reaction is via the half-reaction method.
Electrochemistry
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Balancing Oxidation-Reduction
Equations
This involves treating (on paper only) the
oxidation and reduction as two separate
processes, balancing these half reactions,
and then combining them to attain the
balanced equation for the overall reaction.
Electrochemistry
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Practice – Simple half-reactions
Write half-reactions for a reaction between
Cu2+ and Zn metal.
Overall:
Zn + Cu2+  Zn2+ + Cu
Oxidation:
Zn  Zn2+ + 2 e-’s
Reduction:
Cu2+ + 2 e-’s  Cu
Electrochemistry
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The Half-Reaction Method
(acidic solution)
1. Write the overall unbalanced reaction.
2. (i) Identify oxidized and reduced
substances.
(ii) Write the two incomplete halfreactions.
Electrochemistry
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The Half-Reaction Method
3. Balance each half-reaction.
a. Balance elements other than H and O.
b. Balance O by adding H2O to side
deficient in O atoms.
c. Balance H by adding H+.
d. Balance charge by adding electrons.
4. Multiply the half-reactions by integers
so that the electrons gained and lost
are the same.
Electrochemistry
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The Half-Reaction Method
5. Add the half-reactions, subtracting
things that appear on both sides.
Make sure the equation is balanced
according to mass.
Make sure the equation is balanced
according to charge.
Electrochemistry
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The Half-Reaction Method
Consider the reaction between MnO4− and C2O42− :
1. MnO4− (aq) + C2O42− (aq)  Mn2+ (aq) + CO2 (aq)
Electrochemistry
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The Half-Reaction Method
Secondly, we assign oxidation numbers.
+7
+3
+2
+4
MnO4− + C2O42-  Mn2+ + CO2
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.
Electrochemistry
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Oxidation Half-Reaction
2. (ii) Write the half-reactions.
In this case, begin with oxidation half-reaction:
C2O42−  CO2
To balance the carbon, we add a coefficient
of 2:
C2O42−  2 CO2
Electrochemistry
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Oxidation Half-Reaction
C2O42−  2 CO2
3. The oxygen is now balanced as well.
To balance the charge, we must add 2
electrons to the right side.
C2O42−  2 CO2 + 2 e−
Electrochemistry
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Reduction Half-Reaction
2. (ii) and 3. Reduction half-reaction
MnO4−  Mn2+
The manganese is balanced; to balance the
oxygen, we must add 4 waters to the right
side.
MnO4−  Mn2+ + 4 H2O
Electrochemistry
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Reduction Half-Reaction
MnO4−  Mn2+ + 4 H2O
To balance the hydrogen, we add 8 H+
to the left side.
8 H+ + MnO4−  Mn2+ + 4 H2O
Electrochemistry
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Reduction Half-Reaction
8 H+ + MnO4−  Mn2+ + 4 H2O
To balance the charge, we add 5 e− to
the left side.
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O
Electrochemistry
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3. Combining the Half-Reactions
Now we evaluate the two half-reactions
together:
C2O42−  2 CO2 + 2 e−
5 e−+ 8 H+ + MnO4−  Mn2+ + 4 H2O
To attain the same number of electrons
on each side, we will multiply the first
reaction by 5 and the second by 2. Electrochemistry
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Combining the Half-Reactions
5 C2O42−  10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4−  2 Mn2+ + 8 H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
Electrochemistry
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Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The only thing that appears on both sides are the
electrons. Subtracting them, we are left with:
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2
Electrochemistry
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Sample Exercise 20.2 (p. 849)
Complete and balance the following
equation by the method of halfreactions:
Cr2O72-(aq) + Cl-(aq)  Cr3+(aq) + Cl2(g)
(acidic solution)
Electrochemistry
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Practice Exercise 20.2
Complete and balance the following
oxidation-reduction equations using
the method of half-reactions. Both
reactions occur in acidic solution.
a) Cu(s) + NO3-(aq)  Cu2+(aq) + NO2(g)
b) Mn2+(aq) + NaBiO3(s)  Bi3+(aq) + MnO4-(aq)
Electrochemistry
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Balancing in Basic Solution
• If a reaction occurs in basic solution, one
can balance it as if it occurred in acid.
• Once the equation is balanced, add OH−
to each side to “neutralize” the H+ in the
equation and create water in its place.
• If this produces water on both sides, you
might have to subtract water from each
side.
Electrochemistry
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Sample Exercise 20.3 (p. 850)
Complete and balance this equation for
a redox reaction that takes place in
basic solution:
CN-(aq) + MnO4-(aq)  CNO-(aq) + MnO2(s)
(basic solution)
Electrochemistry
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Practice Exercise 20.3
Complete and balance the following
equations for oxidation-reduction
reactions that occur in basic solution:
a) NO2-(aq) + Al(s)  NH3(aq) + Al(OH)4-(aq)
b) Cr(OH)3(s)+ ClO-(aq) CrO42-(aq) + Cl2(g)
Electrochemistry
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Voltaic Cells
In spontaneous
oxidation-reduction
(redox) reactions,
electrons are
transferred and
energy is released.
Electrochemistry
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Voltaic Cells
• We can use that
energy to do work if
we make the
electrons flow
through an external
device.
• We call such a
setup a voltaic cell.
Electrochemistry
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Voltaic Cells
• A typical cell looks
like this.
• The oxidation
occurs at the anode.
• The reduction
occurs at the
cathode.
Electrochemistry
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Voltaic Cells
Once even one
electron flows from
the anode to the
cathode, the
charges in each
beaker would not be
balanced and the
flow of electrons
would stop.
Electrochemistry
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Voltaic Cells
• Therefore, we use a
salt bridge, usually a
U-shaped tube that
contains a salt
solution, to keep the
charges balanced.
– Cations move toward
the cathode.
– Anions move toward
the anode.
Electrochemistry
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Sample Exercise 20.4 (p. 853)
The following oxidation-reduction reaction is spontaneous:
Cr2O72-(aq) + 14 H+(aq) + 6 I-(aq)  2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
A solution containing K2Cr2O7 and H2SO4 is poured into one
beaker, and a solution of KI is poured into another. A salt bridge
is used to join the beakers. A metallic conductor that will not
react with either solution (such as platinum foil) is suspended in
each solution, and the two conductors are connected with wires
through a voltmeter or some other device to detect an electric
current. The resultant voltaic cell generates an electric current.
Indicate the reaction occurring at the anode, the reaction at
the cathode, the direction of electron and ion migrations,
and the signs of the electrodes.
Electrochemistry
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Practice Exercise 20.4
The two half-reactions in a voltaic cell are
Zn(s)  Zn2+(aq) + 2 eClO3-(aq) + 6 H+(aq)+ 6 e-  Cl-(aq) + 3 H2O(l)
a) Indicate which reaction occurs at the anode
and which at the cathode.
b) Which electrode is consumed in the cell
reaction?
Electrochemistry
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Electrochemistry
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Voltaic Cells
• In the cell, then,
electrons leave the
anode and flow
through the wire to
the cathode.
• As the electrons
leave the anode,
the cations formed
dissolve into the
solution in the
anode
compartment.
Electrochemistry
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Voltaic Cells
• As the electrons
reach the cathode,
cations in the
cathode are
attracted to the now
negative cathode.
• The electrons are
taken by the cation,
and the neutral
metal is deposited
on the cathode.
Electrochemistry
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Electromotive Force (emf)
• Water only
spontaneously flows
one way in a
waterfall.
• Likewise, electrons
only spontaneously
flow one way in a
redox reaction—from
higher to lower
potential energy.
Electrochemistry
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Electromotive Force (emf)
• The potential difference between the
anode and cathode in a cell is called the
electromotive force (emf).
• It is also called the cell potential and is
designated Ecell.
Electrochemistry
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Cell Potential
Cell potential = difference in electrical potential.
Cell potential is measured in volts (V).
J
1V=1
C
One volt = the potential difference required to impart
one joule of energy to a charge of one coulomb.
One Coulomb = 6.25 x 1018 units of charge
1 unit = 1 e- or 1 p+
(derived from the ampere)
Electrochemistry
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Standard Reduction Potentials
Reduction
potentials for
many
electrodes
have been
measured and
tabulated.
Electrochemistry
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Standard Hydrogen Electrode
• Their values are referenced to a standard
hydrogen electrode (SHE).
• By definition, the reduction potential for
hydrogen is 0 V:
2 H+ (aq, 1M) + 2 e−  H2 (g, 1 atm)
Electrochemistry
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Standard Cell Potentials
The cell potential at standard conditions
can be found through this equation:
 = Ered
 (cathode) − Ered
 (anode)
Ecell
Because cell potential is based on
the potential energy per unit of
charge, it is an intensive property.
Electrochemistry
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Cell Potentials
• For the oxidation in this cell,
 = −0.76 V
Ered
• For the reduction,
 = +0.34 V
Ered
Electrochemistry
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Cell Potentials
 = Ered
 (cathode) − Ered
 (anode)
Ecell
= +0.34 V − (−0.76 V)
= +1.10 V
Electrochemistry
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Sample Exercise 20.5 (p. 858)
For the Zn-Cu2+ voltaic cell shown in Figure
20.5, we have
Zn(s) + Cu2+(aq, 1M)  Zn2+(aq, 1M) + Cu(s)
Eocell = 1.10 V
Given that the standard reduction potential of
Zn2+ to Zn is -0.76 V, calculate the Eored for
the reduction of Cu2+ to Cu.
Cu2+(aq, 1M) + 2 e-  Cu(s)
(0.34 V)
Electrochemistry
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Practice Exercise 20.5
A voltaic cell is based on the following halfreactions:
In+(aq)  In3+(aq) + 2 eBr2(l) + 2 e-  2 Br-(aq)
The standard emf for this cell is 1.46 V.
Using the data in Table 20.1 or your AP
Chem packet, calculate Eored for the reduction
of In3+ to In+.
(-0.40 V)
Electrochemistry
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Sample Exercise 20.6 (p. 858)
• Using the standard reduction potentials
listed in Table 20.1 or your AP Chem
packet, calculate the standard emf for
the voltaic cell described in Sample
Exercise 20.4, which is based on the
following reaction:
Cr2O72-(aq) + 14 H+(aq) + 6 I-(aq)
 2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
• (0.79 V)
Electrochemistry
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Practice Exercise 20.6
Using the data in Table 20.1 or your AP
Chem packet, calculate the standard
emf for a cell that employs the following
overall cell reaction:
2 Al(s) + 3 I2(s)  2 Al3+(aq) + 6 I-(aq)
(+2.20 V)
Electrochemistry
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Sample Exercise 20.7 (p. 859)
A voltaic cell is based on the following two standard
half-reactions:
Cd2+(aq) + 2 e-  Cd(s)
Sn2+(aq) + 2 e-  Sn(s)
By using the data in Appendix E or your AP Chem
packet, determine
a) the half-reactions that occur at the
cathode and the anode, and
b) the standard cell potential (0.267 V)
Electrochemistry
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Practice Exercise 20.7
A voltaic cell is based on a Co2+/Co
half-cell and an AgCl/Ag half-cell.
a) What reaction occurs at the anode?
b) What is the standard cell potential?
(1.08 V) (book is wrong)
Electrochemistry
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Oxidizing and Reducing Agents
• The strongest
oxidizers have the
most positive
reduction potentials.
• The strongest
reducers have the
most negative
reduction potentials.
Electrochemistry
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Oxidizing and Reducing Agents
The greater the
difference between
the two, the greater
the voltage of the
cell.
Electrochemistry
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Sample Exercise 20.8 (p. 861)
Using Table 20.1 or your AP Chem
packet, rank the following ions in order
of increasing strength as oxidizing
agents: NO3-(aq), Ag+(aq), Cr2O72-(aq).
Electrochemistry
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Practice Exercise 20.8
Using Table 20.1 or your AP Chem
packet, rank the following species from
the strongest to the weakest reducing
agent: I-(aq), Fe(s), Al(s).
Electrochemistry
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Sample Exercise 20.9 (p. 862)
Using standard reduction potentials
(Table 20.1 or AP Chem packet),
determine whether the following
reactions are spontaneous under
standard conditions:
a) Cu(s) + 2 H+(aq)  Cu2+(aq) + H2(g)
b) Cl2(g) + 2 I-(aq)  2 Cl-(aq) + I2(s)
Electrochemistry
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Practice Exercise 20.9
Using standard reduction potentials
(Appendix E or AP Chem packet), determine
whether the following reactions are
spontaneous under standard conditions:
a) I2(s) + 5 Cu2+(aq) + 6 H2O(l)
 2 IO3-(aq) + 5 Cu(s) + H+(aq)
b) Hg2+(aq) + 2 I-(aq)  Hg(l) + I2(s)
c) H2SO3(aq) + 2 Mn(s) + 4 H+(aq)
 S(s) + 2 Mn2+(aq) + 3 H2O(l)
Electrochemistry
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Where we’ve been
• Electrochemistry –
cell potentials in voltaic
cells under standard
conditions
• Thermodynamics –
Gibbs Free Energy
• Equilibrium combined
with Gibbs Free Energy
Where we’re going
• Apply Gibbs Free Energy
to voltaic cells
• Determine cell potential
of voltaic cells under nonstandard conditions
• Electrolytic cells
• Batteries and Corrosion
(on your own)
• Electrical work
Electrochemistry
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Gibbs Free Energy (DG) remember
• Predicts whether a particular reaction with occur,
involves both enthalpy and entropy
• If DG is negative, the reaction is spontaneous in the
forward direction.
• If DG = 0, the reaction is at equilibrium.
• If DG is positive, the reaction in the forward direction
is nonspontaneous.
• Appendix C - Tables of Thermodynamics Quantities
under standard conditions – 1 M, 1 atm, 25oC.
Electrochemistry
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Free Energy, Equilibrium and
Electrochemistry
1. Combine DGo and K or Q to find DG or K
under nonstandard conditions.
2. Determine DGo of a voltaic cell using E, n
and F. (electrochem)
3. Determine E under non-standard conditions
using the Nernst equation, which includes K
or Q. (electrochem)
Electrochemistry
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Free Energy and Equilibrium
Reminders:
DGo and K apply to standard conditions
(pure solid or liquid, gases at 1 atm and 1M solutions).
DG and Q (equilibrium quotient) apply to any
conditions.
Electrochemistry
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Free Energy and Equilibrium reminder
Under any conditions, standard or
nonstandard, the free energy change can be
found this way:
DG = DG + RT lnQ
(Under standard conditions, all concentrations are 1 M,
so Q = 1 and lnQ = 0; the last term drops out.)
R = 8.314 J/mol.K, T is in K
Electrochemistry
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Free Energy and Equilibrium reminder
• At equilibrium, Q = K, and DG = 0.
• The equation becomes
0 = DG + RT lnK
• Rearranging, this becomes
DG = RT lnK
or,
-DG
K = e RT
Electrochemistry
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Free Energy and Equilibrium reminder
• If DGo < 0, then K >1
when ln K is +  more negative DGo  larger K  products are
favored
• If DGo = 0, then K = 1
• If DGo > 0, then K <1
when ln K is -  more positive DGo  smaller K  reactants
are favored
Electrochemistry
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Free Energy and Redox Reactions
Remember:
Eo = Eored(reduction process) – Eored(oxidation process)
A positive Eo = a spontaneous process
(galvanic or voltaic cell)
A negative Eo = a nonspontaneous process
Electrochemistry
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Electrochemistry and Free Energy
DG for a redox reaction can be found by
using the equation
DG = −nFE
where n is the number of moles of electrons
transferred, and F is a constant, the Faraday.
1 F = 96,485 C = 96,485 J
mol V-mol
Electrochemistry
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Electrochemistry and Free Energy
DG = −nFE
Since n and F are positive, the sign of DG
relies on E.
If E > 0, DG will be <0 and the reaction will be
spontaneous.
(fits with positive Eo  direction of voltaic cell)
Electrochemistry
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Free Energy
Under standard conditions,
DG = −nFE
Electrochemistry
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Sample Exercise 20.10 (p. 864)
a) Use the standard reduction potentials listed
in Table 20.1 or your AP Chem packet to
calculate the standard free-energy change,
DGo, and the equilibrium constant, K, at 298
K for the following reaction:
4 Ag(s) + O2(g) + 4 H+(aq)  4 Ag+(aq) + 2 H2O(l)
(DGo = -170 kJ/mol, K = 9 x 1029)
Electrochemistry
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Sample Exercise 20.10 (p. 864)
b) Suppose the reaction in part (a) were written
as
2 Ag(s) + ½ O2(g) + 2 H+(aq)  2 Ag+(aq) + H2O(l)
What are the values of Eo, DGo and K when
the reaction is written in this way?
(Eo = +0.43 V, DGo = - 83 kJ/mol,
K = 4 x 1014)
Electrochemistry
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Practice Exercise 20.10
For the reaction: 3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH-(aq)
 3 Ni(s) + 2 CrO42-(aq) + 8 H2O(l)
a) What is the value of n for this reaction? (6)
b) Use the data in Appendix E or your AP Chem
packet to calculate DGo for this reaction.
(+87 kJ/mol)
c) Calculate K at T = 298 K (6 x 10-16)
Electrochemistry
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Cell EMF under nonstandard conditions
• A voltaic cell works until E = 0, at which
point equilibrium has been reached.
• The point at which E = 0 is determined
by the concentrations of the species
involved in the redox reaction.
Electrochemistry
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Nernst Equation
(named after Walther Hermann Nernst – Nobel Prize in Chemistry 1920)
• Remember that
DG = DG + RT ln Q
• This means
−nFE = −nFE + RT ln Q
Electrochemistry
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Nernst Equation
Dividing both sides by −nF, we get the
Nernst equation:
RT
ln Q
E = E −
nF
or, using base-10 logarithms,
2.303 RT
log Q
E = E −
nF
Electrochemistry
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Nernst Equation
At room temperature (298 K),
2.303 RT
= 0.0592 V
F
Thus the equation becomes
0.0592
log Q
E = E −
n
Electrochemistry
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Nernst Equation
Zn(s) + Cu2+(aq) D Zn2+(aq) + Cu(s)
If [Cu2+] = 5.0 M and [Zn2+] = 0.050 M,
Ecell = 1.10 V – 0.0592 log 0.050
2
5.0
= 1.16 V
Electrochemistry
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Sample Exercise 20.11 (p. 866)
Calculate the emf at 298 K generated by the
cell described in Sample Exercise 20.4 when
[Cr2O72-] = 2.0 M, [H+] = 1.0 M, [I-] = 1.0 M,
and [Cr3+] = 1.0 x 10-5 M.
Cr2O72-(aq) + 14 H+(aq) + 6 I-(aq)
 2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
(E = +0.89 V)
Electrochemistry
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Practice Exercise 20.11
Calculate the emf generated by the cell
described in the practice exercise
accompanying Sample Exercise 20.6
when [Al3+] = 4.0 x 10-3 M and [I-] =
0.010 M.
i.e. 2 Al(s) + 3 I2(s)  2 Al3+(aq) + 6 I-(aq)
(E = +2.36 V)
Electrochemistry
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Sample Exercise 20.12 (p. 866)
If the voltage of a Zn-H+ cell (like that in
Figure 20.11) is 0.45 V at 25oC when
[Zn2+] = 1.0 M and PH2 = 1.0 atm, what
is the concentration of H+?
([H+] = 5.8 x 10-6 M)
Electrochemistry
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Practice Exercise 20.12
What is the pH of the solution in the
cathode compartment of the cell
pictured in Figure 20.11
when PH2 = 1.0 atm, [Zn2+] in the anode
compartment is 0.10 M, and cell emf is
0.542 V?
(pH = 4.23)
Electrochemistry
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Concentration Cells
• Notice that the Nernst equation implies that a cell
could be created that has the same substance at
both electrodes.
 would be 0, but Q would not.
• For such a cell, Ecell
• Therefore, as long as the concentrations
Electrochemistry
are different, E will not be 0.
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Concentration Cells
• The driving force is the difference in [Ni2+]
• Anode (dilute Ni2+): Ni(s)  Ni2+(aq) + 2 e-
There is very little Ni2+, so this reaction proceeds in the
direction that increases Ni2+, by oxidizing Ni.
• Cathode (concentrated Ni2+): Ni2+(aq) + 2 e-  Ni(s)
There is much Ni2+, so this reaction proceeds in the
direction of decreasing Ni2+, by reducing it to Ni(s).
Electrochemistry
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Sample Exercise 20.13 (p. 870)
A voltaic cell is constructed with two hydrogen electrodes.
Electrode 1 has PH2 = 1.00 atm and an unknown concentration
of H+(aq).
Electrode 2 is a standard hydrogen electrode ([H+] = 1.00 M, PH2
= 1.00 atm).
At 298 K the measured cell voltage is 0.211 V, and the electrical
current is observed to flow from electrode 1 through the external
circuit to electrode 2.
Calculate [H+] for the solution at electrode 1. What is its pH?
(pH = 3.57)
Electrochemistry
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Practice Exercise 20.13
A concentration cell is constructed with two
Zn(s)-Zn2+(aq) half-cells. The first half-cell has
[Zn2+] = 1.35 M, and the second half-cell has
[Zn2+] = 3.75 x 10-4 M.
a) Which half-cell is the anode of the cell?
b) What is the emf of the cell? (0.105 V)
Electrochemistry
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Cell EMF and Chemical Equilibrium
• From the Nernst equation, at equilibrium and 298 K
(E = 0 V and Q = Keq)
0 = Eo – 0.0592 log Keq
n
log Keq =
nEo
0.0592
• Thus, if we know the cell emf, we can calculate the
equilibrium constant (Keq).
Electrochemistry
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Applications of
Oxidation-Reduction
Reactions
Electrochemistry
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Batteries
Electrochemistry
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Lead Battery
• Cathode: PO2 on metal grid in H2SO4
PbO2(s) + HSO4-(aq) + 3 H+(aq) + 2 e-  PbSO4(s)+ 2 H2O(l)
• Anode: Pb
Pb(s)+ HSO4-(aq) PbSO4(s)+ H+(aq)+ 2 e• Overall:
PbO2(s) + Pb(s) + 2 HSO4-(aq) + 2 H+(aq) 
2 PbSO4(s) + 2 H2O(l)
Cell Potential: each cell – about 2 V
Electrochemistry
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Alkaline Batteries
Electrochemistry
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Alkaline Batteries
• Cathode:
MnO2(s) + 2 H2O(l) + 2 e- 
2 MnO(OH)(s) + 2 OH-(aq)
source of “alkaline”
• Anode:
Zn(s)  Zn2+(aq) + 2 eCell potential: about 1.55 V
Electrochemistry
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Nickel–Cadmium, Nickel–Metal–
Hydride, and Lithium–Ion Batteries
• Cathode:
2 NiO(OH)(s) + 2 H2O(l) + 2 e- 
2Ni(OH)2(s) +2 OH-(aq)
Anode:
Cd(s) + 2OH-(aq)  Cd(OH)2(s) + 2eCell potential - about 1.30 V
Cadmium – toxic – difficult disposal
Electrochemistry
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Nickel–Cadmium, Nickel–Metal–
Hydride, and Lithium–Ion Batteries
• Other rechargeable batteries have been
developed:
• NiMH batteries (nickel–metal–hydride).
• Li–ion batteries (lithium–ion batteries).
Electrochemistry
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Hydrogen Fuel Cells
Electrochemistry
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Hydrogen Fuel Cells
• Cathode: reduction of O2
2 H2O(l) + O2(g) + 4 e-  4 OH-(aq)
• Anode: oxidation of H2
2 H2(g) + 4 OH-(aq)  4 H2O(l) + 4 eOverall: O2(g) + 2 H2(g)  2 H2O(l)
Electrochemistry
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Corrosion and…
Electrochemistry
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…Corrosion Prevention
Electrochemistry
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Protection of Underground Pipes
Electrochemistry
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Electrolysis
• Electrolysis reactions =
nonspontaneous reactions that require
an external current in order to force the
reaction to proceed
• take place in electrolytic cells
Electrochemistry
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Electrolysis
• In both voltaic and electrolytic cells,
reduction occurs at the cathode, and
oxidation occurs at the anode.
Electrochemistry
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Electrolysis
• However, in electrolytic cells, electrons are
forced to flow from the anode to the cathode.
• In electrolytic cells the anode is positive and
the cathode is negative.
• In voltaic cells the anode is negative and the
cathode is positive.
Electrochemistry
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Molten NaCl
Electrochemistry
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Electrolysis
• Example: decomposition of molten NaCl.
• Cathode: 2Na+(l) + 2e-  2Na(l) (reduction)
• Anode: 2Cl-(l)  Cl2(g) + 2e- (oxidation)
• Industrially, electrolysis is used to produce
metals like Al.
Electrochemistry
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Electrolysis
• Electrolysis of high-melting ionic
substances requires very high
temperatures.
• Do we get the same products if we
electrolyze an aqueous solution of the
salt?
• Water complicates the issue!
Electrochemistry
© 2009, Prentice-Hall, Inc.
Electrolysis
Example: Consider the electrolysis of NaF(aq):
Na+(aq) + e-  Na(s)
2H2O(l) + 2e-  H2(g) + 2OH-(aq)
Eored = -2.71 V
Eored = -0.83 V
Thus water is more easily reduced the sodium ion.
2F-(aq)  F2(g) + 2e2H2O(l)  O2(g) + 4H+(aq) + 4e-
Eored = +2.87 V
Eored = +1.23 V
Thus it is easier to oxidize water than the fluoride
ion.
Electrochemistry
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Electrolysis with Active Electrodes
Active electrodes: electrodes that take part in
electrolysis.
Example: electroplating.
Anode (nickel strip): Ni(s)  Ni2+(aq) + 2e- (active)
Cathode (steel strip): Ni2+(aq) + 2e-  Ni(s) (inert)
• Ni plates on the inert electrode.
• Electroplating is important in protecting objects from
corrosion.
Electrochemistry
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• Electrolytic cell with an active metal
electrode.
• Nickel dissolves from the anode to form
Ni2+(aq). At the cathode Ni2+(aq) is
reduced and forms a nickel “plate” on
the cathode.
Electrochemistry
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Electroplating
Electrochemistry
© 2009, Prentice-Hall, Inc.
Quantitative Aspects of Electrolysis
How much material can we obtain with electrolysis?
e.g. the reduction of Cu2+ to Cu: Cu2+(aq) + 2e-  Cu(s).
Two mol of electrons will plate 1 mol of Cu.
The charge of one mol of electrons is 96,500 C (1 F).
1 coulomb = amount of charge passing a point in 1 s
when the current is one A.
The amount of Cu can be calculated from the
current (I) and time (t) required to plate: Q = I t
Electrochemistry
© 2009, Prentice-Hall, Inc.
Quantitative Aspects of Electrolysis
Electrochemistry
© 2009, Prentice-Hall, Inc.
Sample Exercise 20.14 (p. 878)
Calculate the number of grams of
aluminum produced in 1.00 hr by the
electrolysis of molten AlCl3 if the
electrical current is 10.0 A.
(3.3.6 g Al)
Electrochemistry
© 2009, Prentice-Hall, Inc.
Practice Exercise 20.14
a)
The half-reaction for formation of magnesium metal
upon electrolysis of molten MgCl2 is
Mg2+ + 2e-  Mg
Calculate the mass of magnesium formed upon
passage of a current of 60.0 A for a period of
4.00 x 103 s.
(30.2 g Mg)
b)
How many seconds would be required to produce
50.0 g of Mg from MgCl2 if the current is 100.0 A?
(3.97 x 103 s)
Electrochemistry
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Electrical Work
Free energy = a measure of the maximum amount of
useful work that can be obtained from a system.
 DG = wmax
Thus
and
DG = -nFE
wmax = -nFE
If Ecell is positive, wmax will be negative.
Work is done by the system on the surroundings.
Electrochemistry
© 2009, Prentice-Hall, Inc.
Electrical Work
• The emf = a measure of the driving force for a
redox process.
• In an electrolytic cell an external source of energy
is required to force the reaction to proceed.
Electrochemistry
© 2009, Prentice-Hall, Inc.
Electrical Work
• w = nFEexternal
• In order to drive the nonspontaneous reaction the
external emf must be greater than Ecell.
• From physics: Work is measured in units of watts:
1 W = 1 J/s
• Electric utilities use units of kilowatt-hours:
 3600s  1 J/s 
6
1 kWh  1000 W1 hour   

  3.6  10 J
 1 hour  1 W 
Electrochemistry
© 2009, Prentice-Hall, Inc.
Sample Exercise 20.15 (p. 880)
Calculate the number of kilowatt-hours
of electricity required to produce 1.0 x
103 kg of aluminum by electrolysis of
Al3+ if the applied emf is 4.50 V.
(1.34 x 104 kWh)
Electrochemistry
© 2009, Prentice-Hall, Inc.
Practice Exercise 20.15
Calculate the number of kilowatt-hours
of electricity required to produce 1.00 kg
of Mg by electrolysis of molten MgCl2 if
the applied emf is 5.00 V. Assume that
the process is 100% efficient.
(11.0 kWh)
Electrochemistry
© 2009, Prentice-Hall, Inc.
Sample Integrative Exercise 20 (p. 880)
The Ksp at 298 K for iron (II) fluoride is 2.4 x 10-6.
a) Write a half-reaction that gives the likely products of
the two-electron reduction of FeF2(s) in water.
b) Use the Ksp value and the standard reduction
potential of Fe2+(aq) to calculate the standard
reduction potential for the half-reaction in part (a).
(-0.606 V)
c) Rationalize the difference in the reduction potential
for the half-reaction in part (a) with that for Fe2+(aq).
Electrochemistry
© 2009, Prentice-Hall, Inc.