Transcript Chapter Eighteen - DePaul University

1
Chapter Eighteen
Electrochemistry
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Chapter Eighteen
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Oxidation–Reduction:
The Transfer of Electrons
Silver metal is formed,
and the solution turns
blue from copper(II)
ions formed.
Electrons from
copper metal are
transferred to
silver ions.
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Half-Reactions
• Review Chapter Four for details of oxidation–reduction
reactions.
• In any oxidation–reduction reaction, there are two halfreactions:
– Oxidation: a species loses electrons to another species.
– Reduction: a species gains electrons from another
species.
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The Half-Reaction Method of
Balancing Redox Equations
1.
2.
3.
4.
5.
6.
Separate a redox equation into two half-equations, one
for oxidation and one for reduction.
Balance the number of atoms of each element in
each half-equation. Usually we balance O and H
atoms last.
Balance each half-reaction for charge by adding
electrons to the left in the reduction half-equation and to
the right in the oxidation half-equation.
Adjust the coefficients in the half-equations so that the
same number of electrons appears in each half-equation.
Add together the two adjusted half-equations to obtain
an overall redox equation.
Simplify the overall redox equation as necessary.
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Redox Reactions in Acidic and in Basic
Solution
• Redox reactions in acidic solution and in basic solution
may be very different from one another.
• If acidic solution is specified, we must add H2O and/or H+
as needed when we balance the number of atoms.
• If basic solution is specified, the final equation may have
OH– and/or water molecules in it.
• A simple way to balance an equation in basic solution:
– Balance the equation as though it were in acidic solution.
– Add as many OH– ions to each side as there are H+ ions in the
equation.
– Combine the H+ and OH– ions to give water molecules on one side,
and simplify the equation as necessary.
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Example 18.1
Permanganate ion, MnO4–, is used in the laboratory as an
oxidizing agent; thiosulfate ion, S2O32–, is used as a reducing
agent. Write a balanced equation for the reaction of these ions
in an acidic aqueous solution to produce manganese(II) ion
and sulfate ion.
Example 18.2
In basic solution, Br2 disproportionates to bromide ions and
bromate ions. Use the half-reaction method to balance the
equation for this reaction:
Br2(l)  Br–(aq) + BrO3–(aq)
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A Qualitative Description of Voltaic
Cells
• A voltaic cell uses a spontaneous redox reaction to produce
electricity.
• A half-cell consists of an electrode (strip of metal or other
conductor) immersed in a solution of ions.
This Zn2+ becomes
a Zn atom.
Both oxidation and
reduction occur at the
electrode surface, and
equilibrium is reached.
This Zn atom leaves the
surface to become a Zn2+ ion.
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Important Electrochemical Terms
• An electrochemical cell consists of two half-cells with the
appropriate connections between electrodes and solutions.
• Two half-cells may be joined by a salt bridge that permits
migration of ions, without completely mixing the solutions.
• The anode is the electrode at which oxidation occurs.
• The cathode is the electrode at which reduction occurs.
• In a voltaic cell, a spontaneous redox reaction occurs and
current is generated.
• Cell potential (Ecell) is the potential difference in volts
between anode and cathode.
• Ecell is the driving force that moves electrons and ions.
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A Zinc–Copper
Voltaic Cell
Positive and negative
ions move through the
salt bridge to equalize
the charge.
… the electrons
produced move
through the wire …
Zn(s) is oxidized to
Zn2+ ions, and …
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… to the Cu(s) electrode,
where they are accepted
by Cu2+ ions to form
more Cu(s).
Reaction: Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)
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Cell Diagrams
•
•
•
•
A cell diagram is “shorthand” for an electrochemical cell.
The anode is placed on the left side of the diagram.
The cathode is placed on the right side.
A single vertical line ( | ) represents a boundary between
phases, such as between an electrode and a solution.
• A double vertical line ( || ) represents a salt bridge or
porous barrier separating two half-cells.
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Example 18.3
Describe the half-reactions and the overall reaction that
occur in the voltaic cell represented by the cell diagram:
Pt(s) | Fe2+(aq), Fe3+(aq) || Cl–(aq) | Cl2(g) | Pt(s)
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Standard Electrode Potentials
• Since an electrode represents
only a half-reaction, it is not
possible to measure the
absolute potential of an
electrode.
• The standard hydrogen
electrode (SHE) provides a
reference for measurement of
other electrode potentials.
• The SHE is arbitrarily assigned
a potential of 0.000 V.
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Standard Electrode Potentials (cont’d)
• The standard electrode potential, E°, is based on the
tendency for reduction to occur at an electrode.
• E° for the standard hydrogen electrode is arbitrarily
assigned a value of 0.000 V.
• All other values of E° are determined relative to the
standard hydrogen electrode.
• The standard cell potential (E°cell) is the difference
between E° of the cathode and E° of the anode.
• E°cell = E°(cathode) – E°(anode)
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Measuring the Standard Potential
of the Cu2+/Cu Electrode
The voltmeter reading
and the direction of
electron flow tell us
that …
… Cu2+ is more easily
reduced than H+, by
0.340 volts.
Standard
hydrogen
electrode
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Cu2+ + 2e  Cu
E° = +0.340 V
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Measuring the Standard Potential
of the Zn2+/Zn Electrode
The voltmeter reading
and the direction of
electron flow tell us
that …
… Zn2+ is harder to
reduce than H+, by
0.763 volts.
Standard
hydrogen
electrode
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Zn2+ + 2e  Zn
E° = – 0.763 V
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Chapter Eighteen
F2 is the strongest
oxidizing agent
F– is the weakest
reducing agent
16
Li is the
strongest
reducing
agent
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Important Points about
Electrode and Cell Potentials
• Standard electrode potentials and cell voltages are
intensive properties; they do not depend on the total
amounts of the species present.
• A “flashlight battery” (D-cell) and a “penlight battery”
(AA cell) produce the same potential—1.5 volts.
• E° does depend on the particular species in the reaction (or
half-reaction).
• As we shall learn later, cell and electrode potentials can
depend on concentration of the species present.
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Example 18.4
Determine E° for the reduction half-reaction
Ce4+(aq) + e–  Ce3+(aq), given that the cell voltage for the
voltaic cell
Co(s) | Co2+(1 M) || Ce4+(1 M), Ce3+(1 M) | Pt(s)
is E°cell = 1.887 V.
Example 18.5
Balance the following oxidation–reduction equation, and
determine E°cell for the reaction.
O2(g) + H+(aq) + I–(aq)  H2O(l) + I2(s)
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Electrode Potentials, Spontaneous
Change, and Equilibrium
• An electrochemical cell does work.
welec = nFEcell
n = number of electrons in the balanced equation
F = 96,485 coulombs per mole.
• The amount of electrical work is also equal to –DG:
DG = –nFEcell
• Under standard conditions:
DG° = –nFE°cell
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Criteria for Spontaneous
Change in Redox Reactions
• If Ecell is positive, the forward reaction is spontaneous.
• If Ecell is negative, the forward reaction is nonspontaneous
(the reverse reaction is _____).
• If Ecell = 0, the system is at equilibrium.
• When a cell reaction is reversed, Ecell and DG change
signs.
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Example 18.6
Will copper metal displace silver ion from aqueous
solution? That is, does the reaction
Cu(s) + 2 Ag+(1 M)  Cu2+(1 M) + 2 Ag(s)
occur spontaneously from left to right?
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The Activity Series Revisited
• In the activity series of metals
(Section 4.4), any metal in the
series will displace a metal
below it from a solution of that
metal’s ions.
• Theoretical basis: The activity
series lists metals in order of
their standard potentials.
• Displacement of a metal from a
solution of its ions by a metal
higher in the series corresponds
to a positive value of Ecell and a
spontaneous reaction.
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Example 18.7
A Conceptual Example
The photograph in Figure 18.11
shows strips of copper and zinc
joined together and then dipped
in HCl(aq). Explain what
happens. That is, what are the
gas bubbles on the zinc and on
the copper, and how did they get
there?
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Equilibrium Constants in Redox
Reactions
• Whereas potential and free energy are related, and free
energy and equilibrium are related, equilibrium and
potential must be related to one another.
DG° = –nFE°cell
and
DG° = –RT ln Keq
therefore
–RT ln Keq = –nFEocell
RT ln Keq
RT
E°cell = ––––––––– = –––– ln Keq
nF
nF
R and F are constant,
therefore at 298 K:
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0.025693 V
E°cell = –––––––– ln Keq
n
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Example 18.8
Calculate the values of ΔG° and Keq at 25 °C for the reaction
Cu(s) + 2 Ag+(1 M)  Cu2+(1 M) + 2 Ag(s)
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Thermodynamics, Equilibrium, and
Electrochemistry: A Summary
From any one of the three
quantities Keq, ΔG°, E°cell, we
can determine the others.
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Effect of Concentrations on Cell
Voltage
• A nonstandard cell differs in potential from a standard cell
(1 M concentrations, 1 atm partial pressures).
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Effect of Concentrations on Cell
Voltage
From the previous relationships
we can show that:
At 25 °C, and converting to
common logarithms:
Ecell = E°cell
RT
– –––– ln Q
nF
Ecell = E°cell
0.0592 V
– ––––––– log Q
n
• This Nernst equation relates a cell voltage for nonstandard
conditions, Ecell, to a standard cell voltage, E°cell, and to the
concentrations of reactants and products expressed through
the reaction quotient, Q.
• We can use the Nernst equation to find cell potential from
concentrations, or we can measure Ecell and determine the
concentration of a species in the cell.
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Example 18.9
Calculate the expected voltmeter reading for the voltaic
cell pictured in Figure 18.13.
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Concentration Cells
• A concentration cell has a
potential determined solely by
the difference in concentrations
of solutes in equilibrium with
identical electrodes.
• Since the two electrodes are
identical, the standard cell
potential is zero.
• To calculate the cell voltage
for the nonstandard
conditions, the Nernst
equation is used.
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pH Measurement
• Consider a concentration cell consisting of a standard
hydrogen electrode, and a hydrogen electrode with
unknown [H+].
• The Nernst equation for such a cell is:
Ecell(in volts) = 0.0592 pH
pH = Ecell /0.0592
• Thus, we can measure the pH of an unknown solution
by making it part of such a concentration cell.
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The pH Meter
In practice, a special pH
electrode is much more
convenient than using
platinum electrodes and
a tank of hydrogen gas!
A stable reference electrode
and a glass-membrane
electrode are contained within
a combination pH electrode.
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The electrode is merely
dipped into a solution, and
the potential difference
between the electrodes is
displayed as pH.
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Batteries: Using Chemical
Reactions to Make Electricity
• We often call any device that stores chemical energy for
later release as electricity a battery.
• Technically, a D, C, or AA “battery” is actually a single
electrochemical cell.
• A battery consists of several cells joined together to
produce higher current or higher voltage.
• A 9-volt “transistor” battery, an automobile battery, and a
rechargable “battery pack” are all true batteries.
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The Dry Cell
• A primary cell employs an
irreversible chemical reaction.
• When the reactants inside the
cell are largely used up, the cell
is “dead.”
• The LeClanché cell or dry cell
(right) is the “ordinary” type of
flashlight “battery.”
• Alkaline cells cost more than
the LeClanché cell but they
have a longer shelf life and
longer service life.
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The Lead–Acid Storage Battery
• A lead–acid storage battery
used in an automobile uses
secondary cells; they are
rechargeable.
• By connecting the cell to an
external electric energy
source, the discharge
reaction is reversed.
Cell reaction:
Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4–(aq)  2 PbSO4(s) + 2 H2O(l)
Charging reaction: 2 PbSO4(s) + 2 H2O(l)  Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4–(aq)
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Other Secondary Cells
• The nickel–cadmium (NiCd) cell uses a cadmium
anode and a cathode containing Ni(OH)2.
• A NiCd cell can be recharged hundreds of times. It
produces 1.2 V (a Leclanché cell produces 1.5 V).
• Nickel–metal hydride cells (NiMH) use a metal alloy
anode that contains hydrogen.
• In use, the anode releases the hydrogen, forming
water. Like the NiCd cell, a NiMH cell produces 1.2 V.
• Lithium-ion cells use a lithium–cobalt oxide or lithium–
manganese oxide material as the anode. The
electrolyte is an organic solvent containing a dissolved
lithium salt.
• Many modern laptop computers and cellular phones
use lithium-ion cells.
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Fuel Cells
• In a fuel cell, the cell reaction is
equivalent to a combustion reaction.
• The reactants are supplied
externally; the cell does not “go
dead” as long as the oxidizing
and reducing agents are provided.
• Fuel cells are generally operated
under nonstandard conditions and
at temperatures considerably
higher than 25 °C.
• H2–O2 fuel cells are seeing use in
some automobiles.
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Corrosion: Metal Loss
Through Voltaic Cells
• In moist air, iron is oxidized to Fe2+, particularly at
scratches, nicks, or dents. These areas are referred to
anodic areas.
• Other regions of the iron serve as cathodic areas, where
the electrons from the anodic areas reduce O2 to OH–.
• Iron(II) ions migrate from the anodic areas to the cathodic
areas where they combine with the hydroxide ions.
• The iron(II) is then further oxidized to iron(III) by
atmospheric oxygen.
• Common rust is Fe2O3 · x H2O.
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Corrosion of an Iron Piling
One way to minimize
rusting is to provide a
different anode reaction.
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Protecting Iron from Corrosion
• The simplest defense against corrosion of iron is to coat it
(with paint or metal) to exclude oxygen from the surface.
• An entirely different approach is to protect iron with a
more active metal.
• Galvanized iron has been coated with zinc.
• The zinc provides an alternative anode reaction; the zinc
corrodes, protecting the iron.
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Cathodic Protection
• In cathodic protection, an iron object to be protected is
connected to a chunk of an active metal.
• The iron serves as the
reduction electrode and
remains metallic. The
active metal is oxidized.
• Water heaters often
employ a magnesium
anode for cathodic
protection.
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Example 18.10
A Conceptual Example
In Figure 18.20, iron nails are placed in a warm colloidal
dispersion of agar in water. Phenolphthalein indicator and
potassium ferricyanide, K3[Fe(CN)6], are also present in the
dispersion. When the faintly yellow dispersion cools and
stands for a few hours, it solidifies into a gel, and soon the
colored regions begin to develop—pink along the middle
part of the nail and blue at both ends. Explain what is
happening in Figure 18.20a.
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Electrolytic Cells
• A voltaic cell corresponds to a spontaneous cell reaction.
• An electrolytic cell corresponds to a nonspontaneous cell
reaction. The reaction is called electrolysis.
• The external source of electricity acts like an “electron
pump.” It pulls electrons away from the anode, where
oxidation takes place, and pushes them toward the
cathode, where reduction takes place.
• The polarities of the electrodes are reversed from those in
the voltaic cell, because now the external source controls
the flow of electrons.
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Electrolysis of Molten
Sodium Chloride
2 NaCl(l)  2 Na(l) + Cl2(g)
The nonspontaneous
reaction is driven by
external potential.
Molten NaCl
(around 1000 °C)
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Predicting Electrolysis Reactions
• In an electrolytic cell, all combinations of cathode and
anode half-reactions give negative values of E°cell.
• The reaction most likely to occur is the one with the least
negative value of E°cell (requires the lowest applied voltage
from the external electricity source). HOWEVER …
• In many half-reactions, particularly those involving gases,
various interactions at electrode surfaces make the required
voltage for electrolysis higher than the voltage calculated
from E° data.
• Overvoltage is the excess voltage above the voltage
calculated from E° values that is required in electrolysis.
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Inert Electrodes
and Active Electrodes
• The electrodes used in the electrolysis of NaCl(l) and
NaCl(aq) are examples of inert electrodes.
• An inert electrode only provides a surface on which the
exchange of electrons can occur.
• In contrast, some electrodes are active electrodes; they
participate directly in a half-reaction.
• Many of the electrodes we saw in voltaic cells were active
electrodes.
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Example 18.11
Predict the electrolysis reaction when AgNO3(aq) is
electrolyzed (a) using platinum electrodes and (b) using a
silver anode and a platinum cathode.
Example 18.12
A Conceptual Example
Two electrochemical cells are connected as shown.
Specifically, the zinc electrodes of the two cells are joined,
as are the copper electrodes. Will there be (a) no current,
(b) a flow of electrons in the direction of the red arrows, or
(c) a flow of electrons in the direction of the blue arrows?
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Quantitative Electrolysis
•
•
•
The unit of electric charge is the coulomb (C), and the
charge carried by one electron is –1.6022 x 10–19 C.
Electric current, expressed in amperes (A), is the rate of
flow of electric charge (C/s).
To calculate the quantitative outcome of an electrolysis
reaction:
1.
2.
3.
4.
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Determine the amount of charge (C)—the product of current
and time.
Convert the amount of charge to moles of electrons.
Use a half-equation to relate moles of electrons to moles of a
reactant or a product.
Convert from moles of reactant or product to the final quantity
desired.
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Example 18.13
We can use electrolysis to determine the gold content of a sample.
The sample is dissolved, and all the gold is converted to Au3+(aq),
which is then reduced back to Au(s) on an electrode of known
mass. The reduction half-reaction is Au3+(aq) + 3e–  Au(s).
What mass of gold will be deposited at the cathode in 1.00 hour
by a current of 1.50 A?
Example 18.14
An Estimation Example
Without doing detailed calculations, determine which of the
following solutions will yield the greatest mass of metal at a
platinum cathode during electrolysis by a 1.50-A electric current
for 30.2 min: CuSO4(aq), AgNO3(aq), or AuCl3(aq).
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Producing Chemicals
by Electrolysis
Electrolysis plays an
important role in the
manufacture and purification
of many substances, including
chlorine, copper, silver,
magnesium, aluminum, lead,
zinc, sodium, fluorine,
titanium, sodium hydroxide,
hydrogen …
Electrolysis of NaCl(aq) is
used to produce H2, Cl2, and
NaOH, all of which have
important industrial uses.
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Electroplating
• Electrolysis can be used to coat
one metal onto another, a process
called electroplating.
• Usually, the object to be
electroplated, such as a spoon, is
cast of an inexpensive metal. It is
then coated with a thin layer of a
more attractive, corrosionresistant, and expensive metal,
such as silver or gold.
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Cumulative Example
An important source of silver is the process in which lead is
purified from lead-containing ores. The percent Ag in a
1.050-g lead sample is determined as follows: The sample is
dissolved in nitric acid to produce 500.0 mL of a solution of
Pb2+(aq) containing a small quantity of Ag+(aq). A strip of
pure Ag(s) is immersed in the solution, and the potential
difference between this half-cell as the cathode and a
standard silver–silver chloride anode is found to be 0.281 V.
What is the mass percent silver in the lead sample?
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Chapter Eighteen