Transcript Chapter 20 Electrochemistry

Chapter 20
Electrochemistry
© 2012 Pearson Education, Inc.
Electrochemical Reactions
(Red-Ox)
In electrochemical reactions,
electrons are transferred from one
species to another.
Electrochemistry
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
(a) Bubbling is caused by water vapor escaping and (b) the exothermic reaction in water produces steam.
(a) Bubbling is caused by water vapor escaping and (b) the endothermic reaction in water produces steam.
(a) Bubbling is caused by hydrogen gas forming and (b) the endothermic reaction in water produces steam.
(a) Bubbling is caused by hydrogen gas forming and (b) the exothermic reaction in water produces steam.
Electrochemistry
Oxidation Numbers
In order to keep
track of what loses
electrons and what
gains them, we
assign oxidation
numbers.
Electrochemistry
© 2012 Pearson Education, Inc.
Oxidation and Reduction
•
•
A species is oxidized when it loses electrons.
(the oxidation number increases)
– Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.
A species is reduced when it gains electrons.
(the oxidation number decreases)
– Here, each of the H+ gains an electron, and they combine to form H2.
Electrochemistry
© 2012 Pearson Education, Inc.
Oxidation and Reduction
• The species reduced is the oxidizing agent.
– H+ oxidizes Zn by taking electrons from it.
• The species oxidized is the reducing agent.
– Zn reduces H+ by giving it electrons.
Electrochemistry
© 2012 Pearson Education, Inc.
Assigning Oxidation Numbers
1.
2.
3.
4.
5.
Elements in their elemental form have an oxidation number of 0.
The oxidation number of a monatomic ion is the same as its charge.
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.
–
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.
The sum of the oxidation numbers in a neutral compound is 0.
The sum of the oxidation numbers in
a polyatomic ion is the charge on
the ion.
Electrochemistry
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
N, +1;
N, +2;
N, +3;
N, +4;
O, –1
O, –2
O, –2
O, –2
Electrochemistry
Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents
The nickel-cadmium (nicad) battery 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.
Practice Exercise
Identify the oxidizing and reducing agents in the reaction
2 H2O(l) + Al(s) + MnO4(aq)  Al(OH)4(aq) + MnO2(s)
Electrochemistry
Balancing Oxidation-Reduction
Equations
Perhaps the easiest way to balance the equation of an oxidation-reduction
reaction is via the half-reaction method.
This method 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
© 2012 Pearson Education, Inc.
The Half-Reaction Method
1.
2.
3.
Assign oxidation numbers to determine what is oxidized and
what is reduced.
Write the oxidation and reduction
half-reactions.
Balance each half-reaction.
a. Balance elements other than H and O.
b. Balance O by adding H2O.
c. Balance H by adding H+.
d. Balance charge by adding electrons.
Electrochemistry
© 2012 Pearson Education, Inc.
The Half-Reaction Method
4.
5.
6.
7.
Multiply the half-reactions by integers so that the electrons
gained and lost are the same.
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
© 2012 Pearson Education, Inc.
The Half-Reaction Method
Consider the reaction between MnO4 and C2O42:
MnO4(aq) + C2O42(aq)  Mn2+(aq) + CO2(aq)
Electrochemistry
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
MnO4- is reduced and the reducing agent is H+.
MnO4- is reduced and the reducing agent is C2O42-.
H+ is reduced and the reducing agent is C2O42-.
H+ is reduced and the reducing agent is MnO4-.
Electrochemistry
The Half-Reaction Method
First, 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
© 2012 Pearson Education, Inc.
Oxidation Half-Reaction
C2O42  CO2
To balance the carbon, we add a coefficient of 2:
C2O42  2CO2
The oxygen is now balanced as well.
To balance the charge, we must add
2 electrons to the right side:
C2O42  2CO2 + 2e
Electrochemistry
© 2012 Pearson Education, Inc.
Reduction Half-Reaction
MnO4  Mn2+
The manganese is balanced; to balance the oxygen, we must
add 4 waters to the right side:
MnO4  Mn2+ + 4H2O
To balance the hydrogen, we add
8H+ to the left side:
8H+ + MnO4  Mn2+ + 4H2O
Electrochemistry
© 2012 Pearson Education, Inc.
Reduction Half-Reaction
8H+ + MnO4  Mn2+ + 4H2O
To balance the charge, we add 5e to the left side:
5e + 8H+ + MnO4  Mn2+ + 4H2O
Electrochemistry
© 2012 Pearson Education, Inc.
Combining the Half-Reactions
Now we evaluate the two half-reactions
together:
C2O42  2CO2 + 2e
5e + 8H+ + MnO4  Mn2+ + 4H2O
To attain the same number of electrons
on each side, we will multiply the first
reaction by 5 and the second by 2:
Electrochemistry
© 2012 Pearson Education, Inc.
Combining the Half-Reactions
5C2O42  10CO2 + 10e
10e + 16H+ + 2MnO4  2Mn2+ + 8H2O
When we add these together, we get:
10e + 16H+ + 2MnO4 + 5C2O42 
2Mn2+ + 8H2O + 10CO2 +10e
Electrochemistry
© 2012 Pearson Education, Inc.
Combining the Half-Reactions
10e + 16H+ + 2MnO4 + 5C2O42 
2Mn2+ + 8H2O + 10CO2 +10e
The only thing that appears on both sides are the
electrons. Subtracting them, we are left with:
16H+ + 2MnO4 + 5C2O42 
2Mn2+ + 8H2O + 10CO2
Electrochemistry
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
Yes, in each half-reaction
Yes, on both sides of the balanced equation for a
redox reaction
No, the electrons do not show in half-reactions and
therefore they do not appear in a balanced equation
for a redox reaction.
No, the electrons cancel in adding half-reactions to
form a balanced equation for a redox reaction.
Electrochemistry
Sample Exercise 20.2 Identifying Oxidizing and Reducing Agents
Complete and balance this equation by the method of half-reactions:
Cr2O72(aq) + Cl(aq)  Cr3+(aq) + Cl2(g)
(acidic solution)
Practice Exercise
Complete and balance the following 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
Balancing in Basic Solution
• If a reaction occurs in a 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
© 2012 Pearson Education, Inc.
Sample Exercise 20.3 Balancing Redox Equations in Basic Solution
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)
Practice Exercise
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
Voltaic Cells
In spontaneous
oxidation-reduction
(redox) reactions,
electrons are
transferred and
energy is released.
Electrochemistry
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
The blue color of Cu2+(aq) lessens as it is reduced to Cu(s).
The blue color of Cu(s) lessens as it is reduced to Cu2+(aq).
The blue color of Zn2+(aq) lessens as it is reduced to Zn(s).
The blue color of Zn(s) lessens as it is reduced to Zn(s).
Electrochemistry
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
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
Cu, because it loses electrons in the chemical reaction.
Cu, because it gains electrons in the chemical reaction.
Zn, because it loses electrons in the chemical reaction.
Zn, because it gains electrons in the chemical reaction.
Electrochemistry
Voltaic Cells
• A typical cell looks like this.
• The oxidation occurs at the
anode.
• The reduction occurs at the
cathode.
• 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
© 2012 Pearson Education, Inc.
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.
• 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.
© 2012 Pearson Education, Inc.
Electrochemistry
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
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
Zn2+(aq) ions formed in the reaction completely migrate through the salt bridge to the right compartment to
maintain electrical charge balance.
Anions from the right compartment migrate through the salt bridge to the left compartment to maintain
electrical charge balance.
NO3-(aq) ions in the salt bridge migrate into the left compartment and Zn 2+(aq) ions migrate into the salt
bridge to maintain electrical charge balance.
Electrochemistry
+
2+
Na (aq) ions in the salt bridge migrate into the left compartment and Zn (aq) ions migrate into the salt
bridge to maintain electrical charge balance.
Sample Exercise 20.4 Describing a Voltaic Cell
The oxidation-reduction reaction
Cr2O72(aq) + 14 H+(aq) + 6 I(aq)  2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
is spontaneous. 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 migration, the direction of ion migration, and the signs of the electrodes.
Practice Exercise
The two half-reactions in a voltaic cell are
Zn(s)  Zn2+(aq) + 2 e
ClO3(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 isElectrochemistry
consumed in the cell reaction? (c) Which electrode is positive?
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.
• 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
© 2012 Pearson Education, Inc.
Cell Potential
Cell potential is measured in volts (V).
J
1V=1
C
Electrochemistry
© 2012 Pearson Education, Inc.
Standard Cell Potential
The cell potential under standard conditions is
denoted E cell
Zn(s) + Cu2+(aq, 1M)  Zn2+(aq,1M) + Cu(s)
E cell
= +1.10 v
Electrochemistry
© 2012 Pearson Education, Inc.
Standard Reduction Potentials
Reduction
potentials for many
electrodes have
been measured
and tabulated.
Electrochemistry
© 2012 Pearson Education, Inc.
A. Yes
B. No
Electrochemistry
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
© 2012 Pearson Education, Inc.
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) + 2e  H2(g, 1 atm)
Electrochemistry
© 2012 Pearson Education, Inc.
Cell Potentials
• For the oxidation in this cell,
 = 0.76 V
Ered
• For the reduction,
 = +0.34 V
Ered
Electrochemistry
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
Na+ migrates into the cathode half-cell to balance electrical charge by balancing the
increase in negative charge of nitrate ions formed in the reaction.
Na+ migrates into the cathode half-cell to balance electrical charge by balancing the
decreasing positive charge of hydrogen ions consumed in the reaction.
Na+ migrates into the cathode half-cell to balance electrical charge by balancing the
decreasing positive charge of zinc ions consumed in the reaction.
Na+ migrates into the cathode half-cell to balance electrical charge by balancing the
increasing positive charge of hydrogen ions formed in the reaction.
Electrochemistry
Cell Potentials
 = Ered
 (cathode)  Ered
 (anode)
Ecell
= +0.34 V  (0.76 V)
= +1.10 V
Electrochemistry
© 2012 Pearson Education, Inc.
A. 1 atm pressure for Cl2(g) and 1 M solution for
Cl–(aq).
–
B. 1 M solution for Cl2(g) and for Cl (aq).
C. 1 atm pressure for Cl2(g) and for Cl–(aq).
D. 1 torr pressure for Cl2(g) and 1 M solution for
Cl–(aq).
Electrochemistry
Sample Exercise 20.5 Calculating
from
For the Zn-Cu2+ voltaic cell shown in Figure 20.5, we have
Given that the standard reduction potential of Zn2+ to Zn(s) is 0.76 V,
calculate the
for the reduction of Cu2+ to Cu:
Cu2+(aq, 1 M) + 2 e  Cu(s)
Practice Exercise
The standard cell potential is 1.46 V for a voltaic
cell based on the following half-reactions:
In+(aq)  In3+(aq) + 2 e
Br2(l) + 2 e  2Br(aq)
Using Table 20.1, calculate
for the reduction of
3+
+
In to In .
Electrochemistry
A. The electrode with the more positive Eredo value is
associated with the cathode.
B. The electrode with the less positive (more negative) Eredo
value is associated with the cathode.
Electrochemistry
Sample Exercise 20.6 Calculating
from
Use Table 20.1 to calculate for the voltaic cell described in Sample Exercise 20.4, which is based on the
reaction Cr2O72(aq) + 14 H+(aq) + 6 I(aq)  2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
Practice Exercise
Using data in Table 20.1, calculate the standard emf for
a cell that employs the overall cell reaction
2 Al(s) + 3 I2(s)  2 Al3+(aq) + 6 I(aq).
Electrochemistry
Sample Exercise 20.7 Determining Half-Reactions at Electrodes and
Calculating Cell Potentials
A voltaic cell is based on the two standard half-reactions
Cd2+(aq) + 2 e  Cd(s)
Sn2+(aq) + 2 e  Sn(s)
Use data in Appendix E to determine (a) which half-reaction occurs at the cathode and which occurs at
the anode and (b) the standard cell potential.
Practice Exercise
A voltaic cell is based on a Co2+/Co half-cell and an AgCl/Ag half-cell.
(a) What half-reaction occurs at the anode? (b) What is the standard cell potential?
Electrochemistry
Oxidizing and Reducing Agents
• The strongest oxidizers
have the most positive
reduction potentials.
• The strongest reducers
have the most negative
reduction potentials.
Electrochemistry
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
A strong oxidizing agent is one that readily gains
electrons and thus has a large positive value of Eredo
for its half-reaction. The larger the positive value of
Eredo the greater is the tendency for the reduction
half-reaction to occur.
A strong oxidizing agent is one that readily gains
electrons and thus has a large negative value of Eredo
for its half-reaction. The larger the negative value of
Eredo the greater is the tendency for the reduction
half-reaction to occur.
A strong oxidizing agent is one that readily loses
electrons and thus has a large positive value of Eredo
for its half-reaction. The larger the positive value Eredo
the greater is the tendency for the reduction halfreaction to occur.
A strong oxidizing agent is one that readily loses
electrons and thus has a large positive value of Eredo
for its half-reaction. The larger the value Eredo the
greater is the tendency for the reduction half-reaction
to occur.
Electrochemistry
Oxidizing and Reducing Agents
The greater the
difference between
the two, the greater
the voltage of the cell.
Electrochemistry
© 2012 Pearson Education, Inc.
Sample Exercise 20.8 Determining Relative Strengths of Oxidizing
Agents
Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents:
NO3 (aq), Ag+ (aq), Cr2O72 (aq).
Practice Exercise
Using Table 20.1, rank the following species from the strongest
to the weakest reducing agent: I (aq), Fe(s), Al(s).
Electrochemistry
Sample Exercise 20.9 Determining Spontaneity
Use Table 20.1 to 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)
Practice Exercise
Using the standard reduction potentials listed in Appendix E, determine which of the following reactions are
spontaneous under standard conditions:
(a) I2(s) + 5 Cu2+(aq) + 6 H2O(l)  2 IO3(aq) + 5 Cu(s) + 12 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
A.
B.
C.
D.
Hg(l) is the stronger reducing agent because Hg(l)
has the more negative Eredo.
Pb(s) is the stronger reducing agent because Pb(s)
has the more negative Eredo.
Hg(l) is the stronger reducing agent because Hg(l)
is a liquid metal and this changes Eredo.
Pb(s) is the stronger reducing agent because Pb(s)
is further right in the periodic table than Hg(l).
Electrochemistry
Free Energy
G for a redox reaction can be found by
using the equation
G = nFE
where n is the number of moles of
electrons transferred, and F is a
constant, the Faraday:
1 F = 96,485 C/mol = 96,485 J/V-mol
Electrochemistry
© 2012 Pearson Education, Inc.
Free Energy
Under standard conditions,
G = nFE
Electrochemistry
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
The number of moles of the oxidizing agent in the balanced chemical equation.
The number of moles of the reducing agent in the balanced chemical equation.
The net number of moles of gas reacting in the balanced chemical equation.
The number of moles of electrons transferred in the balanced chemical equation.
Electrochemistry
Sample Exercise 20.10 Using Standard Reduction Potentials to
Calculate G and K
(a) Use the standard reduction potentials in Table 20.1 to calculate the standard free-energy change,
G,and the equilibrium constant, K, at 298 K for the reaction
(b) Suppose the reaction in part (a) is written
What are the values of E, G, and K when the reaction is written in this way?
Practice Exercise
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? (b) Use the data in Appendix E to calculate G. (c) Calculate K at T = 298 K.
Electrochemistry
Nernst Equation
• Remember that
G = G + RT ln Q
• This means
nFE = nFE + RT ln Q
• Dividing both sides by nF, we get the Nernst
equation:
RT
ln Q
E = E 
nF
• or, using base-10 logarithms,
E = E 
2.303RT
nF
log Q
Electrochemistry
© 2012 Pearson Education, Inc.
Nernst Equation
At room temperature (298 K),
2.303RT
= 0.0592 V
F
Thus, the equation becomes
0.0592
log Q
E = E 
n
Electrochemistry
© 2012 Pearson Education, Inc.
Sample Exercise 20.11 Cell Potential under Nonstandard Conditions
Calculate the emf at 298 K generated by a voltaic cell in which the reaction is
Cr2O72(aq) + 14 H+(aq) + 6 I(aq)  2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
when [Cr2O72] = 2.0 M, [H+] = 1.0 M, [I] = 1.0 M, and [Cr3+] = 1.0  105 M.
Practice Exercise
Calculate the emf generated by the cell described in the practice exercise accompanying Sample
Electrochemistry
Exercise 20.6 when [Al3+] = 4.0  103 M and [I] = 0.0100 M.
Sample Exercise 20.12 Calculating Concentrations in a Voltaic Cell
If the potential of a Zn-H+ cell (like that in Figure 20.9) is 0.45 V at 25 C when [Zn2+] = 1.0 M and
PH2 = 1.0 atm, what is the H+ concentration?
Practice Exercise
What is the pH of the solution in the cathode half-cell in Figure
20.9 when PH2 = 1.0 atm, [Zn2+] in the anode half-cell is 0.10 M,
and the cell emf is 0.542 V?
Electrochemistry
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.
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
The Ni2+(aq) ions and the anions in the salt bridge migrate toward the anode. The NO3-(aq)
ions and the cations in the salt bridge migrate toward the cathode.
The Ni2+(aq) ions and the cations in the salt bridge migrate toward the anode. The NO3-(aq)
ions and the anions in the salt bridge migrate toward the cathode.
The Ni2+(aq) ions and the cations in the salt bridge migrate toward the cathode. The NO3-(aq)
ions and the anions in the salt bridge migrate toward the anode.
The Ni2+(aq) ions and the anions in the salt bridge migrate toward the cathode. The NO3-(aq)
ions and the cations in the salt bridge migrate toward the anode.
Electrochemistry
Sample Exercise 20.13 Determining pH Using a Concentration Cell
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 (PH2 = 1.00 atm, [H+] =
1.00 M). At 298 K the measured cell potential is 0.211 V, and the electrical current is observed to flow
from electrode 1 through the external circuit to electrode 2. Calculate for the solution at electrode 1.What
is the pH of the solution?
Practice Exercise
A concentration cell is constructed with two Zn(s)Zn2+(aq) half-cells. In one half-cell [Zn2+] = 1.35 M,
and in the other [Zn2+] = 3.75  104 M. (a) Which half-cell is the anode? (b) What is the emf of the
cell?
Electrochemistry
Applications of
Oxidation-Reduction
Reactions
Electrochemistry
© 2012 Pearson Education, Inc.
Batteries
Electrochemistry
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
+1
+2
+4
+6
Electrochemistry
Alkaline Batteries
Electrochemistry
© 2012 Pearson Education, Inc.
Hydrogen Fuel Cells
Electrochemistry
© 2012 Pearson Education, Inc.
Corrosion and…
Electrochemistry
© 2012 Pearson Education, Inc.
…Corrosion Prevention
Electrochemistry
© 2012 Pearson Education, Inc.
A.
B.
C.
D.
O2
Fe
Fe2+
H+
Electrochemistry
A.
B.
C.
D.
Al and Cu
Cu and Ni
Al and Zn
Ni and Zn
Electrochemistry
Electrolysis
• Voltaic cells are spontaneous. Electrolytic cells are
non-spontaneous, and work by using an applied
electric current.
• A battery or other source acts as an electron
pump.
Electrochemistry
© 2012 Pearson Education, Inc.
Sample Exercise 20.14 Relating Electrical Charge and Quantity
of Electrolysis
Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if
the electrical current is 10.0 A.
Practice Exercise
(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  103 s. (b)
How many seconds would be required to produce 50.0 g of Mg from MgCl2 if
the current is 100.0 A?
Electrochemistry