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Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 19 Electrochemistry
ONE 19.1 REDOX REACTIONS FORM OF ENERGY THAT HAS TREMENDOUS PRACTICAL SIGNIFICANCE IS ELECTRICAL ENERGY. A 19.2 ELECTROCHEMICAL CELLS DAY WITHOUT ELECTRICITY FROM EITHER THE POWER COMPANY OR BATTERIES IS UNIMAGINABLE IN OUR TECHNOLOG- 19.3 STANDARD ELECTRODE POTENTIALS ICAL SOCIETY. 19.4 SPONTANEITY OF REDOX REACTIONS THE AREA OF CHEMISTRY THAT DEALS WITH THE INTER- CONVERSION OF ELECTRICAL ENERGY AND CHEMICAL ENERGY IS ELEC- 19.5 THE EFFECT OF CONCENTRATION ON
CELL EMF TROCHEMISTRY. 19.6 BATTERIES ELECTROCHEMICAL PROCESSES ARE REDOX REACTIONS IN WHICH THE
ENERGY RELEASED BY A SPONTANEOUS REACTION IS CONVERTED TO ELEC- 19.8 ELECTROLYSIS
TRICITY OR IN WHICH ELECTRICITY IS USED TO DRIVE A NONSPONTANEOUS CHEMICAL REACTION. THIS THE LATTER TYPE IS CALLED ELECTROLYSIS. CHAPTER EXPLAINS THE FUNDAMENTAL PRINCIPLES AND AP- PLICATIONS OF ELECTROCHEMICAL CELLS, THE THERMODYNAMICS OF ELECTROCHEMICAL REACTIONS, AND THE CAUSE AND PREVENTION OF CORROSION BY ELECTROCHEMICAL MEANS. SOME SIMPLE ELECTROLYTIC PROCESSES AND THE QUANTITATIVE ASPECTS OF ELECTROLYSIS ARE ALSO
DISCUSSED. 757 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 758 ELECTROCHEMISTRY 19.1 REDOX REACTIONS Electrochemistry is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy. Electrochemical processes are redox (oxidationreduction) reactions in which the energy released by a spontaneous reaction is converted to electricity or in which electrical energy is used to cause a nonspontaneous
reaction to occur. Although redox reactions were discussed in Chapter 4, it is helpful
to review some of the basic concepts that will come up again in this chapter.
In redox reactions electrons are transferred from one substance to another. The reaction between magnesium metal and hydrochloric acid is an example of a redox reaction:
0 1 Mg(s) 2 0 2HCl(aq) 88n MgCl2(aq) H2(g) Recall that the numbers written above the elements are the oxidation numbers of the
elements. The loss of electrons by an element during oxidation is marked by an increase in the element’s oxidation number. In reduction, there is a decrease in oxidation
number resulting from a gain of electrons by an element. In the above reaction Mg
metal is oxidized and H ions are reduced; the Cl ions are spectator ions. BALANCING REDOX EQUATIONS Equations for redox reactions like the one above are relatively easy to balance. However,
in the laboratory we often encounter more complex redox reactions involving oxoanions such as chromate (CrO2 ), dichromate (Cr2O2 ), permanganate (MnO4 ), nitrate
(NO3 ), and sulfate (SO2 ). In principle we can balance any redox equation using the
procedure outlined in Section 3.7, but there are some special techniques for handling
redox reactions, techniques that also give us insight into electron transfer processes.
Here we will discuss one such procedure, called the ion-electron method. In this approach, the overall reaction is divided into two half-reactions, one for oxidation and
one for reduction. The equations for the two half-reactions are balanced separately and
then added together to give the overall balanced equation.
Suppose we are asked to balance the equation showing the oxidation of Fe2 ions
to Fe3 ions by dichromate ions (Cr2O2 ) in an acidic medium. As a result, the Cr2O2
ions are reduced to Cr3 ions. The following steps will help us balance the equation.
Step 1. Write the unbalanced equation for the reaction in ionic form.
Fe2 Cr2O2 88n Fe3
7 Cr3 Step 2. Separate the equation into two half-reactions.
2 3 Fe2 88n Fe3 Oxidation:
6 3 Reduction: Cr2O2 88n Cr3
7 Step 3. Balance the atoms other than O and H in each half-reaction separately.
The oxidation half-reaction is already balanced for Fe atoms. For the reduction halfreaction we multiply the Cr3 by 2 to balance the Cr atoms.
Cr2O2 88n 2Cr3
7 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.1 REDOX REACTIONS 759 Step 4. For reactions in an acidic medium, add H2O to balance the O atoms and H
to balance the H atoms.
Since the reaction takes place in an acidic medium, we add seven H2O molecules to
the right-hand side of the reduction half-reaction to balance the O atoms:
Cr2O2 88n 2Cr3
7 7H2O To balance the H atoms, we add fourteen H ions on the left-hand side:
Cr2O2 88n 2Cr3
7 14H 7H2O Step 5. Add electrons to one side of each half-reaction to balance the charges. If necessary, equalize the number of electrons in the two half-reactions by multiplying one
or both half-reactions by appropriate coefficients.
For the oxidation half-reaction we write
Fe2 88n Fe3 e We added an electron to the right-hand side so that there is a 2 charge on each side
of the half-reaction.
In the reduction half-reaction there are net twelve positive charges on the lefthand side and only six positive charges on the right-hand side. Therefore, we add six
electrons on the left.
7 6e 88n 2Cr3 7H2O To equalize the number of electrons in both half-reactions, we multiply the oxidation
half-reaction by 6:
6Fe2 88n 6Fe3 6e Step 6. Add the two half-reactions together and balance the final equation by inspection. The electrons on both sides must cancel.
The two half-reactions are added to give
7 14H 6Fe2 6e 88n 2Cr3 6Fe3 7H2O 6e The electrons on both sides cancel, and we are left with the balanced net ionic equation:
7 6Fe2 88n 2Cr3 6Fe3 7H2O Step 7. Verify that the equation contains the same types and numbers of atoms and the
same charges on both sides of the equation.
A final check shows that the resulting equation is “atomically” and “electrically” balanced.
For reaction in a basic medium, we would balance the atoms as we did in step 4
for an acidic medium. Then, for every H ion we would add an equal number of OH
ions to both sides of the equation. Where H and OH appeared on the same side of
the equation, we would combine the ions to give H2O. The following example illustrates the use of this procedure.
EXAMPLE 19.1 Write a balanced ionic equation to represent the oxidation of iodide ion (I ) by permanganate ion (MnO4 ) in basic solution to yield molecular iodine (I2) and manganese(IV) oxide (MnO2). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 760 ELECTROCHEMISTRY Answer Step 1: The unbalanced equation is
MnO4 I 88n MnO2 I2 Step 2: The two half-reactions are
1 Oxidation: I 0 88n I2
7 4 Reduction: MnO4 88n MnO2
Step 3: To balance the I atoms in the oxidation half-reaction, we write
2I 88n I2
Step 4: In the reduction half-reaction, we add two H2O molecules on the right to bal- ance the O atoms:
MnO4 88n MnO2 2H2O To balance the H atoms, we add four H ions on the left:
MnO4 4H 88n MnO2 2H2O Since the reaction occurs in a basic medium and there are four H ions, we
add four OH ions to both sides of the equation:
MnO4 4H 4OH 88n MnO2 2H2O 4OH Combining the H and OH ions to form H2O and canceling 2H2O from
both sides, we write
MnO4 2H2O 88n MnO2 4OH Step 5: Next, we balance the charges of the two half-reactions as follows:
2I 88n I2
MnO4 2H2O 2e 3e 88n MnO2 4OH To equalize the number of electrons, we multiply the oxidation half-reaction
by 3 and the reduction half-reaction by 2:
6I 88n 3I2
2MnO4 4H2O 6e 6e 88n 2MnO2 8OH Step 6: The two half-reactions are added together to give
6I 2MnO4 4H2O 6e 88n 3I2 2MnO2 8OH 6e After canceling the electrons on both sides, we obtain
6I 2MnO4 4H2O 88n 3I2 2MnO2 8OH Step 7: A final check shows that the equation is balanced in terms of both atoms and charges. Similar problems: 19.1, 19.2. PRACTICE EXERCISE Balance the following equation for the reaction in an acidic medium by the ionelectron method:
Fe2 Back Forward Main Menu TOC MnO4 88n Fe3 Study Guide TOC Mn2 Textbook Website MHHE Website 19.2 19.2 ELECTROCHEMICAL CELLS 761 ELECTROCHEMICAL CELLS In Section 4.4 we saw that when a piece of zinc metal is placed in a CuSO4 solution,
Zn is oxidized to Zn2 ions while Cu2 ions are reduced to metallic copper (see Figure
Cu2 (aq) 88n Zn2 (aq) Zn(s) Alphabetically anode precedes
cathode and oxidation precedes
reduction. Therefore, anode is
where oxidation occurs and
cathode is where reduction
Half-cell reactions are similar to
the half-reactions discussed
above. Cu(s) The electrons are transferred directly from the reducing agent (Zn) to the oxidizing
agent (Cu2 ) in solution. However, if we physically separate the oxidizing agent from
the reducing agent, the transfer of electrons can take place via an external conducting
medium. As the reaction progresses, it sets up a constant flow of electrons and hence
generates electricity (that is, it produces electrical work such as driving an electric motor).
An electrochemical cell is the experimental apparatus for generating electricity
through the use of a spontaneous redox reaction. (An electrochemical cell is sometimes referred to as a galvanic cell or voltaic cell, after the scientists Luigi Galvani and
Allessandro Volta, who constructed early versions of this device.) Figure 19.1 shows
the essential components of an electrochemical cell. A zinc bar is immersed in a ZnSO4
solution, and a copper bar is immersed in a CuSO4 solution. The cell operates on the
principle that the oxidation of Zn to Zn2 and the reduction of Cu2 to Cu can be
made to take place simultaneously in separate locations with the transfer of electrons
between them occurring through an external wire. The zinc and copper bars are called
electrodes. This particular arrangement of electrodes (Zn and Cu) and solutions (ZnSO4
and CuSO4) is called the Daniell cell. By definition, the anode in an electrochemical
cell is the electrode at which oxidation occurs and the cathode is the electrode at which
For the Daniell cell, the half-cell reactions, that is, the oxidation and reduction
reactions at the electrodes, are
Zn(s) 88n Zn2 (aq) Zn electrode (anode):
2 Cu electrode (cathode): Cu (aq) 2e 2e 88n Cu(s) Note that unless the two solutions are separated from each other, the Cu2 ions will
react directly with the zinc bar: FIGURE 19.1 An electrochemical cell. The salt bridge (an inverted U tube) containing a KCl
solution provides an electrically
conducting medium between two
solutions. The openings of the U
tube are loosely plugged with
cotton balls to prevent the KCl solution from flowing into the containers while allowing the anions
and cations to move across.
Electrons flow externally from the
Zn electrode (anode) to the Cu
electrode (cathode). Voltmeter
anode Cl– Cu2+
SO4– ZnSO4 solution Forward Main Menu Copper
cathode K+ Salt bridge Zn2+ Back e– TOC 2
SO4– CuSO4 solution Study Guide TOC Textbook Website MHHE Website 762 ELECTROCHEMISTRY Cu2 (aq) Zn(s) 88n Cu(s) Zn2 (aq) and no useful electrical work will be obtained.
To complete the electrical circuit, the solutions must be connected by a conducting medium through which the cations and anions can move from one electrode compartment to the other. This requirement is satisfied by a salt bridge, which, in its simplest form, is an inverted U tube containing an inert electrolyte solution, such as KCl
or NH4NO3, whose ions will not react with other ions in solution or with the electrodes
(see Figure 19.1). During the course of the overall redox reaction, electrons flow externally from the anode (Zn electrode) through the wire and voltmeter to the cathode
(Cu electrode). In the solution, the cations (Zn2 , Cu2 , and K ) move toward the
cathode, while the anions (SO2 and Cl ) move toward the anode. Without the salt
bridge connecting the two solutions, the buildup of positive charge in the anode compartment (due to the formation of Zn2 ions) and negative charge in the cathode compartment (created when some of the Cu2 ions are reduced to Cu) would quickly prevent the cell from operating.
An electric current flows from the anode to the cathode because there is a difference in electrical potential energy between the electrodes. This flow of electric current
is analogous to that of water down a waterfall, which occurs because there is a difference in gravitational potential energy, or the flow of gas from a high-pressure region
to a low-pressure region. Experimentally the difference in electrical potential between
the anode and the cathode is measured by a voltmeter (Figure 19.2) and the reading
(in volts) is called cell voltage. However, two other terms, electromotive force or emf
(E) and cell potential are also used to denote cell voltage. We will see that the voltage
of a cell depends not only on the nature of the electrodes and the ions, but also on the
concentrations of the ions and the temperature at which the cell is operated.
The conventional notation for representing electrochemical cells is the cell diagram. For the Daniell cell shown in Figure 19.1, if we assume that the concentrations
of Zn2 and Cu2 ions are 1 M, the cell diagram is
Zn(s)Zn2 (1 M ) Cu2 (1 M )Cu(s) The single vertical line represents a phase boundary. For example, the zinc electrode
is a solid and the Zn2 ions (from ZnSO4) are in solution. Thus we draw a line between Zn and Zn2 to show the phase boundary. The double vertical lines denote the
FIGURE 19.2 Practical setup
of the electrochemical cell described in Figure 19.1. Note the
U tube (salt bridge) connecting
the two beakers. When the concentrations of ZnSO4 and CuSO4
are 1 molar (1 M) at 25°C, the
cell voltage is 1.10 V. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.3 763 STANDARD ELECTRODE POTENTIALS salt bridge. By convention, the anode is written first, to the left of the double lines and
the other components appear in the order in which we would encounter them in moving from the anode to the cathode. 19.3 The choice of an arbitrary reference for measuring electrode
potential is analogous to choosing
the surface of the ocean as the
reference for altitude, calling it
zero meters, and then referring to
any terrestrial altitude as being a
certain number of meters above
or below sea level.
Standard states are defined in
Table 18.2. STANDARD ELECTRODE POTENTIALS When the concentrations of the Cu2 and Zn2 ions are both 1.0 M, we find that the
voltage or emf of the Daniell cell is 1.10 V at 25°C (see Figure 19.2). This voltage
must be related directly to the redox reactions, but how? Just as the overall cell reaction can be thought of as the sum of two half-cell reactions, the measured emf of the
cell can be treated as the sum of the electrical potentials at the Zn and Cu electrodes.
Knowing one of these electrode potentials, we could obtain the other by subtraction
(from 1.10 V). It is impossible to measure the potential of just a single electrode, but
if we arbitrarily set the potential value of a particular electrode at zero, we can use it
to determine the relative potentials of other electrodes. The hydrogen electrode, shown
in Figure 19.3, serves as the reference for this purpose. Hydrogen gas is bubbled into
a hydrochloric acid solution at 25°C. The platinum electrode has two functions. First,
it provides a surface on which the dissociation of hydrogen molecules can take place:
H2 88n 2H 2e Second, it serves as an electrical conductor to the external circuit.
Under standard-state conditions (when the pressure of H2 is 1 atm and the concentration of the HCl solution is 1 M ), the potential for the reduction of H at 25°C
is taken to be exactly zero:
2H (1 M ) 2e 88n H2(1 atm) E° 0V The superscript “o” denotes standard-state conditions, and E° is the standard reduction potential, or the voltage associated with a reduction reaction at an electrode when
all solutes are 1 M and all gases are at 1 atm. Thus, the standard reduction potential
of the hydrogen electrode is defined as zero. The hydrogen electrode is called the standard hydrogen electrode (SHE).
We can use the SHE to measure the potentials of other kinds of electrodes. For
example, Figure 19.4(a) shows an electrochemical cell with a zinc electrode and a SHE.
In this case the zinc electrode is the anode and the SHE is the cathode. We deduce this
fact from the decrease in mass of the zinc electrode during the operation of the cell,
which is consistent with the loss of zinc to the solution caused by the oxidation reaction:
FIGURE 19.3 A hydrogen
electrode operating under standard-state conditions. Hydrogen
gas at 1 atm is bubbled through
a 1 M HCl solution. The platinum
electrode is part of the hydrogen
electrode. H2 gas at 1 atm Pt electrode
1M HCl Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 764 ELECTROCHEMISTRY Voltmeter
e– e– H2 gas at 1 atm Zn H2 gas at 1 atm Cu
Salt bridge Salt bridge Pt electrode
1M ZnSO4 e– Pt electrode 1M HCl 1M CuSO4 1M HCl Hydrogen electrode
FIGURE 19.4 (a) A cell consisting of a zinc electrode and a hydrogen electrode. (b) A cell conZn(s) 88n Zn2 (aq) 2e
sisting of a copper electrode and
a hydrogen electrode. Both cells
The cell diagram is
are operating under standardZn(s)Zn2 (1 M ) H (1 M )H2(1 atm)Pt(s) state conditions. Note that in (a)
the SHE acts as the cathode, but
As mentioned earlier, the Pt electrode provides the surface on which the reduction takes
in (b) it is the anode.
Zinc electrode place. When all the reactants are in their standard states (that is, H2 at 1 atm, H and
Zn2 ions at 1 M ), the emf of the cell is 0.76 V at 25°C. We can write the half-cell
reactions as follows:
Zn(s) 88n Zn2 (1 M ) Anode (oxidation):
Cathode (reduction): 2H (1 M )
Overall: Zn(s) 2e E°
Zn/Zn2 2e 88n H2(1 atm) 2H (1 M ) 88n Zn2 (1 M ) EH
H2(1 atm) /H2 E°
cell where EZn/Zn2 is the standard oxidation potential, or voltage under standard state
conditions for the oxidation half-reaction, and EH /H2 is the standard reduction potential previously defined for the SHE. The subscript Zn/Zn2 means Zn 88n Zn2
2e , and the subscript H /H2 means 2H
2e 88n H2. The standard emf of the
cell, Ecell, is the sum of the standard oxidation potential and the standard reduction
red (19.1) where the subscripts “ox” and “red” indicate oxidation and reduction, respectively. For
the Zn-SHE cell,
° 0.76 V E°
Zn/Zn2 0 /H2 Thus the standard oxidation potential of zinc is 0.76 V.
We can evaluate the standard reduction potential of zinc, E ° 2
the oxidation half-reaction:
Zn2 (1 M ) 2e 88n Zn(s) /Zn, E° 2
Zn by reversing
/Zn 0.76 V Remember that whenever a half-reaction is reversed, E° changes sign.
The standard electrode potentials of copper can be obtained in a similar fashion,
by using a cell with a copper electrode and a SHE [Figure 19.4(b)]. In this case, the
copper electrode is the cathode because its mass increases during the operation of the
cell, as is consistent with the reduction reaction: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.3 Cu2 (aq) STANDARD ELECTRODE POTENTIALS 765 2e 88n Cu(s) The cell diagram is
Pt(s)H2(1 atm)H (1 M ) Cu2 (1 M )Cu(s) and the half-cell reactions are
Anode (oxidation): H2(1 atm) 88n 2H (1 M )
Cu2 (1 M ) Cathode (reduction): 2 Overall: H2(1 atm) 2e E ° 2/H
H 2e 88n Cu(s) Cu (1 M ) 88n 2H (1 M ) E Cu2
Cu(s) /Cu E°
cell Under standard-state conditions and at 25°C, the emf of the cell is 0.34 V, so we write
cell E H2/H
° E Cu2
° 0 0.34 V /Cu E Cu2
° /Cu Thus the standard reduction potential of copper is 0.34 V and its standard oxidation
potential, E °
Cu/Cu2 , is
For the Daniell cell shown in Figure 19.1, we can now write
Zn(s) 88n Zn2 (1 M ) Anode (oxidation):
2 Cathode (reduction): Cu (1 M )
2 Overall: Zn(s) 2e E°
Zn/Zn2 2e 88n Cu(s)
2 Cu (1 M ) 88n Zn (1 M ) E Cu2
Cu(s) /Cu E°
cell The emf of the cell is†
Zn/Zn2 E Cu2
° 0.76 V 0.34 V /Cu 1.10 V The activity series in Figure 4.15
is based on data given in
This example illustrates how we can use the sign of the emf of a cell to predict
the spontaneity of a redox reaction. Under standard-state conditions for reactants and
products, the redox reaction is spontaneous in the forward direction if the standard emf
of the cell is positive. If it is negative, the reaction is spontaneous in the opposite direction. It is important to keep in mind that a negative E ° does not mean that a recell
action will not occur if the reactants are mixed at 1 M concentrations. It merely means
that the equilibrium of the redox reaction, when reached, will lie to the left. We will
examine the relationships among E ° , G°, and K later in this chapter.
Table 19.1 lists standard reduction potentials for a number of half-cell reactions.
By definition, the SHE has an E° value of 0.00 V. Above the SHE the negative standard reduction potentials increase, and below it the positive standard reduction potentials increase. It is important to know the following points about the table:
The E° values apply to the half-cell reactions as read in the forward (left to right)
• The more positive E° is, the greater the tendency for the substance to be reduced.
For example, the half-cell reaction
• F2(1 atm) 2e 88n 2F (1 M ) E° 2.87 V has the highest positive E° value among all the half-cell reactions. Thus F2 is the
† Some chemists prefer to calculate the standard emf of a cell as follows:
anode where both E°
cathode and E°
anode denote the standard reduction potentials, respectively. Thus for the Daniell cell, we would
0.34 V ( 0.76 V) 1.10 V.
cell Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ELECTROCHEMISTRY TABLE 19.1 Standard Reduction Potentials at 25°C* Increasing strength as oxidizing agent E°(V) Li (aq) e 88n Li(s)
K (aq) e 88n K(s)
Ba2 (aq) 2e 88n Ba(s)
Sr2 (aq) 2e 88n Sr(s)
Ca2 (aq) 2e 88n Ca(s)
Na (aq) e 88n Na(s)
Mg2 (aq) 2e 88n Mg(s)
Be2 (aq) 2e 88n Be(s)
Al3 (aq) 3e 88n Al(s)
Mn2 (aq) 2e 88n Mn(s)
2H2O 2e 88n H2(g) 2OH (aq)
Zn2 (aq) 2e 88n Zn(s)
Cr3 (aq) 3e 88n Cr(s)
Fe2 (aq) 2e 88n Fe(s)
Cd2 (aq) 2e 88n Cd(s)
PbSO4(s) 2e 88n Pb(s) SO2 (aq)
Co2 (aq) 2e 88n Co(s)
Ni2 (aq) 2e 88n Ni(s)
Sn2 (aq) 2e 88n Sn(s)
Pb2 (aq) 2e 88n Pb(s)
2H (aq) 2e 88n H2(g)
Sn4 (aq) 2e 88n Sn2 (aq)
Cu2 (aq) e 88n Cu (aq)
SO2 (aq) 4H (aq) 2e 88n SO2(g) 2H2O
AgCl(s) e 88n Ag(s) Cl (aq)
Cu2 (aq) 2e 88n Cu(s)
O2(g) 2H2O 4e 88n 4OH (aq)
I2(s) 2e 88n 2I (aq)
MnO4 (aq) 2H2O 3e 88n MnO2(s) 4OH (aq)
O2(g) 2H (aq) 2e 88n H2O2(aq)
Fe3 (aq) e 88n Fe2 (aq)
Ag (aq) e 88n Ag(s)
Hg2 (aq) 2e 88n 2Hg(l)
2Hg2 (aq) 2e 88n Hg2 (aq)
NO3 (aq) 4H (aq) 3e 88n NO(g) 2H2O
Br2(l ) 2e 88n 2Br (aq)
O2(g) 4H (aq) 4e 88n 2H2O
MnO2(s) 4H (aq) 2e 88n Mn2 (aq) 2H2O
Cr2O2 (aq) 14H (aq) 6e 88n 2Cr3 (aq) 7H2O
Cl2(g) 2e 88n 2Cl (aq)
Au3 (aq) 3e 88n Au(s)
MnO4 (aq) 8H (aq) 5e 88n Mn2 (aq) 4H2O
Ce4 (aq) e 88n Ce3 (aq)
PbO2(s) 4H (aq) SO2 (aq) 2e 88n PbSO4(s) 2H2O
H2O2(aq) 2H (aq) 2e 88n 2H2O
Co3 (aq) e 88n Co2 (aq)
O3(g) 2H (aq) 2e 88n O2(g) H2O(l)
F2(g) 2e 88n 2F (aq) 3.05
2.87 m77777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777 m77777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777 HALF-REACTION Increasing strength as reducing agent 766 *For all half-reactions the concentration is 1 M for dissolved species and the pressure is 1 atm for gases. These are the
standard-state values. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.3 767 STANDARD ELECTRODE POTENTIALS strongest oxidizing agent because it has the greatest tendency to be reduced. At the
other extreme is the reaction
Li (1 M ) e 88n Li(s) E° 3.05 V which has the most negative E° value. Thus Li is the weakest oxidizing agent because it is the most difficult species to reduce. Conversely, we say that F is the
weakest reducing agent and Li metal is the strongest reducing agent. Under standard-state conditions, the oxidizing agents (the species on the left-hand side of the
half-reactions in Table 19.1) increase in strength from top to bottom and the reducing agents (the species on the right-hand side of the half-reactions) increase in
strength from bottom to top.
• The half-cell reactions are reversible. Depending on the conditions, any electrode
can act either as an anode or as a cathode. Earlier we saw that the SHE is the cathode (H is reduced to H2) when coupled with zinc in a cell and that it becomes the
anode (H2 is oxidized to H ) when used in a cell with copper.
• Under standard-state conditions, any species on the left of a given half-cell reaction will react spontaneously with a species that appears on the right of any halfcell reaction located above it in Table 19.1. This principle is sometimes called the
diagonal rule. In the case of the Cu/Zn electrochemical cell
Zn2 (1 M )
2 Cu (1 M ) 2e 88n Zn(s) E° 2e 88n Cu(s) E° 0.76 V
2 We see that the substance on the left of the second half-cell reaction is Cu and
the substance on the right in the first half-cell reaction is Zn. Therefore, as we saw
earlier, Zn spontaneously reduces Cu2 to form Zn2 and Cu.
• Changing the stoichiometric coefficients of a half-cell reaction does not affect the
value of E° because electrode potentials are intensive properties. This means that
the value of E° is unaffected by the size of the electrodes or the amount of solutions present. For example,
I2(s) 2e 88n 2I (1 M ) E° 0.53 V E° 0.53 V but E° does not change if we multiply the half-reaction by 2:
• 4e 88n 4I (1 M ) Like H, G, and S, E° is a thermodynamic quantity. Therefore, its sign changes but
its magnitude remains the same when we reverse a reaction. As the following examples show, Table 19.1 enables us to predict the outcome of
redox reactions under standard-state conditions, whether they take place in an electrochemical cell, where the reducing agent and oxidizing agent are physically separated
from each other, or in a beaker, where the reactants are all mixed together.
EXAMPLE 19.2 Predict what will happen if molecular bromine (Br2) is added to a solution containing NaCl and NaI at 25°C. Assume all species are in their standard states.
To predict what redox reaction(s) will take place, we need to compare the
standard reduction potentials for the following half-reactions: Answer I2(s) Back Forward Main Menu TOC 2e 88n 2I (1 M ) Study Guide TOC Textbook Website E° 0.53 V MHHE Website 768 ELECTROCHEMISTRY Br2(l ) 2e 88n 2Br (1 M ) E° 1.07 V Cl2(1 atm) 2e 88n 2Cl (1 M ) E° 1.36 V Applying the diagonal rule we see that Br2 will oxidize I but will not oxidize Cl .
Therefore, the only redox reaction that will occur appreciably under standard-state
Oxidation: 2I (1 M ) 88n I2(s) Reduction: Br2(l ) 2e 88n 2Br (1 M ) Overall: 2I (1 M )
Comment Similar problems: 19.14, 19.17. 2e Br2(l ) 88n I2(s) 2Br (1 M ) Note that the Na ions are inert and do not enter into the redox reac- tion.
PRACTICE EXERCISE Can Sn reduce Zn2 (aq) under standard-state conditions?
EXAMPLE 19.3 A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO3)2 solution and a Ag
electrode in a 1.0 M AgNO3 solution. Calculate the standard emf of this electrochemical cell at 25°C.
Table 19.1 gives the standard reduction potentials of the two electrodes: Answer Mg2 (1 M ) 2e 88n Mg(s) E° e 88n Ag(s) E° Ag (1 M ) 2.37 V
0.80 V Applying the diagonal rule, we see that Ag will oxidize Mg:
Mg(s) 88n Mg2 (1 M ) Anode (oxidation):
Cathode (reduction): 2Ag (1 M ) Overall: Mg(s) 2e 2e 88n 2Ag(s) 2Ag (1 M ) 88n Mg2 (1 M ) 2Ag(s) Note that in order to balance the overall equation we multiplied the reduction of
Ag by 2. We can do so because, as an intensive property, E ° is not affected by
this procedure. We find the emf of the cell by using Equation (19.1) and Table 19.1:
2.37 V E Ag
° /Ag 0.80 V 3.17 V
Similar problems: 19.11, 19.12. Comment The positive value of E ° shows that the overall reaction is spontaneous. PRACTICE EXERCISE 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)3 solution? 19.4 SPONTANEITY OF REDOX REACTIONS Our next step is to see how E ° is related to other thermodynamic quantities such as
G° and K. In an electrochemical cell, chemical energy is converted to electrical en- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.4 SPONTANEITY OF REDOX REACTIONS 769 ergy. Electrical energy in this case is the product of the emf of the cell and the total
electrical charge (in coulombs) that passes through the cell:
electrical energy volts coulombs joules
n is the number of moles of electrons exchanged between the reducing agent and the oxidizing
agent in the overall redox
equation. The total charge is determined by the number of moles of electrons (n) that pass through
the circuit. By definition
total charge nF † where F, the Faraday constant, is the electrical charge contained in 1 mole of electrons. Experiments have shown that 1 faraday is equivalent to 96,487 coulombs, or
96,500 coulombs, rounded off to three significant figures. Thus
1F 96,500 C/mol Since
1J 1C 1V we can also express the units of faraday as
A common device for measuring a
cell’s emf is the potentiometer,
which can match the voltage of
the cell precisely without actually
draining any current from it. 96,500 J/V mol The measured emf is the maximum voltage that the cell can achieve. This value
is used to calculate the maximum amount of electrical energy that can be obtained from
the chemical reaction. This energy is used to do electrical work (wele), so
nFEcell The sign convention for electrical
work is the same as that for P-V
work, discussed in Section 6.7. where wmax is the maximum amount of work that can be done. The negative sign on
the right-hand side indicates that the electrical work is done by the system on the surroundings. In Chapter 18 we defined free energy as the energy available to do work.
Specifically, the change in free energy ( G) represents the maximum amount of useful work that can be obtained from a reaction:
G wmax Therefore we can write
G nFEcell (19.2) Both n and F are positive quantities and G is negative for a spontaneous process, so
Ecell must be positive. For reactions in which reactants and products are in their standard states, Equation (19.2) becomes
G° nFE °
cell (19.3) Here again, E ° is positive for a spontaneous process.
Now we can relate E ° to the equilibrium constant (K ) of a redox reaction. In
Section 18.5 we saw that the standard free-energy change G° for a reaction is related
to its equilibrium constant as follows [see Equation (18.10)]:
† Michael Faraday (1791 – 1867). English chemist and physicist. Faraday is regarded by many as the greatest experimental
scientist of the nineteenth century. He started as an apprentice to a bookbinder at the age of 13, but became interested in
science after reading a book on chemistry. Faraday invented the electric motor and was the first person to demonstrate the
principle governing electrical generators. Besides making notable contributions to the fields of electricity and magnetism,
Faraday also worked on optical activity, and discovered and named benzene. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 770 ELECTROCHEMISTRY G° E°
cell RT ln K E°
= nF ∆G° nFE °
cell ll –n
E°e ll Therefore, if we combine Equations (19.3) and (18.10) we obtain Solving for E °
cell When T ∆ G ° = – RT ln K RT ln K (19.4) 298 K, Equation (19.4) can be simplified by substituting for R and F: K (8.314 J/K mol)(298 K)
n(96500 J/V mol) E°
cell FIGURE 19.5 Relationships
among E° , K, and ∆G°.
cell 0.0257 V
n (19.5) Thus, if any one of the three quantities G°, K, or E ° is known, the other two can
be calculated using Equation (18.10), Equation (19.3), or Equation (19.4) (Figure 19.5).
We summarize the relationships among G°, K, and E ° and characterize the sponcell
taneity of a redox reaction in Table 19.2. For simplicity, we will omit the subscript
“cell” in E° and E.
Examples 19.4 and 19.5 apply Equation (19.5).
EXAMPLE 19.4 Calculate the equilibrium constant for the following reaction at 25°C:
Answer 2Cu2 (aq) 34 Sn2 (aq) 2Cu (aq) The two half-reactions for the overall process are
Sn(s) 88n Sn2 (aq) Oxidation:
2 Reduction: 2Cu (aq) In Table 19.1 we find that E ° 2
E° 2e 88n 2Cu (aq) 0.14 V and E ° 2
Cu /Sn 2e E° 2
Sn/Sn E° 2
Cu 0.14 V /Cu 0.15 V. Thus 0.15 V /Cu 0.29 V Equation (19.5) can be written
ln K In the overall reaction we find n 2. Therefore ln K
K Similar problems: 19.21, 19.22. nE°
0.0257 V (2)(0.29 V)
e22.6 6.5 22.6
109 PRACTICE EXERCISE Calculate the equilibrium constant for the following reaction at 25°C:
Fe2 (aq) Back Forward Main Menu TOC 2Ag(s) 34 Fe(s) Study Guide TOC 2Ag (aq) Textbook Website MHHE Website 19.5 TABLE 19.2
Positive THE EFFECT OF CONCENTRATION ON CELL EMF 771 Relationships Among ∆G°, K, and E°
cell REACTION UNDER STANDARD-STATE CONDITIONS Positive
Nonspontaneous. Reaction is spontaneous in the
reverse direction. EXAMPLE 19.5 Calculate the standard free-energy change for the following reaction at 25°C:
Answer 3Ca2 (1 M ) 88n 2Au3 (1 M ) 3Ca(s) First we break up the overall reaction into half-reactions:
2Au(s) 88n 2Au3 (1 M ) Oxidation:
2 Reduction: 3Ca (1 M ) In Table 19.1 we find that E° u3
A 6e 88n 3Ca(s) 1.50 V and E ° a2
C /Au E° u/Au3
A E° 6e 1.50 V E ° a2
C /Ca 2.87 V. Therefore /Ca 2.87 V 4.37 V Now we use Equation (19.3):
The overall reaction shows that n
∆G° 6, so (6 mol)(96,500 J/V mol)( 4.37 V)
2.53 106 J 2.53 Similar problem: 19.24. nFE ° 103 kJ Comment The large positive value of ∆G° tells us that the reaction is not spontaneous under standard-state conditions at 25°C.
PRACTICE EXERCISE Calculate ∆G° for the following reaction at 25°C:
2Al3 (aq) 19.5 3Mg(s) 12 2Al(s) 3Mg2 (aq) THE EFFECT OF CONCENTRATION ON CELL EMF So far we have focused on redox reactions in which reactants and products are in their
standard states, but standard-state conditions are often difficult, and sometimes impossible, to maintain. Nevertheless, there is a mathematical relationship between the
emf of a cell and the concentration of reactants and products in a redox reaction under non-standard-state conditions. This equation is derived below. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 772 ELECTROCHEMISTRY THE NERNST EQUATION Consider a redox reaction of the type
aA bB 88n cC dD From Equation (18.9)
∆G Since ∆G nFE and ∆G° ∆G° RT ln Q nFE°, the equation can be expressed as
nFE Dividing the equation through by nFE° RT ln Q nF, we get
nF E° (19.6) where Q is the reaction quotient (see Section 14.4). Equation (19.6) is known as the
Nernst† equation. At 298 K, Equation (19.6) can be rewritten as
E 0.0257 V
n E° (19.7) At equilibrium, there is no net transfer of electrons, so E 0 and Q K, where K is
the equilibrium constant.
The Nernst equation enables us to calculate E as a function of reactant and product concentrations in a redox reaction. For example, for the Daniell cell in Figure 19.1
Zn(s) Cu2 (aq) 88n Zn2 (aq) Cu(s) The Nernst equation for this cell at 25°C can be written as
Remember that concentrations of
pure solids (and pure liquids)
do not appear in the
expression for Q. E [Zn2 ]
ln [Cu2 ]
2 1.10 V If the ratio [Zn2 ]/[Cu2 ] is less than 1, ln [Zn2 ]/[Cu2 ] is a negative number, so
that the second term on the right-hand side of the above equation is positive. Under
this condition E is greater than the standard emf E°. If the ratio is greater than 1, E is
smaller than E°.
The following example illustrates the use of the Nernst equation.
EXAMPLE 19.6 Predict whether the following reaction would proceed spontaneously as written at
2 given that [Co ] Fe2 (aq) 88n Co2 (aq)
2 0.15 M and [Fe ] Fe(s) 0.68 M. †
Walter Hermann Nernst (1864 – 1941). German chemist and physicist. Nernst’s work was mainly on electrolyte solution
and thermodynamics. He also invented an electric piano. Nernst was awarded the Nobel Prize in Chemistry in 1920 for his
contribution to thermodynamics. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.5 Answer THE EFFECT OF CONCENTRATION ON CELL EMF 773 The half-reactions are
Co(s) 88n Co2 (aq) Oxidation:
2 Reduction: Fe (aq) In Table 19.1 we find that E ° 2
the standard emf is 2e 88n Fe(s) 0.28 V and E ° 2
Fe /Co E° 2e E°
Co/Co2 E° 2
Fe 0.28 V /Fe 0.44 V. Therefore, /Fe ( 0.44 V) 0.16 V From Equation (19.7) we write
E [Co2 ]
n E° 0.16 V 0.0257 V
0.68 0.16 V 0.019 V 0.14 V
Similar problems: 19.29, 19.30. Since E is negative (or ∆G is positive), the reaction is not spontaneous in the direction written.
PRACTICE EXERCISE Will the following reaction occur spontaneously at 25°C, given that [Fe2 ]
and [Cd2 ] 0.010 M?
Cd(s) Fe2 (aq) 88n Cd2 (aq) 0.60 M Fe(s) Now suppose we want to determine at what ratio of [Co2 ] to [Fe2 ] the reaction in Example 19.6 would become spontaneous. We can use Equation (19.7) as follows:
When E 0, Q K. 0.0257 V
n E° We set E equal to zero, which corresponds to the equilibrium situation.
0 [Co2 ]
ln [Fe2 ]
2 0.16 V
ln [Co2 ] 12.5 [Co2 ]
[Fe2 ] e K 4 or 12.5 10 K
6 Thus for the reaction to be spontaneous, the ratio [Co2 ]/[Fe2 ] must be smaller than
4 10 6.
As the following example shows, if gases are involved in the cell reaction, their
concentrations should be expressed in atm. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 774 ELECTROCHEMISTRY EXAMPLE 19.7 Consider the electrochemical cell shown in Figure 19.4(a). In a certain experiment,
the emf (E) of the cell is found to be 0.54 V at 25°C. Suppose that [Zn2 ] 1.0 M
and PH2 1.0 atm. Calculate the molar concentration of H .
Answer The overall cell reaction is
Zn(s) 2H (? M ) 88n Zn2 (1 M ) H2(1 atm) As we saw earlier (p. 764), the standard emf for the cell is 0.76 V. From Equation
(19.7), we write
0.54 V [Zn2 ]PH2
ln [H ]2
2 E° 0.0257 V
2 0.22 V
17.1 2 ln [H ] 17.1
[H ] 1 ln ln [H ]
Similar problems: 19.32. 1 8.6
2 10 4 M PRACTICE EXERCISE What is the emf of a cell consisting of a Cd/Cd2 half-cell and a Pt/H2/H halfcell if [Cd2 ] 0.20 M, [H ] 0.16 M, and PH2 0.80 atm?
Example 19.7 shows that an electrochemical cell whose cell reaction involves H
ions can be used to measure [H ] or pH. The pH meter described in Section 15.3 is
based on this principle, but for practical reasons the electrodes used in a pH meter are
quite different from the SHE and zinc electrode in the electrochemical cell (Figure
19.6). CONCENTRATION CELLS FIGURE 19.6 A glass electrode that is used in conjunction
with a reference electrode in a
pH meter. Because electrode potential depends on ion concentrations, it is possible to construct
a cell from two half-cells composed of the same material but differing in ion concentrations. Such a cell is called a concentration cell.
Consider a situation in which zinc electrodes are put into two aqueous solutions
of zinc sulfate at 0.10 M and 1.0 M concentrations. The two solutions are connected
by a salt bridge, and the electrodes are joined by a piece of wire in an arrangement
like that shown in Figure 19.1. According to Le Chatelier ’s principle, the tendency for
Zn2 (aq) 2e 88n Zn(s) increases with increasing concentration of Zn2 ions. Therefore, reduction should oc- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.6 BATTERIES 775 cur in the more concentrated compartment and oxidation should take place on the more
dilute side. The cell diagram is
Zn(s)Zn2 (0.10 M ) Zn2 (1.0 M )Zn(s) and the half-reactions are
Zn(s) 88n Zn2 (0.10 M ) Oxidation:
2 Reduction: Zn (1.0 M ) 2e 2e 88n Zn(s) Zn2 (1.0 M ) 88n Zn2 (0.10 M ) Overall: The emf of the cell is
E [Zn2 ]dil
ln [Zn2 ]
conc E° where the subscripts “dil” and “conc” refer to the 0.10 M and 1.0 M concentrations,
respectively. E° for this cell is zero (the same electrode and the same type of ions are
1.0 0 0.0296 V 1 mM 1 10 3 M. The emf of concentration cells is usually small and decreases continually during the
operation of the cell as the concentrations in the two compartments approach each other.
When the concentrations of the ions in the two compartments are the same, E becomes
zero, and no further change occurs.
A biological cell can be compared to a concentration cell for the purpose of calculating its membrane potential. Membrane potential is the electrical potential that exists across the membrane of various kinds of cells, including muscle cells and nerve
cells. It is responsible for the propagation of nerve impulses and heart beat. A membrane potential is established whenever there are unequal concentrations of the same
type of ion in the interior and exterior of a cell. For example, the concentrations of K
ions in the interior and exterior of a nerve cell are 400 mM and 15 mM, respectively.
Treating the situation as a concentration cell and applying the Nernst equation, we can
E E° [K ]ex
1 (0.0257 V) ln 15
400 0.084 V or 84 mV where “ex” and “in” denote exterior and interior. Note that we have set E° 0 because
the same type of ion is involved. Thus an electrical potential of 84 mV exists across
the membrane due to the unequal concentrations of K ions. 19.6 BATTERIES A battery is an electrochemical cell, or a series of combined electrochemical cells, that
can be used as a source of direct electric current at a constant voltage. Although the Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 776 ELECTROCHEMISTRY +
Moist paste of
ZnCl2 and NH4Cl
Layer of MnO2
FIGURE 19.7 Interior section
of a dry cell of the kind used in
flashlights and transistor radios.
Actually, the cell is not completely
dry, as it contains a moist electrolyte paste. operation of a battery is similar in principle to that of the electrochemical cells described in Section 19.2, a battery has the advantage of being completely self-contained
and requiring no auxiliary components such as salt bridges. Here we will discuss several types of batteries that are in widespread use.
THE DRY CELL BATTERY The most common dry cell, that is, a cell without a fluid component, is the Leclanché
cell used in flashlights and transistor radios. The anode of the cell consists of a zinc
can or container that is in contact with manganese dioxide (MnO2) and an electrolyte.
The electrolyte consists of ammonium chloride and zinc chloride in water, to which
starch is added to thicken the solution to a pastelike consistency so that it is less likely
to leak (Figure 19.7). A carbon rod serves as the cathode, which is immersed in the
electrolyte in the center of the cell. The cell reactions are
Zn(s) 88n Zn2 (aq) Anode:
Cathode: 2NH4 (aq) 2MnO2(s) 2e 88n Mn2O3(s) 2NH3(aq) H2O(l ) 2MnO2(s) 88n Zn2 (aq) 2NH4 (aq) Overall: Zn(s) 2e
2NH3(aq) H2O(l ) Mn2O3(s) Actually, this equation is an oversimplification of a complex process. The voltage produced by a dry cell is about 1.5 V.
THE MERCURY BATTERY The mercury battery is used extensively in medicine and electronic industries and is
more expensive than the common dry cell. Contained in a stainless steel cylinder, the
mercury battery consists of a zinc anode (amalgamated with mercury) in contact with
a strongly alkaline electrolyte containing zinc oxide and mercury(II) oxide (Figure
19.8). The cell reactions are
Anode: Zn(Hg) Cathode: HgO(s)
Overall: 2OH (aq) 88n ZnO(s) H2O(l ) Zn(Hg) 2e 88n Hg(l ) HgO(s) 88n ZnO(s) H2O(l ) 2e 2OH (aq)
Hg(l ) Because there is no change in electrolyte composition during operation—the overall
cell reaction involves only solid substances—the mercury battery provides a more constant voltage (1.35 V) than the Leclanché cell. It also has a considerably higher capacity and longer life. These qualities make the mercury battery ideal for use in pacemakers, hearing aids, electric watches, and light meters.
Anode (Zn can) THE LEAD STORAGE BATTERY The lead storage battery commonly used in automobiles consists of six identical cells
joined together in series. Each cell has a lead anode and a cathode made of lead dioxide (PbO2) packed on a metal plate (Figure 19.9). Both the cathode and the anode are
immersed in an aqueous solution of sulfuric acid, which acts as the electrolyte. The
cell reactions are
Electrolyte solution containing KOH
and paste of Zn(OH)2 and HgO
FIGURE 19.8 Interior section
of a mercury battery. Back Forward Main Menu Anode: Pb(s) Cathode: SO2
4 PbO2(s) Overall: Pb(s) TOC 4H (aq) PbO2(s) 4H (aq) Study Guide TOC SO2 (aq) 88n PbSO4(s)
(aq) 2e 88n PbSO4(s) 2SO2 (aq) 88n 2PbSO4(s)
2H2O(l ) Textbook Website MHHE Website 19.6 FIGURE 19.9 Interior section
of a lead storage battery. Under
normal operating conditions, the
concentration of the sulfuric acid
solution is about 38 percent by
mass. BATTERIES 777 Removable cap
Anode Cathode – +
H2SO 4 electrolyte Negative plates (lead grills
filled with spongy lead)
Positive plates (lead grills
filled with PbO 2 ) Under normal operating conditions, each cell produces 2 V; a total of 12 V from the
six cells is used to power the ignition circuit of the automobile and its other electrical
systems. The lead storage battery can deliver large amounts of current for a short time,
such as the time it takes to start up the engine.
Unlike the Leclanché cell and the mercury battery, the lead storage battery is
rechargeable. Recharging the battery means reversing the normal electrochemical reaction by applying an external voltage at the cathode and the anode. (This kind of
process is called electrolysis, see p. 783.) The reactions that replenish the original materials are
Anode: PbSO4(s) Cathode: PbSO4(s)
Overall: 2PbSO4(s) 2e 88n Pb(s) SO2 (aq)
4 2H2O(l ) 88n PbO2(s)
2H2O(l ) 88n Pb(s) 4H (aq)
PbO2(s) SO2 (aq)
4 4H (aq) 2e 2SO2 (aq)
4 The overall reaction is exactly the opposite of the normal cell reaction.
Two aspects of the operation of a lead storage battery are worth noting. First, because the electrochemical reaction consumes sulfuric acid, the degree to which the battery has been discharged can be checked by measuring the density of the electrolyte
with a hydrometer, as is usually done at gas stations. The density of the fluid in a
“healthy,” fully charged battery should be equal to or greater than 1.2 g/mL. Second,
people living in cold climates sometimes have trouble starting their cars because the
battery has “gone dead.” Thermodynamic calculations show that the emf of many electrochemical cells decreases with decreasing temperature. However, for a lead storage
battery, the temperature coefficient is about 1.5 10 4 V/°C; that is, there is a decrease in voltage of 1.5 10 4 V for every degree drop in temperature. Thus, even
allowing for a 40°C change in temperature, the decrease in voltage amounts to only
6 10 3 V, which is about
6 10 3 V
12 V 100% 0.05% of the operating voltage, an insignificant change. The real cause of a battery’s appar- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 778 ELECTROCHEMISTRY FIGURE 19.10 A schematic diagram of a solid-state lithium battery. Lithium metal is the anode,
and TiS2 is the cathode. During
operation, Li ions migrate
through the solid polymer electrolyte from the anode to the
cathode while electrons flow externally from the anode to the
cathode to complete the circuit. e– e– Anode Cathode
TiS 2 Li
Li + Solid electrolyte Li Li + + e – TiS 2 + e – –
TiS 2 ent breakdown is an increase in the viscosity of the electrolyte as the temperature decreases. For the battery to function properly, the electrolyte must be fully conducting.
However, the ions move much more slowly in a viscous medium, so the resistance of
the fluid increases, leading to a decrease in the power output of the battery. If an apparently “dead battery” is warmed to near room temperature on a frigid day, it recovers its ability to deliver normal power.
SOLID-STATE LITHIUM BATTERY Unlike the batteries discussed so far, a solid-state battery employs a solid (rather than
an aqueous solution or a water-based paste) as the electrolyte connecting the electrodes.
Figure 19.10 shows a schematic solid-state lithium battery. The advantage in choosing
lithium as the anode is that it has the most negative E ° value, as Table 19.1 shows.
Furthermore, lithium is a lightweight metal, so that only 6.941 g of Li (its molar mass)
are needed to produce 1 mole of electrons. The electrolyte is a polymer material that
permits the passage of ions but not of electrons. The cathode is made of either TiS2 or
V6O13. The cell voltage of a solid-state lithium battery can be as high as 3 V, and it
can be recharged in the same way as a lead storage battery. Although these batteries
do not yet have high reliability and long lifetime, they have been hailed as the batteries of the future.
FUEL CELLS Fossil fuels are a major source of energy, but conversion of fossil fuel into electrical
energy is a highly inefficient process. Consider the combustion of methane:
CH4(g) 2O2(g) 88n CO2(g) 2H2O(l ) energy To generate electricity, heat produced by the reaction is first used to convert water to
steam, which then drives a turbine that drives a generator. An appreciable fraction of
the energy released in the form of heat is lost to the surroundings at each step; even
the most efficient power plant converts only about 40 percent of the original chemical Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.6 FIGURE 19.11 A hydrogenoxygen fuel cell. The Ni and
NiO embedded in the porous
carbon electrodes are
electrocatalysts. e– BATTERIES 779 e– Anode Cathode H2 O2 Porous carbon electrode
containing Ni Porous carbon electrode
containing Ni and NiO Hot KOH solution
2H2 ( g) + 4OH– (aq)
4H2 O(l ) + 4e – Reduction
O2 ( g) + 2H2O(l ) + 4e – 4OH– (aq) energy into electricity. Because combustion reactions are redox reactions, it is more
desirable to carry them out directly by electrochemical means, thereby greatly increasing the efficiency of power production. This objective can be accomplished by a
device known as fuel cell, an electrochemical cell that requires a continuous supply
of reactants to keep functioning.
In its simplest form, a hydrogen-oxygen fuel cell consists of an electrolyte solution, such as potassium hydroxide solution, and two inert electrodes. Hydrogen and
oxygen gases are bubbled through the anode and cathode compartments (Figure 19.11),
where the following reactions take place:
Anode: 2H2(g) Cathode: O2(g)
Overall: 4OH (aq) 88n 4H2O(l ) 2H2O(l )
2H2(g) 4e 4e 88n 4OH (aq)
O2(g) 88n 2H2O(l ) The standard emf of the cell can be calculated as follows, with data from Table 19.1:
0.83 V E°
0.40 V 1.23 V Thus the cell reaction is spontaneous under standard-state conditions. Note that the reaction is the same as the hydrogen combustion reaction, but the oxidation and reduction are carried out separately at the anode and the cathode. Like platinum in the standard hydrogen electrode, the electrodes have a twofold function. They serve as electrical
conductors, and they provide the necessary surfaces for the initial decomposition of
the molecules into atomic species, prior to electron transfer. They are electrocatalysts.
Metals such as platinum, nickel, and rhodium are good electrocatalysts. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 780 ELECTROCHEMISTRY FIGURE 19.12 A hydrogenoxygen fuel cell used in the
space program. The pure water
produced by the cell is consumed
by the astronauts. In addition to the H2-O2 system, a number of other fuel cells have been developed. Among these is the propane-oxygen fuel cell. The half-cell reactions are
Anode: C3H8(g) Cathode: 5O2(g)
Overall: 20H (aq)
C3H8(g) 6H2O(l ) 88n 3CO2(g) 20H (aq) 20e 20e 88n 10H2O(l )
5O2(g) 88n 3CO2(g) 4H2O(l ) The overall reaction is identical to the burning of propane in oxygen.
Unlike batteries, fuel cells do not store chemical energy. Reactants must be constantly resupplied, and products must be constantly removed from a fuel cell. In this
respect, a fuel cell resembles an engine more than it does a battery. However, the fuel
cell does not operate like a heat engine and therefore is not subject to the same kind
of thermodynamic limitations in energy conversion (see the Chemistry in Action essay
on p. 736).
Properly designed fuel cells may be as much as 70 percent efficient, about twice
as efficient as an internal combustion engine. In addition, fuel-cell generators are free
of the noise, vibration, heat transfer, thermal pollution, and other problems normally
associated with conventional power plants. Nevertheless, fuel cells are not yet in widespread use. A major problem lies in the lack of cheap electrocatalysts able to function
efficiently for long periods of time without contamination. The most successful application of fuel cells to date has been in space vehicles (Figure 19.12). 19.7 CORROSION Corrosion is the term usually applied to the deterioration of metals by an electrochemical process. We see many examples of corrosion around us. Rust on iron, tarnish
on silver, and the green patina formed on copper and brass are a few of them (Figure
19.13). Corrosion causes enormous damage to buildings, bridges, ships, and cars. The
cost of metallic corrosion to the U.S. economy has been estimated to be over 100 bil- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.7 CORROSION 781 FIGURE 19.13 Examples of
corrosion: (a) a rusted ship, (b) a
half-tarnished silver dish, and
(c) the Statue of Liberty coated
with patina before its restoration
in 1986. (a) (b) (c) lion dollars a year! This section discusses some of the fundamental processes that occur in corrosion and methods used to protect metals against it.
By far the most familiar example of corrosion is the formation of rust on iron.
Oxygen gas and water must be present for iron to rust. Although the reactions involved
are quite complex and not completely understood, the main steps are believed to be as
follows. A region of the metal’s surface serves as the anode, where oxidation occurs: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 782 ELECTROCHEMISTRY Fe(s) 88n Fe2 (aq) 2e The electrons given up by iron reduce atmospheric oxygen to water at the cathode,
which is another region of the same metal’s surface:
O2(g) 4H (aq) 4e 88n 2H2O(l ) The overall redox reaction is
2Fe(s) O2(g) 4H (aq) 88n 2Fe2 (aq) 2H2O(l ) With data from Table 19.1, we find the standard emf for this process:
E° The positive standard emf means
that the process will occur
red 0.44 V 1.23 V 1.67 V Note that this reaction occurs in an acidic medium; the H ions are supplied in part
by the reaction of atmospheric carbon dioxide with water to form H2CO3.
The Fe2 ions formed at the anode are further oxidized by oxygen:
4Fe2 (aq) O2(g) (4 2x)H2O(l ) 88n 2Fe2O3 xH2O(s) 8H (aq) This hydrated form of iron(III) oxide is known as rust. The amount of water associated with the iron oxide varies, so we represent the formula as Fe2O3 xH2O.
Figure 19.14 shows the mechanism of rust formation. The electrical circuit is completed by the migration of electrons and ions; this is why rusting occurs so rapidly in
salt water. In cold climates, salts (NaCl or CaCl2) spread on roadways to melt ice and
snow are a major cause of rust formation on automobiles.
Metallic corrosion is not limited to iron. Consider aluminum, a metal used to make
many useful things, including airplanes and beverage cans. Aluminum has a much
greater tendency to oxidize than iron does; in Table 19.1 we see that Al has a more
negative standard reduction potential than Fe. Based on this fact alone, we might expect to see airplanes slowly corrode away in rainstorms, and soda cans transformed
into piles of corroded aluminum. These processes do not occur because the layer of insoluble aluminum oxide (Al2O3) that forms on its surface when the metal is exposed
to air serves to protect the aluminum underneath from further corrosion. The rust that
forms on the surface of iron, however, is too porous to protect the underlying metal. FIGURE 19.14 The electrochemical process involved in rust
formation. The H ions are supplied by H2CO3, which forms
when CO2 dissolves in water. Air
Water O2 Rust Fe2+
Fe2+(aq) Back Forward Main Menu Anode
Fe2+(aq) + 2e – Cathode
O2(g) + 4H +(aq) + 4e – 2H2O(l) Fe3+(aq) + e – TOC Study Guide TOC Textbook Website MHHE Website 19.7 CORROSION 783 Coinage metals such as copper and silver also corrode, but much more slowly.
Cu(s) 88n Cu2 (aq)
Ag(s) 88n Ag (aq) 2e
e In normal atmospheric exposure, copper forms a layer of copper carbonate (CuCO3),
a green substance also called patina, that protects the metal underneath from further
corrosion. Likewise, silverware that comes into contact with foodstuffs develops a layer
of silver sulfide (Ag2S).
A number of methods have been devised to protect metals from corrosion. Most
of these methods are aimed at preventing rust formation. The most obvious approach
is to coat the metal surface with paint. However, if the paint is scratched, pitted, or
dented to expose even the smallest area of bare metal, rust will form under the paint
layer. The surface of iron metal can be made inactive by a process called passivation.
A thin oxide layer is formed when the metal is treated with a strong oxidizing agent
such as concentrated nitric acid. A solution of sodium chromate is often added to cooling systems and radiators to prevent rust formation.
The tendency for iron to oxidize is greatly reduced when it is alloyed with certain other metals. For example, in stainless steel, an alloy of iron and chromium, a
layer of chromium oxide forms that protects the iron from corrosion.
An iron container can be covered with a layer of another metal such as tin or zinc.
A “tin” can is made by applying a thin layer of tin over iron. Rust formation is prevented as long as the tin layer remains intact. However, once the surface has been
scratched, rusting occurs rapidly. If we look up the standard reduction potentials, we
find that iron acts as the anode and tin as the cathode in the corrosion process:
Fe(s) 88n Fe2
2 Sn (aq) 2e 2e 88n Sn(s) E°
E° 0.44 V
0.14 V The protective process is different for zinc-plated, or galvanized, iron. Zinc is more
easily oxidized than iron (see Table 19.1):
Zn(s) 88n Zn2 (aq) 2e E° 0.76 V So even if a scratch exposes the iron, the zinc is still attacked. In this case, the zinc
metal serves as the anode and the iron is the cathode.
Cathodic protection is a process in which the metal that is to be protected from
corrosion is made the cathode in what amounts to an electrochemical cell. Figure 19.15
FIGURE 19.15 An iron nail
that is cathodically protected by
a piece of zinc strip does not rust
in water, while an iron nail without such protection rusts readily. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 784 ELECTROCHEMISTRY FIGURE 19.16 Cathodic protection of an iron storage tank
(cathode) by magnesium, a more
electropositive metal (anode).
Since only the magnesium is depleted in the electrochemical
process, it is sometimes called the
sacrificial anode. e– Mg Iron storage tank Mg 2+(aq) + 2e– Oxidation: Mg(s) Reduction: O2( g) + 4H +(aq) + 4e– 2H2 O(l ) shows how an iron nail can be protected from rusting by connecting the nail to a piece
of zinc. Without such protection, an iron nail quickly rusts in water. Rusting of underground iron pipes and iron storage tanks can be prevented or greatly reduced by
connecting them to metals such as zinc and magnesium, which oxidize more readily
than iron (Figure 19.16).
The Chemistry in Action essay on p. 785 shows that dental filling discomfort can
result from an electrochemical phenomenon. 19.8 ELECTROLYSIS In contrast to spontaneous redox reactions, which result in the conversion of chemical
energy into electrical energy, electrolysis is the process in which electrical energy is
used to cause a nonspontaneous chemical reaction to occur. An electrolytic cell is an
apparatus for carrying out electrolysis. The same principles underlie electrolysis and
the processes that take place in electrochemical cells. Here we will discuss three examples of electrolysis based on those principles. Then we will look at the quantitative
aspects of electrolysis.
ELECTROLYSIS OF MOLTEN SODIUM CHLORIDE In its molten state, sodium chloride, an ionic compound, can be electrolyzed to form
sodium metal and chlorine. Figure 19.17(a) is a diagram of a Downs cell, which is used
for large-scale electrolysis of NaCl. In molten NaCl, the cations and anions are the
Na and Cl ions, respectively. Figure 19.17(b) is a simplified diagram showing the
reactions that occur at the electrodes. The electrolytic cell contains a pair of electrodes
connected to the battery. The battery serves as an “electron pump,” driving electrons
to the cathode, where reduction occurs, and withdrawing electrons from the anode,
where oxidation occurs. The reactions at the electrodes are
Cathode (reduction): 2Cl (l ) 88n Cl2(g)
2Na (l ) Overall: 2Na (l) 2e 2e 88n 2Na(l ) 2Cl (l ) 88n 2Na(l ) Cl2(g) This process is a major source of pure sodium metal and chlorine gas. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.8 785 ELECTROLYSIS Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Dental Filling Discomfort
In modern dentistry the material most commonly used
to fill decaying teeth is known as dental amalgam.
(An amalgam is a substance made by combining mercury with another metal or metals.) Dental amalgam
actually consists of three solid phases having stoichiometries approximately corresponding to Ag2Hg3,
Ag3Sn, and Sn8Hg. The standard reduction potentials
for these solid phases are: Hg2 /Ag2Hg3, 0.85 V;
Sn2 /Ag3Sn, 0.05 V; Sn2 /Sn8Hg, 0.13 V.
Anyone who bites a piece of aluminum foil (such
as that used for wrapping candies) in such a way that
the foil presses against a dental filling will probably
experience a momentary sharp pain. In effect, an electrochemical cell has been created in the mouth, with
1.66 V) as the anode, the filling as
the cathode, and saliva as the electrolyte. Contact between the aluminum foil and the filling short-circuits
the cell, causing a weak current to flow between the
electrodes. This current stimulates the sensitive nerve
of the tooth, causing an unpleasant sensation.
Another type of discomfort results when a less
electropositive metal touches a dental filling. For example, if a filling makes contact with a gold inlay in
a nearby tooth, corrosion of the filling will occur. In
this case, the dental filling acts as the anode and the
gold inlay as the cathode. Referring to the E° values FIGURE 19.17 (a) A practical
arrangement called a Downs cell
for the electrolysis of molten
NaCl (m.p. 801°C). The
sodium metal formed at the cathodes is in the liquid state. Since
liquid sodium metal is lighter than
molten NaCl, the sodium floats to
the surface, as shown, and is collected. Chlorine gas forms at the
anode and is collected at the
top. (b) A simplified diagram
showing the electrode reactions
during the electrolysis of molten
NaCl. The battery is needed to
drive the nonspontaneous
reactions. Back Forward Main Menu Gold inlay
O2( g) + 4H+(aq) + 4e– 2H2O(l ) e–
Sn8Hg Sn2+ Dental filling Corrosion of a dental filling brought about by contact with
a gold inlay. for the three phases, we see that the Sn8Hg phase is
most likely to corrode. When that happens, release of
Sn(II) ions in the mouth produces an unpleasant metallic taste. Prolonged corrosion will eventually result in
another visit to the dentist for a replacement filling. Cl2 gas
Anode Liquid Na e–
Cathode Liquid Na Molten NaCl Iron cathode Iron cathode Carbon anode
(a) TOC Study Guide TOC 2Cl– Oxidation
Cl2( g) + 2e – Reduction
2Na+ + 2e –
(b) Textbook Website MHHE Website 786 ELECTROCHEMISTRY Theoretical estimates show that the E° value for the overall process is about
4 V, which means that this is a nonspontaneous process. Therefore, a minimum of
4 V must be supplied by the battery to carry out the reaction. In practice, a higher voltage is necessary because of inefficiencies in the electrolytic process and because of
overvoltage, to be discussed shortly.
ELECTROLYSIS OF WATER Water in a beaker under atmospheric conditions (1 atm and 25°C) will not spontaneously decompose to form hydrogen and oxygen gas because the standard freeenergy change for the reaction is a large positive quantity:
2H2O(l ) 88n 2H2(g) FIGURE 19.18 Apparatus for
small-scale electrolysis of water.
The volume of hydrogen gas generated (left column) is twice that
of oxygen gas (right column). ∆G° O2(g) 474.4 kJ However, this reaction can be induced in a cell like the one shown in Figure 19.18.
This electrolytic cell consists of a pair of electrodes made of a nonreactive metal, such
as platinum, immersed in water. When the electrodes are connected to the battery, nothing happens because there are not enough ions in pure water to carry much of an electric current. (Remember that at 25°C, pure water has only 1 10 7 M H ions and
1 10 7 M OH ions.)
On the other hand, the reaction occurs readily in a 0.1 M H2SO4 solution because
there are a sufficient number of ions to conduct electricity (Figure 19.19). Immediately,
gas bubbles begin to appear at both electrodes. The process at the anode is
2H2O(l ) 88n O2(g) 4H (aq) 4e while at the cathode we have
e 88n 1 H2(g)
2 H (aq) The overall reaction is given by
Anode (oxidation): 2H2O(l ) 88n O2(g) Cathode (reduction): 4[H (aq)
Overall: 4H (aq) 4e 2H2O(l ) 88n 2H2(g) O2(g) E° 1.23 V E° e 88n 1 H2(g)]
2 0.00 V E° 1.23 V Note that no net H2SO4 is consumed.
FIGURE 19.19 A diagram
showing the electrode reactions
during the electrolysis of water. Battery
Cathode Dilute H2SO4 solution Oxidation
2H2O(l) Back Forward Main Menu O2( g) + 4H+(aq) + 4e – TOC Reduction
4H+(aq) + 4e –
2H2( g) Study Guide TOC Textbook Website MHHE Website 19.8 ELECTROLYSIS 787 ELECTROLYSIS OF AN AQUEOUS SODIUM CHLORIDE SOLUTION This is the most complicated of the three examples of electrolysis considered here because aqueous sodium chloride solution contains several species that could be oxidized
and reduced. The oxidation reactions that might occur at the anode are
(1) 2H2O(l ) 88n O2(g) (2) 2Cl (aq) 88n Cl2(g) 4H (aq) 4e 2e Referring to Table 19.1, we find
O2(g) 4H (aq)
Cl2(g) Because Cl2 is more easily reduced than O2, it follows that it
would be more difficult to oxidize
Cl than H2O at the anode. 4e 88n 2H2O(l ) E° 1.23 V 2e 88n 2Cl (aq) E° 1.36 V The standard reduction potentials of (1) and (2) are not very different, but the values
do suggest that H2O should be preferentially oxidized at the anode. (O2 has a smaller
tendency to be reduced than Cl2; therefore, it has a greater tendency to be oxidized.)
However, by experiment we find that the gas liberated at the anode is Cl2, not O2! In
studying electrolytic processes we sometimes find that the voltage required for a reaction is considerably higher than the electrode potential indicates. The overvoltage is
the difference between the electrode potential and the actual voltage required to cause
electrolysis. The overvoltage for O2 formation is quite high. Therefore, under normal
operating conditions Cl2 gas is actually formed at the anode instead of O2.
The reductions that might occur at the cathode are
(3) Na (aq) e 88n Na(s) (4) 2H2O(l) 2e 88n H2(g) (5) 2H (aq) E° 0.83 V E° 2e 88n H2(g) 2.71 V E° 2OH (aq) 0.00 V Reaction (3) is ruled out because it has a very negative standard reduction potential.
Reaction (5) is preferred over (4) under standard-state conditions. At a pH of 7 (as is
the case for a NaCl solution), however, they are equally probable. We generally use
(4) to describe the cathode reaction because the concentration of H ions is too low
(about 1 10 7 M ) to make (5) a reasonable choice.
Thus, the half-cell reactions in the electrolysis of aqueous sodium chloride are
Cathode (reduction): 2Cl (aq) 88n Cl2(g)
2H2O(l ) Overall: 2H2O(l ) 2e 88n H2(g) 2Cl (aq) 88n H2(g) 2e
Cl2(g) 2OH (aq) As the overall reaction shows, the concentration of the Cl ions decreases during electrolysis and that of the OH ions increases. Therefore, in addition to H2 and Cl2, the
useful by-product NaOH can be obtained by evaporating the aqueous solution at the
end of the electrolysis.
Keep in mind the following from our analysis of electrolysis: cations are likely
to be reduced at the cathode and anions are likely to be oxidized at the anode, and in
aqueous solutions water itself may be oxidized and/or reduced. The outcome depends
on the nature of other species present.
The following example deals with the electrolysis of an aqueous solution of sodium
sulfate (Na2SO4). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 788 ELECTROCHEMISTRY EXAMPLE 19.8 An aqueous Na2SO4 solution is electrolyzed, using the apparatus shown in Figure
19.19. If the products formed at the anode and cathode are oxygen gas and hydrogen gas, respectively, describe the electrolysis in terms of the reactions at the electrodes.
Before we look at the electrode reactions, we should consider the following facts: (1) Since Na2SO4 does not hydrolyze in water, the pH of the solution
is close to 7. (2) The Na ions are not reduced at the cathode, and the SO2 ions
are not oxidized at the anode. These conclusions are drawn from the electrolysis of
water in the presence of sulfuric acid and in aqueous sodium chloride solution.
Therefore, the electrode reactions are
Answer The SO2 ion is the conjugate
base of the weak acid HSO4
(Ka 1.3 10 2). However, the
extent to which SO2
hydrolyzes is negligible. Anode: 2H2O(l) 88n O2(g) Cathode: 2H2O(l ) 4H (aq) 2e 88n H2(g) 4e 2OH The overall reaction, obtained by doubling the cathode reaction coefficients and
adding the result to the anode reaction, is
6H2O(l ) 88n 2H2(g) O2(g) 4H (aq) 4OH (aq) If the H and OH ions are allowed to mix, then
4H (aq) 4OH (aq) 88n 4H2O(l ) and the overall reaction becomes
2H2O(l ) 88n 2H2(g) O2(g) PRACTICE EXERCISE An aqueous solution of Mg(NO3)2 is electrolyzed. What are the gaseous products
at the anode and cathode?
Electrolysis has many important applications in industry, mainly in the extraction
and purification of metals. We will discuss some of these applications in Chapter 20. QUANTITATIVE ASPECTS OF ELECTROLYSIS The quantitative treatment of electrolysis was developed primarily by Faraday. He observed that the mass of product formed (or reactant consumed) at an electrode is proportional to both the amount of electricity transferred at the electrode and the molar
mass of the substance in question. For example, in the electrolysis of molten NaCl, the
cathode reaction tells us that one Na atom is produced when one Na ion accepts an
electron from the electrode. To reduce 1 mole of Na ions, we must supply Avogadro’s
number (6.02 1023) of electrons to the cathode. On the other hand, the stoichiometry of the anode reaction shows that oxidation of two Cl ions yields one chlorine molecule. Therefore, the formation of 1 mole of Cl2 results in the transfer of 2 moles of
electrons from the Cl ions to the anode. Similarly, it takes 2 moles of electrons to reduce 1 mole of Mg2 ions and 3 moles of electrons to reduce 1 mole of Al3 ions:
Mg2 Back Forward Main Menu TOC 2e 88n Mg Al3 3e 88n Al Study Guide TOC Textbook Website MHHE Website 19.8 Current
and time Charge
faradays FIGURE 19.20
electrolysis. ELECTROLYSIS Moles of
or oxidized 789 Grams of
or oxidized Steps involved in calculating amounts of substances reduced or oxidized in Therefore
2F 1 mol Mg2 3F 1 mol Al3 where F is the faraday.
In an electrolysis experiment, we generally measure the current (in amperes, A)
that passes through an electrolytic cell in a given period of time. The relationship between charge (in coulombs, C) and current is
1C 1A 1s that is, a coulomb is the quantity of electrical charge passing any point in the circuit
in 1 second when the current is 1 ampere.
Figure 19.20 shows the steps involved in calculating the quantities of substances
produced in electrolysis. Let us illustrate the approach by considering molten CaCl2 in
an electrolytic cell. Suppose a current of 0.452 amperes is passed through the cell for
1.50 hours. How much product will be formed at the anode and at the cathode? In solving electrolysis problems of this type, the first step is to determine which species will
be oxidized at the anode and which species will be reduced at the cathode. Here the
choice is straightforward since we only have Ca2 and Cl ions in molten CaCl2. Thus
we write the half- and overall reactions as
Cathode (reduction): 2Cl (l ) 88n Cl2(g)
Ca2 (l ) Overall: Ca2 (l ) 2e 2e 88n Ca(l ) 2Cl (l ) 88n Ca(l ) Cl2(g) The quantities of calcium metal and chlorine gas formed depend on the number
of electrons that pass through the electrolytic cell, which in turn depends on current
time, or charge:
?C 0.452 A 1.50 h 3600 s
1A s 2.44 103 C Since 1 F 96,500 C and 2 F are required to reduce 1 mole of Ca2 ions, the mass
of Ca metal formed at the cathode is calculated as follows:
? g Ca 2.44 103 C 1F
96,500 C 1 mol Ca
2F 40.08 g Ca
1 mol Ca 0.507 g Ca The anode reaction indicates that 1 mole of chlorine is produced per 2 F of electricity. Hence the mass of chlorine gas formed is
? g Cl2 2.44 103 C 1F
96,500 C 1 mol Cl2
2F 70.90 g Cl2
1 mol Cl2 0.896 g Cl2 The following example applies this approach to the electrolysis in an aqueous
solution. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 790 ELECTROCHEMISTRY EXAMPLE 19.9 A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 hours. Write the half-cell reactions and calculate the volume of gases generated at STP.
Earlier in this chapter (see p. 786) we saw that the half-cell reactions for
this process are Answer Anode (oxidation): 2H2O(l ) 88n O2(g) Cathode (reduction): 4[H (aq) 4H (aq) 2H2O(l ) 88n 2H2(g) Overall: 4e e 88n 1 H2(g)]
O2(g) First we calculate the number of coulombs of electricity that pass through the cell:
?C 1.26 A 7.44 h 3600 s
1A s 3.37 104 C We see that for every mole of O2 formed at the anode, 4 moles of electrons are generated so that
? g O2 3.37 104 C 1F
96,500 C 1 mol O2
4F 32.00 g O2
1 mol O2 2.79 g O2 The volume of 2.79 g O2 at STP is given by
(2.79 g/32.00 g mol 1)(0.0821 L atm/K mol)(273 K)
1.95 L Similarly, for hydrogen we write
? g H2 3.37 104 C 1F
96,500 C 1 mol H2
2F 2.016 g H2
1 mol H2 0.352 g H2 The volume of 0.352 g H2 at STP is given by
(0.352 g/2.016 g mol 1)(0.0821 L atm/K mol)(273 K)
3.91 L Note that the volume of H2 is twice that of O2, which is what we would
expect based on Avogadro’s law (at the same temperature and pressure, volume is
directly proportional to the number of moles of gases). Comment Similar problem: 19.46. PRACTICE EXERCISE A constant current is passed through an electrolytic cell containing molten MgCl2
for 18 hours. If 4.8 105 g of Cl2 are obtained, what is the current in amperes? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website KEY WORDS SUMMARY OF
KEY EQUATIONS • E°
red (19.1) 791 Calculating the standard emf of an electrochemical
Relating free energy change to the emf of the cell.
Relating the standard free energy change to the
standard emf. • ∆G
• ∆G° nFEcell
(19.3) • E°
nF (19.4) Relating the standard emf of the cell to the equilibrium
constant. (19.6) The Nernst equation. For calculating the emf of a cell
under non-standard-state conditions. •E SUMMARY OF FACTS
AND CONCEPTS E° RT
nF 1. Redox reactions involve the transfer of electrons. Equations representing redox processes
can be balanced using the ion-electron method.
2. All electrochemical reactions involve the transfer of electrons and therefore are redox reactions.
3. In an electrochemical cell, electricity is produced by a spontaneous chemical reaction.
Oxidation and reduction take place separately at the anode and cathode, respectively, and
the electrons flow through an external circuit.
4. The two parts of an electrochemical cell are the half-cells, and the reactions at the electrodes are the half-cell reactions. A salt bridge allows ions to flow between the half-cells.
5. The electromotive force (emf) of a cell is the voltage difference between the two electrodes. In the external circuit, electrons flow from the anode to the cathode in an electrochemical cell. In solution, the anions move toward the anode and the cations move toward the cathode.
6. The quantity of electricity carried by 1 mole of electrons is called a faraday, which is
equal to 96,500 coulombs.
7. Standard reduction potentials show the relative likelihood of half-cell reduction reactions
and can be used to predict the products, direction, and spontaneity of redox reactions between various substances.
8. The decrease in free energy of the system in a spontaneous redox reaction is equal to the
electrical work done by the system on the surroundings, or ∆G
9. The equilibrium constant for a redox reaction can be found from the standard electromotive force of a cell.
10. The Nernst equation gives the relationship between the cell emf and the concentrations of
the reactants and products under non-standard-state conditions.
11. Batteries, which consist of one or more electrochemical cells, are used widely as self-contained power sources. Some of the better-known batteries are the dry cell, such as the
Leclanché cell, the mercury battery, and the lead storage battery used in automobiles. Fuel
cells produce electrical energy from a continuous supply of reactants.
12. The corrosion of metals, such as the rusting of iron, is an electrochemical phenomenon.
13. Electric current from an external source is used to drive a nonspontaneous chemical reaction in an electrolytic cell. The amount of product formed or reactant consumed depends
on the quantity of electricity transferred at the electrode. KEY WORDS
Anode, p. 761
Battery, p. 775 Back Forward Cathode, p. 761
Cell voltage, p. 762 Main Menu TOC Corrosion, p. 780
Electrolysis, p. 783 Study Guide TOC Electrolytic cell, p. 784
Electrochemistry, p. 758 Textbook Website MHHE Website 792 ELECTROCHEMISTRY Electrochemical cell, p. 761
Electromotive force (emf),
Faraday, p. 769 Standard emf (E°), p. 764
Standard oxidation potential,
p. 764 Fuel cell, p. 779
Half-cell reaction, p. 761
Nernst equation, p. 772
Overvoltage, p. 787 Standard reduction potential,
p. 763 QUESTIONS AND PROBLEMS
BALANCING REDOX EQUATIONS
Problems 19.1 Balance the following redox equations by the ionelectron method:
H2O (in acidic solu(a) H2O2 Fe2 88n Fe3
NO H2O (in acidic
(b) Cu HNO3 88n Cu2
MnO4 88n CNO
MnO2 (in basic
Br (in basic solution)
(d) Br2 88n BrO3
I2 88n I
S4O2 (in acidic solu(e) S2O2
19.2 Balance the following redox equations by the ionelectron method:
H2O2 88n MnO2 H2O (in basic solu(a) Mn2
Bi (in basic
(b) Bi(OH)3 SnO2 88n SnO2
C2O2 88n Cr3
CO2 (in acidic
Cl 88n Cl2 ClO2 (in acidic solu(d) ClO3
ELECTROCHEMICAL CELLS AND STANDARD EMFs
Review Questions 19.3 Define the following terms: anode, cathode, electromotive force, standard oxidation potential, standard reduction potential.
19.4 Describe the basic features of an electrochemical
cell. Why are the two components of the cell separated from each other?
19.5 What is the function of a salt bridge? What kind of
electrolyte should be used in a salt bridge?
19.6 What is a cell diagram? Write the cell diagram for
an electrochemical cell consisting of an Al electrode
placed in a 1 M Al(NO3)3 solution and a Ag electrode placed in a 1 M AgNO3 solution.
19.7 What is the difference between the half-reactions
discussed in redox processes in Chapter 4 and the
half-cell reactions discussed in Section 19.2? Back Forward Main Menu TOC 19.8 After operating a Daniell cell (see Figure 19.1) for
a few minutes, a student notices that the cell emf
begins to drop. Why?
19.9 Use the information in Table 2.1, and calculate the
19.10 Discuss the spontaneity of an electrochemical reaction in terms of its standard emf (E° ).
Problems 19.11 Calculate the standard emf of a cell that uses the
Mg/Mg2 and Cu/Cu2 half-cell reactions at 25°C.
Write the equation for the cell reaction that occurs
under standard-state conditions.
19.12 Calculate the standard emf of a cell that uses
Ag/Ag and Al/Al3 half-cell reactions. Write the
cell reaction that occurs under standard-state conditions.
19.13 Predict whether Fe3 can oxidize I to I2 under standard-state conditions.
19.14 Which of the following reagents can oxidize H2O
to O2(g) under standard-state conditions? H (aq),
Cl (aq), Cl2(g), Cu2 (aq), Pb2 (aq), MnO4 (aq) (in
19.15 Consider the following half-reactions:
NO3 (aq) 8H (aq)
4H (aq) 5e 88n
3e 88n NO(g) 4H2O(l )
2H2O(l ) Predict whether NO3 ions will oxidize Mn2 to
MnO4 under standard-state conditions.
19.16 Predict whether the following reactions would occur spontaneously in aqueous solution at 25°C.
Assume that the initial concentrations of dissolved
species are all 1.0 M.
(a) Ca(s) Cd2 (aq) 88n Ca2 (aq) Cd(s)
(b) 2Br (aq) Sn2 (aq) 88n Br2(l ) Sn(s)
(c) 2Ag(s) Ni2 (aq) 88n 2Ag (aq) Ni(s)
(d) Cu (aq) Fe3 (aq) 88n
Cu2 (aq) Fe2 (aq) Study Guide TOC Textbook Website MHHE Website 793 QUESTIONS AND PROBLEMS 19.17 Which species in each pair is a better oxidizing agent
under standard-state conditions? (a) Br2 or Au3 , (b)
H2 or Ag , (c) Cd2 or Cr3 , (d) O2 in acidic media or O2 in basic media
19.18 Which species in each pair is a better reducing agent
under standard-state conditions? (a) Na or Li, (b) H2
or I2, (c) Fe2 or Ag, (d) Br or Co2
SPONTANEITY OF REDOX REACTIONS 19.19 Write the equations relating ∆G° and K to the standard emf of a cell. Define all the terms.
19.20 Compare the ease of measuring the equilibrium constant electrochemically with that by chemical means
[see Equation (18.10)].
Problems What is the equilibrium constant for the following
reaction at 25°C?
Mg(s) 19.22 19.23 19.24 19.25 19.26 Zn2 (aq) 34 Mg2 (aq)
Mg2 (aq) 34 Sr2 (aq) Mg(s) is 2.69 1012 at 25°C. Calculate E° for a cell
made up of Sr/Sr2 and Mg/Mg2 half-cells.
Use the standard reduction potentials to find the
equilibrium constant for each of the following reactions at 25°C:
(a) Br2(l ) 2I (aq) 34 2Br (aq) I2(s)
(b) 2Ce4 (aq) 2Cl (aq) 34
Cl2(g) 2Ce3 (aq)
(c) 5Fe (aq) MnO4 (aq) 8H (aq) 34
Mn2 (aq) 4H2O 5Fe3 (aq)
Calculate G° and Kc for the following reactions at
(a) Mg(s) Pb2 (aq) 34 Mg2 (aq) Pb(s)
(b) Br2(l ) 2I (aq) 34 2Br (aq) I2(s)
(c) O2(g) 4H (aq) 4Fe2 (aq) 34
2H2O(l ) 4Fe3 (aq)
(d) 2Al(s) 3I2(s) 34 2Al (aq) 6I (aq)
Under standard-state conditions, what spontaneous
reaction will occur in aqueous solution among the
ions Ce4 , Ce3 , Fe3 , and Fe2 ? Calculate ∆G°
and Kc for the reaction.
Given that E° 0.52 V for the reduction Cu (aq)
e 88n Cu(s), calculate E°, ∆G°, and K for the
following reaction at 25°C:
2Cu (aq) 88n Cu2 (aq) Back Zn(s) The equilibrium constant for the reaction
Sr(s) Forward Main Menu Review Questions 19.27 Write the Nernst equation and explain all the terms.
19.28 Write the Nernst equation for the following
processes at some temperature T:
(a) Mg(s) Sn2 (aq) 88n Mg2 (aq) Sn(s)
(b) 2Cr(s) 3Pb2 (aq) 88n 2Cr3 (aq) 3Pb(s)
Problems Review Questions 19.21 THE EFFECT OF CONCENTRATION ON CELL EMF Cu(s). TOC 19.29 What is the potential of a cell made up of Zn/Zn2
and Cu/Cu2 half-cells at 25°C if [Zn2 ] 0.25 M
and [Cu2 ] 0.15 M?
19.30 Calculate E°, E, and ∆G for the following cell reactions.
(a) Mg(s) Sn2 (aq) 88n Mg2 (aq) Sn(s)
[Mg2 ] 0.045 M, [Sn2 ] 0.035 M
(b) 3Zn(s) 2Cr3 (aq) 88n 3Zn2 (aq) 2Cr(s)
[Cr3 ] 0.010 M, [Zn2 ] 0.0085 M
19.31 Calculate the standard potential of the cell consisting of the Zn/Zn2 half-cell and the SHE. What will
the emf of the cell be if [Zn2 ] 0.45 M, PH2
2.0 atm, and [H ] 1.8 M?
19.32 What is the emf of a cell consisting of a Pb/Pb2
half-cell and a Pt/H2/H half-cell if [Pb2 ]
0.10 M, [H ] 0.050 M, and PH2 1.0 atm?
19.33 Referring to the arrangement in Figure 19.1, calculate the [Cu2 ]/[Zn2 ] ratio at which the following
reaction is spontaneous at 25°C:
Cu(s) Zn2 (aq) 88n Cu2 (aq) Zn(s) 19.34 Calculate the emf of the following concentration
Mg(s)Mg2 (0.24 M ) Mg2 (0.53 M )Mg(s) BATTERIES AND FUEL CELLS
Review Questions 19.35 Explain the differences between a primary electrochemical cell—one that is not rechargeable—and a
storage cell (for example, the lead storage battery),
which is rechargeable.
19.36 Discuss the advantages and disadvantages of fuel
cells over conventional power plants in producing
Problems 19.37 The hydrogen-oxygen fuel cell is described in
Section 19.6. (a) What volume of H2(g), stored at
25°C at a pressure of 155 atm, would be needed to Study Guide TOC Textbook Website MHHE Website 794 ELECTROCHEMISTRY run an electric motor drawing a current of 8.5 A for
3.0 h? (b) What volume (liters) of air at 25°C and
1.00 atm will have to pass into the cell per minute
to run the motor? Assume that air is 20 percent O2
by volume and that all the O2 is consumed in the
cell. The other components of air do not affect the
fuel-cell reactions. Assume ideal gas behavior.
19.38 Calculate the standard emf of the propane fuel cell
discussed on p. 780 at 25°C, given that ∆G° for
propane is 23.5 kJ/mol. 19.47 Considering only the cost of electricity, would it be
cheaper to produce a ton of sodium or a ton of aluminum by electrolysis?
19.48 If the cost of electricity to produce magnesium by
the electrolysis of molten magnesium chloride is
$155 per ton of metal, what is the cost (in dollars)
of the electricity necessary to produce (a) 10.0 tons
of aluminum, (b) 30.0 tons of sodium, (c) 50.0 tons
19.49 One of the half-reactions for the electrolysis of water is CORROSION 2H2O(l ) 88n O2(g) Review Questions 19.39 Steel hardware, including nuts and bolts, is often
coated with a thin plating of cadmium. Explain the
function of the cadmium layer.
19.40 “Galvanized iron” is steel sheet that has been coated
with zinc; “tin” cans are made of steel sheet coated
with tin. Discuss the functions of these coatings and
the electrochemistry of the corrosion reactions that
occur if an electrolyte contacts the scratched surface
of a galvanized iron sheet or a tin can.
19.41 Tarnished silver contains Ag2S. The tarnish can be
removed by placing silverware in an aluminum pan
containing an inert electrolyte solution, such as
NaCl. Explain the electrochemical principle for this
procedure. [The standard reduction potential for the
half-cell reaction Ag2S(s) 2e 88n 2Ag(s)
S2 (aq) is 0.71 V.]
19.42 How does the tendency of iron to rust depend on
the pH of solution?
ELECTROLYSIS 19.50 19.51 19.52 19.53 19.54 Review Questions 19.43 What is the difference between an electrochemical
cell (such as a Daniell cell) and an electrolytic cell?
19.44 What is Faraday’s contribution to quantitative electrolysis?
Problems 19.55 19.45 The half-reaction at an electrode is
Mg2 (molten) 2e 88n Mg(s) Calculate the number of grams of magnesium that
can be produced by supplying 1.00 F to the electrode.
19.46 Consider the electrolysis of molten barium chloride,
BaCl2. (a) Write the half-reactions. (b) How many
grams of barium metal can be produced by supplying 0.50 A for 30 min? Back Forward Main Menu TOC 19.56 4H (aq) 4e If 0.076 L of O2 is collected at 25°C and 755 mmHg,
how many faradays of electricity had to pass through
How many faradays of electricity are required to
produce (a) 0.84 L of O2 at exactly 1 atm and 25°C
from aqueous H2SO4 solution; (b) 1.50 L of Cl2 at
750 mmHg and 20°C from molten NaCl; (c) 6.0 g
of Sn from molten SnCl2?
Calculate the amounts of Cu and Br2 produced in
1.0 h at inert electrodes in a solution of CuBr2 by a
current of 4.50 A.
In the electrolysis of an aqueous AgNO3 solution,
0.67 g of Ag is deposited after a certain period of
time. (a) Write the half-reaction for the reduction of
Ag . (b) What is the probable oxidation half-reaction? (c) Calculate the quantity of electricity used,
A steady current was passed through molten CoSO4
until 2.35 g of metallic cobalt was produced.
Calculate the number of coulombs of electricity
A constant electric current flows for 3.75 h through
two electrolytic cells connected in series. One contains a solution of AgNO3 and the second a solution
of CuCl2. During this time 2.00 g of silver are deposited in the first cell. (a) How many grams of copper are deposited in the second cell? (b) What is the
current flowing, in amperes?
What is the hourly production rate of chlorine gas
(in kg) from an electrolytic cell using aqueous NaCl
electrolyte and carrying a current of 1.500 103 A?
The anode efficiency for the oxidation of Cl is 93.0
Chromium plating is applied by electrolysis to objects suspended in a dichromate solution, according
to the following (unbalanced) half-reaction:
7 e H (aq) 88n Cr(s) H2O(l ) How long (in hours) would it take to apply a Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 19.57 19.58 19.59 19.60 chromium plating 1.0 10 2 mm thick to a car
bumper with a surface area of 0.25 m2 in an electrolytic cell carrying a current of 25.0 A? (The density of chromium is 7.19 g/cm3.)
The passage of a current of 0.750 A for 25.0 min
deposited 0.369 g of copper from a CuSO4 solution.
From this information, calculate the molar mass of
A quantity of 0.300 g of copper was deposited from
a CuSO4 solution by passing a current of 3.00 A
through the solution for 304 s. Calculate the value
of the faraday constant.
In a certain electrolysis experiment, 1.44 g of Ag
were deposited in one cell (containing an aqueous
AgNO3 solution), while 0.120 g of an unknown
metal X was deposited in another cell (containing
an aqueous XCl3 solution) in series with the AgNO3
cell. Calculate the molar mass of X.
One of the half-reactions for the electrolysis of water is
2H (aq) ADDITIONAL PROBLEMS 19.61 For each of the following redox reactions, (i) write
the half-reactions; (ii) write a balanced equation for
the whole reaction, (iii) determine in which direction the reaction will proceed spontaneously under
(a) H2(g) Ni2 (aq) 88n H (aq) Ni(s)
(b) MnO4 (aq) Cl (aq) 88n
Mn2 (aq) Cl2(g) (in acid solution)
(c) Cr(s) Zn2 (aq) 88n Cr3 (aq) Zn(s)
19.62 The oxidation of 25.0 mL of a solution containing
Fe2 requires 26.0 mL of 0.0250 M K2Cr2O7 in
acidic solution. Balance the following equation and
calculate the molar concentration of Fe2 :
7 Fe2 H 88n Cr3 Fe3 19.63 The SO2 present in air is mainly responsible for the
phenomenon of acid rain. The concentration of SO2
can be determined by titrating against a standard
permanganate solution as follows:
5SO2 2MnO4 2H2O 88n
4 2Mn2 4H Calculate the number of grams of SO2 in a sample
of air if 7.37 mL of 0.00800 M KMnO4 solution are
required for the titration. Back Forward 19.64 A sample of iron ore weighing 0.2792 g was dissolved in an excess of a dilute acid solution. All
the Fe(II) was converted to Fe(III) ions. The solution required 23.30 mL of 0.0194 M KMnO4 for
titration. Calculate the percent by mass of iron in
19.65 The concentration of a hydrogen peroxide solution
can be conveniently determined by titration against
a standardized potassium permanganate solution in
an acidic medium according to the following unbalanced equation:
MnO4 Main Menu TOC H2O2 88n O2 Mn2 (a) Balance the above equation. (b) If 36.44 mL of
a 0.01652 M KMnO4 solution are required to completely oxidize 25.00 mL of a H2O2 solution, calculate the molarity of the H2O2 solution.
19.66 Oxalic acid (H2C2O4) is present in many plants and
vegetables. (a) Balance the following equation in
acid solution: 2e 88n H2(g) If 0.845 L of H2 is collected at 25°C and 782 mmHg,
how many faradays of electricity had to pass through
the solution? 795 MnO4 C2O2 88n Mn2
4 CO2 (b) If a 1.00-g sample of H2C2O4 requires 24.0 mL
of 0.0100 M KMnO4 solution to reach the equivalence point, what is the percent by mass of H2C2O4
in the sample?
19.67 Complete the following table. State whether the cell
reaction is spontaneous, nonspontaneous, or at equilibrium.
∆G E CELL REACTION 0
0 19.68 Calcium oxalate (CaC2O4) is insoluble in water.
This property has been used to determine the amount
of Ca2 ions in blood. The calcium oxalate isolated
from blood is dissolved in acid and titrated against
a standardized KMnO4 solution as described in
Problem 19.66. In one test it is found that the calcium oxalate isolated from a 10.0-mL sample of
blood requires 24.2 mL of 9.56 10 4 M KMnO4
for titration. Calculate the number of milligrams of
calcium per milliliter of blood.
19.69 From the following information, calculate the solubility product of AgBr:
AgBr(s) e 88n Ag(s) Ag (aq) Br (aq) e 88n Ag(s) E° 0.07 V E° 0.80 V 19.70 Consider an electrochemical cell composed of the Study Guide TOC Textbook Website MHHE Website 796 ELECTROCHEMISTRY SHE and a half-cell using the reaction Ag (aq)
e 88n Ag(s). (a) Calculate the standard cell potential. (b) What is the spontaneous cell reaction under standard-state conditions? (c) Calculate the cell
potential when [H ] in the hydrogen electrode is
changed to (i) 1.0 10 2 M and (ii) 1.0 10 5 M,
all other reagents being held at standard-state conditions. (d) Based on this cell arrangement, suggest
a design for a pH meter.
19.71 An electrochemical cell consists of a silver electrode
in contact with 346 mL of 0.100 M AgNO3 solution
and a magnesium electrode in contact with 288 mL
of 0.100 M Mg(NO3)2 solution. (a) Calculate E for
the cell at 25°C. (b) A current is drawn from the cell
until 1.20 g of silver have been deposited at the silver electrode. Calculate E for the cell at this stage
19.72 Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of NaCl but fluorine
gas cannot be prepared by electrolyzing an aqueous
solution of NaF.
19.73 Calculate the emf of the following concentration cell
at 25°C: phenolphthalein have been added is electrolyzed using an apparatus like the one shown below: 19.78 Cu(s)Cu2 (0.080 M) Cu2 (1.2 M)Cu(s) 19.74 The cathode reaction in the Leclanché cell is given
2MnO2(s) Zn2 (aq) 2e 88n ZnMn2O4(s) If a Leclanché cell produces a current of 0.0050 A,
calculate how many hours this current supply will
last if there are initially 4.0 g of MnO2 present in
the cell. Assume that there is an excess of Zn2 ions.
19.75 Suppose you are asked to verify experimentally the
electrode reactions shown in Example 19.8. In addition to the apparatus and the solution, you are also
given two pieces of litmus paper, one blue and the
other red. Describe what steps you would take in
19.76 For a number of years it was not clear whether mercury(I) ions existed in solution as Hg or as Hg2 .
To distinguish between these two possibilities, we
could set up the following system:
Hg(l )soln A soln BHg(l ) where soln A contained 0.263 g mercury(I) nitrate
per liter and soln B contained 2.63 g mercury(I) nitrate per liter. If the measured emf of such a cell is
0.0289 V at 18°C, what can you deduce about the
nature of the mercury(I) ions?
19.77 An aqueous KI solution to which a few drops of Back Forward 19.79 Main Menu TOC 19.80 19.81 19.82 Describe what you would observe at the anode and
the cathode. (Hint: Molecular iodine is only slightly
soluble in water, but in the presence of I ions, it
forms the brown color of I3 ions. See Problem
A piece of magnesium metal weighing 1.56 g is
placed in 100.0 mL of 0.100 M AgNO3 at 25°C.
Calculate [Mg2 ] and [Ag ] in solution at equilibrium. What is the mass of the magnesium left? The
volume remains constant.
Describe an experiment that would enable you to
determine which is the cathode and which is the anode in an electrochemical cell using copper and zinc
An acidified solution was electrolyzed using copper
electrodes. A constant current of 1.18 A caused the
anode to lose 0.584 g after 1.52 103 s. (a) What
is the gas produced at the cathode and what is its
volume at STP? (b) Given that the charge of an electron is 1.6022 10 19 C, calculate Avogadro’s
number. Assume that copper is oxidized to Cu2
In a certain electrolysis experiment involving Al3
ions, 60.2 g of Al is recovered when a current of
0.352 A is used. How many minutes did the electrolysis last?
Consider the oxidation of ammonia:
4NH3(g) 3O2(g) 88n 2N2(g) 6H2O(l ) (a) Calculate the ∆G° for the reaction. (b) If this reaction were used in a fuel cell, what would the standard cell potential be?
19.83 An electrochemical cell is constructed by immersing a piece of copper wire in 25.0 mL of a 0.20 M
CuSO4 solution and a zinc strip in 25.0 mL of a 0.20
M ZnSO4 solution. (a) Calculate the emf of the cell
at 25°C and predict what would happen if a small Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS amount of concentrated NH3 solution were added to
(i) the CuSO4 solution and (ii) the ZnSO4 solution.
Assume that the volume in each compartment remains constant at 25.00 mL. (b) In a separate experiment, 25.0 mL of 3.00 M NH3 are added to the
CuSO4 solution. If the emf of the cell is 0.68 V, calculate the formation constant (Kf) of Cu(NH3)2 .
19.84 In an electrolysis experiment a student passes the
same quantity of electricity through two electrolytic
cells, one containing a silver salt and the other a
gold salt. Over a certain period of time, she finds
that 2.64 g of Ag and 1.61 g of Au are deposited at
the cathodes. What is the oxidation state of gold in
the gold salt?
19.85 People living in cold-climate countries where there
is plenty of snow are advised not to heat their
garages in the winter. What is the electrochemical
basis for this recommendation?
19.86 Given that
2 (aq) 2e 88n Hg2 (aq)
2 E° 0.92 V 2e 88n 2Hg(l ) E° 0.85 V calculate ∆G° and K for the following process at
Hg2 (aq) 88n Hg2 (aq)
2 Hg(l ) (The above reaction is an example of a disproportionation reaction in which an element in one oxidation state is both oxidized and reduced.)
19.87 Fluorine (F2) is obtained by the electrolysis of liquid hydrogen fluoride (HF) containing potassium
fluoride (KF). (a) Write the half-cell reactions and
the overall reaction for the process. (b) What is the
purpose of KF? (c) Calculate the volume of F2 (in
liters) collected at 24.0°C and 1.2 atm after electrolyzing the solution for 15 h at a current of
19.88 A 300-mL solution of NaCl was electrolyzed for
6.00 min. If the pH of the final solution was 12.24,
calculate the average current used.
19.89 Industrially, copper is purified by electrolysis. The
impure copper acts as the anode, and the cathode is
made of pure copper. The electrodes are immersed
in a CuSO4 solution. During electrolysis, copper at
the anode enters the solution as Cu2 while Cu2
ions are reduced at the cathode. (a) Write half-cell
reactions and the overall reaction for the electrolytic
process. (b) Suppose the anode was contaminated
with Zn and Ag. Explain what happens to these impurities during electrolysis. (c) How many hours
will it take to obtain 1.00 kg of Cu at a current of
18.9 A? Back Forward Main Menu TOC 797 19.90 An aqueous solution of a platinum salt is electrolyzed at a current of 2.50 A for 2.00 h. As a result, 9.09 g of metallic Pt are formed at the cathode.
Calculate the charge on the Pt ions in this solution.
19.91 Consider an electrochemical cell consisting of a
magnesium electrode in contact with 1.0 M
Mg(NO3)2 and a cadmium electrode in contact with
1.0 M Cd(NO3)2. Calculate E° for the cell, and draw
a diagram showing the cathode, anode, and direction of electron flow.
19.92 A current of 6.00 A passes through an electrolytic
cell containing dilute sulfuric acid for 3.40 h. If the
volume of O2 gas generated at the anode is 4.26 L
(at STP), calculate the charge (in coulombs) on an
19.93 Gold will not dissolve in either concentrated nitric
acid or concentrated hydrochloric acid. However,
the metal does dissolve in a mixture of the acids
(one part HNO3 and three parts HCl by volume),
called aqua regia. (a) Write a balanced equation for
this reaction. (Hint: Among the products are
HAuCl4 and NO2.) (b) What is the function of HCl?
19.94 Explain why most useful electrochemical cells give
voltages of no more than 1.5 to 2.5 V. What are the
prospects for developing practical electrochemical
cells with voltages of 5 V or more?
19.95 A silver rod and a SHE are dipped into a saturated
aqueous solution of silver oxalate, Ag2C2O4, at
25°C. The measured potential difference between
the rod and the SHE is 0.589 V, the rod being positive. Calculate the solubility product constant for
19.96 Zinc is an amphoteric metal; that is, it reacts with
both acids and bases. The standard oxidation potential is 1.36 V for the reaction
Zn(s) 4OH (aq) 88n Zn(OH)2 (aq)
4 2e Calculate the formation constant (Kf) for the reaction
Zn2 (aq) 4OH (aq) 34 Zn(OH)2 (aq)
4 19.97 Use the data in Table 19.1 to determine whether or
not hydrogen peroxide will undergo disproportionation in an acid medium: 2H2O2 88n 2H2O O2.
19.98 The magnitudes (but not the signs) of the standard
electrode potentials of two metals X and Y are
Y2 2e 88n X
2e 88n Y E°
E° 0.25 V
0.34 V where the notation denotes that only the magnitude (but not the sign) of the E° value is shown.
When the half-cells of X and Y are connected, elec- Study Guide TOC Textbook Website MHHE Website 798 ELECTROCHEMISTRY trons flow from X to Y. When X is connected to a
SHE, electrons flow from X to SHE. (a) Are the
E° values of the half-reactions positive or negative?
(b) What is the standard emf of a cell made up of
X and Y?
19.99 An electrochemical cell is constructed as follows.
One half-cell consists of a platinum wire immersed
in a solution containing 1.0 M Sn2 and 1.0 M Sn4 ;
the other half-cell has a thallium rod immersed in a
solution of 1.0 M Tl . (a) Write the half-cell reactions and the overall reaction. (b) What is the equilibrium constant at 25°C? (c) What is the cell voltage if the Tl concentration is increased tenfold?
19.100 Given the standard reduction potential for Au3 in
Table 19.1 and
Au (aq) e 88n Au(s) E° 19.105 19.106 1.69 V answer the following questions. (a) Why does gold
not tarnish in air? (b) Will the following disproportionation occur spontaneously?
3Au (aq) 88n Au3 (aq) 19.101 19.102 19.103 19.104 Back 19.107 2Au(s) (c) Predict the reaction between gold and fluorine
The ingestion of a very small quantity of mercury
is not considered too harmful. Would this statement
still hold if the gastric juice in your stomach were
mostly nitric acid instead of hydrochloric acid?
When 25.0 mL of a solution containing both Fe2
and Fe3 ions is titrated with 23.0 mL of 0.0200 M
KMnO4 (in dilute sulfuric acid), all of the Fe2 ions
are oxidized to Fe3 ions. Next, the solution is
treated with Zn metal to convert all of the Fe3 ions
to Fe2 ions. Finally, 40.0 mL of the same KMnO4
solution are added to the solution in order to oxidize the Fe2 ions to Fe3 . Calculate the molar concentrations of Fe2 and Fe3 in the original solution.
Consider the Daniell cell in Figure 19.1. When
viewed externally, the anode appears negative and
the cathode positive (electrons are flowing from the
anode to the cathode). Yet in solution anions are
moving toward the anode, which means that it must
appear positive to the anions. Since the anode cannot simultaneously be negative and positive, give an
explanation for this apparently contradictory situation.
Lead storage batteries are rated by ampere hours,
that is, the number of amperes they can deliver in
an hour. (a) Show that 1 A h 3600 C. (b) The
lead anodes of a certain lead-storage battery have a
total mass of 406 g. Calculate the maximum theo- Forward Main Menu TOC 19.108
19.109 19.110 retical capacity of the battery in ampere hours.
Explain why in practice we can never extract this
much energy from the battery. (Hint: Assume all of
the lead will be used up in the electrochemical reaction and refer to the electrode reactions on p. 776.)
(c) Calculate E° and ∆G° for the battery.
The concentration of sulfuric acid in the leadstorage battery of an automobile over a period of
time has decreased from 38.0 percent by mass (density 1.29 g/mL) to 26.0 percent by mass
(1.19 g/mL). Assume the volume of the acid remains constant at 724 mL. (a) Calculate the total
charge in coulombs supplied by the battery.
(b) How long (in hours) will it take to recharge the
battery back to the original sulfuric acid concentration using a current of 22.4 amperes.
Consider a Daniell cell operating under non-standard-state conditions. Suppose that the cell’s reaction is multiplied by 2. What effect does this have
on each of the following quantities in the Nernst
equation? (a) E, (b) E°, (c) Q, (d) ln Q, and (e) n?
A spoon was silver-plated electrolytically in a
AgNO3 solution. (a) Sketch a diagram for the
process. (b) If 0.884 g of Ag was deposited on the
spoon at a constant current of 18.5 mA, how long
(in minutes) did the electrolysis take?
Comment on whether F2 will become a stronger oxidizing agent with increasing H concentration.
In recent years there has been much interest in electric cars. List some advantages and disadvantages
of electric cars compared to automobiles with internal combustion engines.
Calculate the pressure of H2 (in atm) required to
maintain equilibrium with respect to the following
reaction at 25°C:
Pb(s) 2H (aq) 34 Pb2 (aq) H2(g) Given that [Pb2 ] 0.035 M and the solution is
buffered at pH 1.60.
19.111 A piece of magnesium ribbon and a copper wire are
partially immersed in a 0.1 M HCl solution in a
beaker. The metals are joined externally by another
piece of metal wire. Bubbles are seen to evolve at
both the Mg and Cu surfaces. (a) Write equations
representing the reactions occurring at the metals.
(b) What visual evidence would you seek to show
that Cu is not oxidized to Cu2 ? (c) At some stage,
NaOH solution is added to the beaker to neutralize
the HCl acid. Upon further addition of NaOH, a
white precipitate forms. What is it?
19.112 The zinc-air battery shows much promise for electric cars because it is lightweight and rechargeable: Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 799 battery every second? Assume that the temperature
is 25°C and the partial pressure of oxygen is 0.21
19.113 Calculate E° for the reactions of mercury with
(a) 1 M HCl and (b) 1 M HNO3. Which acid will
oxidize Hg to Hg2 under standard-state conditions?
Can you identify which test tube below contains
HNO3 and Hg and which contains HCl and Hg? The net transformation is Zn(s) 1 O2(g) 88n
ZnO(s). (a) Write the half-reactions at the zinc-air
electrodes and calculate the standard emf of the battery at 25°C. (b) Calculate the emf under actual operating conditions when the partial pressure of oxygen is 0.21 atm. (c) What is the energy density
(measured as the energy in kilojoules that can be
obtained from one kilogram of the metal) of the zinc
electrode? (d) If a current of 2.1 105 A is to be
drawn from a zinc-air battery system, what volume
of air (in liters) would need to be supplied to the Back Forward Main Menu TOC Answers to Practice Exercises: 19.1 5Fe2
8H 88n 5Fe3
4H2O. 19.2 No. 19.3 0.34 V.
19.4 1.1 10 42. 19.5 ∆G°
4.1 102 kJ. 19.6 Yes.
19.7 0.38 V. 19.8 Anode, O2; cathode, H2. 19.9 2.0 104 A. Study Guide TOC Textbook Website MHHE Website C HEMICAL M YSTERY Tainted Water † T he salesman was persuasive and persistent.
“Do you realize what’s in your drinking water?” he asked Tom. Before
Tom could answer, he continued: “Let me demonstrate.” First he filled a
glass of water from the kitchen faucet, then he produced an electrical device that had a pair of probes and a light bulb. It resembled a standard
conductivity tester. He inserted the probes into the water and the bulb immediately
beamed brightly. Next the salesman poured some water from a jar labeled “distilled
water” into another glass. This time when he inserted the probes into the water, the
bulb did not light.
“Okay, can you explain the difference?” the salesman looked at Tom with a triumphant smile. “Sure,” Tom began to recall an experiment he did in high school long
ago, “The tap water contains minerals that caused. . . .”
“Right on!” the salesman interrupted. “But I’m not sure if you realize how harmful the nation’s drinking water has become.” He handed Tom a booklet entitled The
Miracle of Distilled Water. “Read the section called ‘Heart Conditions Can Result from
Mineral Deposits’,” he told Tom.
“The tap water may look clear, although we know it contains dissolved minerals.
What most people don’t realize is that it also contains other invisible substances that
are harmful to our health. Let me show you.” The salesman proceeded to do another
demonstration. This time he produced a device that he called a “precipitator,” which
had two large electrodes attached to a black box. “Just look what’s in our tap water,”
he said, while filling another large glassful from the faucet. The tap water appeared
clean and pure. The salesman plugged the precipitator into the ac (alternating current)
outlet. Within seconds, bubbles rose from both electrodes. The tap water took on a yellow hue. In a few minutes a brownish scum covered the surface of the water. After 15
minutes the glass of water was filled with a black-brown precipitate. When he repeated
the experiment with distilled water, nothing happened.
Tom was incredulous. “You mean all this gunk came from the water I drink?”
“Where else?” beamed the salesman. “What the precipitator did was to bring out
all the heavy metals and other undesirable substances. Now don’t worry. There is a
remedy for this problem. My company makes a distiller that will convert tap water to
distilled water, which is the only safe water to drink. For a price of $600 you will be
able to produce distilled water for pennies with the distiller instead of paying 80 cents
for a gallon of water from the supermarket.”
Tom was tempted but decided to wait. After all, $600 is a lot to pay for a gadget
that he only saw briefly. He decided to consult his friend Sarah, the chemistry teacher
Adapted with permission from “Tainted Water,” by Joseph J. Hesse, CHEM MATTERS, February, 1988, p. 13. Copyright
1988 American Chemical Society. 800 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website The precipitator with its electrodes immersed in tap water.
Left: Before electrolysis has
started. Right: 15 minutes after
electrolysis commenced. at the local high school, before making the investment. The salesman promised to return in a few days and left the precipitator with Tom so that he could do further testing.
CHEMICAL CLUES 1. 2.
4. 5. After Sarah examined the precipitator, she concluded that it was an electrolytic device consisting of what seemed like an aluminum electrode and an iron electrode.
Since electrolysis cannot take place with ac (why not?), the precipitator must contain a rectifier, a device that converts ac to dc (direct current). Why does the water heat up so quickly during electrolysis?
From the brown color of the electrolysis product(s), deduce which metal acts as
the cathode and which electrode acts as the anode.
Write all possible reactions at the anode and the cathode. Explain why there might
be more than one type of reaction occurring at an electrode.
In analyzing the solution, Sarah detected aluminum. Suggest a plausible structure
for the aluminum-containing ion. What property of aluminum causes it to dissolve
in the solution?
Suggest two tests that would confirm Sarah’s conclusion that the precipitate originated from the electrodes and not from the tap water.
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