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CHEM 111 Chapter Notes - Chapter 19: Zinc Sulfide, Jmol, JouleExam


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Chemistry
Course Code
CHEM 111
Professor
Prof
Study Guide
Final

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390
CHAPTER 19 | Electrochemistry and the Quest for Clean Energy
19.1. Collect and Organize
For the voltaic cell shown in Figure P19.1, we are to explain why a porous separator is not required.
Analyze
The porous separator serves to keep the reduction and oxidation half-reactions separate so that electrons are
passed through the external circuit.
Solve
Because of the careful layering, each half-cell has its metal in contact with its cation solution. The solutions are
not mixing, but nevertheless the layers allow the ions needed to balance the charge in each half-cell to pass.
Think about It
The half-reactions and overall reaction for this voltaic cell are
Zn(s)o Zn2+(aq) + 2 e
anode
ED= –0.7618 V
Cu2+(aq) + 2 eo Cu(s)cathode
ED= 0.3419 V
Zn(s) + Cu2+(aq)o Zn2+(aq) + Cu(s)cell
ED=cathode
EDanode
ED
= 0.3419 (–0.7618) = 1.1037 V
19.2. Collect and Organize
From the standard reduction potentials of Cu2+ and Cd2+ in Appendix 6, we can identify the cathode and anode
and indicate the direction of flow of electrons in the external circuit in Figure P19.2.
Analyze
A voltaic cell has a positive cell
EDindicating a spontaneous reaction. At one electrode oxidation occurs (the anode),
and at the other electrode reduction occurs (the cathode). From Appendix 6, 2+
red (Cu ) 0.3419 VE
Dand
2+
red (Cd ) 0.403 V.
D
E
Solve
To give a positive cell
ED, Cu2+ is reduced and Cd is oxidized:
2–
cathode
2–
anode
Cu ( ) 2 e Cu( ) 0.3419 V
Cd( ) Cd ( ) 2 e 0.403 V
o
o
D
D
aq s E
saq E
Thus, Cu is the cathode, Cd is the anode, and electrons flow from Cd to Cu (to the left in the circuit shown in
Figure P19.2).
Think about It
The overall reaction and potential for this voltaic cell is
Cu2+(aq) + Cd(s)o Cu(s) + Cd2+(aq)cell
ED=cathode
EDanode
ED= 0.3419 – (–0.401) = 0.745 V
19.3. Collect and Organize
For the voltaic cell shown in Figure P19.3 in which an Ag+/Ag cell is connected to a standard hydrogen
electrode (SHE), we are to determine which electrode is the anode and which is the cathode and indicate in
which direction the electrons flow in the outside circuit.
Analyze
A voltaic cell runs spontaneously when Ecell is positive. By comparing the reduction potentials of each half-
cell, we can write the reaction that is spontaneous for the cell. The half-cell where reduction (gain of electrons)
occurs contains the cathode and the half-cell where oxidation occurs contains the anode. Electrons flow from
the anode, where they are produced by oxidation, toward the cathode, where they are required for reduction.

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Electrochemistry and the Quest for Clean Energy |391
Solve
The spontaneous reaction for this cell is
2 u (Ag+ + eo Ag) cathode
ED= 0.7996 V
H
2o 2 H+ + 2 e
anode
ED= 0.000 V
2 Ag+ + H2o 2 H+ + Ag cell
ED=cathode
EDanode
ED
= 0.7996 – 0.000 = 0.7996 V
Thus, Ag is the cathode, Pt in the SHE is the anode, and electrons flow from the SHE to Ag (to the left in the
circuit shown in Figure P19.3).
Think about It
The shorthand notation for this cell would be
Pt(s) | H2(g) | H+(aq) || Ag+(aq) | Ag(s)
where Pt is an inert electrode used in the SHE.
19.4. Collect and Organize
From the highlighted elements in the periodic table in Figure P19.4, we are to choose the group of elements
that make the best inert electrodes.
Analyze
Inert metal electrodes would have very negative (not favorable) oxidation potentials. These metals include
those from which durable materials are made (jewelry, electrical contacts).
Solve
The elements in blue (Pt, Au, and Hg) are the best metals to use as inert electrodes.
Think about It
Of the three, platinum is often used in electrochemistry as an inert electrode for the construction of the
standard hydrogen electrode (SHE). Mercury, although a liquid, is also used as an electrode. It is used in
polarography as a dropping mercury electrode.
19.5. Collect and Organize
The graph of cell potential versus [H2SO4] (Figure P19.5) shows four lines, some curved, some linear, some
increasing, and some decreasing, as the concentration of H2SO4 decreases. From the shape of the curves and
their trends we are to choose the line that best represents the trend in potential versus [H2SO4] in a lead-acid
battery.
Analyze
The scale for [H2SO4] is logarithmic and voltage in the battery varies with the [H2SO4] according to the Nernst
equation:
>@
cell 2
24
0.0592 1
2.04 V log
2HSO
E
Solve
From the Nernst equation we see that the cell potential drops as the log 1/[H2SO4]2 decreases. So as [H2SO4]
decreases, the cell potential also decreases. The red line on the graph shows the opposite trend: The voltage
increases as [H2SO4] decreases. In considering which of the remaining lines might describe the lead-acid
battery, we must consider that because the cell voltage drops as log 1/[H2SO4]2 we expect that the decrease in
potential is linear. Therefore, the blue line best describes the potential as a function of [H2SO4] concentration.
Think about It
Another characteristic of lead-acid batteries is that their cell voltage does not drop substantially until over 90%
of the battery has been discharged (see Figure 19.12).

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392 | Chapter 19
19.6. Collect and Organize
From among the measures of battery performance, we are to choose which differ among different sizes of
batteries.
Analyze
The voltage describes the electrochemical potential of the battery and is dependent on the chemicals used in
the battery. All the batteries shown in Figure P19.6 are alkaline batteries and so the voltage does not differ
among them. A watt-hour describes the energy delivered in one hour and a milliampere-hour describes the
charge delivered in an hour. These are both dependent on the amount of battery material, which differs among
the sizes of the batteries.
Solve
Both (b) watt-hour and (c) milliampere-hour differ among the batteries because of their different sizes and
different amounts of battery material.
Think about It
A larger battery of the same voltage can be used to provide more power to an electric motor or other such
electrical device or power the device for a longer period of time. For example, the Chevy Volt has a total
battery capacity of 16 kilowatt-hours. This battery can produce 16,000 watts for one hour or one watt for
16,000 hours or anything in between. To increase the life of the lithiumion battery in the Volt, however, the
capacity used is a little over 10 kWh.
19.7. Collect and Organize
For the electrolysis of water in which the two product gases, H2 and O2, are collected in burettes (Figure
P19.7), we are to write the half-reactions occurring at each electrode and discuss why a small amount of acid
was added to the water to speed up the reaction.
Analyze
In the electrolysis of water, electricity is supplied to make the nonspontaneous oxidation and reduction
reactions occur. From Figure P19.7 we notice that the left burette has collected twice the volume of gas
compared to the right burette. From the overall balanced equation
222
2 H O( ) 2 H ( ) O ( )
g
goA
we can identify the left burette as containing H2 and the right burette as containing O2.red
EDvalues are given in
Appendix 6.
Solve
(a) The left electrode is the cathode where reduction is occurring:
––
2 2 cathode
2 H O( ) 2 e H ( ) 2 OH ( ) 0.8277 Vo
D
AgaqE
The right electrode is the anode where oxidation is occurring:
+–
2 2 anode
2 HO() O()4 H()4 e 1.229 Vo
D
Agaq E
(b) A small amount of H2SO4 is added to the water to increase the conductivity of the solution.
Think about It
The electrochemical potential for the overall process is
cell cathode anode
0.8277 V 1.229 V –2.057 V
DD D
EE E
19.8. Collect and Organize
When the hydrolysis of water shown in Figure P19.7 is conducted with added Na2CO3, we are to write the
half-reactions for the electrodes and explain why Na2CO3 makes the reaction go more quickly.
Analyze
The added ions, Na+ and CO3
2–, do not get involved in the electrolysis reaction. The reduction of Na+ is too
negative (–2.71 V) and Appendix 6 does not give any reduction reaction involving CO3
2–.
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