One area that is still a significant source of technological transfer is medicine. Over 100 Aboriginal medicines are used today in North American pharmaceutical drugs, including acetylsalicylic acid (Aspirin), which originally came from the willow tree. This is an area of dispute because the mechanisms of some traditional herbs has not been fully understood or accepted by doctors and the medical community. In some cases, Western scientists observed the medicines that have worked, and then patented the medicine, which may force Native people to pay for their own traditional remedies.
14. (a) Some of the requirements for cells used in electronic devices are:
Ɣ long life Ɣ small size
Ɣ constant voltage during their active lives Ɣ reliability for many recharges
Ɣ low expense
(b) The storage capacity of some rechargeable batteries is reduced if they are recharged before being totally drained because the battery then may only discharge to the point where they were recharged previously. Such batteries are sometimes referred to as having a “memory.”
15. Lithiumiodine batteries are commonly used in pacemakers, but some are powered by radioactive isotopes. Conventional pacemaker batteries last from four to eight years, depending on how much the heart uses the pacemaker. A doctor tracks the signals from the pacemaker to determine when the battery is nearing the end of its life. Rechargeable batteries generally hold less energy, and thus would need to be replaced more often.
16. Plastic batteries are unique in having electrodes made of conducting “doped” polymers rather than metallic materials. In some recent plastic cells, the electrolyte is also made of a polymer materialresulting in a “leakproof” battery that contains no liquid. Charging involves ion migration to the electrodes, but the electrodes themselves do not charge. This means that, theoretically, the cells are rechargeable through a very large number of cycles. The cells are very lightweight, can be any shape, and have high-energy density and power density
characteristics, which make them of great interest to the designers of satellites. A significant advantage is that such cells operate through a wide range of temperatures without much change in their electrical properties. Another strong social advantage is that they do not contain environmentally harmful materials. Currently, such cells are expensive and do not store large quantities of energy, although research and development in this area is ongoing.
14.2 VOLTAIC CELLS
Investigation 14.2: A Voltaic Cell (Demonstration) (Pages 623, 659)
Purpose
The purpose of this investigation is to test the design and operation of a voltaic cell used in scientific research.
Problem
What is the design and operation of a voltaic cell?
Evidence
Table 1: Voltage for Various Ag–Cu Cell Designs
Cell design Positive electrode Negative electrode Voltage (V) (a) single electrolyte Ag(s) Cu(s) 0.15 (b) salt bridge Ag(s) Cu(s) 0.45 (c) porous cup Ag(s) Cu(s) 0.46
Ɣ Removing either electrode from the solution, removing the salt bridge, or removing the porous cup immediately produced a reading of 0.00 V. Replacing the removed part restored the original voltmeter reading.
Ɣ No evidence of reaction was visible after several minutes. After several days, long, silvery crystals formed on the silver electrode in cells (b) and (c) and some blue-coloured solution in cell (b) had moved toward the silver side of the cell.
Analysis
On the basis of the evidence gathered, a voltaic cell requires two different electrodes that are both in contact with an electrolyte. All parts, external and internal, must be connected for the cell to operate (to produce a voltage). The salt bridge and porous cup designs functioned in very much the same way, and produced a higher voltage than the design that involved no porous boundary.
Practice (Page 626)
1. A voltaic cell is an arrangement of two half-cells separated by a porous boundary that spontaneously produces electricity.
A half-cell is an electrode–electrolyte combination that forms one-half of a complete cell.
A porous boundary is a barrier that separates two electrolytes while still permitting the movement of ions.
A salt bridge is a U-shaped tube containing an inert electrolyte.
An electrolyte is a solute that forms a solution that conducts an electric current.
An external circuit is an electrical connection that completes the circuit outside of the cell.
(The device that uses electricity is connected in this external circuit.)
An inert electrode is a solid conductor that serves as an anode or a cathode in a voltaic cell, but is chemically unreactive.
2. A cathode is the electrode where reduction of the strongest oxidizing agent occurs.
An anode is the electrode where oxidation of the strongest reducing agent occurs.
3. (a) The reduction half-reaction occurs at the cathode.
(b) The oxidation half-reaction occurs at the anode.
(c) The reaction of the strongest reducing agent occurs at the anode.
(d) The reaction of the strongest oxidizing agent occurs at the cathode.
4. An inert electrode is used in a half-cell where no conducting solid is involved in the half- reaction equation. A carbon, or graphite rod and platinum metal foil are two commonly used inert electrodes.
5. The solution in a salt bridge must be an inert (unreactive) electrolyte; that is, it must contain ions that do not react during the operation of the cell. An example of an inert aqueous electrolyte is sodium sulfate.
6. (a) SOA
Zn(s) | Zn
2+(aq) | | Ag
+(aq) | Ag(s) SRA
cathode 2 [Ag
+(aq) + e
–o Ag(s)]
anode Zn(s) o Zn
2+(aq) + 2 e
–net 2 Ag
+(aq) + Zn(s) o 2 Ag(s) + Zn
2+(aq)
(b) SOA
Al(s) | Al
3+(aq) | | NO
3–(aq), H
+(aq) | Pt(s) SRA
cathode 3 [2 NO
3–(aq) + 4 H
+(aq) + 2 e
–o N
2O
4(g) + 2 H
2O(l)(aq)]
anode 2 [Al(s) o Al
3+(aq) + 3 e
–]
net 2 Al(s) + 6 NO
3–(aq) + 12 H
+(aq) o 2 Al
3+(aq) + 3 N
2O
4(g) + 6 H
2O(l)
7. (a) Cations move toward the cathode and anions move toward the anode.
(b) Ions move to maintain electrical neutrality at each electrode. Why would the solutions at each electrode not remain neutral? In a simple voltaic cell composed of metals and metal ions, reduction removes positive ions (cations) from the solution around the cathode, and oxidation adds positive ions (cations) to the solution around the anode. Anions move toward the anode to balance the excess positive charge in the solution around the anode, while cations move toward the cathode to replace the positive charge being removed from the solution around the cathode.
(c) Not only do the ions in the U-tube move, but the ions in the electrolyte solutions move as well. Since the copper electrode is the anode, it loses electrons and is oxidized (LEO) according to the reaction: Cu(s) o Cu (aq) + 2e
2+ -. The Cu (aq) cation is blue and
2+migrates through the U-tube towards the solution by the positive cathode, Ag(s).
[Alternatively, one half-cell could be placed inside a porcelain cup set inside the beaker containing the other half-cell.]
9. The nickel cathode could be replaced by an inert conducting material such as carbon or platinum. The anode solution could be replaced with an inert aqueous electrolyte such as aqueous potassium sulfate.
Web Activity: Simulation—Voltaic Cells Under Standard Conditions (Page 630)
[No written response is required.]
Investigation 14.3: Testing Voltaic Cells (Pages 631, 660)
Purpose
The purpose of this investigation is to test the predictions of cell potentials and the identity of the electrodes of various cells.
Problem
In cells constructed from various combinations of copper, aluminium, silver, and zinc half-cells, what are the standard cell potentials, and which is the anode and cathode in each case?
Prediction
According to the redox concepts and the table of redox half-reactions:
Cathode (+) Anode(–) Eºcell (V)
Cu(s) Cu2+(aq) Al3+(aq) Al(s) +2.00 Ag(s) Ag+(aq) Cu2+(aq) Cu(s) +0.46 Cu(s) Cu2+(aq) Zn2+(aq) Zn(s) +1.10 Ag(s) Ag+(aq) Al3+(aq) Al(s) +2.46 Zn(s) Zn2+(aq) Al3+(aq) Al(s) +0.90 Ag(s) Ag+(aq) Zn2+(aq) Zn(s) +1.56
Design
Individual metal–metal ion half-cells are constructed. Different combinations are connected with a salt bridge and the electrodes, and the cell potentials are determined. The manipulated variable is the combination of half-cells, and the responding variable is the cell potential. Controlled variables are temperature and electrolyte concentration.
8.
Procedure
1. Clean metal strips with steel wool and rinse with distilled water.
2. Assemble two of the four metal–metal ion half-cells, for example, copper and zinc.
3. Connect the copper half-cell with the zinc half-cell using the salt bridge.
4. Use the voltmeter and connecting wires to determine the cathode and anode of the cell.
5. With the voltmeter connected to the cell, measure the initial voltmeter reading.
6. Remove and rinse the salt bridge.
7. Repeat steps 1 to 6 for the remaining combinations of half-cells.
8. Clean and return the metal strips and recycle the electrolyte solutions.
Evidence/Analysis
Cathode (+) Anode (–) Predicted
potential (V) Measured
potential (V) Percent diff. (%) Cu(s) Cu2+(aq) Al3+(aq) Al(s) +2.00 +0.75 63 Ag(s) Ag+(aq) Cu2+(aq) Cu(s) +0.46 +0.45 2 Cu(s) Cu2+(aq) Zn2+(aq) Zn(s) +1.10 +1.06 4 Ag(s) Ag+(aq) Al3+(aq) Al(s) +2.46 +1.10 55 Zn(s) Zn2+(aq) Al3+(aq) Al(s) +0.90 +0.10 89 Ag(s) Ag+(aq) Zn2+(aq) Zn(s) +1.56 +1.47 6
Evaluation
The design is adequate to answer the problem, with no obvious flaws. The electrodes were easily identified and the cell potentials measured without any difficulties. There are no known
alternatives that would be better. The purity of the metals and the accuracy of the solution concentration are not known, but otherwise, the materials appear adequate. The procedure was generally adequate because the steps were correctly sequenced and sufficient evidence was collected to answer the problem, but the procedure does have some inadequacies. The temperature was controlled, but was not set at the standard value. This should be improved to eliminate some uncertainties in the results. Only one trial was conducted, and this number should be increased. Technological skills appeared adequate and no special skills were required.
On the basis of my evaluation of the experiment, I am quite certain about the cathode and anode, but only moderately certain of the cell potentials. Sources of experimental error or
uncertainty include small measurement uncertainties in the solution preparation and the voltmeter readings, cleanliness of the metal strips, purity of the metals and solutions, and the non-standard conditions of temperature. Without knowing the value of all of these uncertainties and assuming that they will be relatively small (5–10%), the evidence should be of a reasonable quality.
The prediction of the cathode and anode was completely verified in each case. The predicted cell potentials appear to be generally verified for all cells, except those involving aluminium, as the differences are around 5%. However, all measured cell potentials are less than the predicted values. This suggests some small systematic error for all cells. Perhaps this is because the temperature was less than the standard 25 °C and/or the voltmeter was reading consistently low. The drastic percent difference observed with the aluminium half-cells suggests that there is some other unknown factor affecting this particular half-cell. This could be
something about the particular aluminium strip used; perhaps it is an alloynot pure aluminium.
Another possibility is an influence of the oxide coating on the surface of the Al(s) electrode. The redox concepts and redox table are judged acceptable for the majority of cases, because the predictions were generally verified.
The purpose was accomplished to a limited extent. Because of the uncertainties
identified, additional testing with the improvements as noted should be conducted. In particular,
the cells containing the aluminium half-cells need to be redone with new strips and solutions to
identify the source of the large discrepancy.
Practice (Page 631)
10. (a) In a standard cell, each half-cell has all the entities shown in the half-reaction equation at standard ambient temperature and pressure (SATP) conditions. Aqueous solutions have a concentration of 1.0 mol/L.
(b) The standard cell potential, E
ºcell, is the maximum electric potential difference (voltage) of the cell at standard conditions. It represents the difference in ability of the two half-cells to gain electrons.
Eºcell
, = E
ºr, – E
ºrcathode anode
11. (a)
2+ r2+
r
2+ 2+
cell cell
cathode Sn (aq) 2e Sn(s) 0.14 V
anode Cr(s) Cr (aq) 2e 0.91 V
net Sn (aq) + Cr(s) Sn(s) + Cr (aq) ?
0.14 V ( 0.91 V) 0.77 V
E E E E
o q
o q
o q
q
(b)
42 2 3 2 r2+
2 2+ r
4 2 3 2 cell
cell
cathode SO (aq) 4 H (aq) 2 e H SO (s) H O(aq) 0.17 V
anode Co(s) Co (aq) 2 e 0.28 V
net SO (aq) 4 H (aq) Co(s) H SO (s) H O(aq) Co (aq) ? 0.17 V ( 0.28 V) 0.45 V
E E E E
o q
o q
o q
q
(c)
2 2 r2 2 r
2 2 2 cell
cell
cathode O (g) 2 H O(l) + 4 e 4 OH (aq) 0.40 V
anode 2 H (g) + 2 OH (aq) 2 H O(l) 2e 0.83 V
net O (g) 2 H (g) 2 H O(aq) ?
0.40 V ( 0.83 V) 1.23 V
E E E E
o q
ª o º q
¬ ¼
o q
q
Practice (Page 633)
12. (a) Pb(s) | Pb
2+(aq) || Cu
2+(aq) | Cu(s)
anode cathode
cell
0.34 V ( 0.13 V) 0.47 V
Eq
(b) Zn(s) | Zn
2+(aq) || Ni
2+(aq) | Ni(s)
anode cathode
cell
0.26 V ( 0.76 V) 0.50 V
Eq
(c) Pt(s) | H
+(aq), H
2(g) || Fe
3+(aq), Fe
2+(aq) | Pt(s) (or C(s) as anode and cathode)
anode cathode
cell
0.77 V (0.00V) 0.77 V
Eq
13.
3+ r3+
r
3+ 3+
cell
cell r
r
cathode Au (aq) 3 e Au(s) 1.50 V
anode In(s) In (aq) 3e ?
net Au (aq) + In(s) Au(s) + In (aq) 1.84 V
1.84 V 1.50 V 0.34 V
E E E
E E
E
o q
o q
o q
q q
q
14. Changing the reference half-cell does not change the potential difference between any two
half-reactions. Changing the lithium half-cell to 0.00 V means adding 3.04 V to each
reduction potential on the chart.
2+
2+ r
2+ 2+ r
cell cell
cathode Cu (aq) 2 e Cu(s) 3.38 V
anode Zn(s) Zn (aq) 2e 2.28 V
net Cu (aq) + Zn(s) Cu(s) + Zn (aq) ? 3.38 V ( 2.28 V) 1.10 V
E E E E
o q
o q
o q
q
The original reduction potentials would give the following results:
2+
r 2+
r
2+ 2+
cell cell
cathode Cu (aq) 2e Cu(s) 0.34 V
anode Zn(s) Zn (aq) 2 e 0.76 V
net Cu (aq) + Zn(s) Cu(s) + Zn (aq) ? 0.34 V ( 0.76 V) 1.10 V
E E E E
o q
o q
o q
q
15. There are many reasons a predicted cell potential from a table of reduction potentials would be different from measured voltages in the lab. These include:
Ɣ the purity and cleanliness of the electrodes Ɣ the concentration of the electrolytes Ɣ the temperature of the lab
Ɣ the resistance of the wire leads Ɣ the accuracy of the voltmeter
Ɣ the presence of contaminants in the electrolytes
16. If a zinciron cell is run until the potential difference is zero, it means that the concentration of reactants has decreased and the concentration of products has increased to reach an equilibrium—the rate of the forward reaction equals the rate of the reverse reaction.
Lab Exercise 14.A: Developing a Redox Table (Page 633)
Purpose
The purpose of this exercise is to use the concepts and rules of standard cells to develop a redox table.
Problem
What is the table of relative strengths of oxidizing and reducing agents based on measured cell potentials?
Analysis
anode cathode
Pd(s) | Pd
2+(aq) || H
+(aq), Cr
3+(aq), Cr
2O
72–(aq) | C(s)
E°cell= +0.28 V
E°r= ?
E°r= +1.23 V
+0.28 V = +1.23 V – E°
r (anode)Er
°
(anode)= +0.95 V
According to the evidence, the standard reduction potential for the Pd(s) | Pd
2+(aq) half-cell is +0.95 V.
anode cathode
Tl(s) | Tl
+(aq) || Pd
2+(aq) | Pd(s)
E°cell= +1.29 V
E°r= ? E°
r= +0.95 V
+1.29 V = +0.95 V – E°
r (anode)E°r (anode)
= –0.34 V
According to the evidence, the standard reduction potential for the Tl(s) | Tl
+(aq) half-cell is
–0.34 V.
anode cathode
Ti(s) | Ti
2+(aq) || Tl
+(aq) | Tl(s)
E°cell= +1.29 V
E°r= ?
E°r= –0.34 V
+1.29 V = –0.34 V – E°
r (anode) E°r (anode)= –1.63 V
According to the evidence, the standard reduction potential for the Ti(s) | Ti
2+(aq) half-cell is –1.63 V.
According to the evidence, the relative strength of the four oxidizing agents, in decreasing order of E°
r, is:
(SOA) Cr
2O
72–(aq) + 14 H
+(aq) + 6 e
–o Cr
3+(aq) + 7 H
2O(l)
E°r= +1.23 V
Pd
2+(aq) + 2 e
–o Pd(s)
E°r= +0.95 V
Tl
+(aq) + e
–o Tl(s)
E°r= –0.34 V
Ti
2+(aq) + 2 e
–o Ti(s)
E°r= –1.63 V
Mini Investigation: Home Corrosion Experiment (Page 635)
(a) The iron in the steel nail will corrode more in the lemon-lime drink than in the cola because, according to the labels, the lemon-lime soda has more kinds of acids (carbonic acid, malic acid, citric acid) than the cola (carbonic acid and phosphoric acid).
(b) The evidence shows that the nail in the cola corroded more than the one in the lemon-lime soda, but a longer observation period may be necessary to confirm this. The concentrations of the acids are not given on the label so it is possible that the concentration of H
+(aq) played a role. Because H
+(aq) appears above Fe(s) on the table of reduction half-reactions, we expect a spontaneous reaction. It seems likely that the soft drink with the highest concentration of H
+(aq) causes the most corrosion of the iron nail. Another possible explanation might relate to the nature of the acids and their oxidizing ability.
Career Connection: Materials Engineering Technologist (Page 636)
Ɣ Materials engineering technologists work in many areas. For example, they may test, analyze, and inspect metals for corrosion. In addition to performing direct testing for corrosion, materials engineering technologists may also carry out non-destructive testing of materials to test their structural integrity, which can be affected by corrosion.
Ɣ The Northern Alberta Institute of Technology (NAIT) in Edmonton offers a two-year Materials Engineering Technology diploma program. Admission into the program at NAIT requires a high-school diploma or equivalent with English 30 or 33, Math 30 (or equivalent, or 60% in Math 33), and one of: Science 30, Physics 30, or Chemistry 30.
Ɣ Over 1200 Albertans are employed in the Industrial Engineering and Manufacturing Technologists and Technicians occupational group, which is expected to grow 1.7 to 2.7%
each year from 2004 to 2009 in Alberta. It is forecasted that 20 to 40 new positions will be created each year in addition to job openings created by employment turnover.
Ɣ Albertans in the Industrial Engineering and Manufacturing Technologists and Technicians occupational group earned an average salary of $68,200.
Practice (Page 637)
17. For the corrosion of iron to occur, an oxidizing agent (most commonly oxygen and water)
must be present and in contact with the iron.
18. The presence of acidic solutions, electrolytes, mechanical stresses, and contact with less active metals accelerate the corrosion of iron.
19. O
2(g) + 2 H
2O(l) + 4 e
–o 4 OH
–(aq) 2 [Fe(s) o Fe
2+(aq) + 2 e
–] O
2(g) + 2 H
2O(l) + 2 Fe(s) o 2 Fe(OH)
2(s) 20. (a) O
2(g) + 2 H
2O(l) + 4 e
–o 4 OH
–(aq) 2 [Zn(s) o Zn
2+(aq) + 2 e
–]
O
2(g) + 2 H
2O(l) + 2 Zn(s) o 2 Zn(OH)
2(s) (b) O
2(g) + 4 H
+(aq) + 4 e
–o 2 H
2O(l)
2 [Pb(s) o Pb
2+(aq) + 2 e
–]
O
2(g) + 4 H
+(aq) + 2 Pb(s) o 2 H
2O(l) + 2 Pb
2+(aq) (c) H
2S(g) + 2 e
–o H
2(g) + S
2–(s)
2 [Ag(s) o Ag
+(s) + e
–] or 2 Ag(s) + S
2–(s) o Ag
2S(s) + 2 e
–H
2S(g) + 2 Ag(s) o H
2(g) + Ag
2S(s)
21. (a) The painted surface is blistered, providing evidence for damage extending well beyond the break in the paint.
(b) Moisture may be trapped between the steel and the paint, setting up an electrochemical cell. The iron below the paint is oxidized, releasing electrons that travel through the steel to the edge of the crack where both oxygen and water are present to pick up these electrons.
22. Since hydroxide ions (either alone or in combination with other ions) often act as reducing agents, a basic solution might prevent or slow down the corrosion of iron.
23. A zinc coating on iron is better than a tin coating because zinc is more easily oxidized than iron and, thus, the zinc will corrode first. Tin is less active than iron and will promote the oxidation of iron.
24. Two methods of cathodic protection are impressed current and sacrificial anode. These are similar in that both methods force the iron to become the cathode by supplying it with electrons.
Web Activity: Case Study—Galvanizing Steel (Page 637)
In the galvanizing process, the surface of the steel to be galvanized undergoes caustic cleaning, which uses a basic solution to remove oils and other dirt from the surface. Once rinsed, the steel is placed into an acid bath, (called pickling), to remove any mill scale and rust. After it is rinsed again, it is coated in flux, which is a chemical that helps to chemically bond the zinc and steel together, as well as slow any further oxidation of the steel. The steel is then lowered into molten zinc at 850 qC. During this stage, the steel and zinc bond together, forming a layer of zinciron compounds, topped with an outer layer of pure zinc. After it is cooled, the galvanized steel undergoes an inspection process that includes checking the thickness of the coating of zinc.
Comparing a galvanized pipe with a painted pipe: galvanization lasts much longer and requires much less maintenance. A galvanized pipe can last 30 or more years without
maintenance, while painted pipe needs to be retouched every few years and repainted
approximately every 10 years. The zinc coating on the pipe also provides cathodic protection,
which means the zinc can act as the strongest reducing agent instead of the iron in the pipe. This
causes the zinc to react with oxidizing agents instead of the iron, with the added advantage that
the zinc reaction produces zinc oxide, which is a strong, protective coating.
Section 14.2 Questions (Pages 637–638)
1. half-cell: one half of a voltaic cell that consists of one electrode and one electrolyte.
cathode: in a voltaic cell, the positive electrode where reduction occurs.
anode: in a voltaic cell, the negative electrode where oxidation occurs.
cation: a positively charged ion that travels towards the cathode in a voltaic cell.
anion: a negatively charged ion that travels towards the anode in a voltaic cell.
inert electrode: an electrode that does not react itself, but provides a location where a half- reaction can occur.
2. The function of the porous boundary is to keep the two electrolytes separated, but to still allow the movement of ions between the solutions of the two half-cells to maintain neutral solutions. Two common examples of a porous boundary are a salt bridge and a porcelain cup.
3. The external circuit is the electrical connection that connects to each electrode of the two half-cells. The electrons flow along the external circuit from the anode to the cathode in any cell. The internal circuit is the salt bridge or porcelain cup that connects the electrolytes in the two half-cells. Anions flow to the anode and cations flow toward the cathode in the internal circuit.
4. A standard cell is a voltaic cell that has specific composition and conditions under which it operates. Each half-cell has all the entities shown in the half-reaction equation at SATP conditions. Aqueous solutions have a concentration of 1.0 mol/L.
5. (a) The cell potential is predicted from the standard reduction potential values.
E°cell = E°r
(cathode) – E°
r(anode)
(b) This prediction is restricted to cells containing all entities shown in the half-reaction equations and operating at standard conditions (SATP using 1.0 mol/L solutions).
6. A positive cell potential indicates a spontaneous reaction, while a negative cell potential indicates a nonspontaneous reaction.
7. The reactions in voltaic cells are always spontaneous because voltaic cells are designed with two different half-cells each containing the oxidized and reduced entities as shown in the half-reaction equation. Therefore, a spontaneous combination will always exist; the cell potential will always be positive.
8. The standard hydrogen half-cell is an inert platinum electrode immersed in a 1.00 mol/L solution of hydrogen ions, with hydrogen gas at 100 kPa bubbling over the electrode and a cell temperature of 25 °C.
Pt(s) | H
2(g) | H
+(aq)
E°r= 0.00 V
9. A reference half-cell is necessary because it is impossible to determine the reduction potential of a single half-cell. A voltmeter can measure only a potential difference, E°
cell.
10. (a) E°
cell= –0.28 V – (–0.76 V)
E°cell= +0.48 V
The potential of a standard cobalt–zinc cell is +0.48 V.
(b) The theoretical interpretation of this cell potential is that the cobalt(II) ions have a stronger attraction for electrons than do zinc ions. The +0.48 V is a measure of the difference in their abilities to attract electrons.
(c) Some factors that may account for the differences between the experimental and
predicted results are the concentration of the electrolytes, the purity or cleanliness of the electrodes, the temperature of the cells, the accuracy of the voltmeter measuring the potential difference, and the resistance of the wire between the anode and the cathode.
11. (a) SOA: Cu
2+(aq), SRA: Zn(s)
cathode Cu
2+(aq) + 2 e
–o Cu(s)
E°r= +0.34 V
anode Zn(s) o Zn
2+(aq) + 2 e
– E°r= –0.76 V
net Cu
2+(aq) + Zn(s) o Cu(s) + Zn
2+(aq)
E°cell
= +0.34 V – (–0.76 V)
E°cell= +1.10 V
The potential of this cell is +1.10 V.
(b) SOA: Cr
2O
72–(aq), SRA: Sn(s)
cathode: Cr
2O
72–(aq) + 14 H
+(aq) + 6 e
–o 2 Cr
3+(aq) + 7 H
2O(l)
E°r= +1.23 V anode: 3 [Sn(s) o Sn
2+(aq) + 2 e
–]
E°r= –0.14 V net: Cr
2O
72–(aq) + 14 H
+(aq) + 3 Sn(s) o 2 Cr
3+(aq) + 7 H
2O(l) + 3 Sn
2+(aq)
E°cell
= +1.23 V – (–0.14 V)
E°cell= +1.37 V
The potential of this cell is +1.37 V.
12. 2 Ag
+(aq) + X(s) o 2 Ag(s) + X
2+(aq) +1.08 V = +0.80 V – E°
r (anode)E°r (anode)
= +0.80 V – (+1.08 V)
E°r (anode)= –0.28 V
The reduction potential for the X
2+(aq) Ň X(s) half-cell is –0.28 V, which could represent Co
2+(aq)ŇCo(s).
13. A common problem is the rusting of iron in important objects such as bridges, buildings, and
ships. The technological solution of corrosion prevention such as protective coatings and/or
cathodic protection is used to minimize this problem.
Extension
14. (a) Zinc is being used as a sacrificial anode to protect the pipeline. Zinc is a more active metal than iron. A spontaneous electric cell is established, with iron as the cathode and zinc as the anode.
(b) Protective coatings are always used and impressed currents may also be used.
(c) The environmental and safety issues associated with not protecting pipelines come from the possibility of the pipe corroding so badly that it breaks, allowing gas and/or oil to escape into the environment. In addition to the risk of a catastrophic fire, oil and gas leaks can be very harmful to plants and animals. The environmental and safety issues that are associated with protecting pipelines are considerably fewer, although the oxidation of sacrificial anodes can release metal ions into ground water.
15.
ProblemWhat is the total electric potential difference of two cells connected in series?
Prediction
Since the two cells are connected in series, the total electric potential difference of the two cells is equal to the sum of the cell potentials of the two cells.
coppersilver cell
E°cell
= +0.80 V – (+0.34 V)
E°cell= +0.46 V
copperzinc cell
E°cell
