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820 Chapter Seventeen Electrochemistry CHEMICAL IMPACT The Chemistry of Sunken Treasure When the galleon Atocha was destroyed on a reef by a hurricane in 1622, it was bound for Spain carrying ap- proximately 47 tons of copper, gold, and silver from the New World. The bulk of the treasure was silver bars and coins packed in wooden chests. When treasure hunter Mel Fisher salvaged the silver in 1985, corrosion and marine growth had transformed the shiny metal into something that looked like coral. Restoring the silver to its original condition re- quired an understanding of the chemical changes that had occurred in 350 years of being submerged in the ocean. Much of this chemistry we have already considered at var- ious places in this text. As the wooden chests containing the silver decayed, the oxygen supply was depleted, favoring the growth of certain bacteria that use the sulfate ion rather than oxygen as an ox- idizing agent to generate energy. As these bacteria consume sulfate ions, they release hydrogen sulfide gas that reacts with silver to form black silver sulfide: 2Ag1s2 H2S1aq2 ¡ Ag2S1s2 H21g2 Thus, over the years, the surface of the silver became cov- Silver coins and tankards salvaged from the wreck of ered with a tightly adhering layer of corrosion, which for- the Atocha. tunately protected the silver underneath and thus prevented total conversion of the silver to silver sulfide. (Na , Cl , and H2O), only Cl and H2O can be readily oxidized. The half-reactions (writ- ten as oxidization processes) are 2Cl ¡ Cl2 2e 4e e° 1.36 V 2H2O ¡ O2 4H e° 1.23 V Since water has the more positive potential, we would expect to see O2 produced at the anode because it is easier (thermodynamically) to oxidize H2O than Cl . Actually, this does not happen. As the voltage is increased in the cell, the Cl ion is the first to be oxidized. A much higher potential than expected is required to oxidize water. The voltage required in excess of the expected value (called the overvoltage) is much greater for the production of O2 than for Cl2, which explains why chlorine is produced first. The causes of overvoltage are very complex. Basically, the phenomenon is caused by difficulties in transferring electrons from the species in the solution to the atoms on the electrode across the electrode–solution interface. Because of this situation, e° values must be used cautiously in predicting the actual order of oxidation or reduction of species in an electrolytic cell.

17.8 Commercial Electrolytic Processes 821 Another change that took place as the wood decom- Power posed was the formation of carbon dioxide. This shifted the source equilibrium that is present in the ocean, e– e– CO21aq2 H2O1l2 ∆ HCO3 1aq2 H 1aq2 Na+ Coin coated to the right, producing higher concentrations of HCO3 . In OH – with Ag2S turn, the HCO3 reacted with Ca 2 ions present in the sea- H2O water to form calcium carbonate: Anode Cathode Ca2 1aq2 HCO3 1aq2 ∆ CaCO31s2 H 1aq2 As electrons flow, the Ag ions in the silver sulfide are Calcium carbonate is the main component of limestone. reduced to silver metal: Thus, over time, the corroded silver coins and bars became encased in limestone. Ag2S 2e ¡ Ag S2 Both the limestone formation and the corrosion had to As a by-product, bubbles of hydrogen gas from the reduc- be dealt with. Since CaCO3 contains the basic anion CO32 , tion of water form on the surface of the coins: acid dissolves limestone: 2H2O 2e ¡ H21g2 2OH 2H 1aq2 CaCO31s2 ¡ Ca 2 1aq2 CO21g2 H2O1l2 The agitation caused by the bubbles loosens the flakes of Soaking the mass of coins in a buffered acidic bath for sev- metal sulfide and helps clean the coins. eral hours allowed the individual pieces to be separated, and the black Ag2S on the surfaces was revealed. An abrasive These procedures have made it possible to restore the could not be used to remove this corrosion; it would have de- treasure to very nearly its condition when the Atocha sailed stroyed the details of the engraving—a very valuable feature many years ago. of the coins to a historian or a collector—and it would have washed away some of the silver. Instead, the corrosion reac- tion was reversed through electrolytic reduction. The coins were connected to the cathode of an electrolytic cell in a di- lute sodium hydroxide solution as represented in the figure. 17.8 Commercial Electrolytic Processes The chemistry of metals is characterized by their ability to donate electrons to form ions. Because metals are typically such good reducing agents, most are found in nature in ores, mixtures of ionic compounds often containing oxide, sulfide, and silicate anions. The no- ble metals, such as gold, silver, and platinum, are more difficult to oxidize and are often found as pure metals. Production of Aluminum Aluminum is one of the most abundant elements on earth, ranking third behind oxygen and silicon. Since aluminum is a very active metal, it is found in nature as its oxide in an ore called bauxite (named after Les Baux, France, where it was discovered in 1821). Production of aluminum metal from its ore proved to be more difficult than produc- tion of most other metals. In 1782 Lavoisier recognized aluminum to be a metal “whose affinity for oxygen is so strong that it cannot be overcome by any known re- ducing agent.” As a result, pure aluminum metal remained unknown. Finally, in 1854

822 Chapter Seventeen Electrochemistry a process was found for producing metallic aluminum using sodium, but aluminum remained a very expensive rarity. In fact, it is said that Napoleon III served his most honored guests with aluminum forks and spoons, while the others had to settle for gold and silver utensils. The breakthrough came in 1886 when two men, Charles M. Hall in the United States and Paul Heroult in France, almost simultaneously discovered a practical electrolytic process for producing aluminum (see Fig. 17.21). The key factor in the Hall–Heroult process is the use of molten cryolite (Na3AlF6) as the solvent for the aluminum oxide. Electrolysis is possible only if ions can move to the electrodes. A common method for producing ion mobility is dissolving the substance to be electrolyzed in water. This is not possible in the case of aluminum because water is more easily reduced than Al3 , as the following standard reduction potentials show: Al3 3e ¡ Al 2OH e° 1.66 V 2H2O 2e ¡ H2 e° 0.83 V FIGURE 17.21 Thus aluminum metal cannot be plated out of an aqueous solution of Al3 . Charles Martin Hall (1863–1914) was a stu- dent at Oberlin College in Ohio when he Ion mobility also can be produced by melting the salt. But the melting point of solid first became interested in aluminum. One Al2O3 is much too high 12050°C2 to allow practical electrolysis of the molten oxide. A of his professors commented that anyone mixture of Al2O3 and Na3AlF6, however, has a melting point of 1000°C, and the result- who could manufacture aluminum cheaply ing molten mixture can be used to obtain aluminum metal electrolytically. Because of this would make a fortune, and Hall decided to give it a try. The 21-year-old Hall worked in discovery by Hall and Heroult, the price of aluminum plunged (see Table 17.3), and its a wooden shed near his house with an iron use became economically feasible. frying pan as a container, a blacksmith’s forge as a heat source, and galvanic cells Bauxite is not pure aluminum oxide (called alumina); it also contains the oxides of constructed from fruit jars. Using these iron, silicon, and titanium, and various silicate materials. To obtain the pure hydrated alu- crude galvanic cells, Hall found that he mina 1Al2O3 nH2O2, the crude bauxite is treated with aqueous sodium hydroxide. Being could produce aluminum by passing a amphoteric, alumina dissolves in the basic solution: current through a molten Al2O3/Na3AlF6 mixture. By a strange coincidence, Paul Al2O31s2 2OH 1aq2 ¡ 2AlO2 1aq2 H2O1l2 Heroult, a Frenchman who was born and died in the same years as Hall, made the The other metal oxides, which are basic, remain as solids. The solution containing the same discovery at about the same time. aluminate ion (AlO2 ) is separated from the sludge of the other oxides and is acidified with carbon dioxide gas, causing the hydrated alumina to reprecipitate: 2CO21g2 2AlO2 1aq2 1n 12H2O1l2 ¡ 2HCO3 1aq2 Al2O3 nH2O1s2 The purified alumina is then mixed with cryolite and melted, and the aluminum ion is reduced to aluminum metal in an electrolytic cell of the type shown in Fig. 17.22. Because the electrolyte solution contains a large number of aluminum-containing ions, the chemistry is not completely clear. However, the alumina probably reacts with the cryolite anion as follows: Al2O3 4AlF6 3 ¡ 3Al2OF6 2 6F The electrode reactions are thought to be TABLE 17.3 The Price of Aluminum over the Past Century Cathode reaction: AlF63 3e ¡ Al 6F C ¡ 4AlF63 Date Price of Aluminum ($/lb)* Anode reaction: 2Al2OF62 12F CO2 4e 1855 100,000 The overall cell reaction can be written as 1885 100 1890 2 2Al2O3 3C ¡ 4Al 3CO2 1895 0.50 1970 0.30 The aluminum produced in this electrolytic process is 99.5% pure. To be useful as a 1980 0.80 structural material, aluminum is alloyed with metals such as zinc (used for trailer and air- 1990 0.74 craft construction) and manganese (used for cooking utensils, storage tanks, and highway signs). The production of aluminum consumes about 5% of all the electricity used in the *Note the precipitous drop in price after the United States. discovery of the Hall–Heroult process.

17.8 Commercial Electrolytic Processes 823 Electrodes of To external graphite rods power source Carbon dioxide Molten Molten formed at aluminum Al2O3/Na3AlF6 the anodes mixture Carbon-lined Plug iron tank FIGURE 17.22 A schematic diagram of an electrolytic cell for producing aluminum by the Hall–Heroult process. Because molten aluminum is more dense than the mixture of molten cryolite and alumina, it settles to the bottom of the cell and is drawn off periodically. The graphite elec- trodes are gradually eaten away and must be replaced from time to time. The cell operates at a current flow of up to 250,000 A. Electrorefining of Metals Purification of metals is another important application of electrolysis. For example, im- pure copper from the chemical reduction of copper ore is cast into large slabs that serve as the anodes for electrolytic cells. Aqueous copper sulfate is the electrolyte, and thin sheets of ultrapure copper function as the cathodes (see Fig. 17.23). The main reaction at the anode is Cu ¡ Cu2 2e Other metals such as iron and zinc are also oxidized from the impure anode: Zn ¡ Zn2 2e Fe ¡ Fe2 2e FIGURE 17.23 Ultrapure copper sheets that serve as the cathodes are lowered between slabs of im- pure copper that serve as the anodes into a tank containing an aqueous solution of copper sulfate (CuSO4). It takes about four weeks for the anodes to dissolve and for the pure copper to be deposited on the cathodes.

824 Chapter Seventeen Electrochemistry e– Power e– source e– e– e– e– Ag Ag+ Ag Ag+ Anode Cathode (a) (b) FIGURE 17.24 (a) A silver-plated teapot. Silver plating is often used to beautify and protect cutlery and items of table service. (b) Schematic of the electroplating of a spoon. The item to be plated is the cathode, and the anode is a silver bar. Silver is plated out at the cathode: Ag e S Ag. Note that a salt bridge is not needed here because Ag ions are involved at both electrodes. Noble metal impurities in the anode are not oxidized at the voltage used; they fall to the bottom of the cell to form a sludge, which is processed to remove the valuable silver, gold, and platinum. The Cu2 ions from the solution are deposited onto the cathode Cu2 2e ¡ Cu producing copper that is 99.95% pure. Metal Plating Metals that readily corrode can often be protected by the application of a thin coating of a metal that resists corrosion. Examples are “tin” cans, which are actually steel cans with a thin coating of tin, and chrome-plated steel bumpers for automobiles. An object can be plated by making it the cathode in a tank containing ions of the plating metal. The silver plating of a spoon is shown schematically in Fig. 17.24(b). In an actual plating process, the solution also contains ligands that form complexes with the silver ion. By lowering the concentration of Ag in this way, a smooth, even coating of silver is obtained. Electrolysis of Sodium Chloride Addition of a nonvolatile solute lowers the Sodium metal is mainly produced by the electrolysis of molten sodium chloride. Because melting point of the solvent, molten NaCl solid NaCl has a rather high melting point 1800°C2, it is usually mixed with solid CaCl2 to in this case. lower the melting point to about 1600°C2. The mixture is then electrolyzed in a Downs cell, as illustrated in Fig. 17.25, where the reactions are Anode reaction: 2Cl ¡ Cl2 2e Cathode reaction: Na e ¡ Na

17.8 Commercial Electrolytic Processes 825 Cl2 gas Liquid sodium metal Anode Molten mixture of NaCl and CaCl2 Cathode Cathode FIGURE 17.25 Iron Anode Iron The Downs cell for the electrolysis of screen screen molten sodium chloride. The cell is de- signed so that the sodium and chlorine produced cannot come into contact with each other to re-form NaCl. At the temperatures in the Downs cell, the sodium is liquid and is drained off, then cooled, and cast into blocks. Because it is so reactive, sodium must be stored in an inert solvent, such as mineral oil, to prevent its oxidation. Electrolysis of aqueous sodium chloride (brine) is an important industrial process for the production of chlorine and sodium hydroxide. In fact, this process is the second largest consumer of electricity in the United States, after the production of aluminum. Sodium is not produced in this process under normal circumstances because H2O is more easily re- duced than Na , as the standard reduction potentials show: Na e ¡ Na e° 2.71 V 2H2O 2e ¡ H2 2OH e° 0.83 V Hydrogen, not sodium, is produced at the cathode. For the reasons we discussed in Section 17.7, chlorine gas is produced at the anode. Thus the electrolysis of brine produces hydrogen and chlorine: Anode reaction: 2Cl ¡ Cl2 2e Cathode reaction: 2H2O 2e ¡ H2 2OH It leaves a solution containing dissolved NaOH and NaCl. The contamination of the sodium hydroxide by NaCl can be virtually eliminated using a special mercury cell for electrolyzing brine (see Fig. 17.26). In this cell, mercury is the conductor at the cathode, and because hydrogen gas has an extremely high overvoltage with a mercury electrode, Na is reduced instead of H2O. The resulting sodium metal dis- solves in the mercury, forming a liquid alloy, which is then pumped to a chamber where the dissolved sodium reacts with water to produce hydrogen: 2Na1s2 2H2O1l2 ¡ 2Na 1aq2 2OH 1aq2 H21g2 Relatively pure solid NaOH can be recovered from the aqueous solution, and the re- generated mercury is then pumped back to the electrolysis cell. This process, called the chlor–alkali process, was the main method for producing chlorine and sodium hydroxide in the United States for many years. However, because of the environmen- tal problems associated with the mercury cell, it has been largely displaced in the

826 Chapter Seventeen Electrochemistry H2 gas NaOH solution H2O Hg Hg/Na Cl2 gas FIGURE 17.26 Anode The mercury cell for production of chlorine and sodium hydroxide. The large overvolt- Brine Mercury cathode Brine age required to produce hydrogen at a mer- Hg Na in Hg(l) cury electrode means that Na ions are reduced rather than water. The sodium formed dissolves in the liquid mercury and is pumped to a chamber where it reacts with water. chlor–alkali industry by other technologies. In the United States, nearly 75% of the chlor–alkali production is now carried out in diaphragm cells. In a diaphragm cell the cathode and anode are separated by a diaphragm that allows passage of H2O mol- ecules, Na ions, and, to a limited extent, Cl ions. The diaphragm does not allow OH ions to pass through it. Thus the H2 and OH formed at the cathode are kept separate from the Cl2 formed at the anode. The major disadvantage of this process is that the aqueous effluent pumped from the cathode compartment contains a mixture of sodium hydroxide and unreacted sodium chloride, which must be separated if pure sodium hydroxide is a desired product. In the past 30 years, a new process has been developed in the chlor–alkali industry that employs a membrane to separate the anode and cathode compartments in brine elec- trolysis cells. The membrane is superior to a diaphragm because the membrane is imper- meable to anions. Only cations can flow through the membrane. Because neither Cl nor OH ions can pass through the membrane separating the anode and cathode compart- ments, NaCl contamination of the NaOH formed at the cathode does not occur. Although membrane technology is now just becoming prominent in the United States, it is the dominant method for chlor–alkali production in Japan. Key Terms For Review electrochemistry Electrochemistry The study of the interchange of chemical and electrical energy Section 17.1 Employs oxidation–reduction reactions oxidation–reduction (redox) reaction Galvanic cell: chemical energy is transformed into electrical energy by separating reducing agent the oxidizing and reducing agents and forcing the electrons to travel through a oxidizing agent wire oxidation Electrolytic cell: electrical energy is used to produce a chemical change reduction half-reactions Galvanic cell salt bridge Anode: the electrode where oxidation occurs porous disk Cathode: the electrode where reduction occurs galvanic cell anode cathode

For Review 827 cell potential (electromotive force) The driving force behind the electron transfer is called the cell potential (ecell) volt • The potential is measured in units of volts (V), defined as a joule of work voltmeter potentiometer per coulomb of charge: Section 17.2 e1V2 work 1J2 w standard hydrogen electrode charge 1C2 q standard reduction potentials • A system of half-reactions, called standard reduction potentials, can be used to Section 17.3 calculate the potentials of various cells faraday • The half-reaction 2H 2e ¡ H2 is arbitrarily assigned a potential of 0 V Section 17.4 Free energy and work concentration cell The maximum work that a cell can perform is Nernst equation glass electrode wmax qemax ion-selective electrode where emax represents the cell potential when no current is flowing Section 17.5 The actual work obtained from a cell is always less than the maximum because battery energy is lost through frictional heating of the wire when current flows lead storage battery For a process carried out at constant temperature and pressure, the change in free dry cell battery energy equals the maximum useful work obtainable from that process: fuel cell ¢G wmax qemax nF e Section 17.6 corrosion where F (faraday) equals 96,485 C and n is the number of moles of electrons galvanizing transferred in the process cathodic protection Concentration cell Section 17.7 A galvanic cell in which both compartments have the same components but at electrolytic cell electrolysis different concentrations ampere The electrons flow in the direction that tends to equalize the concentrations Section 17.8 Nernst equation Downs cell Shows how the cell potential depends on the concentrations of the cell components: mercury cell chlor–alkali process e e0 0.0591 log Q at 25°C n When a galvanic cell is at equilibrium, e 0 and Q K Batteries A battery consists of a galvanic cell or group of cells connected in series that serve as a source of direct current Lead storage battery • Anode: lead • Cathode: lead coated with PbO2 • Electrolyte: H2SO4(aq) Dry cell battery • Contains a moist paste instead of a liquid electrolyte • Anode: usually Zn • Cathode: carbon rod in contact with an oxidizing agent (which varies depend- ing on the application) Fuel cells Galvanic cells in which the reactants are continuously supplied The H2/O2 fuel cell is based on the reaction between H2 and O2 to form water Corrosion Involves the oxidation of metals to form mainly oxides and sulfides

828 Chapter Seventeen Electrochemistry Some metals, such as aluminum and chromium, form a thin protective oxide coat- ing that prevents further corrosion The corrosion of iron to form rust is an electrochemical process • The Fe2 ions formed at anodic areas of the surface migrate through the mois- ture layer to cathodic regions, where they react with oxygen from the air • Iron can be protected from corrosion by coating it with paint or with a thin layer of metal such as chromium, tin, or zinc; by alloying; and by cathodic protection Electrolysis Used to place a thin coating of metal onto steel Used to produce pure metals such as aluminum and copper REVIEW QUESTIONS 1. What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell. 2. Galvanic cells harness spontaneous oxidation–reduction reactions to produce work by producing a current. They do so by controlling the flow of electrons from the species oxidized to the species reduced. How is a galvanic cell de- signed? What is in the cathode compartment? The anode compartment? What purpose do electrodes serve? Which way do electrons always flow in the wire connecting the two electrodes in a galvanic cell? Why is it necessary to use a salt bridge or a porous disk in a galvanic cell? Which way do cations flow in the salt bridge? Which way do the anions flow? What is a cell potential and what is a volt? 3. Table 17.1 lists common half-reactions along with the standard reduction poten- tial associated with each half-reaction. These standard reduction potentials are all relative to some standard. What is the standard (zero point)? If e° is positive for a half-reaction, what does it mean? If e° is negative for a half-reaction, what does it mean? Which species in Table 17.1 is most easily reduced? Least easily reduced? The reverse of the half-reactions in Table 17.1 are the oxida- tion half-reactions. How are standard oxidation potentials determined? In Table 17.1, which species is the best reducing agent? The worst reducing agent? To determine the standard cell potential for a redox reaction, the standard reduction potential is added to the standard oxidation potential. What must be true about this sum if the cell is to be spontaneous (produce a galvanic cell)? Standard reduction and oxidation potentials are intensive. What does this mean? Summarize how line notation is used to describe galvanic cells. 4. Consider the equation ¢G° nFe°. What are the four terms in this equation? Why does a minus sign appear in the equation? What does the superscript ° indicate? 5. The Nernst equation allows determination of the cell potential for a galvanic cell at nonstandard conditions. Write out the Nernst equation. What are non- standard conditions? What do e, e°, n, and Q stand for in the Nernst equation? What does the Nernst equation reduce to when a redox reaction is at equilib- rium? What are the signs of ¢G° and e° when K 6 1? When K 7 1? When K 1? Explain the following statement: e determines spontaneity, while e° determines the equilibrium position. Under what conditions can you use e° to predict spontaneity? 6. What are concentration cells? What is e° in a concentration cell? What is the driving force for a concentration cell to produce a voltage? Is the higher or the lower ion concentration solution present at the anode? When the anode ion con- centration is decreased and/or the cathode ion concentration is increased, both

Active Learning Questions 829 give rise to larger cell potentials. Why? Concentration cells are commonly used to calculate the value of equilibrium constants for various reactions. For exam- ple, the silver concentration cell illustrated in Fig. 17.9 can be used to determine the Ksp value for AgCl(s). To do so, NaCl is added to the anode compartment until no more precipitate forms. The [Cl ] in solution is then determined some- how. What happens to ecell when NaCl is added to the anode compartment? To calculate the Ksp value, [Ag ] must be calculated. Given the value of ecell, how is [Ag ] determined at the anode? 7. Batteries are galvanic cells. What happens to ecell as a battery discharges? Does a battery represent a system at equilibrium? Explain. What is ecell when a bat- tery reaches equilibrium? How are batteries and fuel cells alike? How are they different? The U.S. space program utilizes hydrogen–oxygen fuel cells to pro- duce power for its spacecraft. What is a hydrogen–oxygen fuel cell? 8. Not all spontaneous redox reactions produce wonderful results. Corrosion is an example of a spontaneous redox process that has negative effects. What hap- pens in the corrosion of a metal such as iron? What must be present for the cor- rosion of iron to take place? How can moisture and salt increase the severity of corrosion? Explain how the following protect metals from corrosion: a. paint b. durable oxide coatings c. galvanizing d. sacrificial metal e. alloying f. cathodic protection 9. What characterizes an electrolytic cell? What is an ampere? When the current applied to an electrolytic cell is multiplied by the time in seconds, what quan- tity is determined? How is this quantity converted to moles of electrons required? How are moles of electrons required converted to moles of metal plated out? What does plating mean? How do you predict the cathode and the anode half-reactions in an electrolytic cell? Why is the electrolysis of molten salts much easier to predict in terms of what occurs at the anode and cathode than the electrolysis of aqueous dissolved salts? What is overvoltage? 10. Electrolysis has many important industrial applications. What are some of these applications? The electrolysis of molten NaCl is the major process by which sodium metal is produced. However, the electrolysis of aqueous NaCl does not produce sodium metal under normal circumstances. Why? What is purification of a metal by electrolysis? Active Learning Questions 3. You want to “plate out” nickel metal from a nickel nitrate solu- tion onto a piece of metal inserted into the solution. Should you These questions are designed to be used by groups of students in class. The use copper or zinc? Explain. questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the 4. A copper penny can be dissolved in nitric acid but not in hy- learning that occurs while students talk to each other about chemical drochloric acid. Using reduction potentials from the book, show concepts. why this is so. What are the products of the reaction? Newer pennies contain a mixture of zinc and copper. What happens to 1. Sketch a galvanic cell, and explain how it works. Look at Figs. 17.1 the zinc in the penny when the coin is placed in nitric acid? Hy- and 17.2. Explain what is occurring in each container and why the drochloric acid? Support your explanations with data from the cell in Fig. 17.2 “works” but the one in Fig. 17.1 does not. book, and include balanced equations for all reactions. 2. In making a specific galvanic cell, explain how one decides on 5. Sketch a cell that forms iron metal from iron(II) while changing the electrodes and the solutions to use in the cell. chromium metal to chromium(III). Calculate the voltage, show

830 Chapter Seventeen Electrochemistry the electron flow, label the anode and cathode, and balance the a. HNO3 e. C6H12O6 i. Na2C2O4 overall cell equation. b. CuCl2 f. Ag j. CO2 6. Which of the following is the best reducing agent: F2, H2, Na, c. O2 g. PbSO4 k. (NH4)2Ce(SO4)3 Na , F ? Explain. Order as many of these species as possible d. H2O2 h. PbO2 l. Cr2O3 from the best to the worst oxidizing agent. Why can’t you order all of them? From Table 17.1 choose the species that is the best 15. Specify which of the following equations represent oxidation– oxidizing agent. Choose the best reducing agent. Explain. 7. You are told that metal A is a better reducing agent than metal B. reduction reactions, and indicate the oxidizing agent, the reducing What, if anything, can be said about A and B ? Explain. 8. Explain the following relationships: ¢G and w, cell potential agent, the species being oxidized, and the species being reduced. and w, cell potential and ¢G, cell potential and Q. Using these relationships, explain how you could make a cell in which both a. CH41g2 H2O1g2 S CO1g2 3H21g2 electrodes are the same metal and both solutions contain the same b. 2AgNO31aq2 Cu1s2 S Cu1NO3221aq2 2Ag1s2 compound, but at different concentrations. Why does such a cell c. Zn1s2 2HCl1aq2 S ZnCl21aq2 H21g2 run spontaneously? d. 2H 1aq2 2CrO42 1aq2 S Cr2O7 2 1aq2 H2O1l2 9. Explain why cell potentials are not multiplied by the coefficients in the balanced redox equation. (Use the relationship between 16. Balance each of the following equations by the half-reaction ¢G and cell potential to do this.) 10. What is the difference between e and e°? When is e equal to method for the pH conditions specified. zero? When is e° equal to zero? (Consider “regular” galvanic cells as well as concentration cells.) a. Cr1s2 NO3 1aq2 S Cr3 1aq2 NO1g2 1acidic2 11. Consider the following galvanic cell: b. Al1s2 MnO4 1aq2 S Al 3 1aq2 Mn 2 1aq2 1acidic2 c. CH3OH1aq2 Ce 4 1aq2 S CO21aq2 Ce 3 1aq2 1acidic2 Zn Ag d. PO33 1aq2 MnO4 1aq2 S PO4 3 1aq2 MnO21s2 1basic2 e. Mg1s2 OCl 1aq2 S Mg1OH221s2 Cl 1aq2 1basic2 f. H2CO1aq2 Ag1NH322 1aq2 S HCO3 1aq2 Ag1s2 NH31aq2 1basic2 Questions 1.0 M Zn2+ 1.0 M Ag+ 17. When magnesium metal is added to a beaker of HCl(aq), a gas is produced. Knowing that magnesium is oxidized and that hy- What happens to e as the concentration of Zn 2 is increased? drogen is reduced, write the balanced equation for the reaction. As the concentration of Ag is increased? What happens to e° How many electrons are transferred in the balanced equation? in these cases? What quantity of useful work can be obtained when Mg is added 12. Look up the reduction potential for Fe3 to Fe 2 . Look up the directly to the beaker of HCl? How can you harness this reac- reduction potential for Fe 2 to Fe. Finally, look up the reduc- tion to do useful work? tion potential for Fe3 to Fe. You should notice that adding the reduction potentials for the first two does not give the potential 18. How can one construct a galvanic cell from two substances, each for the third. Why not? Show how you can use the first two po- having a negative standard reduction potential? tentials to calculate the third potential. 19. The free energy change for a reaction, G, is an extensive property. A blue question or exercise number indicates that the answer to that What is an extensive property? Surprisingly, one can calculate G question or exercise appears at the back of this book and a solution appears from the cell potential, e, for the reaction. This is surprising in the Solutions Guide. because e is an intensive property. How can the extensive property G be calculated from the intensive property e? Review of Oxidation–Reduction Reactions 20. What is wrong with the following statement: The best concen- tration cell will consist of the substance having the most posi- If you have trouble with these exercises, you should review Sections 4.9 and tive standard reduction potential. What drives a concentration 4.10. cell to produce a large voltage? 13. Define oxidation and reduction in terms of both change in oxi- 21. When jump-starting a car with a dead battery, the ground jumper dation number and electron loss or gain. should be attached to a remote part of the engine block. Why? 14. Assign oxidation numbers to all the atoms in each of the fol- 22. In theory, most metals should easily corrode in air. Why? A group lowing. of metals called the noble metals are relatively difficult to corrode in air. Some noble metals include: gold, platinum, and silver. Reference Table 17.1 to come up with a possible reason why the noble metals are relatively difficult to corrode. 23. Consider the electrolysis of a molten salt of some metal. What information must you know to calculate the mass of metal plated out in the electrolytic cell? 24. Although aluminum is one of the most abundant elements on earth, production of pure Al proved very difficult until the late 1800s. At this time, the Hall–Heroult process made it relatively easy to produce pure Al. Why was pure Al so difficult to produce and what was the key discovery behind the Hall–Heroult process?

Exercises 831 Exercises For each galvanic cell, give the balanced cell reaction and de- termine e°. Standard reduction potentials are found in Table 17.1. In this section similar exercises are paired. 34. Give the balanced cell reaction and determine e° for the gal- Galvanic Cells, Cell Potentials, Standard Reduction vanic cells based on the following half-reactions. Standard re- Potentials, and Free Energy duction potentials are found in Table 17.1. a. Cr2O7 2 14H 6e S 2Cr 3 7H2O H2O2 2H 2e S 2H2O 25. Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow and identify the cathode and b. 2H 2e S H2 Al 3 3e S Al anode. Give the overall balanced reaction. Assume that all con- 35. Calculate e° values for the following cells. Which reactions centrations are 1.0 M and that all partial pressures are 1.0 atm. are spontaneous as written (under standard conditions)? Bal- a. Cr 3 1aq2 Cl21g2 ∆ Cr2O7 2 1aq2 Cl 1aq2 b. Cu 2 1aq2 Mg1s2 ∆ Mg 2 1aq2 Cu1s2 ance the reactions. Standard reduction potentials are found in 26. Sketch the galvanic cells based on the following overall reac- tions. Show the direction of electron flow, the direction of ion Table 17.1. a. MnO4 1aq2 I 1aq2 ∆ I21aq2 Mn 2 1aq2 migration through the salt bridge, and identify the cathode and b. MnO4 1aq2 F 1aq2 ∆ F21g2 Mn 2 1aq2 36. Calculate e° values for the following cells. Which reactions are anode. Give the overall balanced reaction. Assume that all con- spontaneous as written (under standard conditions)? Balance the centrations are 1.0 M and that all partial pressures are 1.0 atm. reactions that are not already balanced. Standard reduction po- a. IO3 1aq2 Fe2 1aq2 ∆ Fe3 1aq2 I21aq2 b. Zn1s2 Ag 1aq2 ∆ Zn2 1aq2 Ag1s2 tentials are found in Table 17.1. a. H21g2 ∆ H 1aq2 H 1aq2 27. Calculate e° values for the galvanic cells in Exercise 25. b. Au 3 1aq2 Ag1s2 ∆ Ag 1aq2 Au1s2 28. Calculate e° values for the galvanic cells in Exercise 26. 29. Sketch the galvanic cells based on the following half-reactions. 37. Chlorine dioxide (ClO2), which is produced by the reaction Show the direction of electron flow, show the direction of ion 2NaClO21aq2 Cl21g2 ¡ 2ClO21g2 2NaCl1aq2 migration through the salt bridge, and identify the cathode and has been tested as a disinfectant for municipal water treatment. Using data from Table 17.1, calculate e° and ¢G° at 25°C for anode. Give the overall balanced reaction, and determine e° for the production of ClO2. 38. The amount of manganese in steel is determined by changing it the galvanic cells. Assume that all concentrations are 1.0 M and to permanganate ion. The steel is first dissolved in nitric acid, producing Mn 2 ions. These ions are then oxidized to the deeply that all partial pressures are 1.0 atm. colored MnO4 ions by periodate ion (IO4 ) in acid solution. a. Complete and balance an equation describing each of the a. Cl2 2e S 2Cl e° 1.36 V above reactions. Br2 2e S 2Br e° 1.09 V b. Calculate e° and ¢G° at 25°C for each reaction. b. MnO4 8H 5e S Mn 2 4H2O e° 1.51 V IO4 2H 2e S IO3 H2O e° 1.60 V 30. Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced reaction, and determine e° for 39. Calculate the maximum amount of work that can be obtained from the galvanic cells at standard conditions in Exercise 33. the galvanic cells. Assume that all concentrations are 1.0 M and 40. Calculate the maximum amount of work that can be obtained that all partial pressures are 1.0 atm. from the galvanic cells at standard conditions in Exercise 34. a. H2O2 2H 2e S 2H2O e° 1.78 V O2 2H 2e S H2O2 e° 0.68 V b. Mn 2 2e S Mn e° 1.18 V Fe 3 3e S Fe e° 0.036 V 41. Calculate e° for the reaction 31. Give the standard line notation for each cell in Exercises 25 and 29. CH3OH 1l 2 3 O2 1 g 2 ¡ CO21g2 2H2O1l2 32. Give the standard line notation for each cell in Exercises 26 and 30. 2 33. Consider the following galvanic cells: using values of ¢G°f in Appendix 4. 42. The equation ¢G° nF e° also can be applied to half- reactions. Use standard reduction potentials to estimate ¢G°f for Fe 2 (aq) and Fe3 (aq). ( ¢G°f for e 0.) Au Pt Cd Pt 43. Using data from Table 17.1, place the following in order of increasing strength as oxidizing agents (all under standard con- ditions). 1.0 M Au3+ 1.0 M Cu+ 1.0 M Cd2+ 1.0 M VHO+ 2+ Cd 2 , IO3 , K , H2O, AuCl4 , I2 (a) 1.0 M Cu2+ 1.0 M 1.0 M VO2+ 44. Using data from Table 17.1, place the following in order of increasing strength as reducing agents (all under standard con- ditions). (b) Cu , F , H , H2O, I2, K

832 Chapter Seventeen Electrochemistry 45. Answer the following questions using data from Table 17.1 (all 53. Consider the concentration cell shown below. Calculate the cell under standard conditions). potential at 25°C when the concentration of Ag in the com- a. Is H (aq) capable of oxidizing Cu(s) to Cu 2 (aq)? partment on the right is the following. b. Is Fe3 (aq) capable of oxidizing I (aq)? a. 1.0 M c. Is H2(g) capable of reducing Ag (aq)? b. 2.0 M d. Is Fe 2 (aq) capable of reducing Cr3 (aq) to Cr2 (aq)? c. 0.10 M d. 4.0 10 5 M 46. Consider only the species (at standard conditions) e. Calculate the potential when both solutions are 0.10 M in Ag . For each case, also identify the cathode, the anode, and the di- Na , Cl , Ag , Ag, Zn 2 , Zn, Pb rection in which electrons flow. in answering the following questions. Give reasons for your Ag Ag answers. (Use data from Table 17.1.) a. Which is the strongest oxidizing agent? [Ag+] = 1.0 M b. Which is the strongest reducing agent? c. Which species can be oxidized by SO4 2 (aq) in acid? 54. Consider a concentration cell similar to the one shown in Exer- d. Which species can be reduced by Al(s)? cise 53, except that both electrodes are made of Ni and in the left-hand compartment [Ni2 ] 1.0 M. Calculate the cell po- 47. Use the table of standard reduction potentials (Table 17.1) to tential at 25°C when the concentration of Ni 2 in the compart- pick a reagent that is capable of each of the following oxidations ment on the right has each of the following values. (under standard conditions in acidic solution). a. 1.0 M a. Oxidize Br to Br2 but not oxidize Cl to Cl2 b. 2.0 M b. Oxidize Mn to Mn 2 but not oxidize Ni to Ni2 c. 0.10 M d. 4.0 10 5 M 48. Use the table of standard reduction potentials (Table 17.1) to e. Calculate the potential when both solutions are 2.5 M in Ni 2 . pick a reagent that is capable of each of the following reductions For each case, also identify the cathode, anode, and the direc- (under standard conditions in acidic solution). tion in which electrons flow. a. Reduce Cu2 to Cu but not reduce Cu 2 to Cu . b. Reduce Br2 to Br but not reduce I2 to I . 55. The overall reaction in the lead storage battery is Pb1s2 PbO21s2 2H 1aq2 2HSO4 1aq2 ¡ 49. Hydrazine is somewhat toxic. Use the half-reactions shown be- 2PbSO41s2 2H2O1l2 low to explain why household bleach (a highly alkaline solution Calculate e at 25°C for this battery when [H2SO4] 4.5 M, of sodium hypochlorite) should not be mixed with household that is, [H ] [HSO4 ] 4.5 M. At 25°C, e° 2.04 V for ammonia or glass cleansers that contain ammonia. the lead storage battery. ClO H2O 2e ¡ 2OH Cl e° 0.90 V 56. Calculate the pH of the cathode compartment for the follow- N2H4 2H2O 2e ¡ 2NH3 2OH e° 0.10 V ing reaction given ecell 3.01 V when [Cr 3 ] 0.15 M, [Al 3 ] 0.30 M, and [ Cr2O72 ] 0.55 M. 50. The compound with the formula TlI3 is a black solid. Given the 2Al1s2 Cr2O7 2 1aq2 14H 1aq2 S following standard reduction potentials, 2Al 3 1aq2 2Cr3 1aq2 7H2O1l2 Tl 3 2e ¡ Tl e° 1.25 V 57. Consider the cell described below: I3 2e ¡ 3I e° 0.55 V Zn 0 Zn 2 11.00 M2 0 0 Cu 2 11.00 M2 0 Cu would you formulate this compound as thallium(III) iodide or Calculate the cell potential after the reaction has operated thallium(I) triiodide? long enough for the [Zn 2 ] to have changed by 0.20 mol/L. (Assume T 25°C.) The Nernst Equation 58. Consider the cell described below: 51. A galvanic cell is based on the following half-reactions at 25°C: Al 0 Al 3 11.00 M2 0 0 Pb 2 11.00 M2 0 Pb Ag e ¡ Ag H2O2 2H 2e ¡ 2H2O Predict whether ecell is larger or smaller than e°cell for the fol- lowing cases. a. [Ag ] 1.0 M, [H2O2] 2.0 M, [H ] 2.0 M b. [Ag ] 2.0 M, [H2O2] 1.0 M, [H ] 1.0 10 7 M 52. Consider the concentration cell in Fig. 17.10. If the Fe 2 con- centration in the right compartment is changed from 0.1 M to 1 10 7 M Fe2 , predict the direction of electron flow, and des- ignate the anode and cathode compartments.

Exercises 833 Calculate the cell potential after the reaction has operated a. Calculate the concentration of Ag at the cathode. long enough for the [Al3 ] to have changed by 0.60 mol/L. b. Determine the value of the equilibrium constant for the for- (Assume T 25°C.) mation of Ag(S2O3)2 3 . 59. An electrochemical cell consists of a standard hydrogen elec- Ag 1aq2 2S2O32 1aq2 ∆ Ag1S2O322 3 1aq2 K ? trode and a copper metal electrode. a. What is the potential of the cell at 25°C if the copper electrode 65. Calculate ¢G° and K at 25°C for the reactions in Exercises 25 is placed in a solution in which [Cu 2 ] 2.5 10 4 M? and 29. b. The copper electrode is placed in a solution of unknown [Cu 2 ]. The measured potential at 25°C is 0.195 V. What is 66. Calculate ¢G° and K at 25°C for the reactions in Exercises 26 [Cu 2 ]? (Assume Cu 2 is reduced.) and 30. 60. An electrochemical cell consists of a nickel metal electrode im- 67. An excess of finely divided iron is stirred up with a solution mersed in a solution with 3Ni2 4 1.0 M separated by a porous that contains Cu2 ion, and the system is allowed to come to disk from an aluminum metal electrode. equilibrium. The solid materials are then filtered off, and a. What is the potential of this cell at 25°C if the aluminum electrodes of solid copper and solid iron are inserted into the electrode is placed in a solution in which [Al 3 ] 7.2 remaining solution. What is the value of the ratio [Fe 2 ] [Cu 2 ] 10 3 M? at 25°C? b. When the aluminum electrode is placed in a certain solution in which [Al3 ] is unknown, the measured cell potential at 68. Consider the following reaction: 25°C is 1.62 V. Calculate [Al3 ] in the unknown solution. (Assume Al is oxidized.) Ni 2 1aq2 Sn1s2 S Ni1s2 Sn 2 1aq2 61. An electrochemical cell consists of a standard hydrogen elec- Determine the minimum ratio of [Sn 2 ] [Ni2 ] necessary to trode and a copper metal electrode. If the copper electrode is make this reaction spontaneous as written. placed in a solution of 0.10 M NaOH that is saturated with Cu(OH)2, what is the cell potential at 25°C? (For Cu(OH)2, 69. Under standard conditions, what reaction occurs, if any, when Ksp 1.6 10 19.) each of the following operations is performed? 62. An electrochemical cell consists of a nickel metal electrode im- mersed in a solution with [Ni 2 ] 1.0 M separated by a porous a. Crystals of I2 are added to a solution of NaCl. disk from an aluminum metal electrode immersed in a solution b. Cl2 gas is bubbled into a solution of NaI. with [Al 3 ] 1.0 M. Sodium hydroxide is added to the alu- c. A silver wire is placed in a solution of CuCl2. minum compartment, causing Al(OH)3(s) to precipitate. After d. An acidic solution of FeSO4 is exposed to air. precipitation of Al(OH)3 has ceased, the concentration of OH For the reactions that occur, write a balanced equation and cal- is 1.0 10 4 M and the measured cell potential is 1.82 V. Cal- culate the Ksp value for Al(OH)3. culate e°, ¢G°, and K at 25°C. Al1OH231s2 ∆ Al 3 1aq2 3OH 1aq2 Ksp ? 70. A disproportionation reaction involves a substance that acts as 63. Consider a concentration cell that has both electrodes made of both an oxidizing and a reducing agent, producing higher and some metal M. Solution A in one compartment of the cell con- tains 1.0 M M2 . Solution B in the other cell compartment has lower oxidation states of the same element in the products. a volume of 1.00 L. At the beginning of the experiment 0.0100 mol of M(NO3)2 and 0.0100 mol of Na2SO4 are dissolved in so- Which of the following disproportionation reactions are spon- lution B (ignore volume changes), where the reaction taneous under standard conditions? Calculate ¢G° and K at M2 1aq2 SO4 2 1aq2 ∆ MSO41s2 25°C for those reactions that are spontaneous under standard occurs. For this reaction equilibrium is rapidly established, conditions. whereupon the cell potential is found to be 0.44 V at 25°C. Assume that the process a. 2Cu 1aq2 S Cu 2 1aq2 Cu1s2 1 unbalanced 2 b. 3Fe2 1aq2 S 2Fe 3 1aq2 Fe1s2 M 2 2e ¡ M c. HClO21aq2 S ClO3 1aq2 HClO1aq2 Use the half-reactions: has a standard reduction potential of 0.31 V and that no other redox process occurs in the cell. Calculate the value of Ksp for ClO3 3H 2e ¡ HClO2 H2O e° 1.21 V MSO4(s) at 25°C. HClO2 2H 2e ¡ HClO H2O e° 1.65 V 64. You have a concentration cell in which the cathode has a silver electrode with 0.10 M Ag . The anode also has a silver electrode 71. Consider the galvanic cell based on the following half-reactions: with Ag (aq), 0.050 M S2O32 , and 1.0 10 3 M Ag(S2O3)2 3 . You read the voltage to be 0.76 V. Au 3 3e ¡ Au e° 1.50 V Tl e ¡ Tl e° 0.34 V a. Determine the overall cell reaction and calculate e°cell. b. Calculate ¢G° and K for the cell reaction at 25°C. c. Calculate ecell at 25°C when [Au 3 ] 1.0 10 2 M and [Tl ] 1.0 10 4 M. 72. Consider the following galvanic cell at 25°C: Pt 0 Cr2 10.30 M2, Cr3 12.0 M2 0 0 Co 2 10.20 M2 0 Co

834 Chapter Seventeen Electrochemistry The overall reaction and equilibrium constant value are 85. One of the few industrial-scale processes that produce organic 2Cr 2 1aq2 Co 2 1aq2 ¡ compounds electrochemically is used by the Monsanto Company to produce 1,4-dicyanobutane. The reduction reaction is 2Cr3 1aq2 Co1s2 K 2.79 107 Calculate the cell potential, e, for this galvanic cell and ¢G for 2CH2 “ CHCN 2H 2e ¡ NC¬1CH224¬CN the cell reaction at these conditions. The NC¬(CH2)4¬CN is then chemically reduced using hy- 73. Calculate Ksp for iron(II) sulfide given the following data: drogen gas to H2N¬1CH226¬NH2, which is used in the pro- duction of nylon. What current must be used to produce 150. kg FeS1s2 2e S Fe1s2 S 2 1aq2 e° 1.01 V of NC¬(CH2)4¬CN per hour? 86. A single Hall–Heroult cell (as shown in Fig. 17.22) produces Fe2 1aq2 2e S Fe1s2 e° 0.44 V about 1 ton of aluminum in 24 hours. What current must be used to accomplish this? 74. For the following half-reaction, e° 2.07 V: AlF63 3e ¡ Al 6F 87. It took 2.30 min using a current of 2.00 A to plate out all the sil- ver from 0.250 L of a solution containing Ag . What was the Using data from Table 17.1, calculate the equilibrium constant original concentration of Ag in the solution? at 25°C for the reaction 88. A solution containing Pt4 is electrolyzed with a current of Al 3 1aq2 6F 1aq2 ∆ AlF6 3 1aq2 K ? 4.00 A. How long will it take to plate out 99% of the platinum in 0.50 L of a 0.010 M solution of Pt4 ? 75. Calculate the value of the equilibrium constant for the reaction of zinc metal in a solution of silver nitrate at 25°C. 89. A solution at 25°C contains 1.0 M Cd 2 , 1.0 M Ag , 1.0 M Au3 , and 1.0 M Ni2 in the cathode compartment of an electrolytic 76. The solubility product for CuI(s) is 1.1 10 12. Calculate the cell. Predict the order in which the metals will plate out as the value of e° for the half-reaction voltage is gradually increased. CuI e ¡ Cu I 90. Consider the following half-reactions: Electrolysis IrCl6 3 3e ¡ Ir 6Cl e° 0.77 V PtCl4 2 2e ¡ Pt 4Cl e° 0.73 V 77. How long will it take to plate out each of the following with a PdCl4 2 2e ¡ Pd 4Cl e° 0.62 V current of 100.0 A? a. 1.0 kg Al from aqueous Al3 A hydrochloric acid solution contains platinum, palladium, and b. 1.0 g Ni from aqueous Ni2 iridium as chloro-complex ions. The solution is a constant 1.0 M c. 5.0 mol Ag from aqueous Ag in chloride ion and 0.020 M in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? 78. The electrolysis of BiO produces pure bismuth. How long (Assume that 99% of a metal must be plated out before another would it take to produce 10.0 g of Bi by the electrolysis of a metal begins to plate out.) BiO solution using a current of 25.0 A? 79. What mass of each of the following substances can be produced 91. What reactions take place at the cathode and the anode when each of the following is electrolyzed? in 1.0 h with a current of 15 A? a. molten NiBr2 b. molten AlF3 c. molten MnI2 a. Co from aqueous Co 2 c. I2 from aqueous KI 92. What reactions take place at the cathode and the anode when b. Hf from aqueous Hf 4 d. Cr from molten CrO3 each of the following is electrolyzed? (Assume standard condi- 80. Aluminum is produced commercially by the electrolysis of Al2O3 tions.) in the presence of a molten salt. If a plant has a continuous ca- a. 1.0 M NiBr2 solution b. 1.0 M AlF3 solution pacity of 1.00 million amp, what mass of aluminum can be pro- c. 1.0 M MnI2 solution duced in 2.00 h? 81. An unknown metal M is electrolyzed. It took 74.1 s for a cur- Additional Exercises rent of 2.00 amp to plate out 0.107 g of the metal from a solu- tion containing M(NO3)3. Identify the metal. 93. The saturated calomel electrode, abbreviated SCE, is often used as a reference electrode in making electrochemical measure- 82. Electrolysis of an alkaline earth metal chloride using a current ments. The SCE is composed of mercury in contact with a sat- of 5.00 A for 748 s deposits 0.471 g of metal at the cathode. urated solution of calomel (Hg2Cl2). The electrolyte solution is What is the identity of the alkaline earth metal chloride? saturated KCl. eSCE is 0.242 V relative to the standard hydro- gen electrode. Calculate the potential for each of the following 83. What volume of F2 gas, at 25°C and 1.00 atm, is produced when galvanic cells containing a saturated calomel electrode and the molten KF is electrolyzed by a current of 10.0 A for 2.00 h? given half-cell components at standard conditions. In each case, What mass of potassium metal is produced? At which electrode indicate whether the SCE is the cathode or the anode. Standard does each reaction occur? reduction potentials are found in Table 17.1. 84. What volumes of H2(g) and O2(g) at STP are produced from the electrolysis of water by a current of 2.50 A in 15.0 min?

Additional Exercises 835 a. Cu 2 2e ¡ Cu corroding metal hull of the Monitor. Describe how attaching zinc b. Fe 3 e ¡ Fe 2 to the hull would protect the Monitor from further corrosion. c. AgCl d. Al 3 e ¡ Ag Cl 99. When aluminum foil is placed in hydrochloric acid, nothing hap- e. Ni 2 3e ¡ Al pens for the first 30 seconds or so. This is followed by vigorous 2e ¡ Ni bubbling and the eventual disappearance of the foil. Explain these observations. 94. Consider the following half-reactions: 100. Which of the following statements concerning corrosion is/are Pt 2 2e ¡ Pt e° 1.188 V true? For the false statements, correct them. PtCl4 2 2e ¡ Pt 4Cl e° 0.755 V a. Corrosion is an example of an electrolytic process. NO3 4H 3e ¡ NO 2H2O e° 0.96 V b. Corrosion of steel involves the reduction of iron coupled with the oxidation of oxygen. Explain why platinum metal will dissolve in aqua regia (a mix- c. Steel rusts more easily in the dry (arid) Southwest states than ture of hydrochloric and nitric acids) but not in either concen- in the humid Midwest states. trated nitric or concentrated hydrochloric acid individually. d. Salting roads in the winter has the added benefit of hinder- ing the corrosion of steel. 95. Consider the standard galvanic cell based on the following half- e. The key to cathodic protection is to connect via a wire a metal reactions more easily oxidized than iron to the steel surface to be protected. Cu 2 2e ¡ Cu Ag e ¡ Ag 101. A patent attorney has asked for your advice concerning the mer- its of a patent application that describes a single aqueous gal- The electrodes in this cell are Ag(s) and Cu(s). Does the cell po- vanic cell capable of producing a 12-V potential. Comment. tential increase, decrease, or remain the same when the follow- 102. The overall reaction and equilibrium constant value for a hydrogen–oxygen fuel cell at 298 K is ing changes occur to the standard cell? a. CuSO4(s) is added to the copper half-cell compartment (as- 2H21g2 O21g2 ¡ 2H2O1l2 K 1.28 1083 sume no volume change). a. Calculate e° and ¢G° at 298 K for the fuel cell reaction. b. NH3(aq) is added to the copper half-cell compartment. Hint: b. Predict the signs of ¢H° and ¢S° for the fuel cell reaction. c. As temperature increases, does the maximum amount of work Cu 2 reacts with NH3 to form Cu(NH3)4 2 (aq). c. NaCl(s) is added to the silver half-cell compartment. Hint: obtained from the fuel cell reaction increase, decrease, or re- main the same? Explain. Ag reacts with Cl to form AgCl(s). d. Water is added to both half-cell compartments until the vol- 103. What is the maximum work that can be obtained from a hydrogen–oxygen fuel cell at standard conditions that pro- ume of solution is doubled. duces 1.00 kg of water at 25°C? Why do we say that this is e. The silver electrode is replaced with a platinum electrode. the maximum work that can be obtained? What are the ad- vantages and disadvantages in using fuel cells rather than the Pt2 2e ¡ Pt e° 1.19 V corresponding combustion reactions to produce electricity? 96. A standard galvanic cell is constructed so that the overall cell 104. The overall reaction and standard cell potential at 25°C for the reaction is rechargeable nickel–cadmium alkaline battery is 2Al 3 1aq2 3M1s2 ¡ 3M 2 1aq2 2Al1s2 where M is an unknown metal. If ¢G° 411 kJ for the over- Cd1s2 NiO21s2 2H2O1l2 ¡ e° 1.10 V all cell reaction, identify the metal used to construct the stan- Ni1OH221s2 Cd1OH221s2 dard cell. For every mole of Cd consumed in the cell, what is the maxi- 97. The black silver sulfide discoloration of silverware can be re- mum useful work that can be obtained at standard conditions? moved by heating the silver article in a sodium carbonate solu- tion in an aluminum pan. The reaction is 105. An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is 3Ag2S1s2 2Al1s2 ∆ 6Ag1s2 3S2 1aq2 2Al 3 1aq2 2CO1g2 O21g2 ¡ 2CO21g2 a. Using data in Appendix 4, calculate ¢G°, K, and e° for the above reaction at 25°C. (For Al3 (aq), ¢G°f 480. kJ/mol.) The two half-cell reactions are b. Calculate the value of the standard reduction potential for the following half-reaction: 2e Ag2S1s2 ¡ 2Ag1s2 S 2 1aq2 CO O 2 ¡ CO2 2e O2 4e ¡ 2O 2 98. In 1973 the wreckage of the Civil War ironclad USS Monitor was discovered near Cape Hatteras, North Carolina. (The The two half-reactions are carried out in separate compartments Monitor and the CSS Virginia [formerly the USS Merrimack] fought the first battle between iron-armored ships.) In 1987 in- connected with a solid mixture of CeO2 and Gd2O3. Oxide ions vestigations were begun to see if the ship could be salvaged. It can move through this solid at high temperatures (about 800°C). was reported in Time (June 22, 1987) that scientists were ¢G for the overall reaction at 800°C under certain concentra- considering adding sacrificial anodes of zinc to the rapidly tion conditions is 380 kJ. Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

836 Chapter Seventeen Electrochemistry 106. A fuel cell designed to react grain alcohol with oxygen has the Calculate the Ksp value for Ag2SO4(s). Note that to obtain silver following net reaction: ions in the right compartment (the cathode compartment), ex- cess solid Ag2SO4 was added and some of the salt dissolved. C2H5OH1l2 3O21g2 ¡ 2CO21g2 3H2O1l2 114. A zinc–copper battery is constructed as follows at 25°C: The maximum work 1 mol of alcohol can yield by this process is 1320 kJ. What is the theoretical maximum voltage this cell Zn 0 Zn 2 10.10 M2 0 0 Cu 2 12.50 M2 0 Cu can achieve? 107. Gold is produced electrochemically from an aqueous solution of The mass of each electrode is 200. g. Au(CN)2 containing an excess of CN . Gold metal and oxy- a. Calculate the cell potential when this battery is first con- gen gas are produced at the electrodes. What amount (moles) of O2 will be produced during the production of 1.00 mol of gold? nected. 108. In the electrolysis of a sodium chloride solution, what volume b. Calculate the cell potential after 10.0 A of current has flowed of H2(g) is produced in the same time it takes to produce 257 L of Cl2(g), with both volumes measured at 50.°C and 2.50 atm? for 10.0 h. (Assume each half-cell contains 1.00 L of solution.) 109. An aqueous solution of an unknown salt of ruthenium is elec- c. Calculate the mass of each electrode after 10.0 h. trolyzed by a current of 2.50 A passing for 50.0 min. If 2.618 g d. How long can this battery deliver a current of 10.0 A before Ru is produced at the cathode, what is the charge on the ruthe- nium ions in solution? it goes dead? 110. It takes 15 kWh (kilowatt-hours) of electrical energy to pro- duce 1.0 kg of aluminum metal from aluminum oxide by the 115. A galvanic cell is based on the following half-reactions: Hall–Heroult process. Compare this to the amount of energy necessary to melt 1.0 kg of aluminum metal. Why is it eco- Fe 2 2e ¡ Fe1s2 e° 0.440 V nomically feasible to recycle aluminum cans? (The enthalpy of 2H 2e ¡ H21g2 e° 0.000 V fusion for aluminum metal is 10.7 kJ/mol [1 watt 1 J/s].) where the iron compartment contains an iron electrode and Challenge Problems [Fe 2 ] 1.00 10 3 M and the hydrogen compartment con- tains a platinum electrode, PH2 1.00 atm, and a weak acid, 111. Combine the equations HA, at an initial concentration of 1.00 M. If the observed cell potential is 0.333 V at 25°C, calculate the Ka value for the weak ¢G° nFe° and ¢G° ¢H° T¢S° acid HA. to derive an expression for e° as a function of temperature. De- 116. Consider a cell based on the following half-reactions: scribe how one can graphically determine ¢H° and ¢S° from measurements of e° at different temperatures, assuming that Au 3 3e ¡ Au e° 1.50 V ¢H° and ¢S° do not depend on temperature. What property Fe 3 e ¡ Fe 2 e° 0.77 V would you look for in designing a reference half-cell that would produce a potential relatively stable with respect to temperature? a. Draw this cell under standard conditions, labeling the anode, 112. The overall reaction in the lead storage battery is the cathode, the direction of electron flow, and the concen- Pb1s2 PbO21s2 2H 1aq2 2HSO4 1aq2 ¡ trations, as appropriate. 2PbSO41s2 2H2O1l2 b. When enough NaCl(s) is added to the compartment contain- a. For the cell reaction ¢H° 315.9 kJ and ¢S° 263.5 J/K. ing gold to make the [Cl ] 0.10 M, the cell potential is Calculate e° at 20.°C. Assume ¢H° and ¢S° do not depend observed to be 0.31 V. Assume that Au3 is reduced and as- on temperature. sume that the reaction in the compartment containing gold is b. Calculate e at 20.°C when [HSO4 ] [H ] 4.5 M. c. Consider your answer to Exercise 55. Why does it seem that Au 3 1aq2 4Cl 1aq2 ∆ AuCl4 1aq2 batteries fail more often on cold days than on warm days? Calculate the value of K for this reaction at 25°C. 113. Consider the following galvanic cell: 117. The measurement of pH using a glass electrode obeys the Nernst 0.83V equation. The typical response of a pH meter at 25.00°C is given by the equation Pb Ag emeas eref 0.05916 pH 1.8 M Pb2+ ? M SAOg+42- ? M where eref contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that eref 0.250 V and that emeas 0.480 V. a. What is the uncertainty in the values of pH and [H ] if the uncertainty in the measured potential is 1 mV ( 0.001 V)? b. To what precision must the potential be measured for the un- certainty in pH to be 0.02 pH unit? 118. Zirconium is one of the few metals that retains its structural in- tegrity upon exposure to radiation. For this reason, the fuel rods in most nuclear reactors are made of zirconium. Answer the fol- lowing questions about the redox properties of zirconium based on the half-reaction Ag2SO4 (s) ZrO2 H2O H2O 4e ¡ Zr 4OH e° 2.36 V

Integrative Problems 837 a. Is zirconium metal capable of reducing water to form hydro- c. If the coated silver wire is placed in a solution (at 25°C) in gen gas at standard conditions? which [CrO42 ] 1.00 10 5 M, what is the expected cell potential? b. Write a balanced equation for the reduction of water by zirconium metal. d. The measured cell potential at 25°C is 0.504 V when the coated wire is dipped into a solution of unknown [CrO42 ]. c. Calculate e°, ¢G°, and K for the reduction of water by zir- What is [CrO42 ] for this solution? conium metal. e. Using data from this problem and from Table 17.1, calculate d. The reduction of water by zirconium occurred during the ac- the solubility product (Ksp) for Ag2CrO4. cident at Three Mile Island, Pennsylvania, in 1979. The hy- 123. You have a concentration cell with Cu electrodes and [Cu 2 ] drogen produced was successfully vented and no chemical 1.00 M (right side) and 1.0 10 4 M (left side). explosion occurred. If 1.00 103 kg of Zr reacts, what mass a. Calculate the potential for this cell at 25°C. b. The Cu2 ion reacts with NH3 to form Cu(NH3)42 where of H2 is produced? What volume of H2 at 1.0 atm and 1000.°C the stepwise formation constants are K1 1.0 103, K2 is produced? 1.0 104, K3 1.0 103, and K4 1.0 10 3. Calculate e. At Chernobyl, USSR, in 1986, hydrogen was produced by the the new cell potential after enough NH3 is added to the left cell compartment such that at equilibrium [NH3] 2.0 M. reaction of superheated steam with the graphite reactor core: 124. When copper reacts with nitric acid, a mixture of NO(g) and C1s2 H2O1g2 ¡ CO1g2 H21g2 NO2(g) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the A chemical explosion involving the hydrogen gas did occur equilibrium at Chernobyl. In light of this fact, do you think it was a cor- rect decision to vent the hydrogen and other radioactive gases 2H 1aq2 2NO3 1aq2 NO1g2 ∆ 3NO21g2 H2O1l2 into the atmosphere at Three Mile Island? Explain. Consider the following standard reduction potentials at 25°C: 119. A galvanic cell is based on the following half-reactions: 3e 4H 1aq2 NO3 1aq2 ¡ NO1g2 2H2O1l2 Ag e ¡ Ag1s2 e° 0.80 V e° 0.957 V Cu 2 2e ¡ Cu1s2 e° 0.34 V e 2H 1aq2 NO3 1aq2 ¡ NO21g2 2H2O1l2 In this cell, the silver compartment contains a silver electrode e° 0.775 V and excess AgCl(s) (Ksp 1.6 10 10), and the copper com- partment contains a copper electrode and [Cu 2 ] 2.0 M. a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and NO2 a. Calculate the potential for this cell at 25°C. b. Assuming 1.0 L of 2.0 M Cu 2 in the copper compartment, mixture with only 0.20% NO2 (by moles) at 25°C and 1.00 atm? Assume that no other gases are present and that the calculate the moles of NH3 that would have to be added to change in acid concentration can be neglected. give a cell potential of 0.52 V at 25°C (assume no volume change on addition of NH3). Cu 2 1aq2 4NH31aq2 ∆ K 1.0 1013 Cu1NH324 2 1aq2 Integrative Problems 120. Given the following two standard reduction potentials, M 3 3e S M e° 0.10 V These problems require the integration of multiple concepts to find the M 2 2e S M e° 0.50 V solutions. solve for the standard reduction potential of the half-reaction 125. The following standard reduction potentials have been deter- mined for the aqueous chemistry of indium: M3 e S M2 In 3 1aq2 2e ¡ In 1aq2 (Hint: You must use the extensive property ¢G° to determine In 1aq2 e ¡ In1s2 e° 0.444 V the standard reduction potential.) e° 0.126 V 121. You make a galvanic cell with a piece of nickel, 1.0 M Ni2 (aq), a. What is the equilibrium constant for the disproportionation a piece of silver, and 1.0 M Ag (aq). Calculate the concentra- reaction, where a species is both oxidized and reduced, shown tions of Ag (aq) and Ni2 (aq) once the cell is “dead.” below? 122. A chemist wishes to determine the concentration of CrO4 2 electrochemically. A cell is constructed consisting of a satu- 3In 1aq2 ¡ 2In1s2 In 3 1aq2 rated calomel electrode (SCE; see Exercise 93) and a silver wire coated with Ag2CrO4. The e° value for the following b. What is ¢G°f for In (aq) if ¢G°f 97.9 kJ/mol for half-reaction is 0.446 V relative to the standard hydrogen In3 (aq)? electrode: 126. An electrochemical cell is set up using the following balanced reaction: Ag2CrO4 2e ¡ 2Ag CrO4 2 M a 1aq2 N1s2 ¡ N 2 1aq2 M1s2 a. Calculate ecell and ¢G at 25°C for the cell reaction when Given the standard reduction potentials are: [CrO42 ] 1.00 mol/L. M a ae ¡ M e° 0.400 V b. Write the Nernst equation for the cell. Assume that the SCE N 2 2e ¡ N e° 0.240 V concentrations are constant.

838 Chapter Seventeen Electrochemistry The cell contains 0.10 M N2 and produces a voltage of 0.180 V. a vanadium electrode and V2 at an unknown concentration. The If the concentration of Ma is such that the value of the reaction quotient Q is 9.32 10 3, calculate [Ma ]. Calculate wmax for compartment containing the vanadium (1.00 L of solution) was this electrochemical cell. titrated with 0.0800 M H2EDTA 2 , resulting in the reaction 127. Three electrochemical cells were connected in series so that the H2EDTA 2 1aq2 V 2 1aq2 K? same quantity of electrical current passes through all three cells. In ∆ VEDTA 2 1aq2 2H 1aq2 the first cell, 1.15 g of chromium metal was deposited from a chromium(III) nitrate solution. In the second cell, 3.15 g of os- The potential of the cell was monitored to determine the stoi- mium was deposited from a solution made of Osn and nitrate ions. chiometric point for the process, which occurred at a volume of What is the name of the salt? In the third cell, the electrical charge 500.0 mL of H2EDTA 2 solution added. At the stoichiometric passed through a solution containing X2 ions caused deposition point, ecell was observed to be 1.98 V. The solution was buffered of 2.11 g of metallic X. What is the electron configuration of X? at a pH of 10.00. a. Calculate ecell before the titration was carried out. Marathon Problems b. Calculate the value of the equilibrium constant, K, for the These problems are designed to incorporate several concepts and techniques titration reaction. into one situation. Marathon Problems can be used in class by groups of c. Calculate ecell at the halfway point in the titration. students to help facilitate problem-solving skills. 129. The table below lists the cell potentials for the 10 possible gal- 128. A galvanic cell is based on the following half-reactions: vanic cells assembled from the metals A, B, C, D, and E, and their respective 1.00 M 2 ions in solution. Using the data Cu 2 1aq2 2e ¡ Cu1s2 e° 0.34 V in the table, establish a standard reduction potential table V2 1aq2 2e ¡ V1s2 e° 1.20 V similar to Table 17.1 in the text. Assign a reduction potential In this cell, the copper compartment contains a copper electrode of 0.00 V to the half-reaction that falls in the middle of and [Cu 2 ] 1.00 M, and the vanadium compartment contains the series. You should get two different tables. Explain why, and discuss what you could do to determine which table is correct. E1s2 in E 2 1aq2 A(s) in A2 (aq) B(s) in B2 (aq) C(s) in C2 (aq) D(s) in D2 (aq) D1s2 in D 2 1aq2 C1s2 in C 2 1aq2 0.28 V 0.81 V 0.13 V 1.00 V B1s2 in B 2 1aq2 0.72 V 0.19 V 1.13 V — 0.41 V 0.94 V — 0.53 V — — — — Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.



A60 Answers to Selected Exercises temperatures 45. a. H 803 kJ, S 4 J/K, G 802 kJ; electrons through a wire. This is accomplished by making a galvanic cell which b. H 2802 kJ, S 262 J/K, G 2880. kJ; c. H 416 kJ, separates the reduction reaction from the oxidation reaction in order to control S 209 J/K, G 354 kJ; d. H 176 kJ, S 284 J/K, the flow of electrons through a wire to produce a voltage. 19. An extensive G 91 kJ 47. 5.40 kJ; 328.6 K; ¢G° is negative below 328.6 K. property is one that depends on the amount of substance. The free energy 49. CH4(g) CO2(g) n CH3CO2H(l), H 16 kJ, S 240. J/K, change for a reaction depends on whether 1 mol of product is produced or G 56 KJ; CH3OH(g) CO(g) n CH3CO2H(l), H 173 kJ, S 2 mol of product is produced or 1 million mol of product is produced. This is 278 J/K, G 90. kJ; the second reaction is preferred. It should be run not the case for cell potentials which do not depend on the amount of sub- at temperatures below 622 K. 51. 817 kJ 53. 731 kJ/mol 55. yes stance. The equation that relates ¢G to E is ¢G nFE. It is the n term that 57. 188 kJ 59. a. shifts right; b. no shift since the reaction is at equilib- converts the intensive property E into the extensive property ¢G. The n is the number of mol of electrons transferred in the balanced reaction that ¢G is rium; c. shifts left 61. 8.72; 0.0789 63. 140 kJ 65. 71 kJ/mol associated with. 21. A potential hazard when jump-starting a car is that the 67. H 1.1 105 J/mol; S 330 J/K mol; The major difference in electrolysis of H2O1l2 can occur. When H2O1l2 is electrolyzed, the products are the explosive gas mixture of H21g2 and O21g2. A spark produced during jump the plot is the slope of the line. An endothermic process has a negative slope starting a car could ignite any H21g2 and O21g2 produced. Grounding the jumper cable far from the battery minimizes the risk of a spark nearby the battery for the ln(K) versus 1 T plot, whereas an exothermic process has a positive slope. 69. 447 J/K mol 71. decreases; ¢S will be negative since 2 mol of gaseous reactants form 1 mol of gaseous product. For ¢G to be negative, ¢H must be negative (exothermic). For exothermic reactions, K decreases as where H21g2 and O21g2 could be collecting. 23. You need to know the iden- T increases, so the ratio of the partial pressure of PCl5 to the partial pressure tity of the metal so you know which molar mass to use. You need to know the of PCl3 will decrease. 73. 43.7 K 75. 60 77. a. 1.8 104 J/mol; shifts left; b. 0; no shift since at equilibrium; c. 1.1 104 J/mol; shifts right; oxidation state of metal ion in the salt so the mol of electrons transferred can be determined. And finally, you need to know the amount of current and the d. 0; no shift since at equilibrium; e. 2 103 J/mol; shifts left 79. a. 2.22 time the current was passed through the electrolytic cell. If you know these 105; b. 94.3; c. 0.29 mol ATP 81. S is more favorable for reaction 2 four quantities, then the mass of metal plated out can be calculated. 25. See than for reaction 1, resulting in K2 K1. In reaction 1, seven particles in so- Figure 17.3 of the text for a typical galvanic cell. The anode compartment con- lution form one particle. In reaction 2, four particles form one, which results tains the oxidation half-reaction compounds/ions, and the cathode compart- in a smaller decrease in disorder than for reaction 1. 83. 725 K 85. H ment contains the reduction half-reaction compounds/ions. The electrons flow 286 kJ; G 326 kJ; K 7.22 o1f 0oz5o8n; ePOp3er 3.3 10 41 atm; This partial from the anode to the cathode. For each of the following answers, all solutes pressure represents one molecule 9.5 1017 L of air. Equilib- are 1.0 M and all gases are at 1.0 atm. a. 7H2O(l) 2Cr3 (aq) 3Cl2(g) n rium is probably not maintained under the conditions because the concentra- Cr2O72 (aq) 6Cl (aq) 14H (aq); cathode: Pt electrode; Cl2 bubbled into solution, Cl in solution; anode: Pt electrode; Cr3 , H , and Cr2O72 in solu- tion of ozone is not large enough to maintain equilibrium 87. a. Because tion; b. Cu2 (aq) Mg(s) n Cu(s) Mg2 (aq); cathode: Cu electrode; Cu2 kf A exp a Ea b and kr A exp a 1Ea RT ¢G°2 b, in solution; anode: Mg electrode; Mg2 in solution 27. a. 0.03 V; b. 2.71 V RT 29. See Exercise 25 for a description of a galvanic cell. For each of the fol- kf Ea 1Ea ¢G°2 ¢G° Because kr exp a RT RT b exp a RT b. Because K lowing answers, all solutes are 1.0 M and all gases are at 1.0 atm. In the salt exp a ¢G° b, then K kkfr. b. A catalyst increases the value of the rate con- bridge, cations flow to the cathode and anions flow to the anode. a. Cl2(g) RT 2Br (aq) n Br2(aq) 2Cl (aq), e 0.27 V; cathode: Pt electrode; Cl2(g) stant (increases rate) by lowering the activation energy. For the equilibrium bubbled in, Cl in solution; anode: Pt electrode; Br2 and Br in solution; b. 3H2O(l ) 5IO4 (aq) 2Mn2 (aq) n 5IO3 (aq) 2MnO4 (aq) constant K to remain constant, both kf and kr must increase by the same fac- 6H (aq), e 0.09 V; cathode: Pt electrode; IO4 , IO3 , and H2SO4 (as a tor. Therefore, a catalyst must increase the rate of both the forward and the re- source of H ) in solution; anode: Pt electrode; Mn2 , MnO4 , and H2SO4 in solution 31. 25a. Pt Cr3 (1.0 M), H (1.0 M ), Cr2O72 (1.0 M ) 0 0 Cl2 (1.0 verse reactions. 89. a. 0.333; b. PA 1.50 atm; PB 0.50 atm; c. G atm) Cl (1.0 M) Pt; 25b. Mg 0 Mg2 (1.0 M) 0 0 Cu2 (1.0 M) 0 Cu; 29a. Pt G RT ln(PB PA) 2722 J 2722 J 0 91. at least 7.5 torr 93. 16 g Br (1.0 M), Br2 (1.0 M ) 0 0 Cl2 (1.0 atm) 0 Cl (1.0 M ) 0 Pt; 29b. Pt 0 Mn2 (1.0 M ), MnO4 (1.0 M ), H (1.0 M ) 0 0 IO4 (1.0 M ), H (1.0 M ) IO3 (1.0 95. 61 kJ 97. 4.1 kJ/mol M ), 0 Pt 33. a. Au3 (aq) 3Cu (aq) n 3Cu2 (aq) Au(s), e 1.34 Chapter 17 V; b. 2VO2 (aq) 4H (aq) Cd(s) n Cd2 (aq) 2VO2 (aq) 2H2O(l), e 1.40 V 35. a. 16H 2MnO4 10I S 5I2 2Mn2 8H2O, 13. Oxidation: increase in oxidation number, loss of electrons; reduction: de- e°cell 0.97 V, spontaneous; b. 16H 2MnO4 10F S 5F2 crease in oxidation number, gain of electrons 15. Reactions a, b, and c are 2Mn 2 1.36 V,not spontaneous 37. e 0.41 V, G oxidation–reduction reactions. 8H2O, e°cell Oxidizing Reducing Substance Substance 79 kJ 39. 33a. 388 kJ; 33b. 270. kJ 41. 1.21 V 43. K H2O Agent Agent Oxidized Reduced Cd2 I2 AuCl4 IO3 45. a. no; b. yes; c. yes; d. no 47. a. Cr2O72 , O2, MnO2, IO3 ; b. PbSO4, Cd2 , Fe2 , Cr3 , Zn2 , H2O a. H2O CH4 CH4(C) H2O(H) 49. ClO (aq) 2NH3(aq) n Cl (aq) N2H4(aq) H2O(l ), e cell b. AgNO3 Cu Cu AgNO3(Ag) c. HCl HCl(H) 1.00 V; Because e cell is positive for this reaction, at standard conditions ClO Zn Zn can spontaneously oxidize NH3 to the somewhat toxic N2H4. 51. a. larger; 17. Magnesium is an alkaline earth metal; Mg will oxidize to Mg2 . The ox- b. smaller 53. Electron flow is always from the anode to the cathode. For idation state of hydrogen in HCl is 1. To be reduced, the oxidation state of the cells with a nonzero cell potential, we will identify the cathode, which H must decrease. The obvious choice for the hydrogen product is H21g2 where hydrogen has a zero oxidation state. The balanced reaction is: means the other compartment is the anode. a. 0; b. 0.018 V; compartment Mg1s2 2HCl1aq2 S MgCl21aq2 H21g2. Mg goes from the 0 to the 2 ox- idation state by losing two electrons. Each H atom goes from the 0 to the 1 with [Ag ] 2.0 M is cathode; c. 0.059 V; compartment with [Ag ] 1.0 M oxidation state by gaining one electron. Since there are two H atoms in the is cathode; d. 0.26 V; compartment with [Ag ] 1.0 M is cathode; balanced equation, then a total of two electrons are gained by the H atoms. e. 0 55. 2.12 V 57. 1.09 V 59. a. 0.23 V; b. 1.2 10 5 M 61. 0.16 Hence, two electrons are transferred in the balanced reaction. When the V, copper is oxidized. 63. 1.7 10 30 65. [25] a. G 20 kJ; 1 electrons are transferred directly from Mg to H , no work is obtained. In or- 103; b. G 523 kJ; 5.12 1091; [29] a. G 52 kJ; 1.4 109; der to harness this reaction to do useful work, we must control the flow of b. G 90 kJ; 2 1015 67. 2.5 1026 69. a. no reaction; b. Cl2(g) 2I (aq) S I2(s) 2Cl (aq), e cell 0.82 V; G 160 kJ; K 5.6 1027; c. no reaction; d. 4Fe2 (aq) 4H (aq) O2(g) S

Answers to Selected Exercises A61 4Fe3 (aq) 2H2O(l), e cell 0.46 V; G 180 kJ; K 1.3 1031; 19. a. 29489Cf 188O S 120663Sg 4 01n; b. 120549Rf 21. 690 hours 23. 81Kr is most stable since it has the longest half-life. 73Kr is “hottest” since it decays very 71. a. Au3 (aq) 3Tl(s) S Au(s) 3Tl (aq); e cell 1.84 V; b. G 533 kJ; K 2.52 1093; c. 2.04 V 73. 5.1 10 20 75. 6.19 10 52 rapidly due to its very short half-life. 73Kr, 81s; 74Kr, 34.5 min; 76Kr, 44.4 h; 77. a. 30. hours; b. 33 s; c. 1.3 hours 79. a. 16 g; b. 25 g; c. 71 g; 81Kr, 6.3 105 yr 25. 6.22 mg 32P remains 27. 0.230 29. 26 g d. 4.9 g 81. Bi 83. 9.12 L F2 (anode), 29.2 g K (cathode) 85. 7.44 104 A 87. 1.14 10 2 M 89. Au followed by Ag followed by Ni fol- 31. 2.3 counts per minute per gram of C. No; for a 10.-mg C sample, it would lowed by Cd 91. a. cathode: Ni2 2e S Ni; anode: 2Br S Br2 take roughly 40 min to see a single disintegration. This is too long to wait, and 2e ; b. cathode: Al3 3e S Al; anode: 2F S F2 2e ; c. cathode: Mn2 2e S Mn; anode: 2I S I2 2e 93. a. 0.10 V, SCE is anode; the background radiation would probably be much greater than the 14C activ- b. 0.53 V, SCE is anode; c. 0.02 V, SCE is cathode; d. 1.90 V, SCE is cath- ity. Thus 14C dating is not practical for very small samples. 33. 3.8 109 yr 35. 4.3 106 kg/s 37. 232Pu, 1.715 1014 J/mol; 231Pa, 1.714 1014 ode; e. 0.47 V, SCE is cathode 95. a. decrease; b. increase; c. decrease; J/mol 39. 12C: 1.23 10 12 J/nucleon; 235U: 1.2154 10 12 J/nucleon; since d. decrease; e. same 97. a. G 582 kJ; K 3.45 10102; e 1.01 56Fe is the most stable known nucleus, then the binding energy per nucleon for V; b. 0.65 V; 99. Aluminum has the ability to form a durable oxide coat- 56Fe would be larger than that of 12C or 235U. (See Fig. 18.9 of the text.) ing over its surface. Once the HCl dissolves this oxide coating, Al is exposed 41. 6.01513 amu 43. 2.0 1010 J/g of hydrogen nuclei 45. The to H and is easily oxidized to Al3 . Thus, the Al foil disappears after the ox- Geiger–Müller tube has a certain response time. After the gas in the tube ion- ide coating is dissolved. 101. The claim is impossible. The strongest oxi- izes to produce a “count,” some time must elapse for the gas to return to an dizing agent and reducing agent when combined give e of only about 6 V. electrically neutral state. The response of the tube levels off because, at high 103. wmax 13,200 kJ; the work done can be no larger than the free en- activities, radioactive particles are entering the tube faster than the tube can re- ergy change. If the process were reversible all of the free energy released spond to them. 47. All evolved O2(g) comes from water. 49. 2 neutrons; 4 particles 51. Strontium. Xe is chemically unreactive and not readily in- would go into work, but this does not occur in any real process. Fuel cells corporated into the body. Sr can be easily oxidized to Sr 2 . Strontium is in the are more efficient in converting chemical energy to electrical energy; they same family as calcium and could be absorbed and concentrated in the body are also less massive. Major disadvantage: They are expensive. 105. 0.98 V in a fashion similar to Ca2 . The chemical properties determine where ra- T¢S° ¢nHF°; if we graph e versus T, 107. 0.250 mol 109. 3 111. e° nF dioactive material may be concentrated in the body or how easily it may be excreted. 53. a. unstable; beta production; b. stable; c. unstable; positron we should get a straight line ( y mx b). The slope of the line is equal production or electron capture; d. unstable, positron production, electron to S nF and the y-intercept is equal to H nF. e will have little capture, or alpha production. 55. 49.7 yr 57. 1975 59. 900 g 235U temperature dependence for cell reactions with S close to zero. 61. 7 105 m/s; 8 10 16 J/nuclei; 63. Assuming that (1) the radionu- 113. 9.8 10 6 115. 2.39 10 7 117. a. 0.02 pH units; 6 10 6 clide is long lived enough that no significant decay occurs during the time of M H ; b. 0.001 V 119. a. 0.16 V; b. 8.6 mol 121. 3 Ag 4 4.6 the experiment, and (2) the total activity is uniformly distributed only in the 10 18 M; 3 Ni 2 4 1.5 M 123. a. 0.12 V; b. 0.54 V 125. a. 5.77 10 10; rat’s blood; V 10. mL. 65. a. 12 C; b. 13N, 13C, 14N, 15O, and 15N; 6 b. 12.2 kJ/mol 127. Osmium(IV) nitrate; [Ar]4s13d10 c. 5.950 1011 J/mol 1H 67. 4.3 10 29 69. 29479Bk 1202Ne S 120677Bh 401n; 62.7s; [Rn]7s25f 146d 5 Chapter 18 Chapter 19 1. The characteristic frequencies of energies emitted in a nuclear reaction 1. The gravity of the earth cannot keep H2 in the atmosphere. 3. The acid- ity decreases. Solutions of Be2 are acidic, while solutions of the other M2 suggest that discrete energy levels exist in the nucleus. The extra stability of certain numbers of nucleons and the predominance of nuclei with even num- ions are neutral. 5. The planes of carbon atoms slide easily. Graphite is not bers of nucleons suggest that the nuclear structure might be described by using volatile so the lubricant will not be lost when used in a high vacuum environ- quantum numbers. 3. b-particle production has the net effect of turning a ment. 7. p-type semiconductor 9. For groups 1A–3A, the small size of H neutron into a proton. Radioactive nuclei having too many neutrons typically (as compared to Li), Be (as compared to Mg), and B (as compared to Al) seems undergo b-particle decay. Positron production has the net effect of turning a to be the reason why these elements have nonmetallic properties, while others proton into a neutron. Nuclei having too many protons typically undergo in the groups 1A–3A are strictly metallic. The small size of H, Be, and B also positron decay. 5. The transuranium elements are the elements having more causes these species to polarize the electron cloud in nonmetals, thus forcing protons than uranium. They are synthesized by bombarding heavier nuclei with a sharing of electrons when bonding occurs. For groups 4A–6A, a major dif- neutrons and positive ions in a particle accelerator. 7. ¢E ¢mc2; The key ference between the first and second members of a group is the ability to form difference is the mass change when going from reactants to products. In chem- p bonds. The smaller elements form stable p bonds, while the larger elements ical reactions, the mass change is indiscernible. In nuclear processes, the mass do not exhibit good overlap between parallel p orbitals and, in turn, do not change is discernable. It is the conversion of this discernable mass change into form strong p bonds. For group 7A, the small size of F as compared to Cl is energy that results in the huge energies associated with nuclear processes. used to explain the low electron affinity of F and the weakness of the F¬F 9. Sr-90 is an alkaline earth metal having chemical properties similar to cal- bond. 11. In order to maximize hydrogen bonding interactions in the solid cium. Sr-90 can collect in bones replacing some of the calcium. Once imbed- phase, ice is forced into an open structure. This open structure is why H2O1s2 is less dense than H2O1l2. 13. a. H 207 kJ, S 216 J/K; b. T ded inside the human body, particles can do significant damage. Rn-222 is 958 K 15. a. lithium oxide; b. potassium superoxide; c. sodium peroxide a noble gas so one would expect Rn to be unreactive and pass through the body quickly; it does. The problem with Rn-222 is the rate at which it produces al- 17. a. Li2O(s) H2O(l) S 2LiOH(aq); b. Na2O2(s) 2H2O(l) S pha particles. With a short half-life, the few moments that Rn-222 is in the 2NaOH(aq) H2O2(aq); c. LiH(s) H2O(l ) S H2(g) LiOH(aq); lungs, a significant number of decay events can occur; each decay event d. 2KO2(s) 2H2O(l ) S 2KOH(aq) O2(g) H2O2(aq) 19. 2Li(s) produces an alpha particle which is very effective at causing ionization and 2C2H2(g) S 2LiC2H(s) H2(g); oxidation–reduction 21. a. magnesium can produce a dense trail of damage. 11. a. 2541Cr 10e S 5231V; b. 15313I S carbonate; b. barium sulfate; c. strontium hydroxide 23. CaCO3(s) 01e 15341Xe 13. a. 6381Ga 10e S 3680Zn; b. 6229Cu S 01e 6282Ni; c. 28172Fr S 42He 20885At; d. 12591Sb S 0 15229Te 15. 10 particles; 5 H2SO4(aq) S CaSO4(aq) H2O(l ) CO2(g) 25. In the gas phase, linear 1 e molecules would exist: particles 17. 2563Fe has too many protons. It will undergo positron production, electron capture, and /or alpha-particle production. 2569Fe has too many neu- F Be F trons and will undergo beta-particle production. (See Table 18.2 of the text.)