20. ELECTROCHEMISTRY. Solutions to Exercises

0. ELECTROCHEMISTRY Solutions to Exercises Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiplestep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 0.1 Assign oxidation numbers to the skeleton equation (Step 1). 0 +5 +5 + I + NO 3 IO 3 + NO Separate into two incomplete halfreactions (Step ). Note that iodine is oxidized (increases in oxidation number), and nitrogen is reduced (decreases in oxidation number). I IO 3 NO 3 NO Balance each halfreaction separately. The oxidation halfreaction is not balanced in I, so place a two in front of IO 3 (Step 3a). Then add six H O s to the left side to balance O atoms (Step 3b), and add twelve H + ions to the right side to balance H atoms (Step 3c). Finally, add ten electrons to the right side to balance the charge (Step 3d). The balanced oxidation halfreaction is I + 6H O IO 3 + 1H + + 10e The reduction halfreaction is balanced in N (Step 3a). Add one H O to the right side to balance O atoms (Step 3b), and add two H + ion to the left side to balance H atoms. Finally, add one electron to the left side to balance the charge (Step 3d). (continued)

ELECTROCHEMISTRY 781 The balanced reduction halfreaction is NO 3 + H + + e NO + H O Multiply the reduction halfreaction by five so that, when added, the electrons cancel (Step a). I + 6H O IO 3 10NO 3 I + 10NO 3 + 1H + + 10e + 0H + + 10e 10NO + 10H O + 0H + + 6H O + 10e IO 3 + 10NO + 1H + + 10H O + 10e Simplify the equation by canceling the twelve H + and six H O that appear on both sides. The coefficients do not need to be reduced (Step b). The net ionic equation is I (s) + 10NO 3 (aq) + 8H + (aq) IO 3 (aq) + 10NO (g) + H O(l) 0. After balancing the equation as if it were in acid solution, you obtain the following: H O + ClO ClO + O + H + Add two OH to both sides of the equation (Step 5), and replace the two H + and two OH on the right side with two H O. No further cancellation is required. The balanced equation for the reaction in basic solution is H O + ClO + OH ClO + O + H O 0.3 Silver ion is reduced at the silver electrode (cathode). The halfreaction is Ag + (aq) + e Ag(s) The nickel electrode (anode) is where oxidation occurs. The halfreaction is Ni(s) Ni + (aq) + e (continued)

78 CHAPTER 0 Electron flow in the external circuit is from the nickel electrode (anode) to the silver electrode (cathode). Positive ions will flow in the solution portion of the circuit opposite to the direction of the electrons. A sketch of the cell is given below: Ni (anode) e salt bridge Ag (cathode) + Ni + Ag + Ni Ni + + e Ag + + e Ag 0. The notation for the cell is Zn(s) Zn + (aq) H + (aq) H (g) Pt(s). 0.5 The halfcell reactions are Cd(s) Cd + (aq) + e H + (aq) + e H (g) Summing the halfcell reactions gives the overall cell reaction. Cd(s) + H + (aq) Cd + (aq) + H (g) 0.6 The halfreactions are Zn(s) Zn + (aq) + e Cu + (aq) + e Cu(s) n equals two, and the maximum work for the reaction as written is w max = nfe cell = x 9.65 x 10 C x 1.10 V =.13 x 10 5 V C =.1 x 10 5 J For 6.5 g of zinc metal, the maximum work is 6.5 g Zn x 5 1 mol Zn 65.39 g Zn x.13 x 10 J 1 mol Zn =.13 x 10 =.1 x 10 J

ELECTROCHEMISTRY 783 0.7 The halfreactions and corresponding electrode potentials are as follows Ag + (aq) + e Ag(s) 0.80 V NO 3 (aq) + H + (aq) + e NO(g) + H O(l) 0.96 V The stronger oxidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so NO 3 is the stronger oxidizing agent. 0.8 In this reaction, Cu + is the oxidizing agent on the left side; I is the oxidizing reagent on the right side. The corresponding standard electrode potentials are Cu + (aq) + e Cu(s); E = 0.3 V I (s) + e I (aq); E = 0.5 V The stronger oxidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so I is the stronger oxidizing agent. The reaction is nonspontaneous as written. 0.9 The reduction halfreactions and standard electrode potentials are Zn + (aq) + e Zn(s); E Zn = 0.76 V Cu + (aq) + e Cu(s); E Cu = 0.3 V Reverse the first halfreaction and its halfcell potential to obtain Zn(s) Zn + (aq) + e ; E Zn = 0.76 V Cu + (aq) + e Cu(s); E Cu = 0.3 V Obtain the cell emf by adding the halfcell potentials. E cell = E Cu E Zn = 0.3 V + 0.76 V = 1.10 V 0.10 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Sn + (aq) Sn + (aq) + e E = 0.15 V Hg + (aq) + e Hg + (aq) E = 0.90 V Sn + (aq) + Hg + (aq) Sn + (aq) + Hg + (aq) E cell = 0.75 V (continued)

78 CHAPTER 0 Note that each halfreaction involves two electrons; hence n equals two. Also, E cell equals 0.75 V, and the faraday constant, F, is 9.65 x 10 C. Therefore, G = nfe cell = x 9.65 x 10 C x 0.75 V = 1.75 x 10 5 J = 1. x 10 5 J Thus, the standard freeenergy change is 1. x 10 kj. 0.11 Write the equation with G f 's beneath each substance. Mg(s) + Cu + (aq) Mg + (aq) + Cu(s) G f : 0 65.0 56.0 0 kj Hence, G = Σn G f (products) Σm G f (reactants) = [(56.0) 65.0] kj = 51.0 kj = 5.10 x 10 5 J Obtain n by splitting the reaction into halfreactions. Mg(s) Mg + (aq) + e Cu + (aq) + e Cu(s) Each halfreaction involves two electrons, so n equals two. Therefore, G = nfe cell 5.10 x 10 5 J = x 9.65 x 10 C x E cell Rearrange and solve for E cell. Recall that J = C V. E cell = 5 5.10 x 10 J x 9.65 x 10 C =.699 =.70 V 0.1 The halfreactions and standard electrode potentials are Fe(s) Fe + (aq) + e ; E Fe = 0.1 V Sn + (aq) + e Sn + (s); = 0.15 V The standard emf for the cell is E o + Sn E cell = E Fe = 0.1 V + 0.15 V = 0.56 V E o + Sn (continued)

ELECTROCHEMISTRY 785 Note that n equals two. Substitute into the equation relating E and K. Also K = K c. 0.56 V = 0.059 Solving for K c, you get log K c = 18.91 log K c Take the antilog of both sides: K c = antilog (18.91) = 8.1 x 10 18 = 10 19 0.13 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Zn(s) Zn + (aq) + e E = 0.76 V Ag + (aq) + e Ag(s) E = 0.80 V Zn(s) + Ag + (aq) Zn + (aq) + Ag(s) E cell = 1.56 V Note that n equals two. The reaction quotient is Q = + [Zn ] + [Ag ] = 0.00 (0.0000) = 5.00 x 10 The standard emf is 1.56 V, so the Nernst equation becomes E cell = E cell 0.059 n = 1.56 0.059 log Q log (5.00 x 10 ) = 1.56 0.13909 = 1.09 = 1. V

786 CHAPTER 0 0.1 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Zn(s) Zn + (aq) + e E = 0.76 V Ni + (aq) + e Ni(s) E = 0.3 V Zn(s) + Ni + (aq) Zn + (aq) + Ni(s) E cell = 0.53 V Note that n equals two. The standard emf is 0.53 V, and the emf is 0.3 V, so the Nernst equation becomes E cell = E cell 0.059 n log Q 0.3 V = 0.53 V 0.059 Rearrange and solve for log Q log Q log Q = 0.059 x (0.53 0.3) = 6.18 Take the antilog of both sides Q = + [Zn ] + [Ni ] = antilog (6.18) =.63 x 10 6 Substitute in [Zn + ] = 1.00 M and solve for [Ni + ]. 1.00 M + [Ni ] =.6 x 10 6 [Ni + ] = 3.81 x 10 7 = x 10 7 M 0.15 a. The cathode reaction is K + (l) + e K(l). The anode reaction is Cl (l) Cl (g) + e. b. The cathode reaction is K + (l) + e K(l). The anode reaction is OH (aq) O (g) + H O(g) + e.

ELECTROCHEMISTRY 787 0.16 The species you should consider for halfreactions are Ag + and H O. Two possible cathode reactions are Ag + (aq) + e H O(l) + e Ag(s); E = 0.80 V H (g) + OH (aq); E = 0.83 V Because the silver electrode potential is larger than the reduction potential of water, Ag + is reduced. The only possible anode reaction is the oxidation of water. The expected halfreactions are Ag + (aq) + e Ag(s) H O(l) O (g) + H + (aq) + e 0.17 The conversion of grams of silver to coulombs required to deposit this amount of silver is 0.365 g Ag x 1 mol Ag 107.9 g x 1 mol e 1 mol Ag x 9.65 x 10 C 1 mol e = 36. C The time lapse, 16 min, equals 1.96 x 10 3 s. Thus, Current = charge time = 36. C 1.96 x 10 s =.5188 x 10 =.5 x 10 A 0.18 When the current flows for 1.85 x 10 s, the amount of charge is 0.0565 A x 1.85 x 10 s = 1.05 x 10 3 C Note that four moles of electrons are equivalent to one mol of O. Hence, 1.05 x 10 3 C x 1 mol e 9.65 x 10 C x 1 mol O mol e 3.00 g O x 1 mol O = 0.08665 = 0.0866 g O Answers to Concept Checks 0.1 No sustainable current would flow. The wire does not contain mobile positively and negatively charged species necessary to balance the accumulation of charges in each of the half cells.

788 CHAPTER 0 0. a. Standard reduction potentials are measured against some arbitrarily chosen standard reference halfreaction. Only differences in potentials can be measured. A voltaic cell made from H and I and corresponding solutions will have the same voltage regardless of the choice of the reference cell. If the I /I halfreaction is assigned a value of E = 0.00 V, the H /H + halfreaction must have a voltage of E = 0.5 V to keep the overall voltage the same. b. The voltage of a voltaic cell made from Cu and Zn and corresponding solutions will have the same measured voltage regardless of the choice of the reference halfreaction. c. The calculated voltage is 1.10 V and is the same either way. 0.3 a. Using the standard reduction potentials in Table 0.1, you see the voltage for this cell is positive, which suggests G is negative. b. In order to reduce E cell, change the concentrations in a way that increases the value of Q, where Q = + [Fe ] + [Cu ] For example: Fe(s) Fe + (1.10 M) Cu + (0.50 M) Cu(s) 0. Many of the ions contained in seawater have very high reduction potentials higher than Fe(s). This means spontaneous electrochemical reactions will occur with the Fe(s) causing the iron to form ions and go into solution while, at the same time, the ions in the sea will be reduced and plate out on the surface of the iron. Answers to Review Questions 0.1 A voltaic cell is an electrochemical cell in which a spontaneous reaction generates an electric current (energy). An electrolytic cell is an electrochemical cell that requires electrical current (energy) to drive a nonspontaneous reaction to the right. 0. In both the voltaic and electrolytic cells, the cathode is the electrode at which reduction occurs, and the anode is the electrode at which oxidation occurs. 0.3 The SI unit of electrical potential is the volt (V). 0. The faraday (F) is the magnitude of charge on one mole of electrons; it equals 9.65 x 10 C, or 9.65 x 10 J/V.

ELECTROCHEMISTRY 789 0.5 It is necessary to measure the voltage of a voltaic cell when no current is flowing because the cell voltage exhibits its maximum value only when no current flows. Even if the current flows just for the time of measurement, the voltage drops enough so that what is measured is significantly less than the maximum. 0.6 Standard electrode potentials are defined relative to a standard electrode potential of zero volts (0.00, 0.000 V, etc.) for the H + /H (g) electrode. Because the cell emf is measured using the hydrogen electrode at standard conditions and a second electrode at standard conditions, the cell emf equals the E of the halfreaction at the second electrode. 0.7 The SI unit of energy equals joules equals coulombs x volts. 0.8 The mathematical relationships are as follows: G = nfe cell G = RT ln K Combining these two equations gives ln K = o nfe cell RT 0.9 The first step in the corrosion of iron is Fe(s) + O (g) + H O(l) OH + Fe + The Nernst equation for this reaction is E cell = E cell 0.059 log [OH ] [Fe + ] If the ph increases, the [OH ] increases, and thus E cell becomes more negative (this predicts the reaction becomes less spontaneous). If the ph decreases, the [OH ] decreases, and thus E cell becomes more positive (this predicts the reaction becomes more spontaneous). 0.10 The zinccarbon cell has a zinc can as the anode; the cathode is a graphite rod surrounded by a paste of manganese dioxide and carbon black. Around this is a second paste of ammonium and zinc chlorides. The electrode reactions involve oxidation of zinc metal to zinc(ii) ion and reduction of MnO (s) to Mn O 3 (s) at the cathode. The lead storage battery consists of a spongy lead anode and a lead dioxide cathode, both immersed in aqueous sulfuric acid. At the anode, the lead is oxidized to lead sulfate; at the cathode, lead dioxide is reduced to lead sulfate.

790 CHAPTER 0 0.11 A fuel cell is essentially a battery that does not use up its electrodes. Instead, it operates with a continuous supply of reactants (fuel). An example is the hydrogenoxygen fuel cell in which oxygen is reduced at one electrode to the hydroxide ion, and hydrogen is oxidized at the other electrode to water (H in the +1 oxidation state). Such a cell produces electrical energy in a spacecraft for long periods of time. 0.1 During the rusting of iron, one end of a drop of water exposed to air acts as one electrode of a voltaic cell; at this electrode, an oxygen molecule is reduced by four electrons to four hydroxide ions. Oxidation of metallic iron to iron(ii) ion at the center of the drop of water supplies the electrons, and the center serves as the other electrode of the voltaic cell. Thus, electrons flow from the center of the drop through the iron to the end of the drop. 0.13 When iron or steel is connected to an active metal such as zinc, a voltaic cell is formed with zinc as the anode and iron as the cathode. Any type of moisture forms the electrolyte solution, and the zinc metal is then oxidized to zinc(ii) ion in preference to the oxidation of iron metal. Oxygen is reduced at the cathode to hydroxide ions. If iron or steel is exposed to oxygen while connected to a less active metal such as tin, a voltaic cell is formed with iron as the anode and tin as the cathode, and iron is oxidized to iron(ii) ion rather than tin being oxidized to tin(ii) ion. Thus, exposed iron corrodes rapidly in a tin can. Fortunately, as long as the iron is covered by the tin, it cannot corrode. 0.1 The addition of an ionic species such as strongly ionized sulfuric acid facilitates the passage of current through the solution. 0.15 Sodium metal can be prepared by electrolysis of molten sodium chloride. 0.16 The anode reaction in the electrolysis of molten potassium hydroxide is OH O (g) + H O(g) + e 0.17 The reason different products are obtained is that water instead of Na + is reduced at the cathode during the electrolysis of aqueous NaCl. This is because water has a more positive E (smaller decomposition voltage). At the anode, water instead of chloride ion is oxidized because water has a less positive E (smaller decomposition voltage). 0.18 The Nernst equation for the electrode reaction of Cl (aq) Cl (g) + e is E = 1.36 V 0.059 1 atm log [Cl ] = 1.36 V + 0.059 log [Cl ] This equation implies that E increases as [Cl ] increases. For a sufficiently large [Cl ], Cl will be more readily oxidized than the water solvent.

Answers to Conceptual Problems ELECTROCHEMISTRY 791 0.19 a. Since there is no species present to donate or accept electrons other than zinc, you would expect no change. b. Since there is no species present to donate or accept electrons other than copper, you would expect no change. c. According to the table of standard reduction potentials, the Cu + would undergo reduction, and the Zn would undergo oxidation. You would expect the Zn strip to dissolve as it becomes Zn +, the blue color of the solution to fade as the Cu + becomes Cu, and solid copper precipitate to form. d. According to the table of standard reduction potentials, since Zn + cannot oxidize Cu, you would expect no change. 0.0 You could construct the battery by hooking together, in series, five individual cells, each with E = 1.0 V. 0.1 The Zn is a sacrificial electrode that keeps the hull from undergoing oxidation by the dissolved ions in sea water. Zn works because it is more easily oxidized than Fe. 0. When an electrochemical reaction reaches equilibrium, E cell equals 0, which means no current will flow, and nothing will happen when you turn on the flashlight. This is typically what has occurred when you have a dead battery. 0.3 Since there is more zinc present, the oxidationreduction reactions in the battery will run for a longer period of time. This assumes zinc is the limiting reactant. 0. Any metal that has a more negative standard reduction potential (Mg, Al, etc.) could be used, keeping in mind that group IA metals that fall into this category are too reactive to be of practical use. 0.5 One possible combination using two metals is a cell constructed from a Zn + Zn cathode with a voltage of 0.76 V, and an Al Al 3+ anode with a voltage of 1.66 V. The appropriate halfreactions are Al(s) Al 3+ (aq) + 3 e E ox = 1.66 V Zn + (aq) + e Zn(s) E red = 0.76 V The overall balanced reaction and cell potential is Al(s) + 3 Zn + (aq) Al 3+ (aq) + 3 Zn(s) E cell = 0.90 V

79 CHAPTER 0 0.6 (parts a to f) The completed diagram with all parts labeled is shown below. Al (anode) e.6 V salt bridge Cl K + e Ag (cathode) + Al Al 3+ + 3e oxidation Al 3+ Ag + d. The cell emf is determined as follows. Ag + + e Ag reduction Al 3+ (aq) + 3 e Al(s) E red = 1.66 V Ag + (aq) e Ag(s) E red = 0.80 V E cell = E cathode E anode = 0.80 V (1.66 V) =.6 V g. The species undergoing oxidation is Al(s), and the species undergoing reduction is Ag + (aq). h. The balanced overall reaction is Al(s) + 3 Ag + (aq) Al 3+ (aq) + 3 Ag(s) 0.7 The effect of the various changes on the intensity of the light can be determined using the Nernst equation, E cell = E cell 0.059 log Q n For this reaction, E = 1.10 V, n =, and Q is the reaction quotient. Q = + [Zn ] + [Cu ] If Q > 1, log Q is positive, and the net result will be a decrease in cell potential and a decrease in the intensity of light. If Q < 1, log Q is negative, and the net result will be an increase in cell potential and an increase in the intensity of light. (continued)

ELECTROCHEMISTRY 793 At the start, both [Zn + ] and [Cu + ] are 1.0 M, so Q = 1, and log Q = 0, so E cell = E cell. a. If more CuSO (s) is dissolved in the CuSO solution, [Cu + ] would be greater than one, so Q < 1. The effect is to increase the intensity of light. b. If more Zn(NO 3 ) (s) is dissolved in the Zn(NO 3 ) solution, [Zn + ] would be greater than one, so Q > 1. The effect is to decrease the intensity of light. c. If H O is added to the CuSO solution, [Cu + ] would be less than one, so Q > 1. The effect is to decrease the intensity of light. d. If the salt bridge is removed, the circuit would not be complete, and the cell potential would be zero. No current would flow. 0.8 The most important consideration in battery design is the spontaneity of the cell reaction since this determines the cell voltage. The halfreactions for the reduction of oxygen gas given in Appendix I are: O + H O + e OH E = +0.0 V O + H + + e H O E = +1.3 V These represent the reduction of oxygen under basic and under acidic conditions. Thus, in basic solution, we would require an oxidation halfreaction (anode halfreaction) with a potential greater than 0.0 V to obtain a spontaneous reaction. The density of the reducing agent is the next most important consideration. Hydrogen, the element with the lowest density, should certainly be considered. It does have some drawbacks, however. Because it is a gas, some method of storage needs to be developed. Liquid storage and metal hydride storage have been investigated. Storage of liquid hydrogen presents problems of safety and weight. Storage of hydrogen as a hydride is used in nickelhydride cells presently available for portable computers, cellular phones, etc. (These batteries are rechargeable, but do not use oxygen.) The metal hydride obviously adds weight to the battery. Other elements you might consider are Na, Li, Al, K, Ca, and Zn, which have favorable powertomass ratios. Lithium and sodium might present some disposal problems since they are very reactive metals. Batteries using aluminum with atmospheric oxygen are available.

79 CHAPTER 0 Solutions to Practice Problems Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiplestep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 0.9 In balancing oxidationreduction reactions in acid, the four steps in the text will be followed. For part a, each step is shown. For the other parts, only a summary is shown. a. Assign oxidation numbers to the skeleton equation (Step 1). +6 +3 +3 + Cr O 7 + C O Cr 3+ + CO Separate into two incomplete halfreactions (Step ). Note that carbon is oxidized (increases in oxidation number), and chromium is reduced (decreases in oxidation number). C O Cr O 7 CO Cr 3+ Balance each halfreaction separately. The oxidation halfreaction is not balanced in C, so place a two in front of CO (Step 3a). Finally, add two electrons to the right side to balance the charge (Step 3d). The balanced oxidation halfreaction is C O CO + e The reduction halfreaction is not balanced in Cr, so place a two in front of Cr 3+ (Step 3a). Add seven H O to the right side to balance O atoms (Step 3b), and add fourteen H + ion to the left side to balance H atoms (step 3c). Finally, add six electrons to the left side to balance the charge (Step 3d). The balanced reduction halfreaction is Cr O 7 + 1H + + 6e Cr 3+ + 7H O Multiply the oxidation halfreaction by three so that, when added, the electrons cancel (Step a). 3C O Cr O 7 Cr O 7 6CO + 6e + 1H + + 6e Cr 3+ + 7H O + 3C O + 1H + + 6e Cr 3+ + 6CO + 7H O + 6e (continued)

ELECTROCHEMISTRY 795 The equation does not need to be simplified any further (Step b). The net ionic equation is Cr O 7 + 3C O + 1H + Cr 3+ + 6CO + 7H O b. The two balanced halfreactions are Cu Cu + + e (oxidation) NO 3 + H + + 3e NO + H O (reduction) Multiply the oxidation halfreaction by three and the reduction halfreaction by four, and then add together. Cancel the six electrons from each side. No further simplification is needed. The balanced equation is 3Cu + NO 3 + 8H + 3Cu + + NO + H O c. The two balanced halfreactions are HNO + H O NO 3 + 3H + + e (oxidation) MnO + H + + e Mn + + H O (reduction) Add the two halfreactions together and cancel the two electrons from each side. Also, cancel three H + and one H O from each side. The balanced equation is MnO + HNO + H + Mn + + NO 3 + H O d. The two balanced halfreactions are Mn + + H O MnO PbO + SO + 8H + + 5e (oxidation) + H + + e PbSO + H O (reduction) Multiply the oxidation halfreaction by two and the reduction halfreaction by five, and then add together. Cancel the ten electrons from each side. Also, cancel sixteen H + and eight H O from each side. The balanced equation is 5PbO + Mn + + 5SO + H + 5PbSO + MnO + H O (continued)

796 CHAPTER 0 e. The two balanced halfreactions are HNO + H O NO 3 Cr O 7 + 3H + + e (oxidation) + 1H + + 6e Cr 3+ + 7H O (reduction) Multiply the oxidation halfreaction by three, and then add together. Cancel the six electrons from each side. Also, cancel nine H + and three H O from each side. The balanced equation is 3HNO + Cr O 7 + 5H + Cr 3+ + 3NO 3 + H O 0.30 In balancing oxidationreduction reactions in acid, the four steps in the text will be followed. For part a, each step is shown. For the other parts only a summary is shown. a. Assign oxidation numbers to the skeleton equation (Step 1). + +5 +7 +3 Mn + + BiO 3 MnO + Bi 3+ Separate into two incomplete halfreactions (Step ). Note that manganese is oxidized (increases in oxidation number), and bismuth is reduced (decreases in oxidation number). Mn + MnO BiO 3 Bi 3+ Balance each halfreaction separately. For the oxidation halfreaction, add four H O to the left side to balance O atoms (Step 3b), and add eight H + ion to the right side to balance H atoms (step 3c). Finally, add five electrons to the right side to balance the charge (Step 3d). The balanced oxidation halfreaction is Mn + + H O MnO + 8H + + 5e For the reduction halfreaction, add three H O to the right side to balance O atoms (Step 3b), and add six H + ion to the left side to balance H atoms (step 3c). Finally, add two electrons to the left side to balance the charge (Step 3d). The balanced reduction halfreaction is BiO 3 + 6H + + e Bi 3+ + 3H O Multiply the oxidation halfreaction by two and the reduction halfreaction by five, so that, when added, the electrons cancel (Step a). (continued)

Mn + + 8H O MnO 5BiO 3 + 16H + + 10e + 30H + + 10e 5Bi 3+ + 15H O Mn + + 5BiO 3 + 30H + + 8H O + 10e MnO ELECTROCHEMISTRY 797 + 5Bi 3+ + 16H + + 15H O + 10e The equation can be further simplified by canceling sixteen H + and eight H O from each side (Step b). The net ionic equation is Mn + + 5BiO 3 + 1H + MnO + 5Bi 3+ + 7H O b. The two balanced halfreactions are I + 3H O IO 3 Cr O 7 + 6H + + 6e (oxidation) + 1H + + 6e Cr 3+ + 7H O (reduction) Add the two halfreactions together, and cancel the six electrons from each side. Also, cancel six H + and three H O from each side. The balanced equation is Cr O 7 + I + 8H + Cr 3+ + IO 3 + H O c. The two balanced halfreactions are H SO 3 + H O SO MnO + H + + e (oxidation) + 8H + + 5e Mn + + H O (reduction) Multiply the oxidation halfreaction by five and the reduction halfreaction by two, and then add together. Cancel the ten electrons from each side. Also, cancel sixteen H + and five H O from each side. The balanced equation is MnO + 5H SO 3 Mn + + 5SO + H + + 3H O d. The two balanced halfreactions are Fe + Fe 3+ + e (oxidation) Cr O 7 + 1H + + 6e Cr 3+ + 7H O (reduction) Multiply the oxidation halfreaction by six, and then add together. Cancel the six electrons from each side. The balanced equation is Cr O 7 + 6Fe + + 1H + Cr 3+ + 6Fe 3+ + 7H O (continued)

798 CHAPTER 0 e. The two balanced halfreactions are As + 3H O H 3 AsO 3 + 3H + + 3e (oxidation) ClO 3 + 5H + + e HClO + H O (reduction) Multiply the oxidation halfreaction by four and the reduction halfreaction by three, and then add together. Cancel the twelve electrons from each side. Also, cancel twelve H + and six H O from each side. The balanced equation is As + 3ClO 3 + 3H + + 6H O H 3 AsO 3 + 3HClO 0.31 In balancing oxidationreduction reactions in basic solution, it will first be balanced as if it were in acidic solution; then the extra two steps in the text will be followed. a. The two balanced halfreactions are Mn + + H O MnO + H + + e (oxidation) H O + H + + e H O (reduction) Add the two halfreactions together, and cancel the two electrons from each side. Also, cancel two H + and two H O from each side. The balanced equation in acidic solution is Mn + + H O MnO + H + Now add two OH to each side (Step 5). Simplify by combining the H + and OH to give H O. No further simplification is required (Step 6). The balanced equation in basic solution is Mn + + H O + OH MnO + H O b. The two balanced halfreactions are NO MnO + H O NO 3 + H + + e (oxidation) + H + + 3e MnO + H O (reduction) Multiply the oxidation halfreaction by three and the reduction halfreaction by two, and then add together. Cancel the six electrons from each side. Also, cancel twelve H + and six H O from each side. The balanced equation in acidic solution is MnO + 3NO + H + MnO + 3NO 3 + H O Now add two OH to each side. Simplify by combining the H + and OH to give H O. Then cancel one H O from each side. The balanced equation in basic solution is MnO + 3NO + H O MnO + 3NO 3 + OH (continued)

ELECTROCHEMISTRY 799 c. The two balanced halfreactions are Mn + + H O MnO + H + + e (oxidation) ClO 3 + H + + e ClO + H O (reduction) Multiply the reduction halfreaction by two and then add together. Cancel the two electrons from each side. Also, cancel four H + and two H O from each side. The balanced equation in acidic solution is Mn + + ClO 3 MnO + ClO Since there are no H + on either side, no further simplification is needed. The balanced equation in basic solution is identical to the balanced equation in acidic solution. d. The two balanced halfreactions are NO + H O NO 3 MnO + H + + e (oxidation) + H + + 3e MnO + H O (reduction) Multiply the oxidation halfreaction by three and then add together. Cancel the three electrons from each side. Also, cancel four H + and two H O from each side. The balanced equation in acidic solution is MnO + 3NO + H O MnO + 3NO 3 + H + Now add two OH to each side. Simplify by combining the H + and OH to give H O. Then cancel one H O from each side. The balanced equation in basic solution is MnO + 3NO + OH MnO + 3NO 3 + H O e. The two balanced halfreactions are Cl + 6H O ClO 3 + 1H + + 10e (oxidation) Cl + e Cl (reduction) Multiply the reduction halfreaction by five and then add together. Cancel the ten electrons from each side. Also, divide all the coefficients by two. The balanced equation in acidic solution is 3Cl + 3H O 5Cl + ClO 3 + 6H + Now add six OH to each side. Simplify by combining the H + and OH to give H O. Then cancel three H O from each side. The balanced equation in basic solution is 3Cl + 6OH 5Cl + ClO 3 + 3H O

800 CHAPTER 0 0.3 In balancing oxidationreduction reactions in basic solution, it will first be balanced as if it were in acidic solution, then the extra two steps in the text will be followed. a. The two balanced halfreactions are Cr(OH) CrO + H + + 3e (oxidation) H O + H + + e H O (reduction) Multiply the oxidation halfreaction by two and the reduction halfreaction by three, and then add together. Cancel the six electrons from each side. Also, cancel six H + from each side. The balanced equation in acidic solution is Cr(OH) + 3H O CrO + H + + 6H O Now add two OH to each side. Simplify by combining the H + and OH to give H O. Then combine the H O on the right side into one term. The balanced equation in basic solution is Cr(OH) + 3H O + OH CrO + 8H O b. The two balanced halfreactions are Br + 3H O BrO 3 MnO + 6H + + 6e (oxidation) + H + + 3e MnO + H O (reduction) Multiply the reduction halfreaction by two, and then add together. Cancel the six electrons from each side. Also, cancel six H + and three H O from each side. The balanced equation in acidic solution is MnO + Br + H + MnO + BrO 3 + H O Now add two OH to each side. Simplify by combining the H + and OH to give H O. Then cancel one H O on each side. The balanced equation in basic solution is MnO + Br + H O MnO + BrO 3 + OH c. The two balanced halfreactions are Co + + 3H O Co(OH) 3 + 3H + + e (oxidation) H O + H + + e H O (reduction) Multiply the oxidation halfreaction by two, and then add together. Cancel the two electrons from each side. Also, cancel two H + and two H O from each side. (continued)

ELECTROCHEMISTRY 801 The balanced equation in acidic solution is Co + + H O + H O Co(OH) 3 + H + Now add four OH to each side. Simplify by combining the H + and OH to give H O. Then cancel four H O on each side. The balanced equation in basic solution is Co + + H O + OH Co(OH) 3 d. The two balanced halfreactions are Pb(OH) PbO + H O + e (oxidation) ClO + H + + e Cl + H O (reduction) Add the two halfreactions together, and cancel the two electrons from each side. The balanced equation in acidic solution is Pb(OH) + ClO + H + PbO + Cl + 3H O Now add two OH to each side. Simplify by combining the H + and OH to give H O. Then cancel two H O on each side. The balanced equation in basic solution is Pb(OH) + ClO PbO + Cl + OH + H O e. The two balanced halfreactions are Zn + H O Zn(OH) NO 3 + H + + e (oxidation) + 9H + + 8e NH 3 + 3H O (reduction) Multiply the oxidation halfreaction by four, and then add together. Cancel the eight electrons from each side. Also, cancel nine H + and three H O from each side. The balanced equation in acidic solution is Zn + NO 3 + 13H O Zn(OH) + NH 3 + 7H + Now add seven OH to each side. Simplify by combining the H + and OH to give H O. Then cancel seven H O on each side. The balanced equation in basic solution is Zn + NO 3 + 7OH + 6H O NH 3 + Zn(OH)

80 CHAPTER 0 0.33 a. The two balanced halfreactions are 8H S S 8 + 16H + + 16e (oxidation) NO 3 + H + + e NO + H O (reduction) Multiply the reduction halfreaction by sixteen, and then add together. Cancel the sixteen electrons and sixteen H + from each side. The balanced equation is 8H S + 16NO 3 + 16H + S 8 + 16NO + 16H O b. The two balanced halfreactions are Cu Cu + + e (oxidation) NO 3 + H + + 3e NO + H O (reduction) Multiply the oxidation halfreaction by three and the reduction halfreaction by two, and then add together. Cancel the six electrons from each side. The balanced equation is NO 3 + 3Cu + 8H + NO + 3Cu + + H O c. The two balanced halfreactions are SO + H O SO MnO + H + + e (oxidation) + 8H + + 5e Mn + + H O (reduction) Multiply the oxidation halfreaction by five and the reduction halfreaction by two, and then add together. Cancel the ten electrons from each side. Also, cancel sixteen H + and eight H O from each side. The balanced equation is MnO + 5SO + H O 5SO + Mn + + H + d. The two balanced halfreactions are Sn(OH) 3 + 3H O Sn(OH) 6 + 3H + + e (oxidation) Bi(OH) 3 + 3H + + 3e Bi + 3H O (reduction) Multiply the oxidation halfreaction by three and the reduction halfreaction by two, and then add together. Cancel the six electrons from each side. Also, cancel six H + and six H O from each side. The balanced equation in acidic solution is Bi(OH) 3 + 3Sn(OH) 3 + 3H O 3Sn(OH) 6 + Bi + 3H + Now add three OH to each side. Simplify by combining the H + and OH to give H O. Then cancel three H O on each side. The balanced equation in basic solution is Bi(OH) 3 + 3Sn(OH) 3 + 3OH 3Sn(OH) 6 + Bi

ELECTROCHEMISTRY 803 0.3 a. The two balanced halfreactions are 8H S S 8 + 16H + + 16e (oxidation) Hg + + e Hg (reduction) Multiply the reduction halfreaction by eight, and then add together. Cancel the sixteen electrons from each side. The balanced equation in acidic solution is 8Hg + + 8H S 16Hg + S 8 + 16H + b. The two balanced halfreactions are S + H O SO + 8H + + 8e (oxidation) I + e I (reduction) Multiply the reduction halfreaction by four, and then add together. Cancel the eight electrons from each side. The balanced equation in acidic solution is S + I + H O 8I + SO + 8H + Now add eight OH to each side. Simplify by combining the H + and OH to give H O. Then cancel four H O on each side. The balanced equation in basic solution is S + I + 8OH 8I + SO + H O c. The two balanced halfreactions are Al + H O Al(OH) NO 3 + H + + 3e (oxidation) + 9H + + 8e NH 3 + 3H O (reduction) Multiply the oxidation halfreaction by eight and the reduction halfreaction by three, and then add together. Cancel the twenty four electrons from each side. Also, cancel twenty seven H + and nine H O from each side. The balanced equation in acidic solution is 8Al + 3NO 3 + 3H O 8Al(OH) + 3NH 3 + 5H + Now add five OH to each side. Simplify by combining the H + and OH to give H O. Then cancel five H O on each side. The balanced equation in basic solution is 8Al + 3NO 3 + 5OH + 18H O 8Al(OH) + 3NH 3 (continued)

80 CHAPTER 0 d. The two balanced halfreactions are C O MnO CO + e (oxidation) + H + + 3e MnO + H O (reduction) Multiply the oxidation halfreaction by three and the reduction halfreaction by two, and then add together. Cancel the six electrons from each side. The balanced equation in acidic solution is MnO + 3C O + 8H + MnO + 6CO + H O Now add eight OH to each side. Simplify by combining the H + and OH to give H O. Then cancel four H O on each side. The balanced equation in basic solution is MnO + 3 C O + H O MnO + 6CO + 8OH 0.35 a. The two balanced halfreactions are I + 3H O IO 3 MnO + 6H + + 6e (oxidation) + H + + 3e MnO + H O (reduction) Multiply the reduction halfreaction by two, and then add together. Cancel the six electrons from each side. Also, cancel six H + and three H O from each side. The balanced equation in acidic solution is MnO + I + H + MnO + IO 3 + H O Now add two OH to each side. Simplify by combining the H + and OH to give H O. Then cancel one H O on each side. The balanced equation in basic solution is MnO + I + H O MnO + IO 3 + OH b. The two balanced halfreactions are Cl Cl + e (oxidation) Cr O 7 + 1H + + 6e Cr 3+ + 7H O (reduction) Multiply the oxidation halfreaction by three, and then add together. Cancel the six electrons from each side. The balanced equation is Cr O 7 + 6Cl + 1H + Cr 3+ + 3Cl + 7H O (continued)

c. The two balanced halfreactions are S 8 + 16H O 8SO + 3H + + 3e (oxidation) NO 3 + H + + 3e NO + H O (reduction) ELECTROCHEMISTRY 805 Multiply the oxidation halfreaction by three and the reduction halfreaction by thirty two, and then add together. Cancel the ninety six electrons from each side. Also, cancel ninety six H + and forty eight H O from each side. The balanced equation is 3S 8 + 3NO 3 + 3H + 3NO + SO + 16H O d. The two balanced halfreactions are H O O + H + + e (oxidation) MnO + H + + 3e MnO + H O (reduction) Multiply the oxidation halfreaction by three and the reduction halfreaction by two, and then add together. Cancel the six electrons from each side. Also, cancel six H + from each side. The balanced equation in acidic solution is 3H O + MnO + H + 3O + MnO + H O Now add two OH to each side. Simplify by combining the H + and OH to give H O. Then cancel two H O on each side. The balanced equation in basic solution is 3H O + MnO 3O + MnO + H O + OH e. The two balanced halfreactions are Zn Zn + + e (oxidation) NO 3 + 1H + + 10e N + 6H O (reduction) Multiply the oxidation halfreaction by five, and then add together. Cancel the ten electrons from each side. The balanced equation is 5Zn + NO 3 + 1H + N + 5Zn + + 6H O 0.36 a. The two balanced halfreactions are H O O + H + + e (oxidation) Cr O 7 + 1H + + 6e Cr 3+ + 7H O (reduction) Multiply the oxidation halfreaction by three, and then add together. Cancel the six electrons from each side. Also, cancel six H + from each side. The balanced equation is Cr O 7 + 3H O + 8H + Cr 3+ + 3O + 7H O (continued)

806 CHAPTER 0 b. The two balanced halfreactions are CN + H O CNO + H + + e (oxidation) MnO + H + + 3e MnO + H O (reduction) Multiply the oxidation halfreaction by three and the reduction halfreaction by two, and then add together. Cancel the six electrons from each side. Also, cancel six H + and three H O from each side. The balanced equation in acidic solution is 3CN + MnO + H + 3CNO + MnO + H O Now add two OH to each side. Simplify by combining the H + and OH to give H O. Then, cancel one H O on each side. The balanced equation in basic solution is 3CN + MnO + H O 3CNO + MnO + OH c. The two balanced halfreactions are Cr(OH) CrO + H + + 3e (oxidation) OCl + H + + e Cl + H O (reduction) Multiply the oxidation halfreaction by two and the reduction halfreaction by three, and then add together. Cancel the six electrons from each side. Also, cancel six H + from each side. The balanced equation in acidic solution is Cr(OH) + 3OCl CrO + 3Cl + 3H O + H + Now add two OH to each side. Simplify by combining the H + and OH to give H O. Then combine the H O on the right side into a single term. The balanced equation in basic solution is Cr(OH) + 3OCl + OH CrO + 3Cl + 5H O d. The two balanced halfreactions are SO + H O SO + H + + e (oxidation) Br + e Br (reduction) Add the two halfreactions together. Cancel the two electrons from each side. The balanced equation is Br + SO + H O Br + SO + H + (continued)

ELECTROCHEMISTRY 807 e. The two balanced halfreactions are 8CuS S 8 + 8Cu + + 16e (oxidation) NO 3 + H + + 3e NO + H O (reduction) Multiply the oxidation halfreaction by three and the reduction halfreaction by sixteen, and then add together. Cancel the forty eight electrons from each side. The balanced equation is CuS + 16NO 3 + 6H + Cu + +b 16NO + 3S 8 + 3H O 0.37 Sketch of the cell: Mg (anode) e salt bridge Ni (cathode) + Mg + Ni + Mg Mg + + e Ni + + e Ni 0.38 Sketch of the cell: Ni (anode) e salt bridge Cu (cathode) + Ni + Cu + Ni Ni + + e Cu + + e Cu

808 CHAPTER 0 0.39 Sketch of the cell: Zn (anode) e salt bridge Ag (cathode) + Zn + Ag + Zn Zn + + e Ag + + e Ag 0.0 Sketch of the cell: Fe (anode) e salt bridge Cu (cathode) + Fe + Cu + Fe Fe + + e Cu + + e Cu 0.1 The electrode halfreactions and the overall cell reaction are Anode: Zn(s) + OH (aq) Zn(OH) (s) + e Cathode: Ag O(s) + H O (l) + e Ag(s) + OH (aq) Overall: Zn(s) + Ag O(s) + H O(l) Zn(OH) (s) + Ag(s)

0. The electrode halfreactions and the overall cell reaction are Anode: Zn(s) + OH (aq) Zn(OH) (s) + e ELECTROCHEMISTRY 809 Cathode: HgO(s) + H O(l) + e Hg(l) + OH (aq) Overall: Zn(s) + HgO(s) + H O(l) Zn(OH) (s) + Hg(l) 0.3 Because of its less negative E, Pb + is reduced at the cathode and is written on the right; Ni(s) is oxidized at the anode and is written first, at the left, in the cell notation. The notation is Ni(s) Ni + (aq) Pb + (aq) Pb(s). 0. Because of its less negative E, H + is reduced at the cathode and is written on the right; Al(s) is oxidized at the anode and is written first, at the left, in the cell notation. The notation is Al(s) Al 3+ (aq) H + (aq) H (g) Pt. 0.5 Because of its less negative E, H + is reduced at the cathode and is written on the right; Ni(s) is oxidized at the anode and is written first, at the left, in the cell notation. The notation is Ni(s) Ni + (1 M) H + (1 M) H (g) Pt. 0.6 Because of its less negative E, Fe + is reduced at the cathode and is written on the right; Zn(s) is oxidized at the anode and is written first, at the left, in the cell notation. The notation is Zn(s) Zn + (0.0 M) Fe 3+ (0.30 M) Fe(s). 0.7 The Fe(s), on the left, is the reducing agent. The Ag +, on the right, is the oxidizing agent, gaining just one electron. Multiplying its halfreaction by two to equalize the numbers of electrons and writing both halfreactions gives the overall cell reaction: Ag + (aq) + e Fe(s) Fe + (aq) + e Ag(s) Fe(s) + Ag + (aq) Fe + (aq) + Ag(s) 0.8 The H (g), on the left, is the reducing agent. The Br (l), on the right, is the oxidizing agent. The halfreactions and the overall cell reaction are Br (l) + e H (g) H + (aq) + e Br (aq) H (g) + Br (l) H + (aq) + Br (aq)

810 CHAPTER 0 0.9 The halfcell reactions, the overall cell reaction, and the sketch are Cd(s) Cd + (aq) + e Ni + (aq) + e Ni(s) Cd(s) + Ni + (aq) Cd + (aq) + Ni(s) Cd (anode) e salt bridge Ni (cathode) + Cd + Ni + 0.50 The halfcell reactions, the overall cell reaction, and the sketch are 3 Zn(s) 3Zn + (aq) + 6e Cr 3+ (aq) + 6e Cr(s) 3Zn(s) + Cr 3+ (aq) 3Zn + (aq) + Cr(s) Zn (anode) e salt bridge Cr (cathode) + Zn + Cr 3+

ELECTROCHEMISTRY 811 0.51 The halfcell reactions are 3+ + Fe (aq) + e Fe (aq) + Zn(s) Zn (aq) + e n equals two, and the maximum work for the reaction as written is w = nfe = x 9.65 x 10 C x 0.7 V = 1. 89 x 10 C V max 5 cell 3 5 = 1. 389 x 10 J 3+ Because this is the work obtained by reduction of two mol of Fe, the work for one mol is 5 w max = (1.389 x 10 J)/ mol = 6. 95 x 10 = 6.9 x 10 J = 69 kj 0.5 The halfcell reactions are Cl (aq) + e Cl (aq) + Zn(s) Zn (aq) + e n equals two, and the maximum work for the reaction as written is 5 w = nfe = x 9.65 x 10 C x 0.853 V = 1.66 x 10 C V max cell 5 = 1.66 x 10 J For 0.0 g of zinc, the maximum work is 0.0 g Zn x 5 1 mol Zn 65.39 g Zn x 1.66 x 10 J 1 mol Zn = 5.03 x 10 = = 5.03 x 10 J 0.53 The halfcell reactions are Ag + (aq) + e Ag(s) Ni(s) Ni + (aq) + e n equals two, and the maximum work for the reaction as written is w max = nfe cell = x 9.65 x 10 C x 0.97 V = 1.87 x 10 5 C V 5 = 1.87 x 10 J (continued)

81 CHAPTER 0 For 30.0 g of nickel, the maximum work is 30.0 g Ni x 1molNi 58.69gNi x 5 1.87 x 10 J 1molNi = 9.56 x 10 J = 96 kj 0.5 The halfcell reactions are 3O (g) + 1H + + 1e 6H O(l) Al(s) Al 3+ (aq) + 1e n equals twelve, and the maximum work for the reaction as written is w max = nfe cell = (1)(9.65 x 10 C)(1.15 V) = 1.331 x 10 6 C V = 1.331 x 10 6 J For 50.0 g of aluminum, the maximum work is 50.0 g x 1 mol Al 6.98 g Al x 6 1.331 x 10 J mol Al = 6.169 x 10 5 = = 6.17 x 10 5 J 0.55 The halfreactions and corresponding electrode potentials are as follows NO 3 (aq) + H + (aq) + 3e NO(g) + H O(l) 0.96 V O (g) + H + (aq) + e H O(l) 1.3 V MnO (aq) + 8H + (aq) + 5e Mn + (aq) + H O(l) 1.9 V The order by increasing oxidizing strength is NO 3 (aq), O (g), MnO (aq). 0.56 The halfreactions and corresponding electrode potentials are as follows I (s) + e I (aq) 0.5 V Ag + (aq) + e Ag(s) 0.80 V MnO (aq) + 8H + (aq) + 5e Mn + (aq) + H O(l) 1.9 V The order by increasing oxidizing strength is I (s), Ag + (aq), MnO (aq).

ELECTROCHEMISTRY 813 0.57 The halfreactions and corresponding electrode potentials are as follows Zn + (aq) + e Zn(s) 0.76 V Fe + (aq) + e Fe(s) 0.1 V Cu + (aq) + e Cu + (aq) 0.16 V Zn(s) is the strongest, and Cu + (aq) is the weakest. 0.58 The halfreactions and corresponding electrode potentials are as follows Sn + (aq) + e Sn + (aq) 0.15 V Cu + (aq) + e Cu(s) 0.3 V I (s) + e I (aq) 0.5 V I (s) is the strongest and Sn + is the weakest. 0.59 a. In this reaction, Sn + is the oxidizing agent on the left side; Fe 3+ is the oxidizing reagent on the right side. The corresponding standard electrode potentials are Sn + (aq) + e Sn + (aq) E = 0.15 V Fe 3+ (aq) + e Fe + (aq) E = 0.77 V The stronger oxidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so Fe 3+ is the stronger oxidizing agent. The reaction is nonspontaneous as written. b. In this reaction, MnO is the oxidizing agent on the left side; O is the oxidizing reagent on the right side. The corresponding standard electrode potentials are O (g) + H + (aq) + e H O(l) E = 1.3 V MnO (aq) + 8H + (aq) + 5e Mn + (aq) + H O(l) E = 1.9 V The stronger oxidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so MnO is the stronger oxidizing agent. The reaction is spontaneous as written.

81 CHAPTER 0 0.60 a. The reduction halfreactions and standard potentials are Fe 3+ (aq) + e Fe + (aq) E = 0.77 V Cr O 7 (aq) + 1H + (aq) + 6e Cr + (aq) + 7H O(l) E = 1.33 V The stronger oxidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so Cr O 7 is the stronger oxidizing agent. Thus, Cr O 7 will oxidize iron(ii) ion in acidic solution under standard conditions. b. The reduction halfreactions and standard electrode potentials are Ni + (aq) + e Ni(s) E = 0.3 V Cu + (aq) + e Cu(s) E = 0.3 V The stronger reducing agent is the one involved in the halfreaction with the smaller standard electrode potential, so Ni(s) is the stronger reducing agent. Thus, copper metal will not reduce Ni(II) ion spontaneously. 0.61 The reduction halfreactions and standard electrode potentials are Br (l) + e Br (aq) E = 1.07 V Cl (g) + e Cl (aq) E = 1.36 V F (g) + e F (aq) E =.87 V From these, you see that the order of increasing oxidizing strength is Br, Cl, F. Therefore, chlorine gas will oxidize Br but will not oxidize F. The balanced equation for the reaction is Cl (g) + Br (aq) Cl (aq) + Br (l) 0.6 The reduction halfreactions and standard electrode potentials are Br (l) + e Br (aq) E = 1.07 V Cr O 7 (aq) + 1H + (aq) + 6e Cr 3+ (aq) + 7H O(l) E = 1.33 V MnO (aq) + 8H + (aq) + 5e Mn + (aq) + H O(l) E = 1.9 V From these, you see that the order of increasing oxidizing strength is Br, Cr O 7, MnO. Thus, Cr O 7 will oxidize Br but will not oxidize Mn +. The balanced equation for the reaction is Cr O 7 (aq) + 6Br (aq) + 1H + (aq) Cr 3+ (aq) + 3Br (l) + 7H O(l)

ELECTROCHEMISTRY 815 0.63 The halfreactions and standard electrode potentials are Cr(s) Cr 3+ (aq) + 3e E Cr = 0.7 V Hg + (aq) + e Hg(l) E Hg = 0.80 V Obtain the cell emf by adding the halfcell potentials. E cell = E Hg E Cr = 0.80 V + 0.7 V = 1.5 V 0.6 The halfreactions and standard electrode potentials are Sn(s) Sn + (aq) + e E Sn = 0.1 V Cu + (aq) + e Cu(s) E Cu = 0.3 V Obtain the cell emf by adding the halfcell potentials. E cell = E Cu E Sn = 0.3 V + 0.1 V = 0.8 V 0.65 The halfreactions and standard electrode potentials are Cr(s) Cr 3+ (aq) + 3e E Cr = 0.7 V I (s) + e I (aq) E I = 0.5 V Obtain the cell emf by adding the halfcell potentials. E cell = E Cr E I = 0.5 V + 0.7 V = 1.8 V 0.66 The halfreactions and standard electrode potentials are Al(s) Al 3+ (aq) + 3e E Al = 1.66 V Hg + (aq) + e Hg(l) E Hg = 0.80 V Obtain the cell emf by adding the halfcell potentials. E cell = E Hg E Al = 0.80 V + 1.66 V =.6 V

816 CHAPTER 0 0.67 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: 3Cu(s) 3Cu + (aq) + 6e E = 0.3 V NO 3 (aq) + 8H + (aq) + 6e NO(g) + H O(l) E = 0.96 V 3Cu(s) + NO 3 (aq) + 8H + (aq) 3Cu + (aq) + NO(g) + H O(l) E cell = 0.6 V Note that each halfreaction involves six electrons, hence n = 6. Therefore, G = nfe cell = 6 x 9.65 x 10 C x 0.6 V = 3.58 x 10 5 J = 3.6 x 10 5 J Thus, the standard freeenergy change is 3.6 x 10 kj. 0.68 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Al(s) Al 3+ (aq) + 1e E = 1.66 V 3O (g) + 1H + (aq) + 1e 6H O(l) E = 1.3 V Al(s) + 3O (g) + 1H + (aq) Al 3+ (aq) + 6H O(l) E cell =.89 V Note that each halfreaction involves twelve electrons, hence n = 1. Therefore, G = nfe cell = 1 x 9.65 x 10 C x.89 V = 3.36 x 10 6 J = 3.35 x 10 6 J Thus, the standard freeenergy change is 3.35 x 10 3 kj. 0.69 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: I (aq) I (s) + e E = 0.5 V Cl (g) + e Cl (aq) E = 1.36 V I (aq) + Cl (g) I (s) + Cl (aq) E cell = 0.8 V Note that each halfreaction involves two electrons, hence n =. Therefore, G = nfe cell = x 9.65 x 10 C x 0.8 V = 1.58 x 10 5 J = 1.6 x 10 5 J Thus, the standard freeenergy change is 1.6 x 10 kj.

ELECTROCHEMISTRY 817 0.70 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Na(s) Na + (aq) + e E =.71 V 1/Br (g) + e Br (aq) E = 1.07 V Na(s) + 1/Br (g) Na + (aq) + Br (aq) E cell = 3.78 V Note that each halfreaction involves one electron, hence n = 1. Therefore, G = nfe cell = 1 x 9.65 x 10 C x 3.78 V = 3.67 x 10 5 J = 3.65 x 10 5 J Thus, the standard freeenergy change is 365 kj. 0.71 Write the equation with G f 's beneath each substance. Mg(s) + Ag + (aq) Mg + (aq) + Ag(s) G f : 0 x 77.111 56.01 0 kj Hence, G = Σn G f (products) Σm G f (reactants) = [56.01 x 77.111] kj = 610.3 kj = 6.103 x 10 5 J Obtain n by splitting the reaction into halfreactions. Mg(s) Mg + (aq) + e Ag + (aq) + e Ag(s) Each halfreaction involves two electrons, so n =. Therefore, G = nfe cell 6.103 x 10 5 J = x 9.65 x 10 C x E cell Rearrange and solve for E cell. Recall that J = C V. E cell = 5 6.103 x 10 J x 9.65 x 10 C = 3.1618 = 3.16 V

818 CHAPTER 0 0.7 Write the equation with G f 's beneath each substance. Al(s) + 3Cu + (aq) Al 3+ (aq) + 3Cu(s) G f : 0 3 x 6.98 x (81.) 0 kj Hence, G = Σn G f (products) Σm G f (reactants) = [ x 81. 3 x 6.98] kj = 1157.3 kj = 1.1573 x 10 6 J Obtain n by splitting the reaction into halfreactions. Al(s) Al 3+ (aq) + 3e 3Cu + (aq) + 3e 3Cu(s) Each halfreaction involves three electrons, so n = 3. Therefore, G = nfe cell 1.1573 x 10 6 J = 3 x 9.65 x 10 C x E cell Rearrange and solve for E cell. Recall that J = C V. E cell = 6 1.1573 x 10 J 3 x 9.65 x 10 C = 1.9988 =.00 V 0.73 Write the equation with G f 's beneath each substance: PbO (s) + HSO (aq) + H + (aq) + Pb(s) PbSO (s) + H O(l) G f : 189. (75.87) 0 0 (811.) (37.19) kj Hence, G = [ x 811. + x 37.19 189. x 75.87] kj = 01.9 kj =.019 x 10 5 J Obtain n by splitting the reaction into halfreactions. Pb(s) + HSO (aq) PbSO (s) + H + (aq) + e PbO (s) + HSO (aq) + 3H + (aq) + e PbSO (s) + H O(l) (continued)

Each halfreaction involves two electrons, so n =. Therefore, ELECTROCHEMISTRY 819 G = nfe cell.019 x 10 5 J = x 9.65 x 10 C x E cell Rearrange and solve for E cell. Recall that J = C V. E cell = 5.019 x 10 J x 9.65 x 10 C =.085 =.08 V 0.7 Write the equation with G f 's beneath each substance: 5H C O (aq) + MnO (aq) + 6H + (aq) 10CO (g) + 8H O(l) + Mn + (aq) G f : 5(698) (5.1) 0 10(39.) 8(37.1) (3) kj Hence, G = [10(39.) + 8(37.1) + (3) 5(698) (5.1)] kj = 196.6 kj = 1.966 x 10 6 J Obtain n by splitting the reaction into halfreactions. 5H C O (aq) 10CO (g) + 10H + (aq) + 10e MnO (aq) + 16H + (aq) + 10e Mn + (aq) + 8H O(l) Each halfreaction involves ten electrons, so n = 10. Therefore, G = nfe cell 1.966 x 10 6 J = 10 x 9.65 x 10 C x E cell Rearrange and solve for E cell. Recall that J = C V. E cell = 6 1.966 x 10 J 10 x 9.65 x 10 C =.017 =.0 V 0.75 The halfreactions and standard electrode potentials are Cu(s) Cu + (aq) + e E Cu = 0.3 V Fe 3+ (aq) + e Fe + (aq) = 0.77 V E o 3+ Fe (continued)

80 CHAPTER 0 The standard emf for the cell is o E cell = E E = 0.77 V 0.3 V = 0.3 V 3+ Fe o Cu Note that n =. Substitute into the equation relating E and K. Note that K = K c. 0.3 V = 0.059 Solving for K c, you get log K c = 1.5 log K c Take the antilog of both sides: K c = antilog (1.5) = 3. x 10 1 = 10 1 0.76 The halfreactions and standard electrode potentials are Hg(l) Hg + (aq) + e E Hg = 0.80 V Sn + (aq) + e Sn(s) = 0.1 V The standard emf for the cell is o E o + Sn E cell = E E = 0.1 V 0.80 V = 0.9 V + Sn o Hg Note that n =. Substitute into the equation relating E and K. Note that K = K c. 0.9 V = 0.059 Solving for K c, you get log K c = 31.756 log K c Take the antilog of both sides: K c = antilog (31.756) = 1.75 x 10 3 = 10 3

0.77 The halfreactions and standard electrode potentials are Cu + (aq) Cu + (aq) + e E = 0.16 V Cu + (aq) + e Cu(s) E = 0.5 V The standard emf for the cell is E cell = 0.5 V 0.16 V = 0.36 V ELECTROCHEMISTRY 81 Note that n equals 1. Substitute into the equation relating E and K. Note that K equals K c. 0.36 V = 0.059 1 Solving for K c, you get log K c = 6.081 log K c Take the antilog of both sides: K c = antilog (6.081) = 1. x 10 6 = 1 x 10 6 0.78 The halfreactions and standard electrode potentials are ClO (aq) + OH (aq) ClO 3 (aq) + H O(l) + e E = 0.35 V ClO (aq) + H O(l) + e ClO 3 (aq) + OH (aq) E = 0.17 V The standard emf for the cell is E cell = 0.17 V 0.35 V = 0.18 V Note that n equals two. Substitute into the equation relating E and K. Note that K equals K c. 0.18 V = 0.059 Solving for K c, you get log K c = 6.081 log K c Take the antilog of both sides: K c = antilog (6.081) = 8.9 x 10 7 = 8 x 10 7

8 CHAPTER 0 0.79 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Cr(s) Cr 3+ (aq) + 6e E = 0.7 V 3Ni + (aq) + 6e 3Ni(s) E = 0.3 V Cr(s) + 3Ni + (aq) Cr 3+ (aq) + 3Ni(s) E cell = 0.51 V Note that n equals six. The reaction quotient is Q = 3+ [Cr ] + 3 [Ni ] = 3 (1.0 x 10 ) 3 (1.5) =.96 x 10 7 The standard emf is 0.51 V, so the Nernst equation becomes E cell = E cell 0.059 n = 0.51 0.059 6 log Q log (.96 x 10 7 ) = 0.51 (0.061) = 0.57 = 0.57 V 0.80 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Ni(s) Ni + (aq) + e E = 0.3 V Sn + (aq) + e Sn(s) E = 0.1 V Ni(s) + Sn + (aq) Ni + (aq) + Sn(s) E cell = 0.09 V Note that n equals two. The reaction quotient is Q = + [Ni ] + [Sn ] = 1.0 1.0 x 10 = 1.00 x 10 The standard emf is 0.09 V, so the Nernst equation becomes E cell = E cell 0.059 n = 0.09 0.059 log Q log (1.00 x 10 ) = 0.09 V (0.118 V) = 0.08 = 0.03 V

ELECTROCHEMISTRY 83 0.81 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: 10Br (aq) 5Br (l) + 10e E = 1.07 V MnO (aq) + 16H + (aq) + 10e Mn + (aq) + 8H O(l) E = 1.9 V MnO (aq) + 10Br (aq) + 16H + (aq) Mn + (aq) + 5Br (l) + 8H O(l) E cell = 0. V Note that n equals ten. The reaction quotient is Q = + [Mn ] 10 + 16 [MnO ] [Br ] [H ] = (0.15) 10 (0.010) (0.010) (1.0) 16 =.5 x 10 The standard emf is 0.09 V, so the Nernst equation becomes E cell = E cell 0.059 n log Q = 0. 0.059 10 = 0. (0.133) = 0.876 = 0.9 V log (.5 x 10 ) 0.8 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: 6I (aq) 3I (l) + 6e E = 0.5 V Cr O 7 (aq) + 1H + (aq) + 6e Cr 3+ (aq) + 7H O(l) E = 1.33 V Cr O 7 (aq) + 6I (aq) + 1H + (aq) Cr 3+ (aq) + 3I (s) + 7H O(l) E cell = 0.79 V Note that n equals six. The reaction quotient is Q = 3+ [Cr ] 6 + 1 [CrO 7 ][I ] [H ] = (0.0) 6 1 (0.00)(0.015) (0.50) = 1.1507 x 10 16 The standard emf is 0.79 V, so the Nernst equation becomes E cell = E cell 0.059 n log Q = 0.79 0.059 6 = 0.79 (0.158) = 0.6315 = 0.63 V log (1.1507 x 10 16 )

8 CHAPTER 0 0.83 The overall reaction is Cd(s) + Ni + (aq) Cd + (aq) + Ni(s) Note that n equals two. The reaction quotient is Q = + [Cd ] + [Ni ] = + [Cd ] 1.0 = [Cd + ] The standard emf is 0.170 V, and the emf is 0.0 V, so the Nernst equation becomes E cell = E cell 0.059 n 0.0 = 0.170 0.059 Rearrange and solve for log Q log Q log Q log Q = 0.059 x (0.0 0.170) =.368 Take the antilog of both sides Q = [Cd + ] = antilog (.368) =.31 x 10 3 = 0.00 M The Cd + concentration is 0.00 M. 0.8 The overall reaction is Zn(s) + H + (aq) Zn + (aq) + H (g) Note that n equals two. The reaction quotient is Q = + [Zn ] x PH + [H ] = (1.0)(1.0) + [H ] = 1 + [H ] The standard emf is equal to the standard oxidation potential for Zn, which is 0.76 V. The emf is 0.75 V, so the Nernst equation becomes E cell = E cell 0.059 n log Q (continued)

ELECTROCHEMISTRY 85 0.75 = 0.76 0.059 Rearrange and solve for log Q log Q log Q = 0.059 x (0.75 0.76) = 9.68 Use the properties of logs to simplify log Q as follows: log Q = log Therefore 1 + [H ] = log [H + ] = log [H + ] = (log [H + ]) = x ph log Q = x ph = 9.68 ph = 9.68 =.81 =.8 0.85 a. The cathode reaction is Ca + (l) + e Ca(l). The anode reaction is S (l) S(l) + e. b. The cathode reaction is Cs + (l) + e Cs(l). The anode reaction is OH (l) O (g) + H O(g) + e. 0.86 a. The cathode reaction is Mg + (l) + e Mg(l). The anode reaction is Br (l) Br (l) + e. b. The cathode reaction is Ca + (l) + e Ca(l). The anode reaction is OH (l) O (g) + H O(g) + e. 0.87 a. The species you should consider for halfreactions are Na +, SO, and H O. The possible cathode reactions are Na + (aq) + e Na(s) E =.71 V H O(l) + e H (g) + OH (aq) E = 0.83 V Because the electrode potential for H O is larger (less negative), it is easier to reduce. (continued)

86 CHAPTER 0 The possible anode reactions are H O(l) O (g) + H + (aq) + e SO (aq) S O (aq) + e E = 1.3 V E =.01 V Because the electrode potential for H O is less negative, it is easier to oxidize. The expected halfreactions are H O(l) O (g) + H + (aq) + e E = 1.3 V H O(l) + e H (g) + OH (aq) E = 0.83 V The overall reaction is H O(l) H (g) + O (g) b. The species you should consider for halfreactions are K +, Br, and H O. The possible cathode reactions are K + (aq) + e K(s) E =.9 V H O(l) + e H (g) + OH (aq) E = 0.83 V Because the electrode potential for H O is larger (less negative), it is easier to reduce. The possible anode reactions are H O(l) O (g) + H + (aq) + e Br (aq) Br (l) + e E = 1.3 V E = 1.07 V Because the electrode potential for Br is less negative, it is easier to oxidize. The expected halfreactions are Br (aq) Br (l) + e E = 1.07 V H O(l) + e H (g) + OH (aq) E = 0.83 V The overall reaction is Br (aq) + H O(l) Br (l) + H (g) + OH

ELECTROCHEMISTRY 87 0.88 a. The species you should consider for halfreactions are Cu +, Cl, and H O. The possible cathode reactions are Cu + (aq) + e Cu(s) E = 0.3 V H O(l) + e H (g) + OH (aq) E = 0.83 V Because the electrode potential for Cu + is larger (less negative), it is easier to reduce. The possible anode reactions are H O(l) O (g) + H + (aq) + e Cl (aq) Cl (g) + e E = 1.3 V E = 1.36 V Because the electrode potential for H O is less negative, it is easier to oxidize. The expected halfreactions are H O(l) O (g) + H + (aq) + e E = 1.3 V Cu + (aq) + e Cu(s) E = 0.3 V The overall reaction is Cu + (aq) + H O(l) Cu(s) + H + (aq) + O (g) b. The species you should consider for halfreactions are Cu + and H O. The possible cathode reactions are Cu + (aq) + e Cu(s) E = 0.3 V H O(l) + e H (g) + OH (aq) E = 0.83 V Because the electrode potential for Cu + is larger (less negative), it is easier to reduce. The anode reaction is H O(l) O (g) + H + (aq) + e E = 1.3 V The expected halfreactions are H O(l) O (g) + H + (aq) + e E = 1.3 V Cu + (aq) + e Cu(s) E = 0.3 V The overall reaction is Cu + (aq) + H O(l) Cu(s) + H + (aq) + O (g)

88 CHAPTER 0 0.89 The conversion of grams of aluminum (3.61 kg = 3.61 x 10 3 g) to coulombs required to give this amount of aluminum is 3.61 x 10 3 g x 1molAl 6.98 g x 3mole 1molAl x 9.65 x 10 C 1mole = 3.873 x 10 7 = 3.87 x 10 7 C 0.90 The conversion of grams of chlorine (1.18 kg = 1.18 x 10 3 g) to coulombs required to produce this amount of chlorine is 1.18 x 10 3 1 mol Cl g x 70.91 g x mol e 1 mol Cl x 9.65 x 10 C 1mole = 3.11 x 10 6 = 3.1 x 10 6 C Using the current of 5.00 x 10 A, calculate the time in s: 1 mol Cl Time = 70.91 g 1 mol Cl = 70.91 g = 6.3 x 10 3 = 6. x 10 3 s 0.91 The conversion of coulombs to grams of lithium is 5.00 x 10 3 C x 1 mol e 9.65 x 10 C x 1 mol Li 1 mol e x 6.91 g Li 1 mol Li = 0.3596 = 0.360 g Li 0.9 The conversion of current and time into coulombs is (1 min = 1.36 x 10 s) C = amp x sec = 1.51 A x 1.36 x 10 s =.00 x 10 C The conversion of coulombs to grams of cadmium is.00 x 10 C x 1 mol e 9.65 x 10 C x 1 mol Cd mol e x 11. g Cd 1 mol Cd = 11.66 = 11.7 g Cd

ELECTROCHEMISTRY 89 Solutions to General Problems 0.93 a. The halfreactions and net ionic equation are 3Fe(s) 3Fe + (aq) + 6e NO 3 (aq) + 8H + (aq) + 6e NO(g) + H O(l) 3Fe(s) + NO 3 (aq) + 8H + (aq) 3Fe + (aq) + NO(g) + H O(l) b. The halfreactions and net ionic equation are 3Fe + (aq) 3Fe 3+ (aq) + 3e NO 3 (aq) + H + (aq) + 3e NO(g) + H O(l) 3Fe + (aq) + NO 3 (aq) + H + (aq) 3Fe 3+ (aq) + NO(g) + H O(l) c. Add the results from parts (a) and (b) 3Fe(s) + NO 3 (aq) + 8H + (aq) 3Fe + (aq) + NO(g) + H O(l) 3Fe + (aq) + NO 3 (aq) + H + (aq) 3Fe 3+ (aq) + NO(g) + H O(l) Fe(s) + NO 3 (aq) + H + (aq) Fe 3+ (aq) + NO(g) + H O(l) 0.9 a. The halfreactions and net ionic equation are 3Tl(s) 3Tl + (aq) + 3e NO 3 (aq) + H + (aq) + 3e NO(g) + H O(l) 3Tl(s) + NO 3 (aq) + H + (aq) 3Tl + (aq) + NO(g) + H O(l) b. The halfreactions and net ionic equation are 3Tl + (aq) 3Tl 3+ (aq) + 6e NO 3 (aq) + 8H + (aq) + 6e NO(g) + H O(l) 3Tl + (aq) + NO 3 (aq) + 8H + (aq) 3Tl + (aq) + NO(g) + H O(l) (continued)

830 CHAPTER 0 c. Add the results from parts (a) and (b) 3Tl(s) + NO 3 (aq) + H + (aq) 3Tl + (aq) + NO(g) + H O(l) 3Tl + (aq) + NO 3 (aq) + 8H + (aq) 3Tl + (aq) + NO(g) + H O(l) Tl(s) + NO 3 (aq) + H + (aq) Tl 3+ (aq) + NO(g) + H O(l) 0.95 a. The two balanced halfreactions are 8S S 8 + 16e (oxidation) MnO + H + + 3e MnO + H O (reduction) Multiply the oxidation halfreaction by three and the reduction halfreaction by sixteen, and then add together. Cancel the forty eight electrons from each side. The balanced equation in acidic solution is 16MnO + S + 6H + 16MnO + 3S 8 + 3H O Now add sixty four OH to each side. Simplify by combining the H + and OH to give H O. Then cancel thirty two H O on each side. The balanced equation in basic solution is 16MnO (aq) + S (aq) + 3H O(l) 16MnO (aq) + 3S 8 (aq) + 6OH (l) b. The two balanced halfreactions are HSO 3 IO 3 + H O SO + 3H + + e (oxidation) + 6H + + 6e I + 3H O (reduction) Multiply the oxidation halfreaction by three, and then add together. Cancel the six electrons from each side. Also, cancel six H + and three H O from each side. The balanced equation is IO 3 (aq) + 3HSO 3 (aq) I + 3SO (aq) + 3H + (aq) c. The two balanced halfreactions are Fe(OH) + H O Fe(OH) 3 + H + + e (oxidation) CrO + H + + 3e Cr(OH) (reduction) Multiply the oxidation halfreaction by three, and then add together. Cancel the three electrons and cancel three H + from each side. The balanced equation in acidic solution is CrO (aq) + 3Fe(OH) (s) + H + (aq) + 3H O(l) Cr(OH) (aq) + 3Fe(OH) 3 (s) (continued)

ELECTROCHEMISTRY 831 Now add one OH to each side. Simplify by combining the H + and OH to give H O. Then, combine the H Os on the left side into a single term. The balanced equation in basic solution is CrO (aq) + 3Fe(OH) (aq) + H O(l) Cr(OH) (aq) + 3Fe(OH) 3 (s) + OH (aq) d. The two balanced halfreactions are Cl + H O ClO + H + + e (oxidation) Cl + e Cl (reduction) Add the two halfreactions together. Cancel the two electrons from each side. Divide all the coefficients by a factor of two. The balanced equation in acidic solution is Cl + H O Cl + ClO + H + Now add two OH to each side. Simplify by combining the H + and OH to give H O. Then cancel one H O on each side. The balanced equation in basic solution is Cl (aq) + OH (aq) Cl (aq) + ClO (aq) + H O(l) 0.96 a. The two balanced halfreactions are 8H S S 8 + 16H + + 16e (oxidation) MnO + 8H + + 5e Mn + + H O (reduction) Multiply the oxidation halfreaction by five and the reduction halfreaction by sixteen, and then add together. Cancel the eighty electrons from each side. Also, cancel eighty H + from each side. The balanced equation is 16MnO (aq) + 0H S(aq) + 8H + (aq) 16Mn + (aq) + 5S 8 (aq) + 6H O(l) b. The two balanced halfreactions are Zn Zn + + e (oxidation) NO 3 + 10H + + 8e N O + 5H O (reduction) Multiply the oxidation halfreaction by four, and then add together. The balanced equation is NO 3 (aq) + Zn(s) + 10H + (aq) N O(g) + Zn + (aq) + 5H O(l) (continued)

83 CHAPTER 0 c. The two balanced halfreactions are MnO MnO MnO + e (oxidation) + H + + e MnO + H O (reduction) Multiply the oxidation halfreaction by two, and then add together. Cancel the two electrons from each side. Also, combine the MnO on the left into a single term. The balanced equation in acidic solution is 3MnO + H + MnO + MnO + H O Now add four OH to each side. Simplify by combining the H + and OH to give H O. Then cancel two H O on each side. The balanced equation in basic solution is 3MnO (aq) + H O(l) MnO (s) + MnO (aq) + OH (aq) d. The two balanced halfreactions are Br + 6H O BrO 3 + 1H + + 10e (oxidation) Br + e Br (reduction) Multiply the reduction halfreaction by five, and then add together. Cancel the ten electrons from each side. Also, combine the Br on the left into a single term. The balanced equation in acidic solution is 3Br + 3H O 5Br + BrO 3 + 6H + Now add six OH to each side. Simplify by combining the H + and OH to give H O. Then cancel three H O on each side. The balanced equation in basic solution is 3Br (aq) + 6OH (aq) 5Br (aq) + BrO 3 (aq) + 3H O(l) 0.97 This reaction takes place in basic solution. The skeleton equation is Fe(OH) (s) + O (g) Fe(OH) 3 (s) The two balanced halfreactions are Fe(OH) + H O Fe(OH) 3 + H + + e (oxidation) O + H + + e H O (reduction) Multiply the oxidation halfreaction by four, and then add together. Cancel the four electrons from each side. Also, cancel four H + and two H O from each side. The balanced equation in acidic or in basic solution is Fe(OH) (s) + O (g) + H O(l) Fe(OH) 3 (s)

0.98 This reaction takes place in basic solution. The skeleton equation is SnO (aq) + Bi 3+ (aq) SnO 3 (aq) + Bi(s) The two balanced halfreactions are SnO + H O SnO 3 + H + + e (oxidation) Bi 3+ + 3e Bi (reduction) ELECTROCHEMISTRY 833 Multiply the oxidation halfreaction by three and the reduction reaction by two, and then add together. Cancel the six electrons from each side. The balanced equation in acidic solution is Bi 3+ + 3SnO + 3H O Bi + 3SnO 3 + 6H + Now add six OH to each side. Simplify by combining the H + and OH to give H O. Then cancel three H O on each side. The balanced equation in basic solution is Bi 3+ (aq) + 3SnO (aq) + 6OH (aq) Bi(s) + 3SnO 3 (aq) + 3H O(l) 0.99 The cell notation is Ca(s) Ca + (aq) Cl (aq) Cl (g) Pt(s). The reactions are Anode: Ca(s) Ca + (aq) + e E =.76 V Cathode: Cl (g) + e Cl (aq) E = 1.36 V E cell = 1.36 V +.76 V =.1 V 0.100 The cell notation is Mg(s) Mg + (aq) Ag + (aq) Ag(s). The reactions are Anode: Mg(s) Mg + (aq) + e E =.38 V Cathode: Ag + (aq) + e Ag(s) E = 0.80 V E cell = 0.80 V +.38 V = 3.18 V 0.101 a. In this reaction, Fe 3+ is the oxidizing agent on the left side; Ni + is the oxidizing reagent on the right side. The corresponding standard electrode potentials are Ni + (aq) + e Ni(s) E = 0.3 V Fe 3+ (aq) + e Fe + (aq) E = 0.77 V The stronger oxidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so Fe 3+ is the stronger oxidizing agent. The oxidation of nickel by iron(iii) is a spontaneous reaction. (continued)

83 CHAPTER 0 b. In this reaction, Fe 3+ is the oxidizing agent on the left side; Sn + is the oxidizing reagent on the right side. The corresponding standard electrode potentials are Sn + (aq) + e Sn + (aq) E = 0.15 V Fe 3+ (aq) + e Fe + (aq) E = 0.77 V The stronger oxidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so Fe 3+ is the stronger oxidizing agent. The oxidation of tin(ii) by iron(iii) is a spontaneous reaction. 0.10 a. The halfreactions and the corresponding standard electrode potential values are Cl (g) + e Cl (aq) E = 1.36 V MnO (aq) + 8H + (aq) + 5e Mn + (aq) + H O(l) E = 1.9 V The stronger oxidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so MnO is the stronger oxidizing agent. The oxidation of chloride ion by permanganate ion is a spontaneous reaction. b. The halfreactions and the corresponding standard electrode potential values are Cr O 7 (aq) + 1H + (aq) + 6e Cr 3+ (aq) + 7H O(l) E = 1.33 V Cl (g) + e Cl (aq) E = 1.36 V The stronger oxidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so Cl is the stronger oxidizing agent. The oxidation of chloride ion by dichromate ion is not a spontaneous reaction. 0.103 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Pb(s) Pb + (aq) + e E = 0.13 V H + (aq) + e H (g) E = 0.00 V Pb(s) + H + (aq) Pb + (aq) + H (g) E cell = 0.13 V Note that n equals two. Next, use K sp to calculate [Pb + ]. [Pb + ] = Ksp [SO ] = 8 1.7 x 10 1.0 = 1.7 x 10 8 M (continued)

ELECTROCHEMISTRY 835 The reaction quotient for the cell reaction is Q = + [Pb ] x P H + [H ] = 8 (1.7 x 10 )(1.0) (1.0) = 1.7 x 10 8 The standard emf is 0.13 V, so the Nernst equation becomes E cell = E cell 0.059 n log Q E cell = 0.13 V 0.059 log (1.7 x 10 8 ) = 0.13 V (0.99 V) = 0.3599 = 0.36 V 0.10 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: H (g) H + (aq) + e E = 0.00 V Ag + (aq) + e Ag(s) E = 0.80 V H (g) + Ag + (aq) H + (aq) + Ag(s) E cell = 0.80 V Note that n equals two. Next, use K sp to calculate [Ag + ]. [Ag + ] = Ksp [Cl ] = 10 1.8 x 10 1.0 = 1.8 x 10 10 M The reaction quotient for the cell reaction is Q = + [H ] + H [Ag ] P = (1.0) 10 (1.8 x 10 ) (1.0) = 3.086 x 10 19 The standard emf is 0.80 V, so the Nernst equation becomes E cell = E cell 0.059 n log Q E cell = 0.80 V 0.059 log (3.086 x 10 19 ) = 0.80 V (0.5768 V) = 0.31 = 0. V

836 CHAPTER 0 0.105 a. Note that E = 0.010 V and n equals two. Substitute into the equation relating E and K. E = 0.05 9 log K n 0.010 V = 0.059 Solving for K, you get log K = 0.3378 log K Take the antilog of both sides: K = K c = antilog (0.3378) =.176 =. b. Substitute into the equilibrium expression using [Sn + ] = x, and [Pb + ] = 1.0 M x. K c = [Sn + + ] [Pb ] = x 1.0 x =.176 x = 0.6851 M [Pb + ] = 1.0 M x = 0.319 = 0.3 M 0.106 a. Note that E = 0.030 V and n equals one. Substitute into the equation relating E and K. E = 0.05 9 log K n 0.030 V = 0.059 1 Solving for K, you get log K = 0.5067 log K Take the antilog of both sides: K = K c = antilog (0.5067) = 3.11 = 3. (continued)

ELECTROCHEMISTRY 837 b. After mixing equal volumes, the initial concentrations of Ag + and Fe + are 0.50 M. Substitute into the equilibrium expression using [Fe 3+ ] = x, and [Ag + ] = [Fe + ] = 0.50 M x. K c = 3+ [Fe ] + + [Ag ][Fe ] = x (0.50 x) = 3.11 This equation can be rearranged to give the following quadratic equation. 3.11x.11x + 0.809 = 0 x = (.11) ± (.11) (3.11)(0.809) (3.11) = 0.31 M [Fe + ] = 0.50 x = 0.50 M 0.31 M = 0.686 = 0.7 M 0.107 a. The number of faradays is 1.0 mol Na + x 1 mol e 1 mol Na + x 1 F 1 mol e = 1.0 F The number of coulombs is 1.0 F x 9.65 x 10 C 1 F = 9.65 x 10 = 9.7 x 10 C b. The number of faradays is 1.0 mol Cu + x mol e 1 mol Cu + x 1 F 1 mol e =.0 F The number of coulombs is.0 F x 9.65 x 10 C 1 F = 1.93 x 10 5 = 1.9 x 10 5 C (continued)

838 CHAPTER 0 c. The number of faradays is 1 mol H 1.0 g H O x O 18.01 g H O x mol e 1 mol H O x 1 F 1 mol e = 0.1110 = 0.11 F The number of coulombs is 0.1110 F x 9.65 x 10 C 1 F = 1.07 x 10 = 1.1 x 10 C d. The number of faradays is 1.0 g Cl x 1 mol Cl 35.5 g Cl x mol e mol Cl x 1 F 1 mol e = 0.080 = 0.08 F The number of coulombs is 0.080 F x 9.65 x 10 C 1 F =.7 x 10 3 =.7 x 10 3 C 0.108 a. The number of faradays is 1.0 mol Fe 3+ x 1 mol e 1 mol Fe 3+ x 1 F 1 mol e = 1.0 F The number of coulombs is 1.0 F x 9.65 x 10 C 1 F = 9.65 x 10 = 9.7 x 10 C b. The number of faradays is 1.0 mol Fe 3+ x 3 mol e 1 mol Fe 3+ x 1 F 1 mol e = 3.0 F The number of coulombs is 3.0 F x 9.65 x 10 C 1 F =.895 x 10 5 =.9 x 10 5 C (continued)

ELECTROCHEMISTRY 839 c. The number of faradays is 1.0 g Sn + x + 1 mol Sn 118.7 g Sn + x mol e 1 mol Sn + x 1 F 1 mol e = 0.0168 = 0.017 F The number of coulombs is 0.0168 F x 9.65 x 10 C 1 F = 1.6 x 10 3 = 1.6 x 10 3 C d. The number of faradays is 1.0 g Au 3+ x 1 mol Au 3+ 197.0 g Au 3+ x 3 mol e 1 mol Au 3+ x 1 F 1 mol e = 0.015 = 0.015 F The number of coulombs is 0.015 F x 9.65 x 10 C 1 F = 1.69 x 10 3 = 1.5 x 10 3 C 0.109 From the electrolysis information, calculate the total moles of I formed 65. s x (10.5 x 10 3 A) x 1C 1A s x 1mole 1molI 9.65 x 10 C x mole From the balanced equation = 3.558 x 10 6 mol I 3.558 x 10 6 mol I x 1 mol H AsO 3 1 mol I 3 x 1molAs 1molHAsO 3 3 x 79.gAs 1molAs =.857 x 10 =.83 x 10 g As

80 CHAPTER 0 0.110 From the electrolysis information, calculate the total moles of OH formed. 115 s x (15.6 x 10 3 A) x 1C 1A s x 1mole 9.65 x 10 C x mol OH mol e From the balanced equation = 1.859 x 10 5 mol OH 1.859 x 10 5 mol OH x 1 mol HCHO 1 mol OH 3 5 3 x 90.08 g HC H O 3 5 3 1 mol HCHO 3 5 3 = 1.675 x 10 3 = 1.67 x 10 3 g HC 3 H 5 O 3 0.111 a. The spontaneous chemical reaction and maximum cell potential are Cd(s) Cd + (aq) + e E = 0.0 V Ag + (aq) + e Ag(s) E = 0.80 V Cd(s) + Ag + (aq) Ag(s) + Cd + (aq) E cell = 1.0 V b. Addition of S would greatly decrease the Cd + (aq) concentration and help shift the equilibrium to the right, thus forming more Ag. c. No effect. The size of the electrode makes no difference in the potential. 0.11 a. The spontaneous chemical reaction and maximum cell potential are Fe(s) Fe + (aq) + e E = 0.1 V Cu + (aq) + e Cu(s) E = 0.3 V Fe(s) + Cu + (aq) Cu(s) + Fe + (aq) E cell = 0.75 V b. The Cu + (aq) concentration is greatly decreased shifting the equilibrium to the left and thus forms less Cu. c. No effect. The size of the electrode makes no difference in the potential.

ELECTROCHEMISTRY 81 0.113 Anode: H O(l) O (g) + H + (aq) + e Cathode: Cu + (aq) + e Cu(s) 0.11 Anode: Cl (aq) Cl (g) + e Cathode: H O(aq) + e H (g) + OH (aq) 0.115 a. First, find the moles of silver..8 g Ag x 1 mol Ag 107.9 g Ag = 0.098 mol Ag The number of coulombs is 0.098 mol Ag x 1 mol e 1 mol Ag x 9.65 x 10 C 1 mol e = 18 C Since amp x sec = coul, the number of seconds is time = 18 C 1.50 amp = 1.78 x 103 = 1.8 x 10 3 s b. For the same amount of current, the moles of Cr would be 17 C x 1 mol e 1 mol Cr 9.65 x 10 C x 3 mol e x 5.00 g Cr 1 mol Cr = 0.398 = 0.398 g 0.116 a. First, find the moles of copper..10 g Cu x 1molCu 63.55 g Cu = 0.0330 mol Cu The number of coulombs is 0.0330 mol Cu x mole 1molCu x 9.65 x 10 C 1 mol e = 6377 C (continued)

8 CHAPTER 0 Since amp x sec = coul, the number of seconds is s = 6377 C 1.5 amp = 5.10 x 103 = 5.10 x 10 3 s b. For the same amount of current, the moles of Ag would be 6377 C x 1 mol e 1 mol Ag 9.65 x 10 C x 1 mol e x 107.9 g Ag 1 mol Ag = 7.131 = 7.13 g 0.117 a. Note that 3.50 hours = 1600 s, and the efficiency is 90.0 percent..75 amp x 1600 s x 1 F 9.65 x 10 C x 0.900 = 0.33 = 0.33 F b. The moles of Au that were deposited is mol Au = 1.1 g Au x 1 mol Au 197.0 g = 0.10770 mol Since 1 F = 1 mol e, the ratio of moles of e to moles Au is mol e mol Au = 0.33 0.1077 = 3.00 1 Thus, the ion must be Au 3+. The reaction is Au 3+ + 3e Au 0.118 a. Note that 1.90 hours = 680 s, and the efficiency is 95.0 percent..50 amp x 680 s x 1 F 9.65 x 10 C x 0.950 = 0.1683 = 0.168 F (continued)

ELECTROCHEMISTRY 83 b. The moles of V that were deposited is mol V =.850 g V x 1 mol V 50.9 g V = 0.05598 mol V Since 1 F = 1 mol e, the ratio of moles of e to moles V is mol e mol V = 0.1683 0.05598 = 3.00 1 Thus, the ion must be V 3+. The reaction is V 3+ + 3e V 0.119 a. The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Zn(s) Zn + (aq) + e E = 0.76 V Cu + (aq) + e Cu(s) E = 0.3 V Cu + (aq) + Zn(s) Zn + (aq) + Cu(s) E = 1.10 V Use the Nernst equation to calculate the voltage of the cell. E = E 0.059 n log Q = E 0.059 n + [Zn ] log + [Cu ] Note that n equals two, [Zn + ] = 0.00 M, and [Cu + ] = 0.0100 M. E = 1.10 V 0.059 log [0.00] [0.0100] = 1.10 V 0.03851 V = 1.061 = 1.06 V b. First, calculate the moles of electrons passing through the cell. 1.00 amp x 5 s x 1 mol e 9.65 x 10 C = 0.0033 mol e The moles of Cu deposited are 0.0033 mol e x 1 mol Cu mol e = 0.001166 mol Cu (continued)

8 CHAPTER 0 The moles of Cu remaining in the 1.00 L of solution are 0.0100 0.001166 = 0.00883 = 0.0088 mol Since the volume of the solution is 1.00 L, the molarity of Cu + is 0.0088 M. 0.10 a. The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Fe(s) Fe + (aq) + e E = 0.1 V Cu + (aq) + e Cu(s) E = 0.3 V Fe(s) + Cu + (aq) Fe + (aq) + Cu(s) E = 0.75 V Use the Nernst equation to calculate the voltage of the cell. E = E 0.059 n log Q = E 0.059 n + [Fe ] log + [Cu ] Note that n equals two, [Fe + ] = 0.15 M, and [Cu + ] = 0.036 M. E = 0.75 V 0.059 = 0.7316 = 0.73 V log [0.15] [0.036] b. First, calculate the moles of electrons passing through the cell. 1.5 amp x 335 s x 1 mol e 9.65 x 10 C = 0.0033 9 mol e The moles of Cu deposited are 0.00339 mol e x 1 mol Cu mol e = 0.00170 mol The moles of Cu remaining in the 1.00 L of solution are 0.036 0.00170 = 0.0338 = 0.03 mol Since the volume of the solution is 1.00 L, the molarity of Cu + is 0.03 M.

ELECTROCHEMISTRY 85 0.11 a. Write the cell reaction with the G f s beneath Cd(s) + Co + (aq) Cd + (aq) + Co(s) G f : 0 51.5 77.6 0 kj Hence, G = Σn G f (products) Σm G f (reactants) = [77.6 + 51.5] kj = 6.1 kj =.61 x 10 J b. Next, determine the standard emf for the cell. Note that n equals two. G = nfe.61 x 10 J = (9.65 x 10 C) x E Solve for E to get E =.61 x 10 J (9.65 x 10 C) = 0.135 = 0.135 V The halfreactions and voltages are Cd(s) Cd + (aq) + e E ox = 0.0 V Co + (aq) + e Co(s) E red =? Cd(s) + Co + (aq) Cd + (aq) + Co(s) E = 0.135 V The cell emf is E = E red + (E ox ) 0.135 V = E red + 0.0 V E red = 0.135 V 0.0 V = 0.67 = 0.6 V 0.1 a. Write the cell reaction with the G f s beneath Tl(s) + Pb + (aq) Tl + (aq) + Pb(s) G f : 0.3 (3.) 0 kj Hence, G = [(3.) (.3)] kj = 0.5 kj =.05 x 10 J (continued)

86 CHAPTER 0 b. Next, determine the standard emf for the cell. Note that n equals two. G = nfe.05 x 10 J = (9.65 x 10 C) x E Solve for E to get E =.05 x 10 J (9.65 x 10 C) = 0.0.098 = 0.10 V The halfreactions and voltages are Tl(s) Tl + (aq) + e E ox = E red Pb + (aq) + e Pb(s) E red = 0.13 V Cd(s) + Co + (aq) Cd + (aq) + Co(s) E = 0.098 V The cell emf is E = E red + (E ox ) 0.098 V = 13 V + E red E red = 0.098 V + 0.13 V = 0.3398 = 0.3 V Solutions to CumulativeSkills Problems 0.13 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: 3Zn(s) 3Zn + (aq) + 6e E = 0.76 V Cr 3+ (aq) + 6e Cr(s) E = 0.7 V Cr 3+ (aq) + 3Zn(s) 3Zn + (aq) + Cr(s) E = 0.0 V Note that n equals six. Therefore, G = nfe = 6(9.65 x 10 C)(0.0 V) = 1.158 x 10 J = 11.58 kj (continued)

ELECTROCHEMISTRY 87 Write the cell reaction with the H f s beneath Cr 3+ (aq) + 3Zn(s) 3Zn + (aq) + Cr(s) H f : (1971) 0 3(15.) 0 kj Hence, H = Σn H f (products) Σm H f (reactants) = [3(15.) (1971)] kj = 38.8 = 385 kj Now calculate S. G = H T S 11.58 kj = 38.8 kj 98 K x S Solving for S gives S = 38.8 kj + 11.58 kj 98 K = 11.73 = 11.7 kj/k 0.1 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Cr(s) Cr 3+ (aq) + 6e E = 0.7 V 3Fe + (aq) + 6e 3Fe(s) E = 0.1 V Cr(s) + 3Fe + (aq) Cr 3+ (aq) + 3Fe(s) E = 0.33 V Note that n equals six. Therefore, G = nfe = 6(9.65 x 10 C)(0.33 V) = 1.9107 x 10 5 J = 191.07 kj Write the cell reaction with the H f s beneath Hence, Cr(s) + 3Fe + (aq) Cr 3+ (aq) + 3Fe(s) H f : 0 3(87.9) 3(1971) 0 kj H = Σn H f (products) Σm H f (reactants) = [3(1971) 3(87.9] kj = 3678.3 = 3678 kj (continued)

88 CHAPTER 0 Now calculate S. G = H T S 191.07 kj = 3678.3 kj 98 K x S Solving for S gives S = 3678.3 kj + 191.07 kj 98 K = 11.70 = 11.7 kj/k 0.15 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Br (aq) Br (l) + e E = 1.07 V O (g) + H + (aq) + e H O(l) E = 1.3 V O (g) + H + (aq) + Br (aq) H O(l) + Br (l) E = 0.16 V Now, convert the ph to [H + ] [H + ] = antilog (3.60) =.51 x 10 M Under standard conditions [Br ] = 1 M, and the pressure of O is one atm. Thus, substitute into the Nernst equation, where n equals four. E = E 0.059 n log Q = E 0.059 n log 1 + [H ] [Br ] x P O = 0.16 V 0.059 1 log (.51 x 10 ) = 0.16 V 0.131 V = 0.0531 = 0.05 V Thus, the reaction is nonspontaneous at this [H + ]. 0.16 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Co + (aq) Co 3+ (aq) + e E = 1.8 V O 3 (g) + H + (aq) + e O (g) + H O(l) E =.07 V O 3 (g) + H + (aq) + Co + (aq) O (g) + H O(l) + Co 3+ (aq) E = 0.5 V (continued)

ELECTROCHEMISTRY 89 Now, convert the ph to [H + ] [H + ] = antilog (9.10) = 7.93 x 10 10 M Under standard conditions [Co + ] and [Co 3+ ] = 1 M, and the partial pressures of O and O 3 equals one atm. Thus, substitute into the Nernst equation, where n equals two. E = E 0.059 n log Q = E 0.059 n log [Co 3+ ] x P + + O [H ] [Co ] x P O 3 = 0.5 V 0.059 1 log (7.9 x 10 ) 10 = 0.5 V 0.6936 V = 0.01936 = 0.0 V Thus, the reaction is nonspontaneous at this [H + ]. 0.17 Use the K a to calculate [H+] for the buffer. K a = + [H ][OCN ] [HOCN] = 3.5 x 10 Rearrange, and solve for [H + ], assuming [HOCN] and [OCN ] remain constant in the buffer. Thus, [H + ] = K a = 3.5 x 10 M. In Problem 0.13, E for this cell reaction was found to be 0.16 V. Under standard conditions [Br ] = 1 M, and the pressure of O is one atm. Thus, substitute into the Nernst equation, where n equals four. E = E 0.059 n log Q = E 0.059 n log 1 + [H ] [Br ] x P O = 0.16 V 0.059 1 log (3.5 x 10 ) = 0.16 V 0.059 V = 0.055 = 0.0 V Thus, the reaction is nonspontaneous at this [H + ].

850 CHAPTER 0 0.18 Use the K a to calculate [H+] for the buffer. K a = + [ H ][OCl ] [HOCl] = 3.5 x 10 8 Rearrange, and solve for [H + ] assuming [HOCl] and [OCl ] remain constant in the buffer. Thus, [H + ] = K a = 3.5 x 10 8 M. In Problem 0.1, E for this cell reaction was found to be 0.5 V. Under standard conditions [Co + ] and [Co 3+ ] = 1 M, and the partial pressures of O and O 3 equals one atm. Thus, Substitute into the Nernst equation, where n equals two. E = E 0.059 n log Q = E 0.059 n log [Co 3+ ] x P + + O [H ] [Co ] x P O 3 = 0.5 V 0.059 1 log (3.5 x 10 ) 8 = 0.5 V 0.139 V = 0.19139 = 0.19 V Thus, the reaction is nonspontaneous at this [H + ]. 0.19 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: H (g) H + (aq) + e E = 0.00 V Ag + (aq) + e Ag(s) E = 0.80 V H (g) + Ag + (aq) H + (aq) + Ag(s) E = 0.80 V The standard hydrogen electrode has [H + ] = 1.0 M and the pressure of H equals one atm. Substitute into the Nernst equation, where E = 0.5 V and n equals two. E = E 0.059 n log Q = E 0.059 n log + + [H ] [Ag ] x P H 0.5 V = 0.80 V 0.059 1 + log [Ag ] Using the properties of logs, rearrange to get 0.5 V = 0.80 V + 0.059 log[ag + ] (continued)

ELECTROCHEMISTRY 851 Solve for [Ag + ] log[ag + ] = 0.5 0.80 0.059 = 5.91 [Ag + ] = 10 5.91 = 1. x 10 6 M Finally, determine the solubility product, using [SCN ] = 0.10 M K sp = [Ag + ][SCN ] = (1. x 10 6 )(0.10) = 1. x 10 7 = 1 x 10 7 0.130 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: H (g) H + (aq) + e E = 0.00 V Hg + (aq) + e Hg(l) E = 0.80 V Hg + (aq) + H (g) Hg(l) + H + (aq) E = 0.80 V The standard hydrogen electrode has [H + ] = 1.0 M and the pressure of H equals one atm. Substitute into the Nernst equation, where E = 0.68 V and n equals two. E = E 0.059 n log Q = E 0.059 n log [Hg + [H ] + ] x PH 0.68 V = 0.80 V 0.059 1 log + [Hg ] Using the properties of logs, rearrange to get 0.68 V = 0.80 V + 0.059 Solve for [Hg + ] log[hg + ] log[hg + ] = (0.68 0.80) 0.059 = 17.97 [Hg + ] = 10 17.97 = 1.06 x 10 18 M Finally, determine the solubility product, using [Cl ] = 1.00 M K sp = [Hg + ][Cl ] = (1.06 x 10 18 )(1.00) = 1.06 x 10 18 = 10 18

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