14. See Table 4.2 in Chapter 4 of the text for rules for assigning oxidation numbers.

Transcription

1 CHAPTER SEVENTEEN Review of Oxidation Reduction Reactions 13. Oxidation: increase in oxidation number; loss of electrons Reduction: decrease in oxidation number; gain of electrons 14. See Table 4. in Chapter 4 of the text for rules for assigning oxidation numbers. a. H (1), O (), N (5) b. Cl (1), Cu () c. O (0) d. H (1), O (1) e. Mg (), O (), S (6) f. Ag (0) g. Pb (), O (), S (6) h. O (), Pb (4) i. Na (1), O (), C (3) j. O (), C (4) 4 k. (NH 4) Ce(SO 4) 3contains NH 4 ions and SO 4 ions. Thus, cerium exists as the Ce ion. H (1), N (3), Ce (4), S (6), O () l. O (), Cr (3) 15. The species oxidized shows an increase in oxidation numbers and is called the reducing agent. The species reduced shows a decrease in oxidation numbers and is called the oxidizing agent. The pertinent oxidation numbers are listed by the substance oxidized and the substance reduced. Substance Substance Redox? Ox. Agent Red. Agent Oxidized Reduced a. Yes HO CH4 CH 4 (C, 4 ) HO (H, 1 0) b. Yes AgNO3 Cu Cu (0 ) AgNO 3 (Ag, 1 0) c. Yes HCl Zn Zn (0 ) HCl (H, 1 0) d. No; There is no change in any of the oxidation numbers. 473

2 474 CHAPTER See Chapter 4.10 of the text for rules on balancing oxidationreduction reactions. a. Cr Cr 3 e NO NO 4 H NO NO H O 3 e 4 H NO NO H O Cr Cr 3 e 3 e 4 H NO NO H O H (aq) NO 3(aq) Cr(s) Cr (aq) NO(g) HO(l) b. (Al Al 3 e ) 5 MnO Mn 8 H MnO Mn 4 H O (5 e 8 H MnO Mn 4 H O) Al 5 Al 15 e 15 e 4 H 3 MnO 3 Mn 1 H O H (aq) 3 MnO 4(aq) 5 Al(s) 5 Al (aq) 3 Mn (aq) 1 HO(l) 4 3 c. (Ce e Ce ) 6 CH3OH CO H O CH OH CO 6 H 3 H O CH OH CO 6 H 6 e Ce 6 e 6 Ce H O CH OH CO 6 H 6 e HO(l) CH3OH(aq) 6 Ce (aq) 6 Ce (aq) CO (g) 6 H (aq) d. PO PO (H O PO PO H e ) MnO MnO (3 e 4 H MnO MnO H O) H O 3 PO 3 PO 6 H 6 e 6 e 8 H MnO MnO 4 H O H MnO 4 3 PO 3 3 PO 4 MnO HO Now convert to a basic solution by adding OH to both sides. H OH HO on the reactant side. After converting H to OH, simplify the overall equation by crossing off 1 HO on each side of the reaction. The overall balanced equation is: 3 3 HO(l) MnO 4(aq) 3 PO 3 (aq) 3 PO 4 (aq) MnO (s) OH (aq) e. Mg Mg(OH) OCl Cl

3 CHAPTER HO Mg Mg(OH) H e e H OCl Cl HO H O Mg Mg(OH) H e e H OCl Cl H O OCl (aq) HO(l) Mg(s) Mg(OH) (s) Cl (aq) The final overall reaction does not contain H, so we are done. f. H CO HCO Ag(NH ) Ag NH H O H CO HCO 5 H 4 e (e Ag(NH ) Ag NH ) H O H CO HCO 5 H 4 e 4 e 4 Ag(NH ) 4 Ag 8 NH Questions 4 Ag(NH 3) HO HCO HCO 3 5 H 4 Ag 8 NH3 Convert to a basic solution by adding 5 OH to both sides (5 H 5 OH 5 HO). Then, cross off H O on both sides, which gives the overall balanced equation: 5 OH (aq) 4 Ag(NH 3) (aq) HCO(aq) HCO 3(aq) 3 HO(l) 4 Ag(s) 8 NH 3(aq) 17. In a galvanic cell, a spontaneous redox reaction occurs which produces an electric current. In an electrolytic cell, electricity is used to force a redox reaction that is not spontaneous to occur. 18. The salt bridge allows counter ions to flow into the two cell compartments to maintain electrical neutrality. Without a salt bridge, no sustained electron flow can occur. 19. a. Cathode: The electrode at which reduction occurs. b. Anode: The electrode at which oxidation occurs. c. Oxidation halfreaction: The halfreaction in which electrons are products. In a galvanic cell, the oxidation halfreaction always occurs at the anode. d. Reduction halfreaction: The halfreaction in which electrons are reactants. In a galvanic cell, the reduction halfreaction always occurs at the cathode. 0. a. Purification by electrolysis is called electrorefining. See the text for a discussion of the electrorefining of copper. Electrorefining is possible because of the selectivity of electrode reactions. The anode is made up of the impure metal. A potential is applied so just the metal of interest and all more easily oxidized metals are oxidized at the anode. The metal of interest is the only metal plated at the cathode due to the careful control of the potential applied. The metal ions that could plate out at the cathode in preference to the metal we are purifying will not be in solution, because these metals were not oxidized at the anode. b. A more easily oxidized metal is placed in electrical contact with the metal we are trying to protect. It is oxidized in preference to the protected metal. The protected metal becomes the cathode

4 476 CHAPTER 17 electrode, thus, cathodic protection. 1. As a battery discharges, E cell decreases, eventually reaching zero. A charged battery is not at equilibrium. At equilibrium, E cell = 0 and G = 0. We get no work out of an equilibrium system. A battery is useful to us because it can do work as it approaches equilibrium.. Standard reduction potentials are an intensive property, i.e., they do not depend on how many times the reaction occurs. As long as the concentrations of ions and gases are 1 M or 1 atm, standard reduction potentials and standard oxidation potentials are a constant and not dependent on the coefficients in the balanced equation. 3. Both fuel cells and batteries are galvanic cells. However, fuel cells, unlike batteries, have the reactants continuously supplied and can produce a current indefinitely. 4. Moisture must be present to act as a medium for ion flow between the anodic and cathodic regions. Salt provides ions necessary to complete the electrical circuit in the corrosion process. Together, salt and water make up the salt bridge in this spontaneous electrochemical process. Exercises Three methods discussed in the text to prevent corrosion are galvanizing, alloying and cathodic protection. Galvanizing coats the metal of interest (usually iron) with zinc, which is an easily oxidized metal. Alloying mixes in metals that form durable, effective oxide coatings over the metal of interest. Cathodic protection connects, via a wire, a more easily oxidized metal to the metal we are trying to protect. The more active metal is preferentially oxidized, thus protecting our metal object from corrosion. Galvanic Cells, Cell Potentials, Standard Reduction Potentials, and Free Energy 5. A typical galvanic cell diagram is: The diagram for all cells will look like this. The contents of each halfcell compartment will be identified for each reaction, with all solute concentrations at 1.0 M and all gases at 1.0 atm. For Exercises 17.5 and 17.6, the flow of ions through the salt bridge was not asked for in the questions. If asked, however, cations always flow into the cathode compartment, and anions always flow into the anode compartment. This is required to keep each compartment electrically neutral.

5 CHAPTER a. Table 17.1 of the text lists balanced reduction halfreactions for many substances. For this overall 3 reaction, we need the Cl to Cl reduction halfreaction and the Cr to CrO 7 oxidation halfreaction. Manipulating these two halfreactions gives the overall balanced equation. (Cl e Cl ) HO Cr CrO 7 14 H 6 e 3 7 HO(l) Cr (aq) 3 Cl (g) CrO 7 (aq) 6 Cl (aq) 14 H (aq) The contents of each compartment are: Cathode: Pt electrode; Cl bubbled into solution, Cl in solution 3 Anode: Pt electrode; Cr, H, and CrO 7 in solution We need a nonreactive metal to use as the electrode in each case, since all the reactants and products are in solution. Pt is a common choice. Another possibility is graphite. b. Cu e Cu Mg Mg e Cu (aq) Mg(s) Cu(s) Mg (aq) Cathode: Cu electrode; Cu in solution; Anode: Mg electrode; Mg in solution 6. Reference Exercise 17.5 for a typical galvanic cell diagram. The contents of each halfcell compartment are identified below with all solute concentrations at 1.0 M and all gases at 1.0 atm. a. Reference Table 17.1 for the balanced halfreactions. 5 e 6 H IO 3 1/ I 3 HO 3 (Fe Fe e ) H IO 3 5 Fe 5 Fe 1/ I 3 HO 3 or 1 H (aq) IO 3(aq) 10 Fe (aq) 10 Fe (aq) I (aq) 6 HO(l) Cathode: Pt electrode; IO 3, Iand HSO 4 (H source) in solution. Note: I (s) would make a poor electrode since it sublimes. 3 Anode: Pt electrode; Fe and Fe in solution b. (Ag e Ag) Zn Zn e Zn(s) Ag (aq) Ag(s) Zn (aq)

6 478 CHAPTER 17 Cathode: Ag electrode; Ag in solution; Anode: Zn electrode; Zn in solution 7. To determine E for the overall cell reaction, we must add the standard reduction potential to the standard oxidation potential ( potentials. Remember that used to obtain the overall balanced equation. ). Reference Table 17.1 for values of standard reduction and that standard potentials are not multiplied by the integer 5a. = = 1.36 V (1.33 V) = 0.03 V 5b. = = 0.34 V.37 V =.71 V 8. 6a. = = 1.0 V (0.77 V) = 0.43 V 6b. = = 0.80 V 0.76 V = 1.56 V 9. Reference Exercise 17.5 for a typical galvanic cell design. The contents of each halfcell compartment are identified below with all solute concentrations at 1.0 M and all gases at 1.0 atm. For each pair of halfreactions, the halfreaction with the largest standard reduction potential will be the cathode reaction, and the halfreaction with the smallest reduction potential will be reversed to become the anode reaction. Only this combination gives a spontaneous overall reaction, i.e., a reaction with a positive overall standard cell potential. Note that in a galvanic cell as illustrated in Exercise 17.5, the cations in the salt bridge migrate to the cathode, and the anions migrate to the anode. a. Cl e Cl E = 1.36 V Br Br e E = 1.09 V Cl (g) Br (aq) Br (aq) Cl (aq) = 0.7 V The contents of each compartment are: Cathode: Pt electrode; Cl (g) bubbled in, Cl in solution Anode: Pt electrode; Br and Br in solution b. ( e H IO IO H O) 5 E = 1.60 V (4 H O Mn MnO 8 H 5 e ) E = 1.51 V H 5 IO 4 8 HO Mn 5 IO 3 5 HO MnO 4 16 H = 0.09 V This simplifies to: 3 HO(l) 5 IO 4(aq) Mn (aq) 5 IO 3(aq) MnO 4(aq) 6 H (aq) = 0.09 V Cathode: Pt electrode; IO 4, IO 3, and HSO 4 (as a source of H ) in solution Anode: Pt electrode; Mn, MnO 4 and HSO 4 in solution

7 CHAPTER Reference Exercise 17.5 for a typical galvanic cell design. The contents of each halfcell compartment are identified below, with all solute concentrations at 1.0 M and all gases at 1.0 atm. a. HO H e HO E = 1.78 V HO O H e E = 0.68 V HO (aq) HO(l) O (g) = 1.10 V Cathode: Pt electrode; HO and H in solution Anode: Pt electrode; O (g) bubbled in, HO and H in solution b. 3 (Fe 3 e Fe) E = V (Mn Mn e ) 3 E = 1.18 V 3 Fe (aq) 3 Mn(s) Fe(s) 3 Mn (aq) = 1.14 V 3 Cathode: Fe electrode; Fe in solution; Anode: Mn electrode; Mn in solution 31. In standard line notation, the anode is listed first and the cathode is listed last. A double line separates the two compartments. By convention, the electrodes are on the ends with all solutes and gases towards the middle. A single line is used to indicate a phase change. We also include all concentrations. 3 5a. Pt Cr (1.0 M), CrO 7 (1.0 M), H (1.0 M) Cl (1.0 atm) Cl (1.0 M) Pt 5b. Mg Mg (1.0 M) Cu (1.0 M) Cu 9a. Pt Br (1.0 M), Br (1.0 M) Cl (1.0 atm) Cl (1.0 M) Pt 9b. Pt Mn (1.0 M), MnO 4 (1.0 M), H (1.0 M) IO 4 (1.0 M), H (1.0 M), IO 3 (1.0 M) Pt a. Pt Fe (1.0 M), Fe (1.0 M) IO 3 (1.0 M), H (1.0 M), I (1.0 M) Pt 6b. Zn Zn (1.0 M) Ag (1.0 M) Ag 30a. Pt HO (1.0 M), H (1.0 M) O (1.0 atm) HO (1.0 M), H (1.0 M) Pt 3 30b. Mn Mn (1.0 M) Fe (1.0 M) Fe 33. Locate the pertinent halfreactions in Table 17.1, and then figure which combination will give a positive standard cell potential. In all cases, the anode compartment contains the species with the smallest standard reduction potential. For part a, the copper compartment is the anode, and in part b, the cadmium compartment is the anode.

8 480 CHAPTER 17 3 a. Au 3 e Au E = 1.50 V (Cu Cu e ) 3 E = 0.16 V 3 Au (aq) 3 Cu (aq) Au(s) 3 Cu (aq) = 1.34 V b. (VO H e VO HO) E = 1.00 V Cd Cd e E = 0.40 V VO (aq) 4 H (aq) Cd(s) VO (aq) H O(l) Cd (aq) = 1.40 V 34. a. (H O H e H O) 3 E = 1.78 V Cr 7 H O Cr O 14 H 6 e E = 1.33 V HO (aq) Cr (aq) HO(l) CrO 7 (aq) 8 H (aq) = 0.45 V b. ( H e H ) 3 E = 0.00 V 3 (Al Al 3 e ) E = 1.66 V 3 6 H (aq) Al(s) 3 H (g) Al (aq) = 1.66 V 35. a. Ag e Ag E = 0.80 V Cu Cu e E = 0.34 V Ag Cu Cu Ag = 0.46 V Spontaneous at standard conditions ( > 0). b. Zn e Zn E = 0.76 V Ni Ni e E = 0.3 V Zn Ni Zn Ni = 0.53 V Not spontaneous at standard conditions ( < 0). 36. a. (5 e 8 H MnO Mn 4 H O) E = 1.51 V ( I I e ) 5 E = 0.54 V 4 16 H MnO 4 10 I 5 I Mn 8 HO = 0.97 V Spontaneous b. (5 e 8 H MnO Mn 4 H O) E = 1.51 V ( F F e ) 5 E =.87 V 4 16 H MnO 4 10 F 5 F Mn 8 HO = 1.36 V Not spontaneous 37. Cl e Cl E = 1.36 V (ClO ClO e ) E = V ClO (aq) Cl (g) ClO (aq) Cl (aq) = 0.41 V = 0.41 J/C G = 4 = ( mol e )(96,485 C/mol e )(0.41 J/C) = J = 79 kj

9 CHAPTER a. (4 H NO 3 3 e NO HO) E = 0.96 V (Mn Mn e ) 3 E = 1.18 V 3 Mn(s) 8 H (aq) NO 3(aq) NO(g) 4 HO(l) 3 Mn (aq) =.14 V ( e H IO IO H O) 5 E = 1.60 V (Mn 4 H O MnO 8 H 5 e ) E = 1.51 V IO 4(aq) Mn (aq) 3 HO(l) 5 IO 3(aq) MnO 4(aq) 6 H (aq) = 0.09 V b. Nitric acid oxidation (see above for ): G = 6 = (6 mol e )(96,485 C/mol e )(.14 J/C) = J = 140 kj Periodate oxidation (see above for ): G = (10 mol e )(96,485 C/mol e )(0.09 J/C)(1 kj/1000 J) = 90 kj 39. Since the cells are at standard conditions, w max = G = G =. See Exercise for the balanced overall equations and for. 5 33a. w max = (3 mol e )(96,485 C/mol e )(1.34 J/C) = J = 388 kj 5 33b. w max = ( mol e )(96,485 C/mol e )(1.40 J/C) = J = 70. kj 40. Since the cells are at standard conditions, w max = G = G =. See Exercise for the balanced overall equations and for. 5 34a. w max = (6 mol e )(96,485 C/mol e )(0.45 J/C) =.6 10 J = 60 kj 5 34b. w max = (6 mol e )(96,485 C/mol e )(1.66 J/C) = J = 961 kj 41. CH3OH(l) 3/ O (g) CO (g) HO(l) G = (37) (394) [166] = 70 kj The balanced halfreactions are: HO CH3OH CO 6 H 6 e and O 4 H 4 e HO For 3/ mol O, 6 moles of electrons will be transferred (n = 6). G = nfe, E = = 1.1 J/C = 1.1 V 4. Fe e Fe E = 0.44 V = 0.44 J/C G = nfe = ( mol e )(96,485 C/mol e )(0.44 J/C)(1 kj/1000 J) = 85 kj 85 kj = 0, = 85 kj

10 48 CHAPTER 17 3 We can get two ways. Consider: Fe e Fe E = 0.77 V G = (1 mol e)(96,485 C/mol e )(0.77 J/C) = 74,300 J = 74 kj 3 Fe Fe e Fe Fe e G = 74 kj G = 85 kj 3 Fe Fe 3 e G = 11 kj, = 11 kj/mol 3 or consider: Fe 3 e Fe E = V G = (3 mol e )(96,485 C/mol e )(0.036 J/C) = 10,400 J 10. kj 10. kj = 0, = 10. kj/mol; Roundoff error explains the 1 kj discrepancy. 43. Good oxidizing agents are easily reduced. Oxidizing agents are on the left side of the reduction halfreactions listed in Table We look for the largest, most positive standard reduction potentials to correspond to the best oxidizing agents. The ordering from worst to best oxidizing agents is: K < HO < Cd < I < AuCl 4 < IO3 E (V) Good reducing agents are easily oxidized. The reducing agents are on the right side of the reduction halfreactions listed in Table The best reducing agents have the most negative standard reduction potentials (E ) or the most positive standard oxidation potentials, (= E ). 3 F < Cr < Fe < H < Zn < Li E (V) a. H e H E = 0.00 V; Cu Cu e E = 0.34 V = 0.34 V; No, H cannot oxidize Cu to Cu at standard conditions ( < 0). 3 b. Fe e Fe E = 0.77 V; I I e E = 0.54 V 3 = = 0.3 V; Yes, Fe can oxidize I to I. c. H H e E = 0.00 V; Ag e Ag E = 0.80 V = 0.80 V; Yes, H can reduce Ag to Ag at standard conditions ( > 0). 3 3 d. Fe Fe e E = 0.77 V; Cr e Cr E = 0.50 V 3 = = 1.7 V; No, Fe cannot reduce Cr to Cr at standard conditions. 46. Cl e Cl E = 1.36 V Ag e Ag E = 0.80 V Pb e Pb E = 0.13 V Zn e Zn E = 0.76 V

11 CHAPTER Na e Na E =.71 V a. Oxidizing agents (species reduced) are on the left side of the above reduction halfreactions. Of the species available, Ag would be the best oxidizing agent since it has the largest E value. Note that Cl is a better oxidizing agent than Ag, but it is not one of the choices listed. b. Reducing agents (species oxidized) are on the right side of the reduction halfreactions. Of the species available, Zn would be the best reducing agent since it has the largest E value. c. SO 4 H e H SO H O E = 0.0 V; SO can oxidize Pb and Zn at standard conditions. When SO is coupled with these reagents, is positive d. Al Al 3 e E = 1.66 V; Al can oxidize Ag and Zn at standard conditions since > a. Br Br e E = 1.09 V; Cl Cl e E = 1.36 V; E > 1.09 V to oxidize Br ; E < 1.36 V to not oxidize Cl ; CrO 7, O, MnO, and IO 3 are all possible since when all of these oxidizing agents are coupled with Br, > 0, and when coupled with Cl, < 0 (assuming standard conditions). b. Mn Mn e E = 1.18; Ni Ni e E = 0.3 V; Any oxidizing agent with V > E > 1.18 V will work. PbSO 4, Cd, Fe, Cr, Zn and HO will be able to oxidize Mn but not Ni (assuming standard conditions). 48. a. Cu e Cu E = 0.34 V; Cu e Cu E = 0.16 V; To reduce Cu to Cu but not reduce Cu to Cu, the reducing agent must have a standard oxidation potential ( = E ) between 0.34 V and 0.16 V (so is positive only for the Cu to Cu reduction). The reducing agents (species oxidized) are on the right side of the halfreactions in Table The reagents at standard conditions which have ( = E ) between 0.34 V and 0.16 V are Ag (in 1.0 M Cl ) and H SO 3. b. Br e Br E = 1.09 V; I e I E = 0.54 V; From Table 17.1, VO, Au (in 1.0 M Cl ), NO, ClO, Hg, Ag, Hg, Fe, HO and MnO 4 are all capable at standard conditions of reducing Br to Br but not reducing I to I. When these reagents are coupled with Br, > 0, and when coupled with I, < ClO H O e OH Cl E = 0.90 V NH OH N H H O e E = 0.10 V 3 4 ClO (aq) NH 3(aq) Cl (aq) NH 4(aq) HO(l) = 1.00 V Since is positive for this reaction, then at standard conditions ClO can spontaneously oxidize NH to the somewhat toxic N H Tl e Tl E = 1.5 V 3 I I e E = 0.55 V 3 3 Tl 3 I Tl I 3 = 0.70 V

12 484 CHAPTER 17 3 In solution, Tl can oxidize I to I 3. Thus, we expect TlI 3 to be thallium(i) triiodide. The Nernst Equation 51. HO H e HO E = 1.78 V (Ag Ag e ) E = 0.80 V HO (aq) H (aq) Ag(s) HO(l) Ag (aq) = 0.98 V a. A galvanic cell is based on spontaneous chemical reactions. At standard conditions, this reaction produces a voltage of 0.98 V. Any change in concentration that increases the tendency of the forward reaction to occur will increase the cell potential. Conversely, any change in concentration that decreases the tendency of the forward reaction to occur (increases the tendency of the reverse reaction to occur) will decrease the cell potential. Using Le Chatelier s principle, increasing the reactant concentrations of HO and H from 1.0 M to.0 M will drive the forward reaction further to right (will further increase the tendency of the forward reaction to occur). Therefore, E will be greater than. cell b. Here, we decreased the reactant concentration of H and increased the product concentration of Ag from the standard conditions. This decreases the tendency of the forward reaction to occur which will decrease E as compared to (E < ). cell 7 5. The concentrations of Fe in the two compartments are now 0.01 M and 1 10 M. The driving force for this reaction is to equalize the Fe concentrations in the two compartments. This occurs if the 7 compartment with 1 10 M Fe becomes the anode (Fe will be oxidized to Fe ) and the compartment with the 0.01 M Fe becomes the cathode (Fe will be reduced to Fe). Electron flow, as always for galvanic cells, goes from the anode to the cathode, so electron flow will go from the right 7 compartment ([Fe ] = 1 10 M) to the left compartment ([Fe ] = 0.01 M). 53. For concentration cells, the driving force for the reaction is the difference in ion concentrations between the anode and cathode. In order to equalize the ion concentrations, the anode always has the smaller ion concentration. The general setup for this concentration cell is: Cathode: Ag (x M) e Ag E = 0.80 V Anode: Ag Ag (y M) e E = 0.80 V Ag (cathode, x M) Ag (anode, y M) = 0.00 V cell log Q = log For each concentration cell, we will calculate the cell potential using the above equation. Remember that the anode always has the smaller ion concentration. a. Since both compartments are at standard conditions ([Ag ] = 1.0 M) then = 0 V. No voltage is produced since no reaction occurs. Concentration cells only produce a voltage when the ion concentrations are not equal. b. Cathode =.0 M Ag ; Anode = 1.0 M Ag ; Electron flow is always from the anode to the

13 CHAPTER cathode, so electrons flow to the right in the diagram. log = log = V c. Cathode = 1.0 M Ag ; Anode = 0.10 M Ag ; Electrons flow to the left in the diagram. log = log = V 5 d. Cathode = 1.0 M Ag ; Anode = M Ag ; Electrons flow to the left in the diagram. log = 0.6 V e. Since the ion concentrations are the same, then log ([Ag ] /[Ag ] ) = log (1.0) = 0 and E = 0. No electron flow occurs. anode cathode cell 54. As is the case for all concentration cells, = 0, and the smaller ion concentration is always in the anode compartment. The general Nernst equation for the Ni Ni (x M) Ni (y M) Ni concentration cell is: log Q = log a. Since both compartments are at standard conditions ([Ni ] = 1.0 M), then = 0 V. No electron flow occurs. b. Cathode =.0 M Ni ; Anode = 1.0 M Ni ; Electron flow is always from the anode to the cathode, so electrons flow to the right in the diagram. 3 log = log = V c. Cathode = 1.0 M Ni ; Anode = 0.10 M Ni ; Electrons flow to the left in the diagram. log = V 5 d. Cathode = 1.0 M Ni ; Anode = M Ni ; Electrons flow to the left in the diagram. log = 0.13 V e. Since both concentrations are equal, log (.5/.5) = log 1.0 = 0 and E cell = 0. No electron flow occurs e 8 H MnO 4 Mn 4 HO E = 1.51 V 3 (Fe Fe e ) 5 E = 0.77 V 3 8 H (aq) MnO 4(aq) 5 Fe (aq) 5 Fe (aq) Mn (aq) 4 HO(l) = 0.74 V

14 486 CHAPTER 17 E cell = log Q = 0.74 V ; ph = 4.0 so H = 1 10 M E cell = E cell = 0.74 log (1 10 ) = 0.74 V 0.15 V = 0.59 V = 0.6 V (1 sig. fig. due to concentrations) Yes, E cell > 0 so the reaction will occur as written. 56. n = for this reaction (lead goes from Pb Pb in PbSO 4). E = E log =.04 V log E =.04 V V =.1 V 57. Cu e Cu E = 0.34 V Zn Zn e E = 0.76 V Cu (aq) Zn(s) Zn (aq) Cu(s) = 1.10 V Since Zn is a product in the reaction, the Zn concentration increases from 1.00 M to 1.0 M. This means that the reactant concentration of Cu must decrease from 1.00 M to 0.80 M (from the 1:1 mol ratio in the balanced reaction). log Q = 1.10 V log E cell = 1.10 V log = 1.10 V V = 1.09 V 58. (Pb e Pb) 3 E = 0.13 V 3 (Al Al 3 e ) E = 1.66 V 3 3 Pb (aq) Al(s) 3 Pb(s) Al (aq) = 1.53 V 3 3 From the balanced reaction, when the Al has increased by 0.60 mol/l (Al is a product in the spontaneous reaction), then the Pb concentration has decreased by 3/ (0.60 mol/l) = 0.90 M. E cell = 1.53 V = 1.53 E cell = 1.53 V V = 1.50 V 59. Cu (aq) H (g) H (aq) Cu(s) = 0.34 V 0.00V = 0.34 V; n = mol electrons Since = 1.0 atm and [H ] = 1.0 M: log

15 CHAPTER a. = 0.34 V log = 0.34 V 0.11V = 0.3 V b V = 0.34 V log, log = 4.91, [Cu ] = 10 = M Note: When determining exponents, we will carry extra significant figures Ni (aq) Al(s) Al (aq) 3 Ni(s) = = 1.43 V; n = 6 mol electrons a. E cell = 1.43 V = 1.43 E cell = 1.43 V (0.04 V) = 1.47 V 3 b. 1.6 V = 1.43 V, log [Al ] = [Al ] = 10, [Al ] =.3 10 M 61. Cu (aq) H (g) H (aq) Cu(s) = 0.34 V 0.00 V = 0.34 V; n = Since = 1.0 atm and [H ] = 1.0 M: log Use the K sp expression to calculate the Cu concentration in the cell. 19 Cu(OH) (s) Cu (aq) OH (aq) K sp = = [Cu ] [OH ] 17 From problem, [OH ] = 0.10 M, so: [Cu ] = = M log = 0.34 V log = = 0.16 V Since E cell < 0, the forward reaction is not spontaneous, but the reverse reaction is spontaneous. The Cu electrode becomes the anode and E cell = 0.16 V for the reverse reaction. The cell reaction is: H (aq) Cu(s) Cu (aq) H (g) Ni (aq) Al(s) Al (aq) 3 Ni(s) = 0.3 V 1.66 V = 1.43 V; n = 6 E cell =, 1.8 V = 1.43 V log [Al ] = 39.59, [Al ] = 10, [Al ] = M Al(OH) 3(s) Al (aq) 3 OH (aq) K sp = [Al ] [OH ] ; From the problem, [OH ] 4 = M.

16 488 CHAPTER K sp = ( ) ( ) = See Exercises 17.5, 17.7 and 17.9 for balanced reactions and standard cell potentials. Balanced reactions are necessary to determine n, the moles of electrons transferred. 3 5a. 7 HO Cr 3 Cl CrO 7 6 Cl 14 H = 0.03 V = 0.03 J/C G = 4 = (6 mol e )(96,485 C/mol e )(0.03 J/C) = J = 0 kj E = log Q: At equilibrium, E = 0 and Q = K, so = log K cell cell log K = = 3.05, K = 10 = 1 10 Note: When determining exponents, we will round off to the correct number of significant figures after the calculation is complete in order to help eliminate excessive roundoff error. 5 5b. G = ( mol e )(96,485 C/mol e )(.71 J/C) = J = 53 kj log K = = , K = a. G = ( mol e )(96,485 C/mol e )(0.7 J/C) = J = 5 kj log K = = 9.14, K = b. G = (10 mol e )(96,485 C/mol e )(0.09 J/C) = J = 90 kj log K = = 15.3, K = G = ; log K, log K = 5 6a. G = (10 mol e )(96,485 C/mol e )(0.43 J/C) = J = 410 kj log K = = 7.76, K = 10 = b. G = ( mol e )(96,485 C/mol e )(1.56 J/C) = J = 301 kj log K = = 5.79, K = a. G = ( mol e )(96,485 C/mol e )(1.10 J/C) =.1 10 J = 1 kj log K = = 37.5, K = b. G = (6 mol e )(96,485 C/mol e )(1.14 J/C) = J = 660. kj log K = = , K =

17 CHAPTER a. Possible reaction: I (s) Cl (aq) I (aq) Cl (g) = 0.54 V 1.36 V = 0.8 V This reaction is not spontaneous at standard conditions since < 0. No reaction occurs. b. Possible reaction: Cl (g) I (aq) I (s) Cl (aq) = 0.8 V; This reaction is spontaneous at standard conditions since > 0. The reaction will occur. Cl (g) I (aq) I (s) Cl (aq) = 0.8 V = 0.8 J/C G = 5 = ( mol e )(96,485 C/mol e )(0.8 J/C) = J = 160 kj E = log K, log K = = = 7.75, K = 10 = c. Possible reaction: Ag(s) Cu (aq) Cu(s) Ag (aq) = 0.46 V; No reaction occurs. d. Fe can be oxidized or reduced. The other species present are H, SO 4, HO, and O from air. Only O in the presence of H has a large enough standard reduction potential to oxidize Fe to 3 Fe (resulting in > 0). All other combinations, including the possible reduction of Fe, give negative cell potentials. The spontaneous reaction is: 3 4 Fe (aq) 4 H (aq) O (g) 4 Fe (aq) HO(l) = = 0.46 V G = = (4 mol e )(96,485 C/mol e )(0.46 J/C)(1 kj/1000 J) = 180 kj log K = = 31.13, K = a. Cu e Cu E = 0.5 V Cu Cu e E = 0.16 V Cu (aq) Cu (aq) Cu(s) = 0.36 V; Spontaneous G = = (1 mol e )(96,485 C/mol e)(0.36 J/C) = 34,700 J = 35 kj = log K, log K = = 6.09, K = 10 = b. Fe e Fe E = 0.44 V 3 (Fe Fe e ) E = 0.77 V 3 3 Fe (aq) Fe (aq) Fe(s) = 1.1 V; Not spontaneous c. HClO H e HClO H O E = 1.65 V E = 1.1 V HClO HO ClO 3 3 H e HClO (aq) ClO 3(aq) H (aq) HClO(aq) = 0.44 V; Spontaneous

18 490 CHAPTER 17 G = = ( mol e )(96,485 C/mol e )(0.44 J/C) = 84,900 J = 85 kj log K = = 14.89, K = a. 3 Au 3 e Au E = 1.50 V (Tl Tl e ) 3 E = 0.34 V 3 Au (aq) 3 Tl(s) Au(s) 3 Tl (aq) = 1.84 V 5 b. G = = (3 mol e )(96,485 C/mol e )(1.84 J/C) = J = 533 kj log K = = , K = 10 =.5 10 c. = 1.84 V log = 1.84 log = 1.84 (0.0) =.04 V (Cr Cr e ) Co e Co 3 Cr Co Cr Co 7 = log K = log ( ) = 0.0 V E = E log = 0.0 V log = V 4 G = nfe = ( mol e )(96,485 C/mol e )(0.151 J/C) = J = 9.1 kj 69. CdS e Cd S E = 1.1 V Cd Cd e E = 0.40 V CdS(s) Cd (aq) S (aq) = 0.81 V K sp =? log K = = 7.41, K = 10 = sp sp Al 3 e Al E = 1.66 V 3 Al 6 F AlF 3 e E =.07 V 6

19 CHAPTER Al (aq) 6 F (aq) AlF 6 (aq) = 0.41 V K =? log K = = 0.81, K = 10 = Ag e Ag E = 0.80 V 3 Ag S O Ag(S O ) e E = V Ag (aq) SO 3 (aq) Ag(SO 3) (aq) = 0.78 V K =? For this overall reaction, log K log K = = 13.0, K = 10 = CuI e Cu I =? Cu Cu e E = 0.5 V Electrolysis CuI(s) Cu (aq) I (aq) = 0.5 V 1 For this overall reaction, K = K sp = : 1 = log K sp = log ( ) = 0.71 V = 0.71 V = 0.5, = 0.19 V a. Al 3 e Al; 3 mol e are needed to produce 1 mol Al from Al g Al = s = 30. hours b. 1.0 g Ni = 33 s 3 c. 5.0 mol Ag = s = 1.3 hours 74. The oxidation state of bismuth in BiO is 3 because oxygen has a oxidation state in this ion. Therefore, 3 moles of electrons are required to reduce the bismuth in BiO to Bi(s) g Bi = 554 s = 9.3 min A = = C of charge passed in 1 hour

20 49 CHAPTER 17 4 a C = 16 g Co 4 b C = 5 g Hf 4 c. I I e ; C = 71 g I 6 d. CrO 3(l) Cr 3 O ; 6 mol e are needed to produce 1 mol Cr from molten CrO C = 4.9 g Cr Al is in the 3 oxidation in AlO 3, so 3 mol e are needed to convert Al into Al(s) h = g s = mol M where M = unknown metal Molar mass = ; The element is scandium. Sc forms 3 ions. 78. Alkaline earth metals form ions, so mol of e are transferred to form the metal, M. mol M = 748 s = mol M molar mass of M = = 4.3 g/mol; MgCl was electrolyzed. 79. F is produced at the anode: F F e.00 h = mol e mol e = mol F ; PV = nrt, = 9.1 L F K is produced at the cathode: K e K

21 CHAPTER mol e = 9. g K 80. The halfreactions for the electrolysis of water are: ( e H O H OH ) HO 4 H O 4 e HO(l) H (g) O (g) Note: 4 H 4 OH 4 HO and n = 4 for this reaction as it is written min = mol H At STP, 1 mole of an ideal gas occupies a volume of.4 L (see Chapter 5 of the text) mol H = 0.6 L = 6 ml H mol H = L = 131 ml O Al 3 e Al; 3 mol e are needed to produce Al from Al 4 4 = C/s or a current of A lb Al = 1 10 C of electricity needed 5 5 = 1 10 C/s = 1 10 A min = 138 s; 138 s = mol Ag 3 [Ag ] = mol Ag /0.50 L = M L mol Pt /L = mol Pt 43 To plate out 99% of the Pt, we will produce mol Pt mol Pt = 480 s

22 494 CHAPTER Au 3 e Au E = 1.50 V Ni e Ni E = 0.3 V Ag e Ag E = 0.80 V Cd e Cd E = 0.40 V HO e H OH E = 0.83 V Au(s) will plate out first since it has the most positive reduction potential, followed by Ag(s), which is followed by Ni(s), and finally Cd(s) will plate out last since it has the most negative reduction potential of the metals listed. 86. Species present: Fe, SO 4, H and HO. The possible cathode reactions are: SO SO O 4 4 H e H 3 H E = 0.0 V H e H E = 0.00 V Fe e Fe E = 0.44 V H O e H OH E = 0.83 V Reduction of SO 4 will occur at the cathode since is most positive. The possible anode reactions are: 3 Fe Fe e E = 0.77 V H O O 4 H 4 e E = 1.3 V Oxidation of Fe will occur at the anode since is most positive. 87. Reduction occurs at the cathode, and oxidation occurs at the anode. First, determine all the species present, then look up pertinent reduction and/or oxidation potentials in Table 17.1 for all these species. The cathode reaction will be the reaction with the most positive reduction potential, and the anode reaction will be the reaction with the most positive oxidation potential. a. Species present: Ni and Br ; Ni can be reduced to Ni, and Br can be oxidized to Br (from Table 17.1). The reactions are: Cathode: Ni e Ni E = 0.3 V Anode: Br Br e E = 1.09 V 3 3 b. Species present: Al and F ; Al can be reduced, and F can be oxidized. The reactions are: 3 Cathode: Al 3 e Al E = 1.66 V Anode: F F e E =.87 V c. Species present: Mn and I ; Mn can be reduced, and I can be oxidized. The reactions are: Cathode: Mn e Mn E = 1.18 V Anode: I I e E = 0.54 V 88. These are all in aqueous solutions, so we must also consider the reduction and oxidation of H O in addition to the potential redox reactions of the ions present. For the cathode reaction, the species with

23 CHAPTER the most positive reduction potential will be reduced, and for the anode reaction, the species with the most positive oxidation potential will be oxidized. a. Species present: Ni, Br and HO. Possible cathode reactions are: Ni e Ni E = 0.3 V H O e H OH E = 0.83 V Since it is easier to reduce Ni than HO (assuming standard conditions), Ni will be reduced by the above cathode reaction. Possible anode reactions are: Br Br e E = 1.09 V H O O 4 H 4 e E = 1.3 V Since Br is easier to oxidize than HO (assuming standard conditions), then Br will be oxidized by the above anode reaction. b. Species present: Al, F and H O; Al and H O can be reduced. The reduction potentials are E = 1.66 V for Al and E = 0.83 V for H O (assuming standard conditions). H O will be reduced at the cathode ( H O e H OH ) F and HO can be oxidized. The oxidation potentials are E =.87 V for F and E = 1.3 V for HO (assuming standard conditions). From the potentials, we would predict HO to be oxidized at the anode ( H O O 4 H 4 e ). c. Species present: Mn, I and HO; Mn and HO can be reduced. The possible cathode reactions are: Mn e Mn E = 1.18 V H O e H OH E = 0.83 V Reduction of HO will occur at the cathode since is most positive. I and HO can be oxidized. The possible anode reactions are: I I e E = 0.54 V HO O 4 H 4 e E = 1.3 V Oxidation of I will occur at the anode since is most positive. Additional Exercises 89. The halfreaction for the SCE is: HgCl e Hg Cl = 0.4 V

24 496 CHAPTER 17 For a spontaneous reaction to occur, E cell must be positive. Using the standard reduction potentials in Table 17.1 and the given SCE potential, deduce which combination will produce a positive overall cell potential. a. Cu e Cu E = 0.34 V E cell = = 0.10 V; SCE is the anode. 3 b. Fe e Fe E = 0.77 V E cell = = 0.53 V; SCE is the anode. c. AgCl e Ag Cl E = 0. V E cell = = 0.0 V; SCE is the cathode. 3 d. Al 3 e Al E = 1.66 V E cell = = 1.90 V; SCE is the cathode. e. Ni e Ni E = 0.3 V E cell = = 0.47 V; SCE is the cathode. 90. The potential oxidizing agents are NO 3 and H. Hydrogen ion cannot oxidize Pt under either condition. Nitrate cannot oxidize Pt unless there is Cl in the solution. Aqua regia has both Cl and NO. The overall reaction is: 3 (NO 4 H 3 e NO H O) E = 0.96 V (4 Cl Pt PtCl e ) 3 E = V 1 Cl (aq) 3 Pt(s) NO (aq) 8 H (aq) 3 PtCl (aq) NO(g) 4 H O(l) = 0.1 V Ag (aq) Cu(s) Cu (aq) Ag(s) = V = 0.46 V; A galvanic cell produces a voltage as the forward reaction occurs. Any stress that increases the tendency of the forward reaction to occur will increase the cell potential, while a stress that decreases the tendency of the forward reaction to occur will decrease the cell potential. a. Added Cu (a product ion) will decrease the tendency of the forward reaction to occur, which will decrease the cell potential. b. Added NH 3 removes Cu in the form of Cu(NH 3) 4. Removal of a product ion will increase the tendency of the forward reaction to occur, which will increase the cell potential. c. Added Cl removes Ag in the form of AgCl(s). Removal of a reactant ion will decrease the tendency of the forward reaction to occur, which will decrease the cell potential.

25 CHAPTER d. Q 1 = ; As the volume of solution is doubled, each concentration is halved. = = Q l The reaction quotient is doubled as the concentrations are halved. Since reactions are spontaneous when Q < K and since Q increases when the solution volume doubles, the reaction is closer to equilibrium, which will decrease the cell potential. e. Since Ag(s) is not a reactant in this spontaneous reaction, and since solids do not appear in the reaction quotient expressions, replacing the silver electrode with a platinum electrode will have no effect on the cell potential (Al 3 e Al) E = 1.66 V (M M e ) 3 E =? 3 3 M(s) Al (aq) Al(s) 3 M (aq) = E 1.66 V 3 G = nf, J = (6 mol e )(96,485 C/mol e ), = 0.71 V = E 1.66 V = 0.71 V, E =.37 or E =.37 From table 17.1, the reduction potential for Mg e Mg is.37 V, which fits the data. Hence, the metal is magnesium. 93. a. G = = (480.) 3(86) [3(40.)] = 58 kj From oxidation numbers, n = 6. G = nfe, E = = 1.01 V log K = = , K = 10 = b. e AgS Ag S ) 3 =? 3 (Al Al 3 e ) E = 1.66 V 3 3 AgS(s) Al(s) 6 Ag(s) 3 S (aq) Al (aq) = 1.01 V = 1.66V = 1.01 V 1.66 V = 0.65 V 94. Zn Zn e E = 0.76 V; Fe Fe e E = 0.44 V It is easier to oxidize Zn than Fe, so the Zn would be oxidized, protecting the iron of the Monitor's hull From Exercise 17.7a: 3 Cl (g) Cr (aq) 7 HO(l) 14 H (aq) CrO 7 (aq) 6 Cl (aq) = 0.03 V

26 498 CHAPTER When KCrO 7 and Cl are added to concentrated HSO 4, Q becomes a large number due to [H ] term. The log of a large number is positive. E cell becomes negative, which means the reverse reaction becomes spontaneous. The pungent fumes were Cl (g). 96. Aluminum has the ability to form a durable oxide coating over its surface. Once the HCl dissolves this 3 oxide coating, Al is exposed to H and is easily oxidized to Al, i.e., the Al foil disappears after the oxide coating is dissolved. 97. Consider the strongest oxidizing agent combined with the strongest reducing agent from Table 17.1: F e F E =.87 V (Li Li e ) E = 3.05 V F (g) Li(s) Li (aq) F (aq) = 5.9 V The claim is impossible. The strongest oxidizing agent and reducing agent when combined only give an value of about 6 V. 98. H (g) O (g) HO(l); Oxygen goes from the zero oxidation state to the oxidation state in H O. Since two mol O appear in the balanced reaction, then n = 4 mol electrons transferred. 83 a. = log K = log ( ), = 1.3 V 5 G = nf = (4 mol e )(96,485 C/mol e )(1.3 J/C) = J = 475 kj b. Since mol of gas decrease as reactants are converted into products, then S will be negative (unfavorable). Since the value of G is negative, then H must be negative ( G = H T S ). c. G = w max = H T S. Since S is negative, then as T increases, G becomes more positive (closer to zero). Therefore, w will decrease as T increases. max 99. a. O HO 4 e 4 OH E = 0.40 V (H OH H O e ) E = 0.83 V H (g) O (g) HO(l) = 1.3 V = 1.3 J/C Since standard conditions are assumed, then w max = G for mol HO produced. G = = (4 mol e )(96,485 C/mol e )(1.3 J/C) = 475,000 J = 475 kj 3 For g HO produced, w max is: g HO = 13,00 kj = wmax

27 CHAPTER The work done can be no larger than the free energy change. The best that could happen is that all of the free energy released would go into doing work, but this does not occur in any real process since there is always waste energy in a real process. Fuel cells are more efficient in converting chemical energy into electrical energy; they are also less massive. The major disadvantage is that they are expensive. In addition, H (g) and O (g) are an explosive mixture if ignited; much more so than fossil fuels Cadmium goes from the zero oxidation state to the oxidation state in Cd(OH). Since one mol of Cd appears in the balanced reaction, then n = mol electrons transferred. At standard conditions: 5 w = G = nfe, w = ( mol e )(96,485 C/mol e )(1.10 J/C) =.1 10 J = 1 kj max max 101. (CO O CO e ) O 4 e O CO O CO G = nfe, E = = = 0.98 V 10. In the electrolysis of aqueous sodium chloride, H O is reduced in preference to Na, and Cl is oxidized in preference to H O. The anode reaction is Cl Cl e, and the cathode reaction is H O e H OH. The overall reaction is: HO(l) Cl (aq) Cl (g) H (g) OH (aq). From the 1:1 mol ratio between Cl and H in the overall balanced reaction, if 57 L of Cl (g) are produced, then 57 L of H (g) will also be produced since moles and volume of gas are directly proportional at constant T and P (see Chapter 5 of text) mol e = 50.0 min = mol e mol Ru =.618 g Ru = mol Ru 3 = 3.00; The charge on the ruthenium ions is 3 (Ru 3 e Ru) kwh = = J or kj (Hall process) To melt 1.0 kg Al requires: g Al = kj 3 It is feasible to recycle Al by melting the metal because, in theory, it takes less than 1% of the energy required to produce the same amount of Al by the Hall process.

28 500 CHAPTER 17 Challenge Problems 105. G = nfe = H T S, E = If we graph E vs. T we should get a straight line (y = mx b). The slope of the line is equal to S /nf, and the yintercept is equal to H /nf. From the equation above, E will have a small temperature dependence when S is close to zero a. We can calculate G from G = H T S and then E from G = nfe ; or we can use the equation derived in Exercise For this reaction, n = (from oxidation states). = 1.98 J/C = 1.98 V b. ln Q = 1.98 V ln E 0 = 1.98 V ln = 1.98 V V =.05 V c. From Exercise 17.56, E =.1 V at 5 C. As the temperature decreases, the cell potential decreases. Also, oil becomes more viscous at lower temperatures, which adds to the difficulty of starting an engine on a cold day. The combination of these two factors results in batteries failing more often on cold days than on warm days (Ag e Ag) E = 0.80 V Pb Pb e E = (0.13) Ag Pb Ag Pb = 0.93 V E = E log, 0.83 V = 0.93 V log 3.4 log = = 3.4, = 10, [Ag ] = 0.07 M AgSO 4(s) Ag (aq) SO 4 (aq) K sp = [Ag ] [SO 4 ] Initial s = solubility (mol/l) 0 0 Equil. s s From problem: s = 0.07 M, s = 0.07/ 6 K sp = (s) (s) = (0.07) (0.07/) = a. Zn(s) Cu (aq) Zn (aq) Cu(s) = 1.10 V; E cell = 1.10 V

29 CHAPTER E cell = 1.10 V = 1.10 V V = 1.14 V b h = 1.87 mol Cu produced The Cu concentration will decrease by 1.87 mol/l, and the Zn concentration will increase by 1.87 mol/l. [Cu ] = = 0.63 M; [Zn ] = = 1.97 M E cell = 1.10 V = 1.10 V V = 1.09 V c mol Zn consumed = 1 g Zn; Mass of electrode = = 78 g Zn 1.87 mol Cu formed = 119 g Cu; Mass of electrode = = 319 g Cu d. Three things could possibly cause this battery to go dead: 1. All of the Zn is consumed.. All of the Cu is consumed. 3. Equilibrium is reached (E = 0). cell We began with.50 mol Cu and 00. g Zn 1 mol Zn/65.38 g Zn = 3.06 mol Zn. Cu is the limiting reagent and will run out first. To react all the Cu requires:.50 mol Cu = 13.4 h For equilibrium to be reached: E = 0 = 1.10 V (1.10)/ = K = 10 = This is such a large equilibrium constant that virtually all of the Cu must react to reach equilibrium. So, the battery will go dead in 13.4 hours H e H E = V Fe Fe e E = (0.440V) H (aq) Fe(s) H (g) Fe (aq) = V log Q, where n = and Q = To determine K a for the weak acid, first use the electrochemical data to determine the H concentration in the halfcell containing the weak acid V = V

30 50 CHAPTER = 10 = , [H ] = M Now we can solve for the K a value of the weak acid HA through the normal setup for a weak acid problem. HA H A K a = Initial 1.00 M ~0 0 Equil x x x K = where x = [H ] = M, K = = a 4 7 a 110. a. Nonreactive anions are present in each halfcell to balance the cation charges. 3 3 b. Au (aq) 3 Fe (aq) 3 Fe (aq) Au(s) = = 0.73 V E cell = log Q = 0.73 V 3 Since [Fe ] = [Fe ] = 1.0 M: 0.31 V = 0.73 V , log [Au ] = 1.3, [Au ] = 10 = M Au 4 Cl AuCl ; Since the equilibrium Au concentration is so small, assume [AuCl ] [Au ] 1.0 M, i.e., assume K is large, so the reaction essentially goes to completion o K = 5 =.1 10 ; Assumption good (K is large).

31 CHAPTER a. E = E ph, V = 0.50 V ph cell ref ph = = 3.888; Uncertainty = ± 1 mv = ± V ph = = 3.905; ph = = max min So, if the uncertainty in potential is ± V, the uncertainty in ph is ± or about ± ph units. For this measurement, [H ] = 10 = M. For an error of 1 mv, [H ] = = M. For an error of 1 mv, [H ] = 10 = M. So, the 4 6 uncertainty in [H ] is ± M = ± 6 10 M. b. From part a, we will be within ± 0.0 ph units if we measure the potential to the nearest ± V (1 mv). 11. a. From Table 17.1: HO e H OH E = 0.83 V = 0.83 V.36 V = 1.53 V Yes, the reduction of HO to H by Zr is spontaneous at standard conditions since > 0. b. ( H O e H OH ) Zr 4 OH ZrO H O H O 4 e 3 HO(l) Zr(s) H (g) ZrO HO(s) 5 c. G = nfe = (4 mol e )(96,485 C/mol e )(1.53 J/C) = J = 590. kj E = E log Q; At equilibrium, E = 0 and Q = K. E = log K, log K = = 104, K d kg Zr = mol H mol H = g H 6 V = =.3 10 L H e. Probably yes; Less radioactivity overall was released by venting the H than what would have been released if the H had exploded inside the reactor (as happened at Chernobyl). Neither

32 504 CHAPTER 17 alternative is pleasant, but venting the radioactive hydrogen is the less unpleasant of the two alternatives a. (Ag e Ag) E = 0.80 V Cu Cu e E = 0.34 V Ag (aq) Cu(s) Ag(s) Cu (aq) = 0.46 V log Q where n = and Q = To calculate E, we need to use the K data to determine [Ag ]. cell sp 10 AgCl(s) Ag (aq) Cl (aq) K sp = = [Ag ] [Cl ] Initial s = solubility (mol/l) 0 0 Equil. s s 10 5 K sp = = s, s = [Ag ] = mol/l E cell = 0.46 V log = 0.46 V 0.30 = 0.16 V 13 b. Cu (aq) 4 NH 3(aq) Cu(NH 4) 4 (aq) K = = Since K is very large for the formation of Cu(NH 3) 4,the forward reaction is dominant. At equilibrium, essentially all of the.0 M Cu will react to form.0 M Cu(NH 3) 4. This reaction requires 8.0 M NH 3 to react with all of the Cu in the balanced equation. Therefore, the mol of NH 3 added to 1.0 L solution will be larger than 8.0 mol since some NH 3 must be present at equilibrium. In order to calculate how much NH 3 is present at equilibrium, we need to use the electrochemical data to determine the Cu concentration. log Q, 0.5 V = 0.46 V log.03 3 log =.03, = 10 = [Cu ] = = 10 M (We carried extra significant figures in the calculation.) Note: Our assumption that the.0 M Cu essentially reacts to completion is excellent as only 1 10 M Cu remains after this reaction. Now we can solve for the equilibrium [NH ]. 3

33 CHAPTER K = =, [NH ] = 0.6 M 3 Since 1.0 L of solution is present, then 0.6 mol NH 3remains at equilibrium. The total mol of NH3 added is 0.6 mol plus the 8.0 mol NH 3necessary to form.0 M Cu(NH 3) 4. Therefore, = 8.6 mol NH were added. 3