Homework #3 Chapter 11 Electrochemistry

Homework #3 Chapter 11 Electrochemistry Chapter 4 83. a) Oxidation ½ Reaction Fe + HCl HFeCl 4 Fe + 4HCl HFeCl 4 Fe + 4HCl HFeCl 4 + 3H + Fe + 4HCl HFeCl 4 + 3H + + 3e Reduction ½ Reaction H 2 2H + H 2 2H + + 2e H 2 Balanced Reaction 2(Fe + 4HCl HFeCl 4 + 3H + + 3e ) 3(2H + + 2e H 2 ) 2Fe(s) + 8HCl(aq) 2HFeCl 4 (aq) + 3H 2 (g) b) Oxidization ½ Reaction I I 3 3I I 3 3I I 3 + 2e Reduction ½ Reaction IO 3 I 3 3IO 3 I 3 3IO 3 I 3 + 9H 2 O 3IO 3 + 18H + I 3 + 9H 2 O 3IO 3 + 18H + + 16e I 3 + 9H 2 O Balanced Reaction 3IO 3 + 18H + + 16e I 3 + 9H 2 O 8(3I I 3 + 2e ) 3IO 3 (aq) + 18H + (aq) + 24I (aq) 9I 3 (aq) + 9H 2 O(l) Divide trough by 3 IO 3 (aq) + 6H + (aq) + 8I (aq) 3I 3 (aq) + 3H 2 O(l) c) Oxidation ½ Reaction 4 Cr(NCS) 6 Cr 3+ 2 + NO 3 + CO 2 + SO 4 4 Cr(NCS) 6 Cr 3+ 2 + 6NO 3 + 6CO 2 + 6SO 4 4 Cr(NCS) 6 + 54H 2 O Cr 3+ 2 + 6NO 3 + 6CO 2 + 6SO 4 4 Cr(NCS) 6 + 54H 2 O Cr 3+ 2 + 6NO 3 + 6CO 2 + 6SO 4 + 108H + 4 Cr(NCS) 6 + 54H 2 O Cr 3+ 2 + 6NO 3 + 6CO 2 + 6SO 4 + 108H + + 97e Reduction ½ Reaction Ce 4+ Ce 3+ Ce 4+ + e Ce 3+ 1

Balanced Reaction Cr(NCS) 6 4 + 54H 2 O Cr 3+ + 6NO 3 + 6CO 2 + 6SO 4 2 + 108H + + 97e 97(Ce 4+ + e Ce 3+ ) Cr(NCS) 6 4 (aq)+ 54H 2 O(l) + 97Ce 4+ (aq) Cr 3+ (aq) + 6NO 3 (aq) + 6CO 2 (g) + 6SO 4 2 (aq) + 108H + (aq) + 97Ce 3+ (aq) d) Oxidation ½ Reaction CrI 3 CrO 4 2 + IO 4 CrI 3 CrO 4 2 + 3IO 4 CrI 3 + 16H 2 O CrO 4 2 + 3IO 4 CrI 3 + 16H 2 O CrO 4 2 + 3IO 4 + 32H + CrI 3 + 16H 2 O CrO 4 2 + 3IO 4 + 32H + + 27e Reduction ½ Reaction Cl 2 Cl Cl 2 2Cl Cl 2 + 2e 2Cl Balanced Equation 2(CrI 3 + 16H 2 O CrO 4 2 + 3IO 4 + 32H + + 27e ) 27(Cl 2 + 2e 2Cl ) 2CrI 3 (s) + 32H 2 O(l) + 27Cl 2 (g) 2CrO 4 2 (aq) + 6IO 4 (aq) + 64H + (aq) + 54Cl (aq) The solution is basic not acidic, add 64 OH to both sides 2CrI 3 (s) + 27Cl 2 (g) + 64OH (aq) 2CrO 4 2 (aq) + 6IO 4 (aq) + 32H 2 O(l) + 54Cl (aq) e) Oxidation ½ Reaction Fe(CN) 6 4 Fe(OH) 3 + CO 3 2 + NO 3 Fe(CN) 6 4 Fe(OH) 3 + 6CO 3 2 + 6NO 3 Fe(CN) 6 4 + 39H 2 O Fe(OH) 3 + 6CO 3 2 + 6NO 3 Fe(CN) 6 4 + 39H 2 O Fe(OH) 3 + 6CO 3 2 + 6NO 3 + 75H + Fe(CN) 6 4 + 39H 2 O Fe(OH) 3 + 6CO 3 2 + 6NO 3 + 75H + + 61e Reduction ½ Reaction Ce 4+ Ce(OH) 3 Ce 4+ + 3H 2 O Ce(OH) 3 Ce 4+ + 3H 2 O Ce(OH) 3 + 3H + Ce 4+ + 3H 2 O + e Ce(OH) 3 + 3H + Balance Reaction Fe(CN) 6 4 + 39H 2 O Fe(OH) 3 + 6CO 3 2 + 6NO 3 + 75H + + 61e 61(Ce 4+ + 3H 2 O + e Ce(OH) 3 + 3H + ) Fe(CN) 6 4 (aq) + 222H 2 O(l) + 61Ce 4+ (aq) Fe(OH) 3 (s) + 6CO 3 2 (aq) + 6NO 3 (aq)+ 258H + (aq) + 61Ce(OH) 3 (s) The solution is basic not acidic, add 258OH to both sides Fe(CN) 6 4 (aq) + 258OH (aq) + 61Ce 4+ (aq) Fe(OH) 3 (s) + 6CO 3 2 (aq) + 6NO 3 (aq)+ 61Ce(OH) 3 (s) + 36H 2 O(l) 2

Chapter 11 3. Reactions of Interest Ni 2+ + 2e Ni E =0.23 V Cu + + e Cu E =0.52 V Cu 2+ + 2e Cu E =0.34 V Zn 2+ + 2e Zn E =0.76 V In order to plate out Ni you need the Ni reaction to be the reduction ½ reaction (cathode). In addition, you also need the cell to be galvanic (E >0). The E cell of the nickel/copper cell is 0.75 V or 0.57 V depending on the ion of copper that is used. Therefore, neither of these cells would be a galvanic cell, resulting in copper not being an appropriate material. The E cell of the nickel/zinc cell is 0.53 V. Therefore, the nickel/zink cell would plate out Ni. 10. E is the reaction potential at the standard state. The standard state is 1 M concentration of aqueous solutions or 1 atm pressure of gasses. By convention E is set to 0 for hydrogen s reduction ½ reaction (2H + + 2e H 2 ). E is the reaction potential not at the standard state. E is 0 when the cell is at equilibrium. 19. a) Reactions of interest 2 Cr 2 O 7 + 14H + + 6e 2Cr 3+ + 7H 2 O E = 1.33 V Cl 2 + 2e 2Cl E = 1.36 V 3(Cl 2 + 2e 2Cl ) E = 1.36 V 2Cr 3+ 2 + 7H 2 O Cr 2 O 7 + 14H + + 6e E = 1.33 V 3Cl 2 (g) + 2Cr 3+ (aq) + 7H 2 O(l) 6Cl (aq) + Cr 2 O 2 7 (aq) + 14H + (aq) E =1.36 V1.33 V=0.03V 3

b) Reactions of Interest Cu 2+ + 2e Cu E = 0.34 V Mg 2+ + 2e Mg E =2.37 V Cu 2+ + 2e Cu E = 0.34 V Mg Mg 2+ + 2e E =2.37 V Mg(s) + Cu 2+ (aq) Cu(s) + Mg 2+ (aq) E = 0.34 V + 2.37 V = 2.71 V c) Reactions of Interest IO 3 + 6H + + 5e ½I 2 +3H 2 O E =1.20 V Fe 3+ + e Fe 2+ E = 0.77 V IO 3 + 6H + + 5e ½I 2 +3H 2 O 5(Fe 2+ Fe 3+ + e ) IO 3 (aq) + 6H + (aq) + 5Fe 2+ (aq) ½I 2 (s) +3H 2 O(l) + 5Fe 3+ (aq) E = 1.20 V E =0.77 V E =1.20 V0.77V= 0.43 V Note: I 2 does not conduct electricity; therefore, the Pt electrode is used d) Reactions of Interest Ag + + e Ag E = 0.80 V Zn 2+ + 2e Zn E = 0.76 V 2(Ag + + e Ag) E = 0.80 V Zn Zn 2+ + 2e E = 0.76 V 2Ag + (aq) + Zn(s) 2Ag(s) + Zn 2+ (aq) E = 0.80 V + 0.76 V = 1.56 V 4

21. a) Cl 2 + 2e 2Cl E = 1.36 V 2Br Br 2 + 2e E =1.09 V Cl 2 (g) + 2Br (aq) 2Cl (aq) + Br 2 (aq) E =1.36 V 1.09 V= 0.27 V Note: Br 2 is a liquid at room temperature b) 5(IO 3 + 2H + + 2e IO 3 +H 2 O) 2(Mn 2+ + 4H 2 O MnO 4 + 8H + + 5e ) 5IO 3 (aq)+2mn 2+ (aq)+3h 2 O(l)5IO 3 (aq)+2mno 4 (aq)+6h + (aq) E = 1.60 V E = 1.51 V E = 1.60V 1.51V=0.09V 5

c) H 2 O 2 +2H + + 2e 2H 2 O E = 1.78 V H 2 O 2 O 2 + 2H + + 2e E = 0.68 V 2H 2 O 2 (aq) O 2 (g) + 2H 2 O(l) E = 1.78 V 0.68 V = 1.10 V d) 2(Fe 3+ + 3e Fe) E = 0.036V 3(Mn Mn 2+ + 2e ) E = 1.18 V 2Fe 3+ (aq) + 3Mn(s) 2Fe(s) + Mn 2+ (aq) E = 0.036 V + 1.18 V = 1.14 V 22. a) For a galvanic cell E must be positive therefore Cu + must be oxidized given: Au 3+ + 3e Au E = 1.50V Cu 2+ + e Cu + E = 0.16 V Au 3+ + 3e Au E = 1.50V 3(Cu + Cu 2+ +e ) E = 0.16 V Au 3+ (aq) + 3Cu + (aq) Au(s) + 3Cu 2+ (aq) E = 1.34 V 6

b) For a galvanic cell E must be positive therefore Cd must be oxidized given: + VO 2 + 2H + + e VO 2+ + H 2 O E = 1.00V Cd 2+ +2e Cd E = 0.40 V + 2(VO 2 + 2H + + e VO 2+ + H 2 O) E = 1.00V Cd Cd 2+ +2e E = 0.40 V 2VO + 2 (aq) +4H + (aq) + Cd(s) 2VO 2+ (aq) +2H 2 O(l) + Cd 2+ (aq) E = 1.40 V 24. a) Cu 2+ + 2e Cu E = 0.34 V In order for the cell to be a galvanic cell Cu 2+ must be reduced, therefore, SCE is oxidized and at the anode. E cell = 0.34 V + 0.242 V = 0.10 V b) Fe 3+ + e Fe 2+ E = 0.77 V In order for the cell to be a galvanic cell, Fe 3+ must be reduced, therefore, SCE is oxidized and at the anode. E cell = 0.77 V + 0.242 V = 0.53 V c) AgCl + e Ag + Cl E = 0.22 V In order for the cell to be a galvanic cell, Ag must be oxidized, therefore, SCE is reduced and at the cathode. E cell = 0.242 V 0.22 V = 0.02 V d) Al 3+ + 3e Al E = 1.66 V In order for the cell to be a galvanic cell, Al must be oxidized, therefore, SCE is reduced and at the cathode. E cell = 0.242 V + 1.66 V = 1.90 V e) Ni 2+ + 2e Ni E = 0.23 V In order for the cell to be a galvanic, cell Ni must be oxidized, therefore, SCE is reduced and at the cathode. E cell = 0.242 V + 0.23 V = 0.47 V 25. a) Cu Cu 2+ +2e E = 0.34 V 2H + + 2e H 2 E = 0.0 V No H + cannot oxidize Cu b) 2I I 2 +2e E =0.54 V Fe 3+ + e Fe 2+ E = 0.77 V Yes Fe 3+ is capable of oxidizing I if it is going to Fe 2+ 2I I 2 +2e E =0.54 V Fe 3+ + 3e Fe E = 0.04 V No Fe 3+ cannot oxidize I if it is going to Fe 7

c) Ag + + e Ag E =0.80 V H 2 2H + +2e E = 0.0 V Yes H 2 is capable of reducing Ag + d) Cr 3+ + e Cr 2+ E = 0.50 V Fe 2+ Fe 3+ + e E =0.77 V No Fe 2+ is not capable of reducing Cr 3 + to Cr 2+ 26. The oxidizing agent is the species that is reduced. Therefore, the best oxidizing reagent is the species that has the largest E value for the reduction ½ reaction. K + < H 2 O < Cd 2+ < I 2 < AuCl 4 < IO 3 27. The reducing agent is the species that is oxidized. Therefore, the best reducing agent is the species that has the smallest E value for the reduction ½ reaction. F < H 2 O < I 2 < Cu + < H < K 28. Choices Cl 2 + 2e 2Cl Ag + + e Ag Pb 2+ + 2e Pb E = 1.36 V E = 0.80 V E = 0.13 V Zn 2+ + 2e Zn E = 0.76 V Na + + e Na E = 2.71 V Underlined are possible answers a) The oxidizing agent is the species that is being reduced or species in the reduction reaction. Therefore Cl, Ag, Pb, and Zn can be eliminated because if they were on the reactants side of the reaction the reaction would be an oxidation reaction, therefore, they can only be reducing agents. Out of the remaining species the best oxidizing agent is the one with the largest E value in the reduction ½ reactions. Ag + b) The reducing agent is a species that is being oxidized or the species in the oxidizing reaction. Therefore Ag +, Zn 2+, and Na + can be eliminated because if you flip the reaction to make it an oxidation reaction they are on the product side of the reaction and not the reactants, therefore, they can only be oxidizing agents. Out of the remaining species the best oxidizing agent is the one with the most negative E value in the reduction ½ reactions (the E value will become positive when the reaction is flipped). Zn 2 c) SO 4 + 4H + + 2e H 2 SO 3 + H 2 O E = 0.20 V 2 In order for SO 4 to oxidize a species the species has to be in an oxidation reaction; therefore, Ag +, Zn 2+, and Na + can be eliminated because they are on the reactant side of reduction reactions. Of the remaining species the E value of the reduction ½ reaction must be smaller than 0.20 V. Pb and Zn 8

d) Al 3+ + 3e Al E = 1.66 V Reaction of interest Al Al 3+ + 3e E = 1.66 V In order for Al to reduce a species it must be in a reduction reaction. Therefore, Cl, Ag, Pb, and Zn can be eliminated because if they were on the reactants side of the reaction it would be an oxidation reaction. Of the remaining species the E value of the reduction ½ reaction must be greater than 1.66 V. Zn 2+ and Ag + 29. a) Br 2 + 2e 2Br E = 1.09 V Cl 2 + 2e 2Cl E =1.36 V Reactions of interest 2Br Br 2 + 2e E = 1.09 V 2Cl Cl 2 + 2e E =1.36 V In order to oxidize Br and not Cl the E value of the reduction ½ reaction must be greater than 1.09 V but smaller than 1.36 V. 2 Cr 2 O 7 + 14H + + 6e 2Cr 3+ + 7H 2 O E = 1.33 V O 2 + 4H + + 4e 2H 2 O E = 1.23 V MnO 2 + 4H + + 4e Mn 2+ + 2H 2 O E = 1.21 V IO 3 + 6H + + 5 e ½I 2 + 3H 2 O E = 1.20 V Therefore, Cr 2 O 2 7, O 2, MnO 2, and IO 3 could oxidize Br to Br 2 but not oxidize Cl to Cl 2. b) Mn 2+ + 2e Mn E = 1.18 V Ni 2+ + 2e Ni E = 0.23 V Reactions of interest Mn Mn 2+ + 2e E = 1.18 V Ni Ni 2+ + 2e E = 0.23 V In order to oxidize Mn and not Ni the E value of the reduction ½ reaction must be greater than 1.18 V but smaller than 0.23 V. PbSO 4 + 2e 2 Pb + SO 4 E = 0.35 V Cd 2+ +2e Cd E = 0.40 V Fe 2+ + 2e Fe E = 0.44 V Cr 3+ + e Cr 2+ E = 0.50 V Cr 3+ + e Cr E = 0.73 V Zn 2+ + 2e Zn E =0.76 V 2H 2 O + 2e H 2 + 2OH E = 0.83 V Therefore, PbSO 4, Cd 2+, Fe 2+, Cr 3+, Zn 2+, and H 2 O are capable of oxidizing Mn to Mn 2+ but not oxidizing Ni to Ni 2+. 9

30. a) Cu 2+ + 2e Cu E = 0.34 V Cu 2+ + e Cu + E = 0.16 V In order to reduce Cu 2+ to Cu but not to Cu + the E values of the reduction ½ reaction must be greater than 0.16 V and less than 0.34 V. The species also must be on the product side of the reduction reaction. HgCl 2 + 2e 2Hg + 2Cl E = 0.27 V AgCl + e Ag + 4Cl E = 0.22 V 2 SO 4 + 4H + + 2e H 2 SO 4 + H 2 O E = 0.20 V Therefore, Hg/Cl, Ag/Cl, and H 2 SO 3 are capable of reducing Cu 2+ to Cu but not to Cu +. b) Br 2 + 2e 2Br E = 1.09 V I 2 + 2e 2I E =0.54 V In order to reduce Br 2 to Br but not I 2 to I the E values of the reduction ½ reaction must be great than 0.54 V and less than 1.09 V. The species also must be on the product side of the reduction reaction. + VO 2 + 2H + + e VO 2+ + H 2 O E = 1.00V AuCl 4 + 3e Au + 4Cl E = 0.99V NO 3 + 4H + + 3e NO + 2H 2 O E = 0.96V ClO 2 + e ClO 2 E =0.954 V 2Hg 2+ + 2e 2+ Hg 2 E =0.91 V Ag + + e Ag E = 0.80 V 2Hg 2+ + 2e 2Hg E =0.80 V Fe 3+ + 2e Fe 2+ E = 0.77 V O 2 + 2H + + 2e H 2 O 2 E =0.68 V MnO 4 + e 2 MnO 4 E =0.56 V Therefore, VO 2+,Au/Cl, NO, ClO 2, Hg 2+ 2, Ag, Hg, Fe 2+, H 2 O 2, and MnO 4 are capable of reducing Br 2 to Br but not I 2 to I. 42. a) 2(ClO 2 ClO 2 +e ) E =0.95 V Cl 2 + 2e 2Cl E = 1.36 V 2ClO 2 (aq) + Cl 2 (g) 2ClO 2 (g) + 2Cl (aq) E =1.36 V 0.95 V = 0.41 V Note: The Na + in the equation is just there as a spectator ion 296,485 79,.. 7.410 10

b) Assume acidic conditions Reduction ½ Reaction ClO 2 Cl ClO 2 Cl + 2H 2 O ClO 2 + 4H + Cl +2H 2 O ClO 2 + 4H + + 5e Cl +2H 2 O Oxidation ½ Reaction ClO 2 ClO 3 ClO 2 + H 2 O ClO 3 ClO 2 + H 2 O ClO 3 + 2H + ClO 2 + H 2 O ClO 3 + 2H + + e Balanced Reaction ClO 2 + 4H + + 5e Cl +2H 2 O 5(ClO 2 + H 2 O ClO 3 + 2H + + e ) 6ClO 2 (g) + 3H 2 O(l) Cl (aq) + 5ClO 3 (aq) + 6H + (aq) 43. a) Reaction 1: Unbalanced Mn(s) + NO 3 (aq) NO(g) + Mn 2+ (aq) Oxidation ½ reaction Mn Mn 2+ Mn Mn 2+ + 2e Reduction ½ reaction (The problem told you that you have nitric acid as a reactant. To determine what species NO 3 forms, use the standard reduction potentials) NO 3 NO NO 3 NO + 2H 2 O NO 3 + 4H + NO + 2H 2 O NO 3 + 4H + + 3e NO + 2H 2 O Balanced Reaction 3(Mn Mn 2+ + 2e ) 2(NO 3 + 4H + + 3e NO + 2H 2 O) 3Mn(s) + 2NO 3 (aq) +8H + (aq) 3Mn 2+ (aq) + 2NO(g) + 4H 2 O(l) Reaction 2: Unbalanced Mn 2+ (aq)+ IO 4 (aq) MnO 4 (aq) + IO 3 (aq) Oxidation ½ reaction Mn 2+ MnO 4 Mn 2+ + 4H 2 O MnO 4 Mn 2+ + 4H 2 O MnO 4 + 8H + Mn 2+ + 4H 2 O MnO 4 + 8H + + 5e Reduction ½ Reaction IO 4 IO 3 IO 4 IO 3 + H 2 O IO 4 + 2H + + 2e IO 3 + H 2 O IO 4 + 2H + + 2e IO 3 + H 2 O 11

Balanced Reaction 2(Mn 2+ + 4H 2 O MnO 4 + 8H + + 5e ) 5(IO 4 + 2H + + 2e IO 3 + H 2 O) 2Mn 2+ (aq) + 5IO 4 (aq) +3H 2 O(l) 2MnO 4 (aq) + 5IO 3 (aq) + 6H + (aq) b) Reaction 1 3(Mn Mn 2+ + 2e ) E = 1.18 V 2(NO 3 + 4H + + 3e NO + 2H 2 O) E = 0.96 V 3Mn(s) + 2NO 3 (aq) +8H + (aq) 3Mn 2+ (aq) + 2NO(g) + 4H 2 O(l) E cell = 1.18 V + 0.96 V = 2.14 V 696,485 2.14 1.2410 1,240,... The K number is most likely too large for your calculator Reaction 2 2(Mn 2+ + 4H 2 O MnO 4 + 8H + + 5e ) E = 1.51 V 5(IO 4 + 2H + + 2e IO 3 + H 2 O) E = 1.60 V 2Mn 2+ (aq) + 5IO 4 (aq) +3H 2 O(l) 2MnO 4 (aq) + 5IO 3 (aq) + 6H + (aq) E cell = 1.60 V + 1.51 V = 0.09 V 10 96,485 0.09 910,.. 210 90 44. a) 8.3145 298 1.28 10 4.7410 474 Need to find number of e transferred 2H 2 (g) + O 2 (g) 2H 2 O(l) H 0 +1 1 e since H 2 total of 2 e O 02 2 e since O 2 total of 4e Therefore, 4 e total were transferred in reaction 8.3145 298 496,485 1.28 10 1.22 b) Two gases go to a liquid, therefore, the positional probability decreases causing ΔS to be negative. If ΔS is negative the only way for ΔG to be negative is if ΔH is negative. 12

c) ΔG is the maximum possible work that can be done. Increasing the temperature increases the TΔS term, therefore, since this term is +, less possible work can be done. 45. If E vs. T was plotted on a graph, the intercept would equal and the slope of the line would equal. The smaller the term is the smaller the temperature dependence of E. Therefore, cells that have small ΔS terms are relatively temperature independent. 47. a) Cu + Cu 2+ + e E = 0.16 V Cu + + e Cu E = 0.52 V 2Cu + Cu 2+ + Cu E = 0.16 V + 0.52 V = 0.36 V This reaction is spontaneous 196,485,.. 1.210 35 b) 2(Fe 2+ Fe 3+ +e ) E = 0.44 V Fe 2+ +2e Fe E = 0.77 V 3Fe 2+ 2Fe 3+ + Fe E =0.44 V+ 0.77 V=1.21 V This reaction is not spontaneous c) HClO 2 + H 2 O ClO 3 + 3H + + 2e E =1.21 V HClO 2 + 2H + + 2e HClO + H 2 O E =1.65 V 2HClO 2 (aq) ClO 3 (aq) + H + (aq) + HClO(aq) E = 1.21 V + 1.65 V =0.44 V 296,485 85,.. 7.610 55. a) For galvanic cell E must be positive Au 3+ +3e Au E = 1.50 V 3(Tl Tl + + e ) E = 0.34 V Au 3+ (aq) + 3Tl(s) Au(s) + 3Tl + (aq) E = 1.50 V + 0.34 V = 1.84 V 13

b) 396,485 1.84 5.32 10 532 2.2610 1.84 8.3145 298 1.0 10 2.04 396,485 1.0 10 c) 56. 2(Cr 2+ Cr 3+ + e ) Co 2+ + 2e Co 2Cr 2+ + Co 2+ 2Cr 3+,.. 8.3145 298 2.79 10 0.220 296,485 0.220 8.3145 2 96,485 298 2.0 296,485 0.30 0.151 0.20 0.151 29,100 29.1 59. Given Al 3+ + 3e Al E = 1.66 V Pb 2+ + 2e Pb E = 0.13 V 2(Al Al 3+ + 3e ) E = 1.66 V 3(Pb 2+ + 2e Pb) E = 0.13 V 2Al(s) + 3Pb 2+ (aq) 2Al 3+ (aq) + 2Pb(s) E = 1.66 V + 0.13 V= 1.53 V Calculate the final Pb 2+ concentration: Pb 2+ Al 3+ Initial 1.00 1.00 Change 3x +2x=0.60 Final 1.003x 1.00+0.60=1.60 x must equal 0.30 M, therefore, the Pb 2+ concentration will change by 0.90 M and the final Pb 2+ concentration equals 1.000.90=0.10 M. Calculate E of cell 1.53 8.3145 298 1.60 696,485 0.10 1.50 14

65. Since the same material is on both sides of the cell E = 0 In order to be a galvanic cell the concentration on the anode side of Ni 2+ must be smaller than the concentration of Ni 2+ on the cathode side Ni 2+ + 2e Ni a) 0.. 0.0., No electron flow. b) 0.,. 0.0089. Anode is on the left (1.0 M) the cathode is on the right (2.0 M) and electrons flow from left to right on the diagram. c) 0.,. 0.030. Anode is on the right (0.1 M) the cathode is on the left (1.0 M) and electrons flow from right to left on the diagram. d) 0.,. 0.13. Anode is on the right (4.0 10 5 M) the cathode is on the left (1.0 M) and electrons flow from right to left on the diagram. e) 0. No electron flow. 72. a) Al 3+ + 3e Al 1.0 b) Ni 2+ + 2e Ni 1.0 c) Ag + + e Ag 5.0,.,... 0.0. 110 1.110 29 0.034., 33. 5.0., 4800 1.3. 73.,. a) Co 2+ + 2e Co 0.56. 0.56 17 15

b) Hf 4+ + 4e Hf 0.56 c) 2I I 2 +2e. 25 0.56. 71 d) CrO 3 + 6H + + 6e Cr + 3H 2 O 0.56. 4.9 74. The question wants you to calculate the molarity of the initial solution We know the volume so we need to calculate the mole of Ag +. The ½ reaction that we are interested in is Ag + +e Ag If we know the moles of electrons we can get the moles of Ag +. Calculate the moles of electrons. 2.00 2.30 0.00286 96,485 Calculate the moles of Ag + 0.00286 0.00286 Calculate molarity 0.00286 0.0114 0.250 75. The question asked you to calculate time to plate out 10 g Bi. To calculate the time use Therefore, need to determine the number of e. To do this you must find a relationship between e and Bi. Balance the equation. BiO + Bi BiO + Bi + H 2 O BiO + + 2H + Bi + H 2 O BIO + + 2H + + 3e Bi + H 2 O Note: If you balanced it in basic conditions you would also come out with 3 electrons Determine the number of e 10.0 0.144 Determine time. 0.144 96,485 556 25.0 81. For electrolysis reaction E cell is a) Cathode reaction Ni 2+ + 2e Ni E = 0.23 V Anode reaction 2Br Br 2 + 2e E = 1.09 V 16

b) Cathode reaction Al 3+ + 3e Al E = 1.66 V Anode reaction 2F F 2 + 2e E = 2.87 V c) Cathode reaction Mn 2+ + 2e Mn E = 1.18 V Anode reaction 2I I 2 + 2e E = 0.54 V For the aqueous solution we must also consider the reactions with H 2 O d) Cathode reaction Ni 2+ + 2e Ni E = 0.23 V 2H 2 O + 2e H 2 + 2OH E = 0.83 V Since it is easier to reduce Ni 2+ than H 2 O the nickel reaction will occur at the cathode. Anode reaction 2Br Br 2 + 2e E = 1.09 V H 2 O O 2 + 4H + + 4e E = 1.23 V Since it is easier to oxidize B than H 2 O the bromine reaction will occur at the anode. e) Cathode reaction Al 3+ + 3e Al E =1.66 V 2H 2 O + 2e H 2 + 2OH E = 0.83 V Since it is easier to reduce H 2 O than Al 3+ the water reaction will occur at the cathode. Anode reaction 2F F 2 + 2e E = 2.87 V H 2 O O 2 + 4H + + 4e E = 1.23 V Since it is easier to oxidize H 2 O than F the water reaction will occur at the anode. f) Cathode reaction Mn 2+ + 2e Mn E = 1.18 V 2H 2 O + 2e H 2 + 2OH E = 0.83 V Since it is easier to reduce H 2 O than Mn 2+ the water reaction will occur at the cathode. Anode reaction 2I I 2 + 2e E =0.54 V H 2 O O 2 + 4H + + 4e E =1.23 V Since it is easier to oxidize I than H 2 O the iodine reaction will occur at the anode. 84. The question wants you to determine the charge on the ruthenium. The reaction that we are interested in is: Ru n+ + ne Ru. Therefore, we should be able to identify n by comparing the moles of Ru to the moles of e. Calculate the moles of Ru 2.618 0.0259. Calculate the moles of e 2.50 50.0 0.0777 96,486 17

Because the ratio between the moles of e and moles of Ru is 3:1 the charge on the ruthenium must be 3+. 92. The question wants you to calculate the volume of O 2 and H 2 gas produced in the electrolysis of water. The question gives you the time and the current. With this information you can calculate the mole of electrons. 2.50 15.0 0.0233 96,486 You need to find a relationship between the moles of electrons and the moles of O 2 and H 2, to do this look at the equations for the electrolysis of water. 2(2H + + 2e H 2 ) 2H 2 O O 2 + 4H + + 4e 2H 2 O 2H 2 + O 2 Calculate the moles of H 2 0.0233 0.0117 Calculate the volume of H 2 0.0117 0.08206 273.15 0.262 1.00 Calculate moles of O 2 0.0233 0.00583 Calculate the volume of O 2 0.00583 0.08206 273.15 0.131 1.00 98. A battery is a galvanic cell. When a battery is new there are many more reactants than products in the battery and therefore, Q is small. As the battery is used it turns reactants into product and Q gets larger. Since as the battery is used up E gets smaller and smaller until it is 0. A battery is not a system at equilibrium otherwise E = 0. Fuel cells and batteries are both galvanic cells. The difference between them is that batteries are enclosed and therefore, eventually reach equilibrium. For fuel cells the reactants are continually supplied and therefore, never reach equilibrium. A hydrogen oxygen fuel cell is a fuel cell that you need to supply hydrogen and oxygen to get out voltage. 100. The reaction that goes on in a hydrogen oxygen fuel cell 2H 2 (g) + O 2 (g) 2H 2 O(l) O 2 + 4H + + 4e 2H 2 O E = 1.23 V 2(H 2 2H + + 2e ) E = 0.0 V O 2 (g) + 2H 2 (g) 2H 2 O(l) E = 1.23 V + 0.0 V = 1.23 V Because standard conditions 18

496,485 1.23 4.7510 475 This is the w max for 2 moles of H 2 O. Need to find the work mass for 1.00 kg of H 2 O 1.00 13,200. The work done can be no larger than 13,200 kj. Usually the max work is not done and some of the energy is lost to heat. Fuel cells are more efficient in converting chemical energy into electrical energy than combustion reaction. The main disadvantage of fuel cells is their cost. 19