Fundamentals of Analytical Chemistry by Skoog et. al. 8th ed. Chapter 18 answers

Fundamentals of Analytical Chemistry: 8th ed.Chapter 18 Chapter 18 18-1 (a) Oxidation is a process in which a species loses one or more electrons. (b) An oxidizing agent is an electron acceptor. (c) A salt bridge is a device that provides electrical contact but prevents mixing of dissimilar solutions in an electrochemical cell. (d) A liquid junction is the interface between dissimilar liquids. A potential develops across the interface. (e) The Nernst equation relates the potential to the concentrations (strictly, activities) of the participants in an electrochemical reaction. 18-2 (a) The electrode potential is the potential of an electrochemical cell in which a standard hydrogen electrode acts as the reference electrode on the left and the half-cell of interest is on the right as written in cell notation. (b) The formal potential of a half-reaction if the potential of the system (measured against the standard hydrogen electrode) when the concentration of each solute participating in the half-reaction has a concentration of exactly one molar and the concentrations of all other constituents of the solution are carefully specified. (c) The standard electrode potential for a half-reaction is the potential of a cell consisting of the half-reaction of interest on the right and a standard hydrogen electrode on the left as written in cell notation. The activities of all of the participants in the half-reaction are specified as having a value of unity. The additional specification that the standard hydrogen electrode is the reference electrode implies that the standard potential for the half-reaction is always a reduction potential. (d) A liquid-junction potential is the potential that develops across the interface between Fundamentals of Analytical Chemistry: 8th ed. Chapter 18 two dissimilar solutions. (e) An oxidation potential is the potential of an electrochemical cell in which the cathode is a standard hydrogen electrode and the half-cell of interest acts as anode. 18-3 (a) Reduction is the process whereby a substance acquires electrons; a reducing agent is a supplier of electrons. (b) A galvanic cell is one in which a spontaneous electrochemical reaction occurs and is thus a source of energy. The reaction in an electrolytic cell is forced in a nonspontaneous direction through application of an external source of electrical energy. (c) The anode of an electrochemical cell is the electrode at which oxidation occurs. The cathode is the electrode at which reduction occurs. (d) In a reversible cell, alteration of the direction of the current simply causes a reversal in the electrochemical process. In an irreversible cell, reversal of the current results in a different reaction at one or both of the electrodes. (e) The standard electrode potential is the potential of an electrochemical cell in which the standard hydrogen electrode acts as the reference electrode on the left and all participants in the right-hand electrode process have unit activity. The formal potential differs in that the molar concentrations of the reactants and products are unity and the concentrations of other species in the solution are carefully specified. 18-4 The first standard potential is for a solution that is saturated with I2 and has an I2 (aq) activity significantly less than one. The second potential if for a hypothetical half-cell in which the I2 (aq) activity is unity. Such a half-cell, if it existed, would have a greater potential because the driving force for the reduction would be greater at the higher I2 The second half-cell potential. Only under these conditions is the hydrogen activity constant so that the electrode potential is constant and reproducible. Chapter 18 concentration. is nevertheless useful for calculating electrode potentials for solutions that are undersaturated in I2. 18-5 It is necessary to bubble hydrogen through the electrolyte in a hydrogen electrode in order to keep the solution saturated with the gas.)  18-7 (a) 2 Fe 3 Sn 2  2 Fe 2 Sn 4 (b) Cr ( s ) 3Ag   Cr 3 3Ag( s )  (c) 2 NO 3 Cu( s ) 4 H   2 NO 2 ( g ) 2 H 2 O Cu 2  (d) 2 MnO 4 5H 2SO3  2 Mn 2 5SO 4 (e) Ti 3 Fe(CN ) 6 3 2 4 H  3H 2 O H 2 O  TiO 2 Fe(CN ) 6 (f) H 2 O 2 2Ce 4  O 2 ( g ) 2Ce 3 2 H  (g) 2Ag( s ) 2I  Sn 4  2AgI ( s ) Sn 2 (h) UO 2 2 Zn( s ) 4 H   U 4 Zn 2 2 H 2 O 4 2 H  .250 V. although hypothetical. 18-6 The potential in the presence of base would be more negative because the nickel ion activity in this solution would be far less than 1 M. and the electrode potential would be significantly more negative.Fundamentals of Analytical Chemistry: 8th ed.72 V. Consequently the driving force for the reduction if Ni (II) to the metallic state would also be far less. whereas the  standard electrode potential for Ni 2 2e   Ni( s ) is –0. (In fact the standard electrode potential for the reaction Ni(OH ) 2 2e   Ni( s ) 2OH  has a value of –0. Cr ( s )  Cr 3 3e    (c) Oxidizing agent NO3-. Fe 3 e   Fe 2  Reducing agent Sn2+. H 2SO 3 H 2 O  SO 4  2 4 H  2e  3 (e) Oxidizing agent Fe(CN)63-. NO 3 2 H  e   NO 2 ( g ) H 2 O  Reducing agent Cu. Sn 4 2e   Sn 2  4 . Ag  e  Ag( s )  Reducing agent Cr. Fe (CN ) 6 e   Fe (CN ) 6  Reducing agent Ti3+. Ce 4 e   Ce 3  Reducing agent H2O2.  Chapter 18  (i) 5HNO 2 2 MnO 4 H   5NO 3 2 Mn 2 3H 2 O   (j) H 2 NNH 2 IO 3 2 H  2Cl   N 2 ( g ) ICl 2 3H 2 O 18-8 (a) Oxidizing agent Fe3+. H 2 O 2   O 2 ( g ) 2 H  2e  (g) Oxidizing agent Sn4+.Fundamentals of Analytical Chemistry: 8th ed. MnO 4 8H  5e   Mn 2 4 H 2 O  Reducing agent H2SO3. Cu( s )  Cu 2 2e    (d) Oxidizing agent MnO4-. Ti 3 H 2 O  TiO 2 2 H  e   (f) Oxidizing agent Ce4+. Sn 2  Sn 4 2e   (b) Oxidizing agent Ag+. Ag( s ) I   AgI( s ) e   (h) Oxidizing agent UO22+. Zn( s )  Zn 2 2e    (i) Oxidizing agent MnO4-. IO 3 6H  2Cl  4e   ICl 2 3H 2 O    Reducing agent H2NNH2. Chapter 18 Reducing agent Ag. HNO 2 H 2 O  NO 3 3H  2e     (j) Oxidizing agent IO3-. UO 2 2 4 H  2e   U 4 2 H 2 O  Reducing agent Zn. MnO 4 8H  5e   Mn 2 4 H 2 O   Reducing agent HNO2.Fundamentals of Analytical Chemistry: 8th ed. H 2 NNH 2 18-9 N 2 ( g ) 4 H  4e    (a) MnO 4 5VO 2 11H 2 O  Mn 2 5V(OH ) 4 2 H  (b) I 2 H 2S( g )  2I  S( s ) 2 H  (c) Cr2 O 7 2 3U 4 2 H   2Cr 3 3UO 2 2 H 2 O (d) 2Cl  MnO 2 ( s ) 4 H   Cl 2 ( g ) Mn 2 2 H 2 O  (e) IO 3 6H  5I   3I 2 H 2 O   (f) IO 3 2I  6H  6Cl   3ICl 2 3H 2 O (g) HPO 3 2  3 2 MnO 4 3OH   PO 4 2 MnO 4  (h) SCN  BrO 3 H 2 O  SO 4 2 2 HCN Br  H   (i) V 2 2V(OH ) 4 2 H   3VO 2 5H 2 O  2 H 2 O (j) 2 MnO 4 3Mn 2 4OH   5MnO 2 ( s ) 2 H 2 O . H 2S( g )  S( s ) 2 H  2e   (c) Oxidizing agent Cr2O72-. I    1 I 2 3H 2 O 2 1 I 2 e  2    (f) Oxidizing agent IO 3 . Cr2 O 7 2 14 H  6e   2Cr 3 7 H 2 O  Reducing agent U4+. I  2Cl   ICl 2 2e    (g) Oxidizing agent MnO4-. IO 3 6H  5e    Reducing agent I-. I 2 (ag ) 2e   2I   Reducing agent H2S. BrO 3 6 H  6e   Br  3H 2 O  . 2Cl   Cl 2 ( g ) 2e     (e) Oxidizing agent IO 3 . VO 2 3H 2 O  V(OH ) 4 2 H  e   (b) Oxidizing agent I2.Fundamentals of Analytical Chemistry: 8th ed. MnO 4 e   MnO 4  Reducing agent HPO32-. HPO 3 2 3OH   PO 4   2 3 2 H 2 O 2e  (h) Oxidizing agent BrO3-. Chapter 18  18-10 (a) Oxidizing agent MnO4-. U 4 2 H 2 O  UO 2  2 4 H  2e  (d) Oxidizing agent MnO2. MnO 2 ( s ) 4 H  2e   Mn 2 2 H 2 O  Reducing agent Cl-. MnO 4 8H  5e   Mn 2 4 H 2 O   Reducing agent VO2+. IO 3 6 H  2Cl  4e   ICl 2 3H 2 O   Reducing agent I-. 256 Zn 2 2e   Zn( s )  -0. MnO 4 4 H  3e   MnO 2 ( s ) 2 H 2 O  Reducing agent Mn2+. SCN  4 H 2 O  SO 4  2 HCN 7 H  6e   (i) Oxidizing agent V(OH)4+.36 AgBr ( s ) e   Ag( s ) Br   0.25 Tl 3 2e   Tl   3  Tl   Tl 3 2e   (b).Fundamentals of Analytical Chemistry: 8th ed.073 V 3 e   V 2  -0. Mn 2 2 H 2 O  MnO 2 ( s ) 4 H  2e   18-11 (a) AgBr ( s ) e   Ag( s ) Br   V 2  V 3 e   Tl 3 2e   Tl   Fe (CN ) 6 V 3 e   V 2  Zn( s )  Zn 2 2e   3 Fe (CN ) 6 e   Fe (CN ) 6  S2O8 2 2e   2SO 4  4 2 2 2e   2SO 4  3 Fe(CN ) 6 e  AgBr ( s ) e   Ag( s ) Br   E 2. Chapter 18 Reducing agent SCN-. V 2 H 2 O  VO 2 2 H  2e    (j) Oxidizing agent MnO4-.01 2 1. V(OH ) 4 2 H  e   VO 2 3H 2 O  Reducing agent V2+.763 . (c) S2O8 4  Fe(CN ) 6 e   Fe(CN ) 6  4 0. Chapter 18 18-12 (a) Sn( s )  Sn 2 2e   2 H  2e   H2 (g)  Ag  e   Ag( s )  Fe 2  Fe 3 e   Sn 4 2e   Sn 2  H 2 (g)  2 H  2e   Fe 3 e   Fe 2  Sn 2  Sn 4 2e   Sn 2 2e   Sn( s )  Co( s )  Co 2 2e   (b).00 Sn 2 2e   Sn( s )  -0.771 Sn 4 2e   Sn 2  0.277 18-13 (a) 0. (c) E Ag  e   Ag( s )  0.154 2 H  2e   H 2 (g)  0.337  log 0.0440  .Fundamentals of Analytical Chemistry: 8th ed.136 Co 2 2e   Co( s )  -0.297 V 2 0.799 Fe 3 e   Fe 2  0.0592  1  ECu 0. 521  log 0.128 0.8 10 20 [Cu 2][OH ]2 0.190 V (c) K Cu ( OH )2 4.337  0.62 10  2 3 4 4 [Cu ][ NH 3 ] 11 0.521  log K  1 1 [Cu ]   CuCl  0.337 0.521  log(3.33 1016 ) 20  2 2 4.0592  4 [ NH 3 ]4   ECu 0.0592 [Cl ]   ECu 0.290 0.337  log 2 0.048 V .0592  0.0592  0. Chapter 18 (b) K CuCl 1.521  log  0.03 109 )   2 0 .337  log log(3.Fundamentals of Analytical Chemistry: 8th ed.0400  0.0592 0.152 V (d) 2 [Cu( NH ) ] 4 5.521 0.0592  1  0.0592 5.0592  0.0592  [OH ]2 ECu 0.0592  1  0.95 105 ) 7  1 1.337  log log(6.337  log 2 0.62 1011  0.489 0.9 10 7 [Cu ][Cl ] 0.0592 0.337  log [Cu( NH ) 2]  2 2 [Cu ]  3 4   4 0. 0250 2   0.331 0.337  log K 2 2 [Cu ]   Cu ( OH )2     2 0.337 0.0750  0.9 10  1  0.0592  1  0.337  0.8 10  0. 0 10  0.0592  1  E Zn 0.33 1011 )  16  2 2 3.337  log  [CuY 2] 2 2 [Cu ]        0.337  log 0.341 1.0100  0.763  log K 2 2 [Zn ]   Zn ( OH )2     2 0.763  log log(3.33 0.0592 2.763  0.0592  0.763  log 0.3 1018  2. Chapter 18 (e) [CuY 2] 4 K CuY  3.763 0.0250 0.337  log(1.0592  1  0.337 0.00 10 3  0.0592  0.0250 0.0592  0.76 10  2 3 4 4 [Zn ][ NH 3 ] 8 . 00  10 2   0.3 1010  0.0592 4 K CuY 2cT ECu 0.3 1010 [Cu 2]cT [CuY 2] 4.0592  1  0.90 10 2   4.Fundamentals of Analytical Chemistry: 8th ed.0600  (b) K Zn ( OH )2 3.337  log 2 0.007 V 18-14 (a) 0.763  log 2 0.6 10 9  6.0592  [OH ]2 E Zn 0.799 V 2 0.00 10 3 cT  2.0 10 16 [Zn 2][OH ]2 0.4 1011 ) 3   2 4 .10 V (c) 2 [Zn( NH ) ] 4 7. 477 1.00  E E o  log 2 2 0.763  log [Zn( NH ) 2]  2 [ Zn ] 2   3 4   4 0.2 10 2  3.0592 4 K ZnY 2cT E Zn 0.763 0.0592  0.3 1016 ) 3   2 5 .0592  1.763 0.250 0. E Zn Chapter 18 0.2 1016  1.763  0.7 1015 [ Zn 2]cT [ZnY 2] 5.0100 2   0.45 10 2   5.0592  1  0.763  log(1.00 10 3  0.00 10 3 cT  4.01 V (d) [ZnY 2] 4 K ZnY  5.0100 0.763  log log(3.0395 0.763  log  [ZnY 2] 2 2 [ Zn ]        0. 00  10 2   0.24 V 18-15 2 H  2e   H2 (g)   pH  0.76 108  0.913  0.0592  4 [ NH 3 ]4   0.913  .763  log 0.0395 0.0592 7.0592  0.00 0.763  log 2 0.121 0.0100 12 0.0592  1.763  log 2 0.7 1015  0.00  log 2 2   2 0.0592  1.0100   2 From Table 10-2 H  0.121 V E 0.0100 12 0.0592  1  0.03 108 )   2 0.0592 0.251 1.00  log  2 2   [ H ]   2 2 a    H H    The ionic strength of the solution is given by 1 0.Fundamentals of Analytical Chemistry: 8th ed.00 0. 960 0.00 0.154 ( 0.920 mmol Fe 3  1.1568 mmol FeCl 3 1 mmol Fe 3 mmol Fe 3consumed   25.210 V E Pt 0.0586 2 0.0  E Pt 0.73 ( 0.960 mmol Sn 4 3 2 mmol Fe mmol Sn 2remaining 2.0592  1.30 mmol Sn 2 mL mmol SnCl 2 0.00 L 3.359 V  0.0592  0.0592  0.30 1.920 mmol Fe 3 mL mmol FeCl 3 1 mmol Sn 4 mmol Sn 4formed 3.154  log 7.0918 mmol SnCl 2 1 mmol Sn 2 mmol Sn 2consumed   25.1492   0.340 mmol Sn 2 0.149 0.50 10 3  E Pt 0.Fundamentals of Analytical Chemistry: 8th ed.100  (e) 2 Fe 3 Sn 2  2 Fe 2 Sn 4  0.50 10 2  0.154  log 0.00 L 2.0592 2. 0263   (b) 0.359  log 2   2 0.340 / 50.0    .355 V E Pt 0.78 V E Pt 0.177 V  2 1.359 0.73  log  2 0 .0353  0.051) 0.198 V 2   (c)  0.000  log 2   6  2 1.00 10   (d) VO 2 2 H  e   V 3 H 2 O  E o 0.023) 0. 18-16 PtCl 4 2 2e   Pt ( s ) 4Cl   Chapter 18 E o 0.73 V (a) 4 0.960 / 50.044) 0.0592   0.154 ( 0. 29 V 0. Chapter 18  (f) V(OH ) 4 V 3  2VO 2 2 H 2 O   0.0592 log 0.174 mmol VO 2 3 mmol V  mmol V(OH ) 4 remaining 2.00 0.08 mmol V(OH ) 4 mL  0.174  / 75.993 / 75.0400  E Pt 0.087 0.00845  (c) pH 5.022 0.749 V  2  0.0592 log 2   0.1996  .0592 log 0.36 0.000  log 2 6   2 2 .0789 1.771 0.00   18-17 (a) 0.86 V E Pt 1.771 0.894 V E Pt 1.55 [ H 3O ] 2.00 0.106 0.000 0.00 0.08 1.00566  (b)  0.Fundamentals of Analytical Chemistry: 8th ed.139 0.329 V E Pt 0.0813  E Pt 0.00 0.087 mmol V 3 mL mmol V2 (SO 4 ) 3 2 mmol VO 2 mmol VO 2formed 1.00 L 2.0592  1.0832 mmol V(OH ) 4  mmol V(OH ) 4 consumed  25.329 0.00 1.1000 0.00 L 1.01087 mmol V2 (SO 4 ) 3 2 mmol V 3 mmol V 3consumed   50.087 mmol V 3  2.82 10 6  0.068 0. 82  10    (d)   0.00 0.993 mmol V(OH ) 4     2.0800 0.36 0.0592 log 2   0. 0607 mmol Ce(SO 4 ) 2 1 mmol Ce 4 mmol Ce 4consumed   50.16 0.011) 0.359 0.100 0.00  E Ni 0.314 mmol V 3  0.00 L 5.85 mmol V(OH ) 4    3.00 0.Fundamentals of Analytical Chemistry: 8th ed.00 18-18 (a) 0.68 0.0592 1.04 0.04 mmol Fe 3 mmol Fe 2remaining 5.250  log 0.00 L mL mmol Ce(SO 4 ) 2 3.00 mmol Fe 2 mL mmol FeCl 2 2 mmol Fe 3formed mmol Ce 4consumed 3.250 0.04 / 100.314 mmol V(OH ) 4 mL 0.04 mmol Ce 4 0.0  (f)  V(OH ) 4 V 3  2 VO 2 2 H 2 O   0.00 3.030 0.965 / 100.0943  .69 V 3.00 L 0.16 mmol V 3 2 mmol VO 2 mmol VO 2formed 0.0628 mmol V(OH ) 4   mmol V(OH ) 4 consumed  50.00 L mL mmol V2 (SO 4 ) 3 4.359 0.280 V anode 2 0.196 mmol Fe 2 1. Chapter 18 (e) 0.628 mmol VO 2 mmol V 3  mmol V(OH ) 4 remaining 4.628 / 75.68 ( 0.100 mmol FeCl 2 1 mmol Fe 2 mmol Fe consumed   50.0   E Pt 0.0592 log 2    0.194 V E Pt 0.85 / 75.165 0.0592 log 0.314 3.0832 mmol V2 (SO 4 ) 3 2 mmol V 3 mmol V 3consumed   25. 237 V anode E Pt 0.50 10  780 / 760 (d) 0.771 0.Fundamentals of Analytical Chemistry: 8th ed.1628  (d) E Pt 0.0922 0.1439   E Ag 0.017 0.0897  (b) E Pt 0.015 0. Chapter 18 (b) E Ag 0.984 0.756 V cathode 0.090 V anode (c)  0.003 V cathode EO2 1.0592 log 0.306 V cathode 2 0.171 V cathode 2 0.031 0.00753 0.0592 0.009 V anode   2 18-19 (a) 0.00 0.229 0.0592  0.229  log 4 4   4  1.0592 log 0.771 0.237 0.0592  1.337 0.061) 0.00 1.185 0.337  log 0.226 1.026 0.0592  1.151 0.017 0.154  log 0.00  log 4 2   2 1.0944  E Pt 0.054) 0.131 V anode (c)  0.350  (e)  0.017) 0.154 ( 0.185 ( 0.0906  .0592 log 0.00  ECu 0.0592 log 0.00 10  0.151 ( 0.12140. 5 10 14   K sp  18-21 2 Ni 2 4e  2 Ni( s )  E o 0.0592  1.0827 0.799 0. Chapter 18 (e) 2  0.  Thus.7 10 13 K sp 4 [ P2 O 7 ]  0.00  0.799  log  0 .0592  1.0592  1.31 ( 0.390 V   2 2 1.250 4 [ Ni 2]2 [ P2 O 7 ] 1.0699   E Ag 0. E = Eo for Ni 2 P2 O 7 ( s ) 4e  2 Ni( s ) P2 O 7 .250  log   4 [ Ni ] 4 K    sp  4 When [P2O74-] =1.150  log    4 4 1. 0.779 2 [Ag ]2 [SO 3 ] 1.439 V .250  log 2 2 0.31 0.00.00  E 0.00  0.  Thus.0592    E 0.799  log 0.799  log 0.Fundamentals of Analytical Chemistry: 8th ed.24 V anode   18-20 2Ag  2e  2Ag( s )  E o 0. 0.0592  1  0.189 0.00  E 0.5 10 14 K sp 2 [SO 3 ]  0.073) 0.00.7 10 13   K sp  0.0592 log  0.0592    E 0. 799  log  [ Ag ]2   K  2 2    sp  2 When [SO32-] =1.250 0.409 0.0592  1  0.0592  1.250  log 0. E = Eo for Ag 2SO 3 ( s ) 2e  2Ag( s ) SO 3 . 0.0592  1  0.126  log  K  6 6 [ Pb ]  sp   When [AsO42-] =1.0592  1.475 V 0.00  E 0.0592  1.  Thus. E = Eo for Pb 3  AsO 4  ( s ) 4e  3Pb( s ) 2AsO 4 .0592   E 0.0592  1  0.126 2 [ Pb 2]3 [AsO 4 ]2 4.349 0.00  E 0.0592  1.628 0.00.336  log 0.0592  [S2]   E 0. E = Eo for Tl 2S( s ) 2e  2Tl ( s ) S 2.1 10 36 K sp 2 [ AsO 4 ]2  0.336  log 22  K  2 2 6 10  sp   0.336 [Tl ]2 [S 2] 6 10 22 K sp 0.336  log K  2 2 [Tl ]   sp  When [S2-] =1. 18-22 2Tl  2e  2Tl ( s )  Chapter 18 E o 0.336  log  2 0.126  log 2 3 0.00.00  0.Fundamentals of Analytical Chemistry: 8th ed.126  log 0. 0.336 0.00  0. 2  2 Thus.0592  1.96 V 18-23 3Pb 3 6e  3Pb( s )  E o 0.1 10   sp  0.126  log 36  K  6 6 4.126 0.0592  1  18-24 E 0.2 1016 [Zn 2][ Y 4] .763  log 2  2 [ Zn ]  [ZnY 2] 3. E E o FeY  1.763 0.153 0.771 0.00  7.2 1016   E 0.62 1011    0.2 10  [ NH 3 ]2  5.Fundamentals of Analytical Chemistry: 8th ed.00   1.153 0.0592 log 10  1.1 10    18-26  2 [Cu( NH 3 ) 2 ] [Cu( NH 3 ) 2 ] [Cu ]  and [Cu 2]  2 10 [ NH 3 ]  7.100 V E 0.00.0592 log [Cu 2]     [Cu( NH 3 ) 2 ]  5.00.25 V 2 1.0592 [ Y 4] 3.0592 log 2 10    [ Cu ( NH ) ] 7 .763  log 2   2  [ ZnY ]  When [Y4-] = [ZnY2-] = 1.771 0.62 1011  0.2 1016   E 0.0592  1. 2  10 3 2   When [Cu(NH3)2+] = [Cu(NH3)22+] = 1. E E o Cu ( NH 3 )22   1. E E o ZnY 2   0.1 1014  [ Fe 2]  E 0.13 V E 0.2 10   .3 10 25  0.0592 log 14  [ FeY ] 2.489 1.153 0.00  3.64 0.0592 log 14   1.3 10 25   0.771 0.00  5.3 10 25  [ Y 4] 2. Chapter 18  0.00   18-25 [ FeY ] [ FeY 2] 2 [ Fe 3]  4 and [ Fe ]  [ Y ] 1.153 0.0592 log [ Fe 3]     [ FeY 2]  1.763  log  0.053 0.771 0.00.1 10   When [FeY2-] = [FeY-] = 1.62 1011   [Cu ]  E 0.00  2. 00 1.00 40.400 0.50 1.764 0.500 0.50 5.V 0.812 0.676 0.00 0.00 20.781 0.617 0.753 0.25 1.785 0.200 0.01 0.250 0.100 0.854 0.771 o o We also assume 25 C 3+ 2+ 3+ E=E -0.771 0.0592log([Fe ]/[Fe ]) 2+ 2+ [Fe ]/[Fe ] 3+ [Fe ]/[Fe ] E.0075 0.800 0.593 0.025 0.00 400.830 0.0025 0.571 0.653 0.005 1000.V Note: We use the Nernst equation in column C to calculate the potentials from 0.00 75.00 Spreadsheet Documentation B6=1/A6 C6=$B$3-0. A 1 2 3 4 5 6 7 8 3+ 18-27 Fe /Fe 2+ Chapter 18 B C D half-cell potentials o E.010 0.704 0.00 200.795 0.00 100.00 10.00 13.001 0.667 0.Fundamentals of Analytical Chemistry: 8th ed.75 2.05 0.712 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 0.635 9 10 11 12 13 14 0.00 2.00 0.33 100.040 0.33 10.882 0.100 133.075 0.00 0.33 1.0592*LOG10(B6) 4.694 0.750 1.645 0.889 .777 0.00 1.013 0.735 0.00 25. 00 13.00 20.V 0.564 1.571 1.724 1.00 10.00 133.759 1.00 100.693 1.7 o [Ce4+]/[Ce3+] o 3+ 4+ E=E -(0.50 5.500 0.00 400.546 1.0592)log([Ce ]/[Ce ]) [Ce3+]/[Ce4+] E.010 1.040 0.682 1.00 4.783 1.250 100.750 1.800 0.641 1.100 0.605 1.33 10.741 1.667 0.Fundamentals of Analytical Chemistry: 8th ed.00 Spreadsheet Documentation B6=1/A6 C6=$B$3-0.400 0.01 0.623 1.075 0.710 1.001 0.00 40.00 75.V 4 We also assume1 M HClO4 25 C 5 6 7 8 9 C Note: We use the Nernst equation in column C to calculate the potentials from 1.200 0.00 25.633 1.0075 1000.00 1.005 0.00 200.00 0.664 16 17 18 19 20 21 0.05 0.811 1. Chapter 18 18-28 A 1 2 4+ 18-28 Ce /Ce 3+ B D half-cell potentials o 3 E.00 1.574 10 11 12 13 14 15 0.00 1.706 1.013 0.0592*LOG10(B6) 0.33 1.522 1.50 1.818 .714 22 23 24 25 26 27 28 29 30 2.33 1.25 1.700 1.582 1.025 0.100 0.0025 0.75 2. 850 E.900 0.700 1. Chapter 18 Plot for Probelm 18-27 0.600 0.Fundamentals of Analytical Chemistry: 8th ed.600 1.950 0.650 0.500 0 20 40 60 4+ 3+ [Ce ]/[Ce ] .750 0.700 0. V 1.550 0 20 40 60 80 100 120 80 100 120 [Fe3+]/[Fe2+] Plot of Problem 18-28 1.750 1.800 0.550 1.650 1. V 0.850 1.800 E. 30 0.50 0.30 0.785 1.70 1.12 1.777 1.676 0.00 0.40 0.712 0.00 2.57 0.66 0.75 0. Eo.12 1.500 -0.01 -2.60 0.62 0.25 0.56 0.250 -0.735 0.30 1.30 1.12 1.0025 -2.0075 -2. 18-28 and log ratio D E F o 3+ Note: We calculate the logarithm of concentration E .71 1.0025 -2.854 25. V (Fe ) 0.753 0.617 0.24 1.50 0.60 0.69 1.71 2.0592*LOG10(A6) F6=$B$4+0.00 0.40 1.00 1.645 0.7 3+ 2+ 3+ 2+ 4+ 3+ 4+ 3+ [Fe ]/[Fe ] Log([Fe ]/[Fe ]) E.812 5.00 1.500 -0.18 0.V [Ce ]/[Ce ] Log([Ce ]/[Ce ]) E.10 0.63 0.00 0.001 -3.750 -0.00 0.76 25.18 1.882 75.55 0.075 -1.00 1.70 1.00 1.025 -1.78 75.12 0.40 0.74 10.00 0.52 0.005 -2.60 1.593 0.88 1.30 1.694 0.00 1.25 0.100 -1.40 1.0592*LOG10(D6) .01 -2.60 1.10 1.100 -1.12 0.00 1.12 0.00 1.750 -0.60 0.00 1.795 2.71 1.830 10.88 0.001 -3.771 4+ ratio in column B.Fundamentals of Analytical Chemistry: 8th ed.00 2.250 -0.764 0.704 0.771 1.05 -1. Chapter 18 18-29 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A B C 18-29 Plots for Problems 18-27.00 0.81 100.00 0.24 0.889 100.72 5.075 -1.61 0.60 1.00 1.05 -1. V (Ce ) 1.82 Spreadsheet Documentation B6=LOG10(A6) E6=LOG10(D6) C6=$B$3+0.70 0.75 0.005 -2.68 0.635 0.00 0.00 0.781 1.00 0.653 0.00 1.0075 -2.00 1.64 0.30 0.50 0.V 0.025 -1.50 0.00 1.58 0. Plot of Problem 18-27 0.00 1.00 3. V 1.950 0.00 2. Chapter 18 A plot of potential versus logarithm of the concentration ratio is a straight line.00 3+ 1.00 3.600 0.00 -1.70 1.Fundamentals of Analytical Chemistry: 8th ed.65 1.750 0.650 0.700 0.900 0.00 -2.85 1.00 -3.55 1.550 0.00 2+ Log([Fe ]/[Fe ]) Plot of Problem 18-28 1.00 0.850 E.60 1.80 E.00 0.50 -4. V 0.800 0.500 -4.00 2.00 -3.00 -2.00 -1.00 4+ 3+ Log([Ce ]/[Ce ]) .75 1.