SOLUCIONÁRIO LIVRO FUNDAMENTOS DA QUÍMICA ANALÍTICA SKOOG 8ª-EDIÇÃO - CAPÍTULO 18

Prévia do material em texto

Fundamentals of Analytical Chemistry: 8
th
 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 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. 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
(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 concentration. The second half-cell potential, although hypothetical, 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. Only under these conditions is the hydrogen activity constant so that the 
electrode potential is constant and reproducible. 
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. Consequently the driving force for the reduction if Ni (II) to the 
metallic state would also be far less, and the electrode potential would be significantly more negative. (In 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
fact the standard electrode potential for the reaction 


  OH2)(Nie2)OH(Ni 2 s
 has a value of –0.72 
V, whereas the standard electrode potential for 
)(Nie2Ni2 s

 
 is –0.250 V.) 
18-7 (a) 
  4223 SnFe2SnFe2
 
(b) 
)(Ag3CrAg3)(Cr 3 ss  
 
(c) 
  2223 CuOH2)(NO2H4)(CuNO2 gs
 
(d) 
OH3H4SO5Mn2SOH5MnO2 2
2
4
2
324 

 
(e) 
  H2)CN(FeTiOOH)CN(FeTi
4
6
2
2
3
6
3
 
(f) 
  H2Ce2)(OCe2OH 32
4
22 g
 
(g) 
  24 Sn)(AgI2SnI2)(Ag2 ss
 
(h) 
OH2ZnUH4)(ZnUO 2
242
2 
 s
 
(i) 
OH3Mn2NO5HMnO2HNO5 2
2
342 

 
(j) 
OH3ICl)(NCl2H2IONNHH 222322 
 g
 
18-8 (a) Oxidizing agent Fe3+; 


  23 FeeFe
 
Reducing agent Sn2+; 


  e2SnSn 42
 
(b) Oxidizing agent Ag+;
)(AgeAg s

 
 
Reducing agent Cr; 


  e3Cr)(Cr 3s
 
(c) Oxidizing agent NO3
-; 
OH)(NOeH2NO 223  
 g
 
Reducing agent Cu; 


  e2Cu)(Cu 2s
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
(d) Oxidizing agent MnO4
-; 
OH4Mne5H8MnO 2
2
4 



 
Reducing agent H2SO3; 


  e2H4SOOHSOH
2
4232
 
(e) Oxidizing agent Fe(CN)6
3-; 


 
4
6
3
6 )CN(Fee)CN(Fe
 
Reducing agent Ti3+; 


  eH2TiOOHTi 22
3
 
(f) Oxidizing agent Ce4+; 


  34 CeeCe
 
Reducing agent H2O2; 


  e2H2)(OOH 222 g
 
(g) Oxidizing agent Sn4+; 


  24 Sne2Sn
 
Reducing agent Ag; 


  e)(AgII)(Ag ss
 
(h) Oxidizing agent UO2
2+; 
OH2Ue2H4UO 2
42
2 



 
Reducing agent Zn; 


  e2Zn)(Zn 2s
 
(i) Oxidizing agent MnO4
-; 
OH4Mne5H8MnO 2
2
4 



 
Reducing agent HNO2; 


  e2H3NOOHHNO 322
 
(j) Oxidizing agent IO3
-; 
OH3ICle4Cl2H6IO 223 



 
Reducing agent H2NNH2; 


  e4H4)(NNNHH 222 g
 
18-9 (a) 
  H2)OH(V5MnOH11VO5MnO 4
2
2
2
4
 
(b) 
  H2)(SI2)(SHI 22 sg
 
(c) 
OHUO3Cr2H2U3OCr 2
2
2
342
72 

 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
(d) 
OH2Mn)(ClH4)(MnOCl2 2
2
22 
 gs
 
(e) 
OHI3I5H6IO 223 

 
(f) 
OH3ICl3Cl6H6I2IO 223 

 
(g) 
OH2MnO2POOH3MnO2HPO 2
2
4
3
44
2
3 

 
(h) 
  HBrHCNSOOHBrOSCN
2
423
 
(i) 
OH5VO3H2)OH(V2V 2
2
4
2  

 
(j) 
OH2)(MnO5OH4Mn3MnO2 22
2
4 
 s
 
18-10 (a) Oxidizing agent MnO4
-; 
OH4Mne5H8MnO 2
2
4 



 
Reducing agent VO2+; 


  eH2)OH(VOH3VO 42
2
 
(b) Oxidizing agent I2; 


 I2e2)(I2 ag
 
Reducing agent H2S; 


  e2H2)(S)(SH2 sg
 
(c) Oxidizing agent Cr2O7
2-; 
OH7Cr2e6H14OCr 2
32
72 



 
Reducing agent U4+; 


  e2H4UOOH2U
2
22
4
 
(d) Oxidizing agent MnO2; 
OH2Mne2H4)(MnO 2
2
2 


s
 
Reducing agent Cl-;

  e2)(ClCl2 2 g
 
(e) Oxidizing agent 

3IO
; 
OH3I
2
1
e5H6IO 223  

 
Reducing agent I-; 


  eI
2
1
I 2
 
(f) Oxidizing agent 

3IO
; 
OH3ICle4Cl2H6IO 223 



 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
Reducing agent I-; 


  e2IClCl2I 2
 
(g) Oxidizing agent MnO4
-; 


 
2
44 MnOeMnO
 
Reducing agent HPO3
2-; 


  e2OH2POOH3HPO 2
3
4
2
3
 
(h) Oxidizing agent BrO3
-; 
OH3Bre6H6BrO 23 



 
Reducing agent SCN-; 


  e6H7HCNSOOH4SCN
2
42
 
(i) Oxidizing agent V(OH)4
+; 
OH3VOeH2)OH(V 2
2
4 



 
Reducing agent V2+; 


  e2H2VOOHV 22
2
 
(j) Oxidizing agent MnO4
-; 
OH2)(MnOe3H4MnO 224  
 s
 
Reducing agent Mn2+; 


  e2H4)(MnOOH2Mn 22
2 s
 
18-11 (a) 


  Br)(Age)(AgBr ss
 


  eVV 32
 


  Tle2Tl3
 


  e)CN(Fe)CN(Fe
3
6
4
6
 


  23 VeV
 


  e2Zn)(Zn 2s
 


 
4
6
3
6 )CN(Fee)CN(Fe
 


  Br)(Age)(AgBr ss
 


 
2
4
2
82 SO2e2OS
 


  e2TlTl 3
 
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
 
(b), (c) E 


 
2
4
2
82 SO2e2OS
 
2.01 


  Tle2Tl3
 
1.25 


 
4
6
3
6 )CN(Fee)CN(Fe
 
0.36 


  Br)(Age)(AgBr ss
 
0.073 


  23 VeV
 
-0.256 
)(Zne2Zn2 s

 
 
-0.763 
 
18-12 (a) 


  e2Sn)(Sn 2s
 
)(He2H2 2 g
 
 
)(AgeAg s

 
 


  eFeFe 32
 


  24 Sne2Sn
 


  e2H2)(H2 g
 


  23 FeeFe
 


  e2SnSn 42
 
)(Sne2Sn2 s

 
 


  e2Co)(Co 2s
 
 
(b), (c) E 
)(AgeAg s

 
 0.799 


  23 FeeFe
 0.771 


  24 Sne2Sn
 0.154 
)(He2H2 2 g
 
 0.00 
)(Sne2Sn2 s

 
 -0.136 
)(Coe2Co2 s

 
 -0.277 
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
18-13 (a) 
V297.0
0440.0
1
log
2
0592.0
337.0Cu 





E
 
(b) 
V190.0331.0521.0
)1095.3log(
1
0592.0
521.0
109.1
0750.0
log
1
0592.0
521.0
]Cl[
log
1
0592.0
521.0
]Cu[
1
log
1
0592.0
521.0
]Cl][Cu[109.1
5
7
CuCl
Cu
7
CuCl



























K
E
K
 
(c) 
2220
)OH(Cu ]OH][Cu[108.42
 K
 
 
V152.0489.0337.0
)1033.3log(
2
0592.0
337.0
108.4
0400.0
log
2
0592.0
337.0
]OH[
log
2
0592.0
337.0
]Cu[
1
log
2
0592.0
337.0
16
20
2
)OH(Cu
2
2Cu
2




























 K
E
 
(d) 
4
3
2
2
4311
4
]NH][Cu[
])NH(Cu[
1062.5



 
 
V048.0290.0337.0
)1003.6log(
2
0592.0
337.0
0250.0
128.01062.5
log
2
0592.0
337.0
])NH(Cu[
]NH[
log
2
0592.0
337.0
]Cu[
1
log
2
0592.0
337.0
9
411
2
43
4
34
2Cu








 








 








E
 
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
(e) 
   
    0250.01000.41090.2
1000.4]CuY[
103.2103.6106.3
]Cu[
]CuY[
32
T
32
10189
CuY4
T
2
2








c
K
c
 
 
V007.033.0337.0
)104.1log(
2
0592.0
337.0
1000.4
0250.0103.2
log
2
0592.0
337.0
]CuY[
log
2
0592.0
337.0
]Cu[
1
log
2
0592.0
337.0
11
3
10
2
TCuY4
2Cu
2















 









 cK
E
 
18-14 (a) 
V799.0
0600.0
1
log
2
0592.0
763.0Zn 





E
 
(b) 
2216
)OH(Zn ]OH][Zn[100.32
 K
 
 
V10.1341.0763.0
)1033.3log(
2
0592.0
763.0
100.3
0100.0
log
2
0592.0
763.0
]OH[
log
2
0592.0
763.0
]Zn[
1
log
2
0592.0
763.0
11
16
2
)OH(Zn
2
2Zn
2




























 K
E
 
(c) 
4
3
2
2
438
4
]NH][Zn[
])NH(Zn[
1076.7



 
 
V01.1251.0763.0
)1003.3log(
2
0592.0
763.0
0100.0
250.01076.7
log
2
0592.0
763.0
])NH(Zn[
]NH[
log
2
0592.0
763.0
]Zn[
1
log
2
0592.0
763.0
8
48
2
43
4
34
2Zn








 








 








E
 
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
(d) 
   
    0395.01000.51045.4
1000.5]ZnY[
107.1102.3102.5
]Zn[
]ZnY[
32
T
32
15162
ZnY4
T
2
2








c
K
c
 
 
V24.1477.0763.0
)103.1log(
2
0592.0
763.0
1000.5
0395.0107.1
log
2
0592.0
763.0
]ZnY[
log
2
0592.0
763.0
]Zn[
1
log
2
0592.0
763.0
16
3
15
2
TZnY4
2Zn
2















 









 cK
E
 
18-15 
)(He2H2 2 g
 
 
    



















 2
H
22
H
Ho
]H[
00.1
log
2
0592.0
00.0
p
log
2
0592.0
2
a
EE
 
The ionic strength of the solution  is given by 
     0100.010100.010100.0
2
1 22 
 
From Table 10-2 
   
V121.0121.000.0
913.00100.0
00.1
log
2
0592.0
00.0
913.0
22
H








 
E
 
18-16 
V73.0Cl4)(Pte2PtCl o
2
4 


 Es
 
(a) 
 
V78.0)051.0(73.0
0263.0
1492.0
log
2
0592.0
73.0
4
Pt 







E
 
(b) 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
V198.0)044.0(154.0
1050.7
1050.2
log
2
0592.0
154.0
2
3
Pt 










E
 
(c) 
 
V355.0
1000.1
00.1
log
2
0592.0
000.0
26
Pt 










E
 
(d) 
V359.0OHVeH2VO o2
32  

 E
 
 
   
V210.0149.0359.0
100.00353.0
20586.0
log
2
0592.0
359.0
2Pt








E
 
(e) 


  4223 SnFe2SnFe2
 


  2
2
2
22 Snmmol30.2L00.25
SnClmmol
Snmmol1
mL
SnClmmol0918.0
consumedSnmmol
 











22
4
3
4
34
3
3
3
33
Snmmol340.0960.130.2remainingSnmmol
Snmmol960.1
Femmol2
Snmmol1
Femmol920.3formedSnmmol
Femmol920.3L00.25
FeClmmol
Femmol1
mL
FeClmmol1568.0
consumedFemmol
 
 
 
V177.0)023.0(154.0
0.50/960.1
0.50/340.0
log
2
0592.0
154.0Pt 





E(f) 
OH2VO2V)OH(V 2
23
4 



 



 4
4
4 )OH(Vmmol08.2L00.25
mL
)OH(Vmmol0832.0
consumed)OH(Vmmol
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 











44
2
3
2
32
3
342
3
3423
)OH(Vmmol993.0087.108.2remaining)OH(Vmmol
VOmmol174.2
Vmmol
VOmmol2
Vmmol087.1formedVOmmol
Vmmol087.1L00.50
)SO(Vmmol
Vmmol2
mL
)SO(Vmmol01087.0
consumedVmmol
 
 
  
V86.0139.000.1
1000.000.75/993.0
00.75/174.2
log0592.000.1
2Pt






E
 
18-17 (a) 
V29.0068.036.0
00566.0
0813.0
log0592.036.0Pt 





E
 
(b) 
 
V749.0022.0771.0
200845.0
0400.0
log0592.0771.0Pt 






E
 
(c) 
6
3 1082.2]OH[55.5pH
 
 
 
V329.0329.0000.0
1082.2
00.1
log
2
0592.0
000.0
26
Pt 










E
 
(d) 
  
V894.0106.000.1
0800.01996.0
0789.0
log0592.000.1
2Pt






E
 
(e) 





4
24
4
244
Cemmol04.3
L00.50
)SO(Cemmol
Cemmol1
mL
)SO(Cemmol0607.0
consumedCemmol 








22
343
2
2
2
22
Femmol196.004.300.5remainingFemmol
Femmol04.3consumedCemmolformedFemmol
Femmol00.5L00.50
FeClmmol
Femmol1
mL
FeClmmol100.0
consumedFemmol
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
V69.0)011.0(68.0
0.100/04.3
0.100/965.1
log0592.068.0Pt 





E
 
(f) 
OH2VO2V)OH(V 2
23
4 



 



 4
4
4 )OH(Vmmol314.0L00.50
mL
)OH(Vmmol0628.0
consumed)OH(Vmmol
 












44
2
3
2
32
3
342
3
3423
)OH(Vmmol85.3314.016.4remaining)OH(Vmmol
VOmmol628.0
Vmmol
VOmmol2
Vmmol314.0formedVOmmol
Vmmol16.4
L00.25
)SO(Vmmol
Vmmol2
mL
)SO(Vmmol0832.0
consumedVmmol
 
  
V194.0165.0359.0
100.000.75/628.0
00.75/85.3
log0592.0359.0
2Pt






E
 
18-18 (a) 
anodeV280.0030.0250.0
0943.0
00.1
log
2
0592.0
250.0Ni 





E
 
 
(b) 
  anodeV090.0)061.0(151.00922.0log0592.0151.0Ag E
 
(c) 
  
cathodeV003.1226.0229.1
1050.1760/780
00.1
log
4
0592.0
229.1
44
O2












E
 
(d) 
cathodeV171.0)017.0(154.0
350.0
0944.0
log
2
0592.0
154.0Pt 





E
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
(e) 
 
anodeV009.0026.0017.0
00753.0
1439.0
log0592.0017.0
2
Ag 







E
 
18-19 (a) 
cathodeV306.0031.0337.0
0897.0
00.1
log
2
0592.0
337.0Cu 





E
 
(b) 
  anodeV131.0)054.0(185.01214.0log0592.0185.0Pt E
 
(c) 
 
anodeV237.0237.000.0
1000.1
984.0
log
2
0592.0
00.0
24
Pt 










E
 
(d) 
cathodeV756.0015.0771.0
0906.0
1628.0
log0592.0771.0Pt 





E
 
(e) 
 
anodeV24.0)073.0(31.0
0827.0
0699.0
log0592.031.0
2
Ag 







E
 
18-20 
779.0)(Ag2e2Ag2 o 

 Es
 



















sp
2
3
2
sp
142
3
2
]SO[
log
2
0592.0
799.0
]Ag[
1
log
2
0592.0
799.0E
105.1]SO[]Ag[
K
K
 
When [SO3
2-] =1.00, E = Eo for 


 
2
332 SO)(Ag2e2)(SOAg ss
. 
Thus, 
V390.0409.0799.0
105.1
00.1
log
2
0592.0
799.0
00.1
log
2
0592.0
799.0
14
sp

















K
E
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
18-21 
250.0)(Ni2e4Ni2 o2 

 Es
 



















sp
4
72
22
sp
134
72
22
]OP[
log
4
0592.0
250.0
]Ni[
1
log
4
0592.0
250.0
107.1]OP[]Ni[
K
E
K
 
When [P2O7
4-] =1.00, E = Eo for 


 
4
72722 OP)(Ni2e4)(OPNi ss
. 
Thus, 
V439.0189.0250.0
107.1
00.1
log
4
0592.0
150.0
00.1
log
4
0592.0
250.0
13
sp


















K
E
 
18-22 
336.0)(Tl2e2Tl2 o 

 Es
 



















sp
2
2
sp
2222
]S[
log
2
0592.0
336.0
]Tl[
1
log
2
0592.0
336.0
106]S[]Tl[
K
E
K
 
When [S2-] =1.00, E = Eo for 


  22 S)(Tl2e2)(STl ss
. 
Thus, 
V96.0628.0336.0
106
00.1
log
2
0592.0
336.0
00.1
log
2
0592.0
336.0
22
sp


















K
E
 
18-23 
126.0)(Pb3e6Pb3 o3 

 Es
 



















sp
22
4
32
sp
3622
4
32
]AsO[
log
6
0592.0
126.0
]Pb[
1
log
6
0592.0
126.0
101.4]AsO[]Pb[
K
E
K
 
When [AsO4
2-] =1.00, E = Eo for 
  

 
2
4243
AsO2)(Pb3e4)(AsOPb ss
. 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
Thus, 
V475.0349.0126.0
101.4
00.1
log
6
0592.0
126.0
00.1
log
6
0592.0
126.0
36
sp


















K
E
 
18-24 







 ]Zn[
1
log
2
0592.0
763.0
2
E
 
16
42
2
102.3
]Y][Zn[
]ZnY[


 
 





 



]ZnY[
102.3]Y[
log
2
0592.0
763.0
2
164
E
 
When [Y4-] = [ZnY2-] = 1.00, 
 2ZnY
oEE
 
 
V25.1489.0763.0
00.1
102.300.1
log
2
0592.0
763.0
16





 
E
 
18-25 
   144
2
2
254
3
101.2]Y[
]FeY[
]Fe[and
103.1]Y[
]FeY[
]Fe[










 
 
  




















14
252
3
2
101.2]FeY[
103.1]FeY[
log0592.0771.0
]Fe[
]Fe[
log0592.0771.0E
 
When [FeY2-] = [FeY-] = 1.00, 
 FeY
oEE
 
 
 
V13.064.0771.0
101.200.1
103.100.1
log0592.0771.0
14
25








E
 
18-26 
   1123
2
232
102
3
23
1062.5]NH[
])NH(Cu[
]Cu[and
102.7]NH[
])NH(Cu[
]Cu[








 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
 
  





















102
23
11
23
2
102.7])NH(Cu[
1062.5])NH(Cu[
log0592.0153.0
]Cu[
]Cu[
log0592.0153.0E
 
When [Cu(NH3)2
+] = [Cu(NH3)2
2+] = 1.00, 
 2
23 )NH(Cu
oEE
 
 
 
V100.0053.0153.0
102.700.1
1062.500.1
log0592.0153.0
10
11








E
 
 A B C D 
1 18-27 Fe
3+
/Fe
2+
 half-cell potentials 
2 
3 E
o
, V 0.771 
Note: We use the Nernst 
equation in column C to 
calculate the potentials from 
4 We also assume25
o
C E=E
o
-0.0592log([Fe
2+
]/[Fe
3+
]) 
5 [Fe
3+
]/[Fe
2+
] [Fe
2+
]/[Fe
3+
] E,V 
6 0.001 1000.00 0.593 
7 0.0025 400.00 0.617 
8 0.005 200.00 0.635 
9 0.0075 133.33 0.645 
10 0.01 100.00 0.653 
11 0.025 40.00 0.676 
12 0.05 20.00 0.694 
13 0.075 13.33 0.704 
14 0.100 10.00 0.712 
15 0.250 4.00 0.735 
16 0.500 2.00 0.753 
17 0.750 1.33 0.764 
18 1.00 1.00 0.771 
19 1.25 0.800 0.777 
20 1.50 0.667 0.781 
21 1.75 0.571 0.785 
22 2.50 0.400 0.795 
23 5.00 0.200 0.812 
24 10.00 0.100 0.830 
25 25.00 0.040 0.854 
26 75.00 0.013 0.882 
27 100.00 0.010 0.889 
28 Spreadsheet Documentation 
29 B6=1/A6 
30 C6=$B$3-0.0592*LOG10(B6) 
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
18-28 
 A B C D 
1 18-28 Ce
4+
/Ce
3+
 half-cell potentials 
2 
3 E
o
, V 1.7 
Note: We use the Nernst 
equation in column C to 
calculate the potentials from 
4 We also assume1 M HClO4 25
o
C E=E
o
-(0.0592)log([Ce
3+
]/[Ce
4+
]) 
5 [Ce
4+
]/[Ce
3+
] [Ce
3+
]/[Ce
4+
] E,V 
6 0.001 1000.00 1.522 
7 0.0025 400.00 1.546 
8 0.005 200.00 1.564 
9 0.0075 133.33 1.574 
10 0.01 100.00 1.582 
11 0.025 40.00 1.605 
12 0.05 20.00 1.623 
13 0.075 13.33 1.633 
14 0.100 10.00 1.641 
15 0.250 4.00 1.664 
16 0.500 2.00 1.682 
17 0.750 1.33 1.693 
18 1.00 1.00 1.700 
19 1.25 0.800 1.706 
20 1.50 0.667 1.710 
21 1.75 0.571 1.714 
22 2.50 0.400 1.724 
23 5.00 0.200 1.741 
24 10.00 0.100 1.759 
25 25.00 0.040 1.783 
26 75.00 0.013 1.811 
27 100.00 0.010 1.818 
28 Spreadsheet Documentation 
29 B6=1/A6 
30 C6=$B$3-0.0592*LOG10(B6) 
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
Plot for Probelm 18-27
0.550
0.600
0.650
0.700
0.750
0.800
0.850
0.900
0.950
0 20 40 60 80 100 120
[Fe3+]/[Fe2+]
E,
 V
 
Plot of Problem 18-28
1.500
1.550
1.600
1.650
1.700
1.750
1.800
1.850
0 20 40 60 80 100 120
[Ce
4+
]/[Ce
3+
]
E,
 V
 
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
18-29 
 A B C D E F 
1 18-29 Plots for Problems 18-27, 18-28 and log ratio 
2 
3 E
o
, V (Fe
3+
) 0.771 Note: We calculate the logarithm of 
concentration ratio in column B. 4 Eo, V (Ce
4+
) 1.7 
5 [Fe
3+
]/[Fe
2+
] Log([Fe
3+
]/[Fe
2+
]
) 
E,V [Ce
4+
]/[Ce
3+
] Log([Ce
4+
]/[Ce
3+
]
) 
E,V 
6 0.001 -3.00 0.593 0.001 -3.00 1.52 
7 0.0025 -2.60 0.617 0.0025 -2.60 1.55 
8 0.005 -2.30 0.635 0.005 -2.30 1.56 
9 0.0075 -2.12 0.645 0.0075 -2.12 1.57 
10 0.01 -2.00 0.653 0.01 -2.00 1.58 
11 0.025 -1.60 0.676 0.025 -1.60 1.61 
12 0.05 -1.30 0.694 0.05 -1.30 1.62 
13 0.075 -1.12 0.704 0.075 -1.12 1.63 
14 0.100 -1.00 0.712 0.100 -1.00 1.64 
15 0.250 -0.60 0.735 0.250 -0.60 1.66 
16 0.500 -0.30 0.753 0.500 -0.30 1.68 
17 0.750 -0.12 0.764 0.750 -0.12 1.69 
18 1.00 0.00 0.771 1.00 0.00 1.70 
19 1.25 0.10 0.777 1.25 0.10 1.71 
20 1.50 0.18 0.781 1.50 0.18 1.71 
21 1.75 0.24 0.785 1.75 0.24 1.71 
22 2.50 0.40 0.795 2.50 0.40 1.72 
23 5.00 0.70 0.812 5.00 0.70 1.74 
24 10.00 1.00 0.830 10.00 1.00 1.76 
25 25.00 1.40 0.854 25.00 1.40 1.78 
26 75.00 1.88 0.882 75.00 1.88 1.81 
27 100.00 2.00 0.889 100.00 2.00 1.82 
28 Spreadsheet Documentation 
29 B6=LOG10(A6) E6=LOG10(D6) 
30 C6=$B$3+0.0592*LOG10(A6) F6=$B$4+0.0592*LOG10(D6) 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 18 
A plot of potential versus logarithm of the concentration ratio is a straight line. 
Plot of Problem 18-27
0.500
0.550
0.600
0.650
0.700
0.750
0.800
0.850
0.900
0.950
-4.00 -3.00 -2.00 -1.00 0.00 1.00 2.00 3.00
Log([Fe
3+
]/[Fe
2+
])
E,
 V
 
Plot of Problem 18-28
1.50
1.55
1.60
1.65
1.70
1.75
1.80
1.85
-4.00 -3.00 -2.00 -1.00 0.00 1.00 2.00 3.00
Log([Ce
4+
]/[Ce
3+
])
E,
 V