19 Applications of Standard Electrode Potentials
(1) Calculating thermodynamic cell potentials
(2) Calculating equilibrium constants for redox reactions
(3) Constructing redox titration curves
19A Calculating Potentials of Electrochemical Cells
Ecell = Eright – Eleft
Ex. 19-1 Calculate the thermodynamic potential of the following cell and the free
energy change associated with the cell reaction.
Cu|Cu2+(0.0200 M)∥Ag+(0.0200 M)|Ag
Ag+ + e- → Ag(s)
Cu2+ + 2e- → Cu(s)
E Ag + /Ag = 0.799 − 0.0592 log
E Cu 2 + /Cu = 0.337 −
Eº = 0.799 V
Eº = 0.337 V
1
= 0.6984V
0.0200
0.0592
1
log
= 0.2867V
0.0200
2
Ecell = EAg+ - ECu2+ = 0.6984 - 0.2867 = + 0.412 V
ΔG for Cu(s) + 2Ag+ → Cu2+ + 2Ag(s)
ΔG = -nFEcell = -2 × 96485 C × 0.412 V = -79500J (18.99 Kcal)
Ex. 19-2 Calculate the thermodynamic potential of the cell
Ag|Ag+(0.0200 M)∥Cu2+(0.0200 M)|Cu
Ecell = ECu2+ - EAg+ = 0.2867 - 0.6984 = – 0.412 V
Ex. 19-3 Calculate the potential of the following cell and indicate whether it is
galvanic or electrolytic (see Fig. 19-1).
Pt|UO22+(0.0150 M),U4+(0.200 M), H+(0.0300 M)∥
Fe2+(0.0100 M), Fe3+(0.0250 M),|Pt
Eº = 0.771 V
Fe3+ + e- → Fe2+
2+
+
4+
UO2
+ 2H2O Eº = 0.334 V
+ 4H + 2e → U
E cathode
E° anode
[Fe 2+ ]
0.0100
0
.
771
0
.
0592
log
= 0.771 − 0.0592 log
=
−
0.0250
[Fe 3+ ]
= 0.771-(-0.0236) = 0.7946 V
0.200
[ U 4+ ]
0.0592
0.0592
= 0.334 −
= 0.334 −
log
log
2+
+ 4
2
2
0.0150 × 0.0300 4
[ UO 2 ][H ]
= 0.334 - (0.2136) = 0.1204 V
127
Ecell = Ecathode – Eanode = 0.7946 – 0.1204 = +0.674 V
( positive sign → galvanic)
Fig. 19-2 Cell without liquid junction for ex. 19-4
Fig. 19-1 Cell for ex. 19-3
Ex. 19-4 Calculate the theoretic potential for
Ag|AgCl(sat'd), HCl(0.0200 M) | H2(0.800 atm), Pt
2H+
+
AgCl(s)
2e+
H2(g)
→
e-
→
E°cathode = 0.000 −
Eº = 0.000 V
+Cl-
Ag(s)
Eº = 0.222 V
0.0592
0.800
= −0.0977V
log
2
2
0.0200
Eanode = 0.222 – 0.0592 log 0.0200 = 0.3226 V
Ecell = – 0.0977 – 0.3226 = – 0.420 V
The negative sign means that the cell reaction
2H+
+
Ag(s)
Cl-
+
→
H2(g)
+
AgCl(s)
is nonspontaneous, and thus the cell is electrolytic.
Ex. 19-5 Calculate the potential for the following cell using (a) concentration and (b)
activities: Zn|ZnSO4(x M), PbSO4(sat’d)|Pb,
where x = 5.00 × 10-4, 2.00 × 10-3, 1.00 × 10-2 and 5.00 × 10-2,
(a) In a neutral solution, [SO42-] = cZnSO4 = x = 5.00 × 10-4,
+
PbSO4
Zn2+
+
2e2e-
⇔
⇔
Pb(s) + SO42Zn(s)
Eº = -0.350 V
Eº = -0.763 V
128
E PbSO3/Pb = −0.350 −
E Zn 2+ /Zn = −0.763 −
0.0592
log(5.00 × 10 −4 ) = −0.252V
2
0.0592
1
log
= −0.860V
2
5.00 × 10 − 4
Ecell = Eright – Eleft = -0252 – (-0.860) = +0.608 V
(b) Calculate activity coefficient for Zn2+ and SO421
2
μ = (5.00 × 10 −4 × 2 2 + 5.00 × 10 −4 × 2 2 ) = 2.00 × 10 −3
− log γ SO 2 − =
4
0.51× 2 2 2.00 × 10 −3
1 + 3.3 × 0.4 2.00 × 10 −3
E PbSO 4 /Pb = −0.350 −
E Zn 2 + /Zn = −0.763 −
= 8.61 × 10 − 2 , γ SO 2- = 0.820, γ Zn 2 + = 0.825
4
0.0592
log(0.820 × 5.00 × 10 −4 ) = −0.250V
2
0.0592
1
log
= −0.863V
0.825 × 5.00 ×10 −4
2
Ecell = Eright – Eleft = -0250 – (-0.863) = + 0.613 V
Table 19-1 Effect of Ionic Strength on the Potential of a Galvanic Cell
E, V, based on E, V, Experimental
E, V, based on
μ, Ionic
[ZnSO4], M
Concentration
Activity
Values
Strength
-4
5.00 × 10
2.00 × 10-3
1.00 × 10-2
2.00 × 10-2
5.00 × 10-2
2.00 × 10-3
8.00 × 10-3
4.00 × 10-2
8.00 × 10-2
2.00 × 10-1
0.608
0.573
0.531
0.513
0.490
0.613
0.582
0.550
0.537
0.521
0.611
0.583
0.553
0.542
0.529
Ex. 19-6 Calculate the potential required to initiate deposition of Cu from a solution
that is 0.010 M in CuSO4 and contains sufficient H2SO4 to give a pH of 4.00.
Eº = +1.229 V
O2(g) + 4H+ + 4e- ⇔ 2H2O
2+
+ 2e ⇔ Cu(s)
Eº = +0.337 V
Cu
0.0592
1
E Cu 2 + /Cu = +0.337 −
log
= +0.278V
2
0.010
E O 2 /H 2 O = +1.229 −
1
0.0592
1
0.0592
=
−
= +0.992V
log
1
.
229
log
pO2 × [H + ]4
1 atm × 10- 4
4
4
Ecell = Eright – Eleft = +0.278 – 0.992 = -0.714 V
⇒ The cell reaction Cu2+ + 2H2O ⇔ O2(g) + 4H+ + Cu(s) is nonspontaneous
and that to cause Cu to be deposited, we must apply a cathode potential more
negative than -0.714 V
129
19B Determining Standard Potentials Experimentally
Ex. 19-7 Pt,H2(1.00 atm)|HCl(3.215×10-3 M), AgCl(sat’d)|Ag
Eº = +0.52053 V
Calculate the Eº for the half-reaction AgCl(s) + e ⇔ Ag(s) +Cl0
E right = E AgCl
− 0.0592 log(γ Cl- )(c HCl )
1
Eleft = EH0 + /H − 0.0592 log
H+ + e- ⇔ ½ H2(g)
2
p H22
(γ H + )(cHCl )
Ecell = Eright – Eleft
1
⎡
⎤
2
p
H
0
0
2
⎢
⎥
= [ E AgCl − 0.0592 log(γ Cl− )(γ HCl )] − EH + /H − 0.0592 log
2
⎢
(γ H + )(cHCl ) ⎥
⎣
⎦
(γ + )(cHCl )
0
= E AgCl
− 0.0592 log(γ Cl− )(γ HCl ) − 0.000 − 0.0592 log H 1
p H22
Ecell = 0.52053 = E
E
0
AgCl
0
AgCl
− 0.0592 log
= 0.52053 + 0.0592 log
(γ H + )(γ Cl− )(cHCl ) 2
1
p H22
(0.945)(0.939)(3.215 × 10 −3 ) 2
1.00
1
= 0.2223 ≈ 0.222 V
2
19C Calculating Redox Equilibrium Constants
Cu(s)
+
2Ag+
Cu2+ +
→
2Ag(s)
K eq =
[Cu 2 + ]
[Ag + ]2
Cu|Cu2+(x M)∥Ag+(y M)|Ag
at chemical equilibrium
Ecell = 0 = Eright – Eleft = EAg – ECu
or
Eright = Eleft = EAg = ECu
Eox1 = Eox2= Eox3 = Eox4
E 0Ag −
0.0592
1
0.0592
1
log
log
= E 0Cu −
+ 2
2
2
[Ag ]
[Cu 2+ ]
2Ag+ +
E
0
Ag
−E
0
Cu
2e-
→
2Ag(s)
Eº = 0.799V
0.0592
1
0.0592
1
0.0592
[Cu 2+ ]
=
log
−
log
=
log
2
[ Ag + ]2
2
[Cu 2+ ]
2
[Ag + ]2
0.0592
log K eq
=
2
130
2 (E oAg − E oCu )
[Cu 2 + ]
= log
= log K eq
[Ag + ] 2
0.0592
ln K eq
0
ΔG 0 nFE cell
=−
=
RT
RT
→
n (E oB − E oA )
log K eq =
0.0592
at 25°C log K eq
0
n (E oright − E oleft )
nE cell
=
=
0.0592
0.0592
Ex. 19-8 Calculate the Keq for the reaction Cu(s) + 2Ag+ → Cu2+ + 2Ag(s)
log K eq = log
[Cu 2 + ]
[Ag + ]2
2 (0.799 − 0.337)
= 15.61
0.0592
=
Keq = antilog 15.61 = 4.1 × 1015
Ex. 19-9 Calculate the Keq for the reaction 2Fe3+ + 3I- → 2Fe2+ + I32Fe3+ +
I3-
E
+
Fe 3+ /Fe 2+
2e2e-
→
→
= E°
2Fe2+
0.771 V
3I-
Fe 3+ /Fe 2+
0.536 V
0.0592
[Fe 2 + ]2
log
−
2
[ Fe3+ ]2
0.0592
[ I - ]3
E - - = E° - - −
log
I 3 /I
I 3 /I
2
[ I3- ]
E
at equilibrium
E°
Fe3+ /Fe 2 +
Fe3+ /Fe 2 +
=E
I3- /I -
0.0592
[ Fe 2 + ]2 =
0.0592
[I - ]3
−
log
E° - - −
log
I 3 /I
2
2
[Fe3+ ]2
[I3- ]
2(E °
Fe3+ /Fe 2 +
− E °- )
I3
0.0592
= log
[Fe 2 + ]2
[Fe3+ ]2
− log
[I - ]3
[I3- ]
= log
[Fe 2 + ]2 [I3- ]
[Fe3+ ]2 [I - ]3
2(E ° 3+ 2 + − E °- - )
2+ 2 [Fe ] [I3 ]
Fe /Fe
I3 / I
=
log
3+ 2 - 3
0.0592
[Fe
log K eq =
2(E°
Fe
] [I ]
3+
/Fe
2+
− E °-
I3 / I-
)
=
0.0592
Keq = antilog 7.94 = 8.7 × 107
2(0.771 − 0.536)
= 7.94
0.0592
131
Ex. 19-10 Calculate the Keq for the reaction
2MnO4- + 3Mn2+ + 2H2O → 5MnO2(s) + 4H+
2MnO4- + 8H+ + 6e- →2MnO2(s) + 4H2O
Eº = 1.695 V
3MnO2(s) + 12H+ + 6e- → 3Mn2+ + 6H2O
Eº = 1.23 V
EMnO4-/MnO2 = EMnO2/Mn2+
1.695 -
0.0592
1
0.0592
[Mn 2 + ]3
log
log
= 1.23 − log
6
6
[MnO-4 ]2 [H + ]8
[H + ]12
6(1.695 - 1.23)
[H + ]12
[H + ]12
1
= log
+ log
= log
0.0592
[MnO -4 ] 2 [H + ] 8
[Mn 2 + ] 3
[MnO -4 ] 2 [Mn 2 + ] 3 [H + ] 8
log K eq = log
[H + ]4
[MnO -4 ]2 [Mn 2 + ]3
= 47.1
Keq = antilog 47.1 = 1 × 1047
19D Constructing Redox Titration Curves
19D-1 Electrode Potentials during Redox Titrations
Fe2+ + Ce4+ → Fe3+ + Ce3+
ECe4+/Ce3+ = EFe3+/Fe2+ = Esystem = EIn
SHE∥Ce4+, Ce3+, Fe3+, Fe2+ | Pt
Most end point in oxidation/reduction titrations are based on the rapid changes in
Esystem that occur at or near chemical equivalence.
Before the equivalence point, Esystem calculations are easiest to make using the
Nernst equation for the analyte. Beyond the equivalence point, the Nernst equation
for the titrant is more convenient.
Equivalence-point Potentials
at the equivalence point
[Fe 2 + ]
[Ce3+ ]
o
o
and E eq = E 3+ − 0.0592 log
E eq = E 4 + − 0.0592 log
4+
3+
Ce
Fe
[Ce
[Fe
]
[Ce3+ ][Fe 2 + ]
o
o
= E o 4 + + E o 3+
2E eq = E 4 + + E 3+ − 0.0592 log
Ce
Fe
Ce
Fe
[Ce 4 + ][Fe3+ ]
3+
3+
2+
4+
([Fe ] = [Ce ], [Fe ] = [Ce ])
E eq =
132
Eo
Ce 4 +
+ Eo
Fe3+
2
]
Ex. 19-11 Obtain an expression for the equivalence-point potential in the titration of
0.0500 M U4+ with 0.1000 M Ce4+. Assume both solutions are 1.0 M in H2SO4.
U4+
+
2Ce4+
UO22+ +
Ce4+
4H+
e-
+
+ 2H2O →
2e-
+
→
→
Ce3+
UO22+
U4+
+
2Ce3+
+ 2H2O
+ 4H+
Eº = 0.334 V
(formal potential) Eº' = 1.44 V
0.0592
E eq = E ° 2 + −
log
UO 2
2
[U 4+ ]
[ UO 22 + ][H + ]4
[Ce3+ ]
0.0592
°'
log
E eq = E 4 + −
4+
Ce
1
[Ce
(a)
(b)
]
[ U 4 + ][Ce3+ ]
2a + b → 3E eq = 2E ° 2 + + E °' 4 + − 0.0592 log
UO 2
Ce
[ UO 2 + ][Ce 4 + ][H + ]4
2
[U4+] = [Ce4+]/2
at equivalence
E eq =
=
2E °
UO 22 +
+ E °'
Ce 4 +
3
2E °
UO 22 +
+ E °'
Ce 4 +
3
and
[UO22+] = [Ce3+]/2
−
2[Ce 4 + ][Ce3+ ]
0.0592
log
3
2[Ce3+ ][Ce 4 + ][H + ]4
−
0.0592
1
log
3
[H + ]4
pH dependent
19D-2 The Titration Curves
Titration of 50.00 mL of 0.0500 M Fe2+ with 0.1000 M Ce4+ in a medium that is 1.0
M in H2SO4 at all times.
Ce4+
+
Fe2+ →
Ce4+ + e- → Ce3+
Fe3+ + e- → Fe2+
Initial Potential
Fe3+
+
Ce3+
Eº' = 1.44 V (1M H2SO4)
Eº' = 0.68 V (1M H2SO4)
Potential After the Addition of 5.00 mL of Cerium(IV)
[Fe3+ ] =
5.00 × 0.100
0.500
− [Ce 4 + ] =
− [Ce 4 + ]
50.00 + 5.00
55.00
[ Fe 2 + ] =
50.00 × 0.0500 − 5.00 × 0.100
2.00
+ [Ce 4 + ] =
+ [Ce 4 + ]
55.00
55.00
[Fe3+] ≒ 0.500/55.00
E system = +0.68 −
and
[Fe2+] ≒ 2.00/55.00
0.0592
2.00 / 55.00
log
= 0.64 V
1
0.500 / 55.00
133
Equilvalence-Point Potential
E 0'
E eq = Ce
4+
+ E 0' 3 +
Fe
2
=
1.44 + 0.68
= 1.06 V
2
Potential After the Addition of 25.10 mL of Cerium(IV)
[Ce3+ ] =
25.00 × 0.100
2.500
− [Fe2 + ] ≈
75.10
75.10
[Ce 4 + ] =
25.10 × 0.1000 − 50.00 × 0.0500
0.010
+ [Fe 2 + ] ≈
75.10
75.10
E = +1.44 −
0.0592
[Ce 3+ ]
0.0592
2.500 / 75.10
log
= 1.44 −
log
= 1.30 V
1
1
0.010 / 75.10
[Ce 4 + ]
Table 19-2 Electrde Potential versus SHE in Titrations with 0.100 M Ce4+
Reagent Volume, mL
5.00
15.00
20.00
24.00
24.90
Potential, vs, SHE
50.00 mL of 0.0500 M Fe2+
50.00 mL of 0.02500 M U4+
0.64
0.316
0.69
0.339
0.72
0.352
0.76
0.375
0.82
0.405
25.00
1.06
25.10
26.00
30.00
1.30
1.36
1.40
Equivalence Point
0.703
1.30
1.36
1.40
Oxidation/reduction curves are independent of the concentration of the
reactants except when the solution is very dilute.
Fig. 19-3 Titration curves for
0.1000M Ce4+ titration.
A: Titration of 50.00 mL of
0.05000 M Fe2+.
B: Titration of 50.00 mL of
0.02500 M U4+.
134
19D-3 Effect of Variables on Redox Titration Curves
Reactant concentration
titration curves are usually independent of analyte and reagent conc.
Completeness of the Reaction
completeness of the reaction↑ → change in Esystem in the equivalence-point
region ↑
Fig. 19-5
Effect of titrant electrode potential on reaction
completeness. The standard electrode potential
for the analyte (EA0) is 0.200V; starting with
curve A, standard electrode potentials for the
titrant (ET0) are 1.20, 1.00, 0.80, 0.60 and 0.40,
respectively. Both analyte and titrant undergo a
one-electron change.
19E Oxidation/Reduction Indicators
*Specific Indicators: react with one of the participants in the titration to produce a
color.
starch: form a deep blue complex with iodine
KSCN: form a red Fe(III)/thiocyanate complex with Fe(III)
*General Redox Indicators: respond to the potential of the system.
Inox +
ne-
a color change
→
0
E = EIn
−
ox /In red
Inred
Inox →
[In ]
0.0592
log red
n
[In ox ]
Inred
0.0592
[In red ] 1
[In red ]
0
E = EIn
±
≤
→
≥ 10
n
[In ox ] 10
[In ox ]
For a typical indicator to undergo a useful transition in color, the titrant must cause
a change of 0.118/n V in the potential of the system.
For many indicators, n is 2, so a change of 0.059 V is sufficient.
Iron(II) Complexes of Orthophenanthrolines
(phen)3Fe3+ +
pale blue
e- ⇔
(phen)3Fe2+
red
135
Table 19-3 Selected Oxidation/Reduction Indicators
Color
Transition
Indicator
Potential, V
Oxidized
Reduced
5-Nitro-1,10Pale blue
Red-violet
+1.25
phenanthroline iron(II)
complex
2,3’-Diphenylamine
Blue-violet Colorless
+ 1.12
dicarboxylic acid
1,10-Phenanthroline
Pale blue
Red
+ 1.11
iron(II) complex
5-Methyl-1,10Pale blue
Red
+ 1.02
phenanthroline iron(II)
complex
Erioglaucin A
Blue-red
Yellow-green
+ 0.98
Diphenylamine
Red-violet Colorless
+ 0.85
sulfonic acid
Diphenylamine
Violet
Colorless
+ 0.76
p-Ethoxychrysoidine Yellow
Red
+ 0.76
Methylene blue
Blue
Colorless
+ 0.53
Indigo tetrasulfonate Blue
Colorless
+ 0.36
Phenosafranine
Red
Colorless
+ 0.28
N
Conditions
1M H2SO4
7-10 M H2SO4
1M H2SO4
1M H2SO4
0.5M H2SO4
Dilute acid
Dilute acid
Dilute acid
1M acid
1M acid
1M acid
N
2+
Fe
N
N
3
1,10-phenanthroline
NO2
ferroin (phen)3Fe2+
N
H3C
N
N
N
5-nitro-1,10- phenanthroline
5-methyl-1,10- phenanthroline
136