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Chapter 13 &14
Electrochemistry and
Electrode Potentials
Oxidation: a loss of electrons to an oxidizing agent
Reduction: a gain of electrons from a reducing agent
16A Principles Reduction-oxidation reaction
(redox reaction)Ox1 + Red2 Red1 + Ox2
An oxidizing substance:Ma+ + ne- M(a-n)+
An reducing substance: Ma+ M(a-n)+ + ne-
16B Electrochemical Cells(1) Galvanic (Voltaic) cell:
a chemical reaction spontaneously occurs to produce electrical energy.
Ex: lead storage battery
(2) Electrolytic cell:electrical energy is used to force a nonsponta
neous chemical reaction to occur.Ex: electrolysis of water
the anode: oxidation occurs
the cathode: reduction occurs
Salt bridge: allows charge transfer through the solutions but prevents mixing of the solutions.
Fig 16-2
AgeAg:cathode the
e2CuCu:anode thereaction-half
CuAg2Cu2Ag
-
2
2
Cu: reducing agent
Ag+: oxidizing agent
Electrode potential: the tendency of the ions to give off or take on electrons.
Normal Hydrogen Electrode (NHE)or Standard Hydrogen Electrode (SHE)
2H+ + 2e- = H2 or H+ + e- = 1/2H2
Eo ( 標準電位 ) = 0.000 V
Table 16.1( 於 1953, the 17th IUPAC meeting 決定半反應以「還原反應」來表示 )
Potential are dependent on [con]. & temp.
Standard reduction potential: activity=1
The more positive the electrode potential, the greater the tendency of the oxidized form to be reduced.
The more negative the electrode, the greater the tendency of the reduced form to be oxidized.
Related to free energy
-nFεΔG
:condition standard
nFεΔG
ΔG w& nFq & qε w
q
w-ε
(C) charge
(J)work (V) emf
max
The Nernst Equation
lnK RTnFEΔG
EEE
FeCuFeCu
ΔG calculate
0rxn
0ox
0red
0rxn
(aq)2
(s)(s)(aq)2
Ex:Predict whether 1M HNO3 will dissolve
gold metal to form 1M Au3+?
0.54Vεεε
-1.50Vε 3eAuAu
0.96Vε
O2HNO3e4HNO
0ox
0re
0cell
0ox
3
0re
23
Cell representation
anodesolutioncathode
ionfor titratpoint end sharp 0.3~0.2E
0.617V0.1540.771
E E E :Ex
E-EE-EE-EE
Pt|)(CFe ),(CFe||)(CSn ),(CSn|Pt
Pt|)(CCe ),(CCe||)(CFe ),(CFe|Pt :Ex
cell
Sn,SnFe,Fecell
-anodecathodeleftrightcell
42
33
24
12
43
34
23
12
2423
The Nernst Equation & [C] effect
Activities should be used in the Nernst equation. We will use concentrations here because titrations de
al with large potential changes, and the errors are small by doing so.
Table 16-1, 每個離子之活性( activity )皆為 1 ,叫「標準還原電位」
Dependence of the cell potential on [C]
a
b
[Ox][Red]
lognF
2.3026RTEE
bRedneaOx
E is the reduction potential at the specified concentrations n: the number of electrons involved in the half-reactionR: gas constant (8.3143 V coul deg-1mol-1)T: absolute temperatureF: Faraday constant (96,487 coul eq-1)
at 25°C 2.3026RT/F=0.05916
ex:[C] & Ecell
standard conditions: [C]=1M what if [C]≠1M?
a)[Al3+]=2.0M, [Mn2+]=1.0M Ecell<0.48Vb)[Al3+]=1.0M, [Mn2+]=3.0M Ecell>0.48V
0.48VE
3Mn2Al3Mn2Al 0cell
(s)(aq)3
(aq)2
(s)
0E mequilibriuat
][I][Fe
][I][Felog
2
0.059EE
][I
][Ilog
2
0.059E
][Fe
][Felog
2
0.059E
0EEE
I2Fe3I2Fe :17.6Ex
cell
3233
22
I,IFe,Fe
3
3
I,I
23
22
Fe,Fe
I,IFe,Fecell
32-3
323
3
23
323
After 兩個半反應 reached eq., the cell voltage necessarily becomes zero and the reaction is complete.
Ex: One beaker contains a solution of 0.020 M KMnO4, 0.005 M M
nSO4, and 0.500 M H2SO4; and a second beaker contains 0.150
M FeSO4 and 0.0015 M Fe2 (SO4)3. The 2 beakers are connecte
d by a salt bridge and Pt electrodes are placed one in each. The electrodes are connected via a wire with a voltmeter in between.
What would be the potential of each half-cell (a) before reaction and (b) after reaction?
What would be the measured cell voltage (c) at the start of the reaction and (d) after the reaction reaches eq.?
Assume H2SO4 to be completely ionized and equal volumes in e
ach beaker.
5Fe+2 + MnO4- + 8H+ = 5Fe+3 + Mn+2 + 4H2O
Pt/Fe+2(0.15 M), Fe+3(0.003 M)//MnO4-(0.02 M), Mn+2(0.005 M), H+(1.00 M)/Pt
(a) EFe = EoFe(III)/Fe(II) – (0.059/1) log [Fe+2]/[Fe+3]
= 0.771 – 0.059 log (0.150)/(0.0015 × 2) = 0.671 V
EMn = EoMnO4-/Mn+2 – (0.059/5)log [Mn+2]/[MnO4
-][H+]8
= 1.51 – 0.059/5 log (0.005)/(0.02)(1.00) 8 = 1.52 V
(b) At eq., EFe = EMn, 可以含鐵之半反應來看,先找出平衡時兩個鐵離子的濃度,得EFe = 0.771 – 0.059 log (0.05)/(0.103) = 0.790 V
(c) Ecell = EMn - EFe = 1.52 – 0.671 = 0.849 V
(d) At eq., EFe = EMn, 所以 Ecell = 0 V
16C-7 Limitation to use E0
The sources of differences: For Fe3+ + e- Fe2+
(1) Use [conc] vs ax (activities) Fe(II)/ Fe(III) = 0.4/0.18 at = 0.1M
(2) Other equilibria:complexes Fe(III) with Cl-, SO4
-2 are more stable than those of Fe(II).
16C-7 Formal PotentialEx: Ce4+ + e- Ce3+ E°=1.6V
with H+A- E°≠1.61V則酸之陰離子會與之起某種程度的螯合,而影響他們的活性
再是標準還原電位 Eo 改叫形式電位 Eo`
Formal potential: (E°’)The standard potential of a redox couple with the oxidized and r
educed forms at 1M concentrations and with the solution conditions specified.
Ex: Ce4+/Ce3+ in 1M HCl E°’=1.28V
Dependence of potential on pH
0.059pHE'E
]AsO[H
]AsO[Hlog
2
0.059-0.059pHE
]AsO[H
]AsO[Hlog
2
0.059-]0.059log[HE
][H ]AsO[H
]AsO[Hlog
2
0.059-EE
OHAsOH2e2HAsOH :Ex
43
33
43
33
243
33
233-
43
Many redox reactions involved protons, and their potentials are influenced greatly by pH.
Dependence of potential on complexation:
][FeCl
][Fe0.059log-0.70E
constant[HCl] If
4ClFeeFeCl
FeCl iscomplex theIf
0.70V'E HCl 1MIn
0.771VE /FeFe :Ex
-4
2
2-4
-4
23
Complexing one ion reduces its effective concentration, which changes the potential.
In effect, we’ve stabilized the Fe+3 by complexing it, make it more difficult to reduce.
Ex: Systems involving ppt Calculate Ksp for AgCl at 25℃ & E =0.58V
Ag Ag
1MAgNO 3(aq)
1M NaCl(s)
& AgCl(s)
Sol:
1.0Msp
0
ClAgK
1.0
Aglog
1
0.05920
0.58Vε
(why?) 0ε
Ex 17-4: Calculate the cell potential for Ag AgCl(sat’d), HCl(0.0200 M) H∣ ∣ 2(0.800 atm), Pt
Sol: 2H+ + 2e- H2(g) E0H+/H2
= 0.000 V
AgCl(s) + e- Ag(s) E0AgCl/Ag = 0.222 V
32
Hright
)0200.0(
800.0log
2
0592.0
]H[log
2
0592.0000.0E
2
P
= -0.0977 V
Eleft = 0.222 – 0.0592 log[Cl-] = 0.222 – 0.0592 log 0.0200
= 0.3226 V
Ecell = Eright – Eleft = -0.0977 – 0.3226 = -0.420 V
2H+ + 2Ag(s) H2(g) + 2AgCl(s)
17B Calculating Redox Equilibrium Constants
Ex 17-6: Calculate the equilibrium constant for the reaction
2Fe3+ + 3I- 2Fe2+ + I3-
Sol: 2Fe3+ + 2e- 2Fe2+ E0 = 0.771 V
I3- + 2e- 3I- E0 = 0.536 V
23
22
/FeFe0
/FeFe ]Fe[
]Fe[log
2
0592.0EE 23
23
][I
]I[log
2
0592.0EE
3
3
/II0
/II 33
]I[
]I[log
]Fe[
]Fe[log
0.0592
)E-2(E
]I[
]I[log
2
0592.0E
]Fe[
]Fe[log
2
0592.0E
EE
3
3
23
22/II
0/FeFe
0
3
3
/II0
23
22
/FeFe0
/II/FeFe
--3
23
323
323
3233
22
]I[]Fe[
]I[][Felog
0592.0
)EE(2
]I[]Fe[
]I[][Felog
/II0
/FeFe0
3233
223
23
7eq
/II0
/FeFe0
eq
107.894.7loganti
94.70592.0
)536.0771.0(2
0592.0
)EE(2log 3
23
K
K
Example 17-7
Calculate the equilibrium constant for the reaction
2MnO4- + 3Mn2+ + 2H2O 5MnO2(s) + 4H+
Sol: 2MnO4- + 8H+ +6e- 2MnO2(s) + 4H2O E0 = +1.695 V
3MnO2(s) + 12H+ + 6e- 3Mn2+ + 6H2O E0 = +1.23 V
EMnO4-/MnO2
= EMnO2/Mn2+
12
32
824 ]H[
]Mn[log
6
0592.023.1
]H[]MnO[
1log
6
0592.0695.1
83224
12
]H[]Mn[]MnO[
]H[log
0592.0
)23.1695.1(6
eq3224
4
log]Mn[][MnO
]H[log1.47 K
47eq 1011.47loganti K
32
12
824 ]Mn[
]H[log
]H[]MnO[
1log
0592.0
)23.1695.1(6
17C Constructing Redox Titration Curves
Example 17-8
Obtain an expression for the equivalence-point potential in the titration of 0.0500 M U4+ with 0.1000 M Ce4+. Assume
that both solutions are 1.0 M in H2SO4.
U4+ + 2Ce4+ + 2H2O UO22+ + 2Ce3+ + 4H+
Sol: UO22+ + 4H+ + 2e- → U4+ + 2H2O E0 = 0.334 V
Ce4+ + e- Ce3+ E0' = 1.44 V422
4
/UUO0
eq]][HUO[
]U[log
2
0592.0EE 42
2
]Ce[
]Ce[log
1
0592.0EE
4
3
/CeCe0
eq34
422
4
/UUO0
eq]H][UO[
]U[log0592.0E2E2 42
2
4422
34
/CeCe0
/UUO0
eq]H][Ce][UO[
]Ce][[Ulog0592.0EE2E3 3442
2
2
]Ce[]U[
44
2
]Ce[]UO[
32
2
443
34/CeCe
0/UUO
0
eq ]H][[Ce]Ce[2
]Ce][Ce[2log
3
0592.0
3
EE2E
34422
4
/CeCe0
/UUO0
]H[
1log
3
0592.0
3
EE2 34422
Electrode Potential versus SHE in Titrations with 0.100 M Ce4+
Potential, V vs. SHE
Reagent Volume, mL
50.00 mL of 0.0500 M Fe2+
50.00 mL of 0.02500 M U4+
5.00 0.64 0.316
15.00 0.69 0.339
20.00 0.72 0.352
24.00 0.76 0.375
24.90 0.82 0.405
25.00 1.06 ←Equivalence →
Point
0.703
25.10 1.30 1.30
26.00 1.36 1.36
30.00 1.40 1.40
Table 17-1
Note: H2SO4 concentration is such that [H+] = 1.0 M throughout.
17D Oxidation/Reduction Indicators Self-indication:
If the titrant is highly colored, this color may be used to detect the end point.
Ex : MnO4- Mn2+
purple faint pink
Starch indicator:This indicator is used for titrations invol
ving iodineStarch + I2 dark-blue color complex
Redox Indicators:These are highly colored dyes that are weak r
educing or oxidizing agents that can be oxidized or reduced
Oxind + ne- Redind
ind
ind0indind Ox
Redlog
n
0.059EE
10
1
]In[
]In[
ox
red 10
]In[
]In[
ox
red
A potential equal to 2×(0.059/n)V is required for a sharp color change
n = 1 0.12Vn = 2 0.060V
The redox indicator reaction must be rapid and reversible.
Table 17.2Ex:
(1) Ferroin: [tris(1,10-phenanthroline)ion(II) sulfate]
for titrations with cerium(IV)(2) Starch/Iodine soln.
18B Reducing AgentsThiosulfate: stable to air oxidation Iron(II): E0 = 0.771V for titration of cerium(IV), chromium(V
I), vanadium(V) indicator:
ferroin or diphenylamine sulfonate.
18C Oxidizing AgentsPotassium permanganate (KMnO4)
E0=1.51In neutral solution: MnO4
-MnO2
In acid solution: MnO4-Mn2+
Autocatalytic decomposition: Standardization: Na2C2O4
5H2C2O4+2MnO4-+6H+
10CO2+2Mn2++8H2O
Cerium (IV): Ce4+ / H2SO4: E0 = 1.44V; Ce4+ / HClO4: E0 = 1.70V
1) the rate of oxidation of chloride ion is slow2) is stable in H2SO4
3) (NH4)2Ce(NO3)6 can be obtained as a primary standard.
indicator: Ferroin
Potassium dichromate: K2Cr2O7
a slightly weaker oxidizing agent than KMnO4 primary standard
Cr2O72- Cr3+
E0 = 1.33~1.00V in 1M HCl