chapter 14 electrode potentials. 14-1 redox chemistry & electricity oxidation: a loss of...
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Chapter 14
Electrode Potentials
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14-1 Redox Chemistry & Electricity
Oxidation: a loss of electrons to an oxidizing agentReduction: a gain of electrons from a reducing agent
Reduction-Oxidation reaction (redox reaction)
ex:
Half-reactions:re: Fe3+ + e- Fe2+
ox: V2+ V3+ + e-
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1. Chemistry & Electricity
• Electrochemistry: the study of the interchange of chemical & electrical energy.
• Electric charge (q) is measured in coulombs(C). • The magnitude of the charge of a single electron (or proto
n) is 1.602×10 - 19 C. A mole of electrons therefore has a charge of (1.602×10 - 19 C)(6.022×1023 /mol)= 9.649×104 C/mol, which is called the Faraday constant, F.
Example at p.310
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2. Electric current is proportional to the rate of a redox reaction
I (ampere; A) = electric current
= a flow of 1 coulomb per second = 1C/s
Example at p. 310:
Sn4+ + 2e- Sn2+
at a constant rate of 4.24 mmole/h.
How much current flows into the solution?
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3. Voltage & Electrical Work
wire q I E
hose H2O VH2O PH2O
The difference in electric potential between two points measures the work that is needed (or can be done) when electrons move from one point to another.
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Ask yourself at p.312
• Consider the redox reaction
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14.2 Galvanic Cells
Chemical reaction spontaneously occurs to produce electrical energy.
Ex: lead storage battery
When the oxidizing agent & reducing agent are physically separated, e transfer through an external wire.
generates electricity.
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A cell in action
Electrodes: the redox rxn occur
anode: oxidation occur
cathode: reduction occur
Salt bridge: connect two solns.
External wire
cathodeanode e
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Cell representation: Line Notation
Example :: Interpreting Line Diagrams of Cells
Figure 14-4 Another galvanic cell.
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14-3 Standard PotentialsCell potential ( Ecell)
a)The voltage difference between the electrodes.
electromotive force (emf)
b)can be measured by voltmeter.
c)emf of a cell depends on
The nature of the electrodes & [ions]
Temp.
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14-3 Standard Potentials
S.H.E. (standard hydrogen electrode )
It is impossible to measure Ecell of a half-rxn directly, need a reference rxn.
standard hydrogen electrode:
1atmP 1M,H
0E )( H2e2H
2H
0red cell2
gaq)(
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The standard reduction potential (E0) for each half-cell is measured by an experiment shown in idealized form in Fig.14-6.
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Table 14-1 & Appendix C
( 於 1953, the 17th IUPAC meeting 決定半反應以「還原反應」來表示 )
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Standard Reduction Potentials for reaction
0.34V
0CuCu
0.76V
0ZnZn
1.10V
0cell
(s)(aq2
(aq)2
(s)
0.76V
0ZnZn
0
0HH
0.76V
0cell
2(g)(aq)2
(s)(aq)
0red
0ox
0cell
22
22
EEE
Cu)ZnCuZn :rxn
EEE
HZnZn 2H:rxn
EEE
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0.34V0.77VEEE
0.77VE 2Fe2e2Fe
0.34VEE 2eCuCu
0.77VE FeeFe:red
FeCuCue F
0ox
0re
0cell
0re
23
0re
0ox
2
0re
23
(aq)2
(aq)2
(s)(aq)3
2
rxn:
Standard Reduction Potentials for reaction
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• AgCl (s) + e- Ag (s) + Cl-
0.222 V
0.197 V in saturated KCl (formal potentional)
E0 = 0.222V
S.H.E.║ Cl- (aq, 1M) | AgCl (s) | Ag(s)
E0’ (formal potential) = 0.197 V (in saturated KCl)
S.H.E.║ KCl (aq, saturated) | AgCl (s) | Ag(s)
Formal potential
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Formal Potential
Ex: Ce4+ + e- Ce3+ E°=1.6V
with H+A- E°≠1.61V
Formal potential: (E°’)
The potential for a cell containing a [reagent] ≠1M.
Ex: Ce4+/Ce3+ in 1M HCl E°’=1.28V
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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
14-4 The Nernst EquationThe net driving force for a reaction is expressed by the Nernst eqn.
Nernst Eqn for a Half-Reaction
where
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a
b
[A]
[B]log
n
0.05916EE
bBneaA
E = E0 when [A] = [B] = 1M
Q (Reaction quotient ) =1 E = E0
Where, Q = [B]b / [A]a
Nernst equation for a half-reaction at 25ºC
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[C] & Ecell
standard conditions: [C]=1M
what if [C]≠1M?
(ex)
a)[Al3+]=2.0M, [Mn2+]=1.0M Ecell<0.48V
b)[Al3+]=1.0M, [Mn2+]=3.0M Ecell>0.48V
0.48VE
3Mn2Al3MnAl 2
:ε
0cell
(s)(aq)3
(aq)2
(s)
0cell
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Dependence of potential on pH
0.059pHE'E
]AsO[H
]AsO[Hlog
2
0.0592-0.0592pHE
]AsO[H
]AsO[Hlog
2
0.0592-]H0.0592log[E
][H ]AsO[H
]AsO[Hlog
2
0.0592-EE
OHAsOH2e2HAsOH :Ex
43
33
43
33
243
33
233-
43
Many redox reactions involved protons, and their potentials are influenced greatly by pH.
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Nernst Equation for a Complete Reaction Equation for a Complete Reaction
• 1. Write reduction half-reactions for both half-cells and find E0 for each in Appendix C.
• 2. Write Nernst equation for the half-reaction in the right half-cell.
• 3. Write Nernst equation for the half-reaction in the left half-cell.
• 4. Fine the net cell voltage by subtraction: E=E +- E - .
• 5. To write a balanced net cell reaction.
P.321
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Nernst Equation for a complete reaction
]3.6
1log 18 [log
2
0.0592-201.1
]10[3.6
[0.01]log
2
0.0592--0.402)](-799.0[
][Ag
][Cdlog
2
0.0592-EE
2
210
2
2
Example at p. 321 Rxn: 2Ag+ (aq) + Cd (s) Ag (s) + Cd 2+ (aq)
2Ag+ + 2e- Ag (s) E0+ = 0.799
Cd 2+ + 2e- Cd (s) E0- = -0.402
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Electrons Flow Toward More Positive Potential• Electrons always flow from left to right in a diagra
m like Figure 14-7.
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14-5 E0 and the Equilibrium Constant
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Qlog n
0.059 E
]][Cl[Fe
][Felog
n
0.059EE
]log[Cln
0.059E
][Fe
][Felog
n
0.059E
(EEEEE
VE AgClFeAgFe
VE ClAgeAgCl
VE FeeFe
3
2
3
2
AgAgCl,Fe,Fecell
23
--
2-3
23
)(
))(
.)s()s(
.)s()s(
.
0
5490
2220
7710
14-5 E0 and the Equilibrium Constant
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Klog n
0.05916E
Klog n
0.05916E0
Qlog n
0.05916EE
• E = 0 and Q = K
• E0 > 0 K > 1, • E0 < 0, K < 1
At equilibrium
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Ex:• One beaker contains a solution of 0.020 M KMnO4, 0.005
M MnSO4, 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 connected 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 each beaker.
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Ans:
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
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Determinea) e- flow direction?
b) anode? cathode?
c) E =? at 25℃
Concentration Cells
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Ag Ag
1MAgNO 3(aq)
1M NaCl(s)
& AgCl(s)
1.0Msp
0
ClAgK
1.0
Aglog
1
0.05920
0.58Vε
0?ε
(ex) Calculate Ksp for AgCl at 25℃ ε=0.58V
soln:
Ex: Systems involving ppt
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14-6 Reference Electrodes
Indicator electrode: responds to analyte concentration
Reference electrode: maintains a fixed potential
]log[Cl0592.0E][Fe
][Felog0592.0EEEE
3
2
cell
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• Silver-Sliver Chloride• AgCl + e- Ag(s) +Cl-
E0 = 0.222 V
E (saturated KCl) = 0.197 V
• Calomel • Hg2Cl2 + 2e- 2Hg(l) +2Cl-
E0 = 0.268 V
E (saturated KCl) = 0.241 V saturated calomel electrode (S.C.E.)
Reference Electrodes
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Voltage conversion between different reference scales
• The potential of A ?
?