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Chapter 20. Electrochemistry. Chapter Goals. Electrical Conduction Electrodes Electrolytic Cells The Electrolysis of Molten Potassium Chloride The Electrolysis of Aqueous Potassium Chloride The Electrolysis of Aqueous Potassium Sulfate Faraday’s Law of Electrolysis - PowerPoint PPT Presentation
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2
Chapter Goals
1. Electrical Conduction
2. Electrodes
Electrolytic Cells
3. The Electrolysis of Molten Potassium Chloride
4. The Electrolysis of Aqueous Potassium Chloride
5. The Electrolysis of Aqueous Potassium Sulfate
6. Faraday’s Law of Electrolysis
7. Commercial Applications of Electrolytic Cells
3
Chapter GoalsVoltaic or Galvanic Cells
8. The Construction of Simple Voltaic Cells
9. The Zinc-Copper Cell
10. The Copper-Silver Cell
Standard Electrode Potentials
11. The Standard Hydrogen Electrode
12. The Zinc-SHE Cell
13. The Copper-SHE Cell
14. Standard Electrode Potentials
15. Uses of Standard Electrode Potentials
4
Chapter Goals16. Standard Electrode Potentials for Other Half-
Reactions
17. Corrosion
18. Corrosion Protection
Effect of Concentrations (or Partial Pressures) on Electrode Potentials
19. The Nernst Equation
20. Using Electrohemical Cells to Determine Concentrations
21. The Relationship of Eocell to Go and K
5
Chapter GoalsPrimary Voltaic Cells
22. Dry Cells
Secondary Voltaic Cells
23. The Lead Storage Battery
24. The Nickel-Cadmium (Nicad) Cell
25. The Hydrogen-Oxygen Fuel Cell
6
Electrochemistry Electrochemical reactions are oxidation-reduction
reactions. The two parts of the reaction are physically separated.
The oxidation reaction occurs in one cell. The reduction reaction occurs in the other cell.
There are two kinds electrochemical cells.
1. Electrochemical cells containing in nonspontaneous chemical reactions are called electrolytic cells.
2. Electrochemical cells containing spontaneous chemical reactions are called voltaic or galvanic cells.
7
Electrical Conduction Metals conduct electric currents well in a
process called metallic conduction. In metallic conduction there is electron flow
with no atomic motion. In ionic or electrolytic conduction ionic motion
transports the electrons. Positively charged ions, cations, move toward the
negative electrode. Negatively charged ions, anions, move toward the
positive electrode.
8
Electrodes
The following convention for electrodes is correct for either electrolytic or voltaic cells:
The cathode is the electrode at which reduction occurs.
The cathode is negative in electrolytic cells and positive in voltaic cells.
The anode is the electrode at which oxidation occurs.
The anode is positive in electrolytic cells and negative in voltaic cells.
9
Electrodes
Inert electrodes do not react with the liquids or products of the electrochemical reaction.
Two examples of common inert electrodes are graphite and platinum.
10
Electrolytic Cells
Electrical energy is used to force nonspontaneous chemical reactions to occur.
The process is called electrolysis. Two examples of commercial electrolytic
reactions are:1. The electroplating of jewelry and auto parts.
2. The electrolysis of chemical compounds.
11
Electrolytic Cells
Electrolytic cells consist of:1. A container for the reaction mixture.
2. Two electrodes immersed in the reaction mixture.
3. A source of direct current.
12
The Electrolysis of Molten Potassium Chloride Liquid potassium is produced at one
electrode. Indicates that the reaction K+
() + e- K() occurs at this electrode.
Is this electrode the anode or cathode? Gaseous chlorine is produced at the other
electrode. Indicates that the reaction 2 Cl- Cl2(g) + 2 e-
occurs at this electrode. Is this electrode the anode or cathode?
13
The Electrolysis of Molten Potassium Chloride
Diagram of this electrolytic cell.
Porous barrier
e-
K+ + e- K()
cathode reaction2Cl- Cl2 (g) + 2e- anode reaction
Generator-source of DC
- electrode + electrode
e-
molten KCl
14
The Electrolysis of Molten Potassium Chloride The nonspontaneous redox reaction that occurs is:
K 2 Cl K 2 Cl 2 reaction Cell
K e K2reaction Cathode
e 2 Cl Cl 2 reaction Anode
g2-
-
-2(g)
-
15
The Electrolysis of Molten Potassium Chloride In all electrolytic cells, electrons are forced
to flow from the positive electrode (anode) to the negative electrode (cathode).
16
The Electrolysis of Aqueous Potassium Chloride In this electrolytic cell, hydrogen gas is produced at one electrode.
The aqueous solution becomes basic near this electrode. What reaction is occurring at this electrode? You do it!You do it!
Gaseous chlorine is produced at the other electrode. What reaction is occurring at this electrode? You do it!You do it!
These experimental facts lead us to the following nonspontaneous electrode reactions:
ed!electrolyz isr that wateNote ion.spectator a is K
OH 2Cl H OH 2 Cl 2 reaction Cell
OH 2 H e 2 OH 2 reaction Cathode
e 2 Cl Cl 2 reaction Anode
-g2g22
-
-g2
-2
-2(g)
-
17
The Electrolysis of Aqueous Potassium Chloride
2 H2O + 2e- H2 (g) + 2 OH-
cathode reaction2Cl- Cl2 (g) + 2e-
anode reaction
Cell diagram Battery, a source of direct currente- flow
- electrode + electrode
e- flow
aqueous KCl
Cl2 gasH2 gas
+ pole of battery- pole of battery
18
The Electrolysis of Aqueous Potassium Sulfate In this electrolysis, hydrogen gas is produced at one
electrode. The solution becomes basic near this electrode. What reaction is occurring at this electrode?
You do it! Gaseous oxygen is produced at the other electrode
The solution becomes acidic near this electrode. What reaction is occurring at this electrode?
You do it! These experimental facts lead us to the following
electrode reactions:
19
The Electrolysis of Aqueous Potassium Sulfate
O 2H OH 2 isreaction overall The
OH 4 4H O H 2 OH 6reaction Cell
)2OH H2e- OH 2(2reaction Cathode
e 4 H 4 O OH 2 reaction Anode
2(g)2(g)2
OH 4
-2(g)2(g)2
-2(g)2
-2(g)2
2
20
The Electrolysis of Aqueous Potassium Sulfate
2 H2O + 2e- H2 (g) + 2 OH-
cathode reaction2H2O O2 (g) + 4H+ + 4e-
anode reaction
Cell diagram Battery, a source of direct currente- flow
- electrode + electrode
e- flow
aqueous K2SO4
O2 gasH2 gas
+ pole of battery- pole of battery
21
Electrolytic Cells
In all electrolytic cells the most easily reduced species is reduced and the most easily oxidized species is oxidized.
22
Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Faraday’s Law - The amount of substance undergoing chemical
reaction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the electrolytic cell.
A faraday is the amount of electricity that reduces one equivalent of a species at the cathode and oxidizes one equivalent of a species at the anode.
-23 e 106.022y electricit offaraday 1
23
Counting Electrons: Coulometry and Faraday’s Law of Electrolysis A coulomb is the amount of charge that passes a
given point when a current of one ampere (A) flows for one second.
1 amp = 1 coulomb/second
coulombs 487,96e 106.022faraday 1 -23
24
Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Faraday’s Law states that during electrolysis, one
faraday of electricity (96,487 coulombs) reduces and oxidizes, respectively, one equivalent of the oxidizing agent and the reducing agent. This corresponds to the passage of one mole of electrons
through the electrolytic cell.
-23
-23
e 106.022 of lossagent reducing of equivalent 1
e 106.022 ofgain agent oxidizing of equivalent 1
25
Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Example 21-1: Calculate the mass of palladium produced by
the reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes.
g 106 2(96,500) g 106
mol 1 mol 2 mol 1
Pd 2e +Pd :Cathode 0-+2
Cathode: Pd + 2e Pd
1 mol 2 mol 1 mol
106 g 2(96,500) 106 g
3.20 amp = 3.20
g = 30.0 min60 smin
Cs
g Pd2 96,500 C
g Pd
2+ - 0
Cs
?.
.3 20 106
316
26
Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Example 21-2: Calculate the volume of oxygen
(measured at STP) produced by the oxidation of water in example 21-1.
C 96,5004 L 22.4
mol 4 mol 4 mol 1 mol 2
4e+ 4H + O OH 2 :Anode
STP
-+g22
2STP2STP
2STP2STP
STP
-+g22
O mL 334or O L 334.0
C 96,5004
O L 4.22
s
C 20.3
min
s 60min 30.0 O L ?
C 96,5004 22.4L
mol 4 mol 4 mol 1 mol 2
4e+ 4H + O OH 2 :Anode
27
Commercial Applications of Electrolytic CellsElectrolytic Refining and Electroplating of Metals
Impure metallic copper can be purified electrolytically to 100% pure Cu. The impurities commonly include some active metals plus
less active metals such as: Ag, Au, and Pt. The cathode is a thin sheet of copper metal
connected to the negative terminal of a direct current source.
The anode is large impure bars of copper.
28
Commercial Applications of Electrolytic Cells The electrolytic solution is CuSO4 and H2SO4 The impure Cu dissolves to form Cu2+. The Cu2+ ions are reduced to Cu at the cathode.
Anode impure Cu Cu 2e
Cathode very pure Cu 2e Cu
Net rxn. No net rxn.
s0
aq2
aq2+
s0
29
Commercial Applications of Electrolytic Cells Any active metal impurities are oxidized to cations that are more difficult
to reduce than Cu2+.
This effectively removes them from the Cu metal.
metals active
otherfor forth so And
2eFeFe
2eZnZn20
20
30
Commercial Applications of Electrolytic Cells The less active metals are not oxidized and
precipitate to the bottom of the cell. These metal impurities can be isolated and
separated after the cell is disconnected. Some common metals that precipitate include:
Ag, Au, Pt, Pd
Se, Te
31
Voltaic or Galvanic Cells
Electrochemical cells in which a spontaneous chemical reaction produces electrical energy.
Cell halves are physically separated so that electrons (from redox reaction) are forced to travel through wires and creating a potential difference.
Examples of voltaic cells include:
batteries calculator andComputer
batteries Flashlight
batteries Auto
32
The Construction of Simple Voltaic Cells Voltaic cells consist of two half-cells which
contain the oxidized and reduced forms of an element (or other chemical species) in contact with each other.
A simple half-cell consists of: A piece of metal immersed in a solution of its ions. A wire to connect the two half-cells. And a salt bridge to complete the circuit, maintain
neutrality, and prevent solution mixing.
34
The Zinc-Copper Cell
Cell components for the Zn-Cu cell are:1. A metallic Cu strip immersed in 1.0 M copper (II) sulfate.
2. A metallic Zn strip immersed in 1.0 M zinc (II) sulfate.
3. A wire and a salt bridge to complete circuit The cell’s initial voltage is 1.10 volts
35
The Zinc-Copper Cell
V10.1Eith reaction w sspontaneou a is This
CuZnCu Znreaction cell Overall
Cue2Cu reaction Cathode
e2Znn Zreaction Anode
0cell
0+2+20
0+2
+20
In all voltaic cells, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode).
36
The Zinc-Copper Cell There is a commonly used short hand notation for
voltaic cells. The Zn-Cu cell provides a good example.
Zn/Zn2+(1.0 M) || Cu2+(1.0 M)/Cu
species (and concentrations)in contact with electrode surfaces
electrode surfaces
salt bridge
37
The Copper - Silver Cell
Cell components:1. A Cu strip immersed in 1.0 M copper (II) sulfate.
2. A Ag strip immersed in 1.0 M silver (I) nitrate.
3. A wire and a salt bridge to complete the circuit.
The initial cell voltage is 0.46 volts.
38
The Copper - Silver Cell
V46.0Eith reaction w sspontaneou a is This
Ag 2+CuAg 2+Cu reaction cell Overall
Age+Ag2 reaction Cathode
2eCuCu reaction Anode
0cell
0+2+0
0-+
20
Compare the Zn-Cu cell to the Cu-Ag cell The Cu electrode is the cathode in the Zn-Cu cell. The Cu electrode is the anode in the Cu-Ag cell.
Whether a particular electrode behaves as an anode or as a cathode depends on what the other electrode of the cell is.
39
The Copper - Silver Cell
These experimental facts demonstrate that Cu2+ is a stronger oxidizing agent than Zn2+.
In other words Cu2+ oxidizes metallic Zn to Zn2+. Similarly, Ag+ is is a stronger oxidizing agent than Cu2+.
Because Ag+ oxidizes metallic Cu to Cu 2+. If we arrange these species in order of increasing strengths, we see that:
agent reducing asstrength
Ag>Cu>Zn
agent oxidizing asstrength
Ag<Cu<Zn
000
++2+2
40
Standard Electrode Potential
To measure relative electrode potentials, we must establish an arbitrary standard.
That standard is the Standard Hydrogen Electrode (SHE). The SHE is assigned an arbitrary voltage of 0.000000… V
41
The Zinc-SHE Cell
For this cell the components are:1. A Zn strip immersed in 1.0 M zinc (II) sulfate.
2. The other electrode is the Standard Hydrogen Electrode.
3. A wire and a salt bridge to complete the circuit. The initial cell voltage is 0.763 volts.
42
The Zinc-SHE Cell
V 763.0E H+Zn2H+Znreaction Cell
V 0.000 He 2+H 2reaction Cathode
V 0.763 e 2+Zn Znreaction Anode
E
0cellg2
+2+0
g2-+
-+20
0
The cathode is the Standard Hydrogen Electrode. In other words Zn reduces H+ to H2.
The anode is Zn metal. Zn metal is oxidized to Zn2+ ions.
43
The Copper-SHE Cell
The cell components are:1. A Cu strip immersed in 1.0 M copper (II) sulfate.
2. The other electrode is a Standard Hydrogen Electrode.
3. A wire and a salt bridge to complete the circuit. The initial cell voltage is 0.337 volts.
44
The Copper-SHE Cell
V 337.0E Cu+2HCu+H reaction Cell
V 0.337 Cu2e+Cureaction Cathode
V 0.000 e 2+H 2 H reaction Anode
E
0cell
0++2g2
0-+2
-+g2
0
In this cell the SHE is the anode The Cu2+ ions oxidize H2 to H+.
The Cu is the cathode. The Cu2+ ions are reduced to Cu metal.
45
Uses of Standard Electrode Potentials Electrodes that force the SHE to act as an anode are assigned
positive standard reduction potentials. Electrodes that force the SHE to act as the cathode are assigned
negative standard reduction potentials. Standard electrode (reduction) potentials tell us the tendencies of
half-reactions to occur as written. For example, the half-reaction for the standard potassium electrode
is:
V -2.925=E K e K 00
The large negative value tells us that this reaction will occur only under extreme conditions.
46
Uses of Standard Electrode Potentials Compare the potassium half-reaction to fluorine’s half-
reaction:
The large positive value denotes that this reaction occurs readily as written.
Positive E0 values denote that the reaction tends to occur to the right. The larger the value, the greater the tendency to occur to
the right. It is the opposite for negative values of Eo.
F + 2 e 2 F E = +2.87 V20 - - 0
47
Uses of Standard Electrode Potentials Use standard electrode potentials to predict
whether an electrochemical reaction at standard state conditions will occur spontaneously.
Example 21-3: Will silver ions, Ag+, oxidize metallic zinc to Zn2+ ions, or will Zn2+ ions oxidize metallic Ag to Ag+ ions?
Steps for obtaining the equation for the spontaneous reaction.
48
Uses of Standard Electrode Potentials1. Choose the appropriate half-reactions from a table of standard
reduction potentials.
2. Write the equation for the half-reaction with the more positive E0 value first, along with its E0 value.
3. Write the equation for the other half-reaction as an oxidation with its oxidation potential, i.e. reverse the tabulated reduction half-reaction and change the sign of the tabulated E0.
4 Balance the electron transfer.5 Add the reduction and oxidation half-reactions and their potentials.
This produces the equation for the reaction for which E0cell is positive,
which indicates that the forward reaction is spontaneous.
49
Uses of Standard Electrode Potentials
V +1.5662=E Zn+2AgZn+2Agreaction Cell
V) (-0.7628- )e 2 + Zn1(Zn Oxidation
V 0.7994+ )Age+2(Ag Reduction
E
0cell
+200+
-+2
0-+
0
50
Electrode Potentials for Other Half-Reactions Example 21-4: Will permanganate ions,
MnO4-, oxidize iron (II) ions to iron (III) ions,
or will iron (III) ions oxidize manganese(II) ions to permanganate ions in acidic solution?
Follow the steps outlined in the previous slides.
Note that E0 values are not multiplied by any stoichiometric relationships in this procedure.
51
Electrode Potentials for Other Half-Reactions Example 21-4: Will permanganate ions, MnO4
-, oxidize iron (II) ions to iron (III) ions, or will iron (III) ions oxidize manganese(II) ions to permanganate ions in acidic solution?
Thus permanganate ions will oxidize iron (II) ions to iron (III) and are reduced to manganese (II) ions in acidic solution.
V +0.74=E 5Fe+O4HMn8HFe5MnO reaction Cell
V) 0.771(- )e + Fe5(Fe Oxdation
V 1.51 + O)4HMn5e+8H1(MnO Reduction
E
0cell
+32
22-4
-+32
22--
4
0
52
Electrode Potentials for Other Half-Reactions Example 21-5: Will nitric acid, HNO3, oxidize
arsenous acid, H3AsO3, in acidic solution? The reduction product of HNO3 is NO in this reaction.
You do it!
V +0.38=E 6HAsO3H+OH2NO2HAsOH32NO
6HAsO3H+O4H2NOOH38HAsOH32NO reaction Cell
V) 0.58(- )2e + 2HAsOHOHAsO3(H Oxidation
V 0.96 + O)2HNO3e+4H2(NO Reduction
E
0cell43233
-3
432233-3
-43233
2--
3
0
53
Corrosion
Metallic corrosion is the oxidation-reduction reactions of a metal with atmospheric components such as CO2, O2, and H2O.
points. exposedat rapidly occursreaction The
reaction overall OFe 2O 3+Fe 4 3202
0
54
Corrosion Protection
Some examples of corrosion protection.1 Plate a metal with a thin layer of a less active (less
easily oxidized) metal.
"Tin plate" for steel.
55
Corrosion Protection
2. Connect the metal to a sacrificial anode, a piece of a more active metal.
anodes. lsacrificia as
exterior on the Zn and Mg
have hulls ship and pipes Soil
56
Corrosion Protection
3. Allow a protective film to form naturally.
foil. Al ofexterior on film
nt transparehard, a forms OAl
OAl 2O 3+Al 4
32
3202
0
57
Corrosion Protection
4 Galvanizing, the coating of steel with zinc, provides a more active metal on the exterior.
rust. tobegins Fe before oxidized
bemust Zn ofcoat thin The
ZincSteel
58
Corrosion Protection
5. Paint or coat with a polymeric material such as plastic or ceramic.
Steel bathtubs are coated with ceramic.
59
Effect of Concentrations (or Partial Pressures) on Electrode PotentialsThe Nernst Equation Standard electrode potentials, those compiled
in appendices, are determined at thermodynamic standard conditions.
Reminder of standard conditions. 1.00 M solution concentrations
1.00 atm of pressure for gases
All liquids and solids in their standard thermodynamic states.
Temperature of 250 C.
60
The Nernst Equation The value of the cell potentials change if conditions
are nonstandard. The Nernst equation describes the electrode
potentials at nonstandard conditions. The Nernst equation is:
61
The Nernst Equation
quotientreaction =Q
e mol J/V 487,96
VCJ 1)e C/mol (96,487=faraday the=F
ed transferrelectrons ofnumber =n
Kin etemperatur=TK mol
J 8.314=constant gas universal=R
conditions standardunder potentialE
interest ofcondition under potential=E
Q log nF
2.303RT-E=E
-
.-
0
0
62
The Nernst Equation
Substitution of the values of the constants into the Nernst equation at 25o C gives:
Q log n
0.0592-E=E Thus
0592.0e mol J/V 96,487
K 298314.8303.2
F
RT 2.303
0
-K mol
J
63
The Nernst Equation
For this half-reaction:
The corresponding Nernst equation is:
V +0.153=ECueCu 0-+2
E = E -
0.05921
log Cu
Cu0
+
2
64
The Nernst Equation Substituting E0 into the above expression gives:
If [Cu2+] and [Cu+] are both 1.0 M, i.e. at standard conditions, then E = E0 because the concentration term equals zero.
E = 0.153 V -
0.05921
log Cu
Cu
+
2
E = 0.153 V -0.0592
1 log
11
66
The Nernst Equation
Example 21-6: Calculate the potential for the Cu2+/ Cu+ electrode at 250C when the concentration of Cu+ ions is three times that of Cu2+ ions.
3Cu
Cu 3
Cu
Cu=Q
Cue +Cu
+2
+2
+2
+
-+2
67
The Nernst Equation
Example 21-6: Calculate the potential for the Cu2+/ Cu+ electrode at 250C when the concentration of Cu+ ions is three times that of Cu2+ ions.
3 log 1
0.0592-V 0.153=E
Q log 1
0.0592-E=E 0
68
The Nernst Equation
Example 21-6: Calculate the potential for the Cu2+/ Cu+ electrode at 250C when the concentration of Cu+ ions is three times that of Cu2+ ions.
V 0.125=E
V 0.0282-0.153=E
V 0.4770.0592-0.153V=E
3 log 1
0.0592-V 0.153=E
Q log 1
0.0592-E=E 0
69
The Nernst Equation
Example 21-7: Calculate the potential for the Cu2+/Cu+ electrode at 250C when the Cu+ ion concentration is 1/3 of the Cu2+ ion concentration.
You do it!
333.0Cu
Cu 31
Cu
Cu=Q
Cue +Cu
+2
+2
+2
+
-+2
70
The Nernst Equation
Example 21-7: Calculate the potential for the Cu2+/Cu+ electrode at 250C when the concentration of Cu+ ions is 1/3 that of Cu2+ ions.
31 log
1
0.0592-V 0.153=E
Q log 1
0.0592-E=E 0
71
The Nernst Equation
V 0.181=E
V 0.0282+0.153=E
V 0.477-0.0592-0.153V=E
31 log
1
0.0592-V 0.153=E
Q log 1
0.0592-E=E 0
72
The Nernst Equation
Example 21-8: Calculate the electrode potential for a hydrogen electrode in which the [H+] is 1.0 x 10-3 M and the H2 pressure is 0.50 atmosphere.
5
232+
H
2-+
100.5Q
100.1
0.50
H
P=Q
He 2 + H 2
2
73
The Nernst Equation
Example 21-8: Calculate the electrode potential for a hydrogen electrode in which the [H+] is 1.0 x 10-3 M and the H2 pressure is 0.50 atmosphere.
V 168.0E
699.52
0.05920E
105.0 log 2
0.0592-E=E 50
74
The Nernst Equation
The Nernst equation can also be used to calculate the potential for a cell that consists of two nonstandard electrodes.
Example 21-9: Calculate the initial potential of a cell that consists of an Fe3+/Fe2+ electrode in which [Fe3+]=1.0 x 10-2 M and [Fe2+]=0.1 M connected to a Sn4+/Sn2+ electrode in which [Sn4+]=1.0 M and [Sn2+]=0.10 M . A wire and salt bridge complete the circuit.
75
The Nernst Equation
Calculate the E0 cell by the usual procedure.
V 62.0E SnFe 2SnFe 2reaction Cell
V 0.15- e 2+SnSn1 Oxidation
V 0.771 Fee+Fe2 Reduction
E
0cell
+4+2+2+3
-+4+2
+2+3
0
76
The Nernst Equation
Substitute the ion concentrations into Q to calculate Ecell.
E = E -0.0592
2Q
= 0.62 V -0.0592
2
Fe Sn
Fe Sn
cell cell0
2 4
3 2
log
log
2
2
E = 0.62 V -0.0592
2cell log. .
. .
010 10
10 10 010
2
2 2
77
The Nernst Equation
V 0.53=E
V 09.062.0E
00.32
0.0592-V 0.62E
10.0100.1
0.110.0log
2
0.0592-V 0.62=E
cell
cell
cell
22
2
cell
78
Relationship of E0cell to G0 and K From previous chapters we know the relationship of
G0 and K for a reaction.
K log RT 303.2G
or
lnK -RTG
0
0
79
Relationship of E0cell to G0
and K The relationship between G0 and E0
cell is also a simple one.
e ofnumber n
e mol J/V 96,487 F where
E F-n G
-
-
0cell
0
80
Relationship of E0cell to G0
and K Combine these two relationships into a single
relationship to relate E0cell to K.
RT
E Fn K ln
or
lnK RT -E Fn -
0cell
0cell
81
Relationship of E0cell to G0
and K Example 21-10: Calculate the standard Gibbs
free energy change, G0 , at 250C for the following reaction.
Cu PbCuPb 22
82
Relationship of E0cell to G0
and K1. Calculate E0
cell using the appropriate half-reactions.
V 463.0E CuPbPbCu
V 0.126-- e 2PbPb
V 0.337 Cue 2Cu
E
0cell
022
2
02
0
83
Relationship of E0cell to G0
and K2. Now that we know E0
cell , we can calculate G0 .
V 463.0e mol V
J 500,96e mol 2G
E Fn G
-0
0cell
0
kJ -89.4G
)rxn" of mol"(per J 1094.8G
V 463.0e mol V
J 500,96e mol 2G
E Fn G
0
40
-0
0cell
0
84
Relationship of E0cell to G0
and K Example 21-11: Calculate the thermodynamic equilibrium constant for the reaction in example 21-10 at 250C.
2
215
40
0
Cu
Pb105K 36.1 K ln
K 298K molJ 8.314 -
molJ1094.8
RT
G K ln
K ln RT G
85
Relationship of E0cell to G0
and K Example 21-12: Calculate the Gibbs Free
Energy change, G and the equilibrium constant at 250C for the following reaction with the indicated concentrations.
MM 0.50 Zn Ag2 0.30 Ag 2 Zn 2
86
Relationship of E0cell to G0
and K1. Calculate the standard cell potential E0
cell.
V 562.1E ZnAg 2 ZnAg 2
V 0.763-- 2eZnZn
V 0.799 Age Ag2
E
0cell
200
-20
0-
0
87
Relationship of E0cell to G0
and K2. Use the Nernst equation to calculate Ecell for
the given concentrations.
Q log 2
0592.0EE
0.748=5.6 log
6.530.0
50.0
Ag
Zn=Q
0cellcell
22+
+2
88
Relationship of E0cell to G0
and K
V 5.6 log 2
0592.0-V 562.1E
Q log 2
0592.0EE
0.748=5.6 log
6.530.0
50.0
Ag
Zn=Q
cell
0cellcell
22+
+2
89
Relationship of E0cell to G0
and K
V 1.540=E
V 0.022-1.562=E
V 5.6 log 2
0592.0-V 562.1E
Q log 2
0592.0EE
0.748=5.6 log
6.530.0
50.0
Ag
Zn=Q
cell
cell
cell
0cellcell
22+
+2
90
Relationship of E0cell to G0 and K Ecell = +1.540 V, compared to E0cell = +1.562 V. We can use this information to calculate G.
The negative G tells us that the reaction is spontaneous.
kJ/mol -297=G
rxn. of molper J 1097.2G
V 540.1e mol V
J 500,96e mol 2G
E Fn -=G
5
--
cell
91
Relationship of E0cell to G0
and K Equilibrium constants do not change with reactant concentration.
We can use the value of E0cell at 250C to get K.
2+
+252
0cell
0cell
Ag
Zn106=K
8.52K log
298314.8303.2
562.1500,962K log
RT 2.303
E Fn -=K log
K log RT 2.303 E Fn -
92
Primary Voltaic Cells
As a voltaic cell discharges, its chemicals are consumed.
Once the chemicals are consumed, further chemical action is impossible.
The electrodes and electrolytes cannot be regenerated by reversing current flow through cell. These cells are not rechargable.
93
One example of a dry cell is flashlight, and radio, batteries.
The cell’s container is made of zinc which acts as an electrode.
A graphite rod is in the center of the cell which acts as the other electrode.
The space between the electrodes is filled with a mixture of:
1. ammonium chloride, NH4Cl
2. manganese (IV) oxide, MnO2
3. zinc chloride, ZnCl24. and a porous inactive solid.
The Dry Cell
94
The Dry Cell As electric current is produced, Zn dissolves and
goes into solution as Zn2+ ions. The Zn electrode is negative and acts as the anode.
95
The Dry Cell
The anode reaction is:
The graphite rod is the positive electrode (cathode). Ammonium ions from the NH4Cl are reduced at the cathode.
Zn Zn + 2 e0 2+ -
2 NH 2 e 2 NH H4+ -
3 g 2 g
96
The Dry Cell
The cell reaction is:
g2g3+2+
40
g2g3-+
4
-+20
H NH 2 ZnNH 2 + Znreaction Cell
H NH 2 e 2 + NH 2 reaction Cathode
e 2+Zn Zn reaction Anode
97
The Dry Cell
The other components in the cell are included to remove the byproducts of the reaction.
MnO2 prevents H2 from collecting on graphite rod.
At the anode, NH3 combines with Zn2+ to form a soluble complex and removing the Zn2+ ions from the reaction.
H MnO 2 MnO(OH)2 g 2 s s 2
Zn 4 NH Zn NH2+3 3
42
99
The Dry Cell
Alkaline dry cells are similar to ordinary dry cells except that KOH, an alkaline substance, is added to the mixture.
Half reactions for an alkaline cell are:
V 5.1E
MnO(OH) 2OHZnOH 2 MnO 2+ Znreaction Cell
OH 2 +MnO(OH) 2 e 2 +OH 2 MnO 2reaction Cathode
e 2OH ZnOH 2 +n Z reaction Anode
0cell
ss22s20s
-s
-2s2
s2-0
s
100
The Dry Cell
Alkaline dry cells are similar to ordinary dry cells except that KOH, an alkaline substance, is added to the mixture.
Half reactions for an alkaline cell are:
101
Secondary Voltaic Cells
Secondary cells are reversible, rechargeable. The electrodes in a secondary cell can be regenerated by the
addition of electricity. These cells can be switched from voltaic to electrolytic cells.
One example of a secondary voltaic cell is the lead storage or car battery.
102
The Lead Storage Battery
In the lead storage battery the electrodes are two sets of lead alloy grids (plates).
Holes in one of the grids are filled with lead (IV) oxide, PbO2. The other holes are filled with spongy lead. The electrolyte is dilute sulfuric acid.
104
The Lead Storage Battery
As the battery discharges, spongy lead is oxidized to lead ions and the plate becomes negatively charged.
The Pb2+ ions that are formed combine with SO42- from
sulfuric acid to form solid lead sulfate on the Pb electrode.
discharge. duringpost battery negative theis anode The
reaction. anode theis This
e 2 PbPb -+20s
Pb SO PbSO2+42-
4 s
105
The Lead Storage Battery
The net reaction at the anode during discharge is:
Electrons are produced at the Pb electrode. These electrons flow through an external circuit
(the wire and starter) to the PbO2 electrode. PbO2 is reduced to Pb2+ ions, in the acidic solution. The Pb2+ ions combine with SO4
2- to form PbSO4 and coat the PbO2 electrode.
PbO2 electrode is the positive electrode (cathode).
Pb SO PbSO + 2 es0
42-
4 s-
106
The Lead Storage Battery
The cell reaction for a discharging lead storage battery is:
As the cell discharges, the cathode reaction is:
OH 2+PbSO 2 SO+H 4 + PbO Pb reaction Cell
OH 2+PbSOe 2+SO+H 4+PbOreaction Cathode
e 2 PbSO SO Pb reaction Anode
2s4-2
4+
s2s
2s4--2
4+
s2
-s4
-24s
PbO + 4 H + SO + 2 e PbSO + 2 H O2 s+
42- -
4 s 2
107
The Lead Storage Battery
The cell reaction for a discharging lead storage battery is:
As the cell discharges, the cathode reaction is:
PbO + 4 H + SO + 2 e PbSO + 2 H O2 s+
42- -
4 s 2
108
The Lead Storage Battery
What happens at each electrode during recharging? At the lead (IV) oxide, PbO2, electrode, lead ions are oxidized
to lead (IV) oxide.
The concentration of the H2SO4 decreases as the cell discharges.
Recharging the cell regenerates the H2SO4.
--2
4+
s22s4 e 2 SO H 4 PbO OH 2 PbSO
charge 2s4
discharge-2
4+
s2s O2HPbSO 2SO+4H+PbOPb
109
The Lead Storage Battery
What happens at each electrode during recharging? At the lead (IV) oxide, PbO2, electrode, lead ions are oxidized
to lead (IV) oxide.
The concentration of the H2SO4 decreases as the cell discharges.
Recharging the cell regenerates the H2SO4.
--2
4+
s22s4 e 2 SO H 4 PbO OH 2 PbSO
110
The Nickel-Cadmium (Nicad) Cell
Nicad batteries are the rechargeable cells used in calculators, cameras, watches, etc.
As the battery discharges, the half-reactions are:
V4.1E
Ni(OH)+Cd(OH) OH 2 + NiO Cdrxn Cell
OH 2+Ni(OH)e 2+OH 2+NiO Cathode
e 2 Cd(OH) OH 2 Cd Anode
0
s2s22s2s
-s2
-2s2
-2
-s
111
The Hydrogen-Oxygen Fuel Cell Fuel cells are batteries that must have their
reactants continuously supplied in the presence of appropriate catalysts.
A hydrogen-oxygen fuel cell is used in the space shuttle The fuel cell is what exploded in Apollo 13.
Hydrogen is oxidized at the anode. Oxygen is reduced at the cathode.
112
The Hydrogen-Oxygen Fuel Cell
Notice that the overall reaction is the combination of hydrogen and oxygen to form water. The cell provides a drinking water supply for the astronauts
as well as the electricity for the lights, computers, etc. on board.
Fuel cells are very efficient. Energy conversion rates of 60-70% are common!
OH 2 O H 2 reaction Cell
OH 4e 4OH 2O reaction Cathode
e 2OH 2OH 2 H2 reaction Anode
2g2g2
-(aq)
-)(2g2
-)(2
-(aq)g2
113
Synthesis Question
What are the explosive chemicals in the fuel cell that exploded aboard Apollo 13?
114
Synthesis Question
The Apollo 13 fuel cells contained hydrogen and oxygen. Both are explosive, especially when mixed. The oxygen tank aboard Apollo 13 exploded.
115
Group Question
Some of the deadliest snakes in the world, for example the cobra, have venoms that are neurotoxins. Neurotoxins have an electrochemical basis. How do neurotoxins disrupt normal chemistry and eventually kill their prey?