Chapter 21
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Chapter Twenty One
Electrochemistry
1
Chapter Goals 1. 2.
4. 5. 6. 7. 8.
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 Commercial Applications of Electrolytic Cells
2
Chapter Goals 2. 3. 4.
6. 7. 8. 9. 10.
Voltaic or Galvanic Cells The Construction of Simple Voltaic Cells The Zinc-Copper Cell The Copper-Silver Cell Standard Electrode Potentials The Standard Hydrogen Electrode The Zinc-SHE Cell The Copper-SHE Cell Standard Electrode Potentials Uses of Standard Electrode Potentials 3
Chapter Goals 1.
2. 3.
Standard Electrode Potentials for Other HalfReactions Corrosion Corrosion Protection Effect of Concentrations (or Partial Pressures) on Electrode Potentials The Nernst Equation Using Electrohemical Cells to Determine Concentrations The Relationship of Eocell to ∆Go and K
4
Chapter Goals Primary Voltaic Cells 2.
4. 5. 6.
Dry Cells Secondary Voltaic Cells The Lead Storage Battery The Nickel-Cadmium (Nicad) Cell The Hydrogen-Oxygen Fuel Cell
5
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. Electrochemical cells containing in nonspontaneous chemical reactions are called electrolytic cells. Electrochemical cells containing spontaneous chemical reactions are called voltaic or galvanic cells. 6
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. 7
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. 8
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.
9
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. 2.
The electroplating of jewelry and auto parts. The electrolysis of chemical compounds.
10
Electrolytic Cells Electrolytic cells consist of:
1. 2. 3.
A container for the reaction mixture. Two electrodes immersed in the reaction mixture. A source of direct current.
11
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 eoccurs at this electrode. Is this electrode the anode or cathode? 12
The Electrolysis of Molten Potassium Chloride Diagram of this electrolytic cell. eGenerator-source eof DC - electrode + electrode
molten KCl K+ + e- → K() cathode reaction
Porous barrier
2Cl- → Cl2 (g) + 2eanode reaction 13
The Electrolysis of Molten Potassium Chloride
The nonspontaneous redox reaction that occurs is:
Anode reaction
2 Cl → Cl 2(g) + 2 e -
(
+
Cathode reaction 2 K + e → K ( ) -
-
)
Cell reaction 2 Cl - + 2 K + → Cl 2 ( g ) + 2 K ( )
14
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).
15
The Electrolysis of Aqueous Potassium Chloride In this electrolytic cell, hydrogen gas is produced at Anode reaction 2 Cl → Cl 2(g) + 2 e one electrode.
The aqueous solution becomes basic near this electrode. Cathode reaction 2 H 2 O + 2 e → H 2( g ) + 2 OH What reaction is occurring at this electrode? You do it!
Cell reaction 2 Cl is+produced 2 H 2 O →atHthe + Cl 2electrode. Gaseous chlorine 2 ( g ) other ( g ) + 2 OH -
+
What reaction is occurring at this electrode? You do it!
-
K is a spectator ion. Note that wate r is electrolyz ed! These experimental facts lead us to the following nonspontaneous electrode reactions:
16
The Electrolysis of Aqueous Potassium Chloride Cell diagram e- flow
- pole of battery + pole of battery Battery, a source e- flow of direct current
- electrode H2 gas
+ electrode Cl2 gas
aqueous KCl 2 H2O + 2e- → H2 (g) + 2 OH2Cl- → Cl2 (g) + 2ecathode reaction anode reaction
17
The Electrolysis of Aqueous Potassium Sulfate
In this electrolysis, hydrogen gas is produced at one electrode.
Gaseous oxygen is produced at the other electrode
The solution becomes basic near this electrode. What reaction is occurring at this electrode? You do it! 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: 18
The Electrolysis of Aqueous Potassium Sulfate Anode reaction
2 H 2 O → O 2(g) + 4 H + + 4 e -
Cathode reaction 2(2 H 2 O + 2e- → H 2(g) + 2OH - ) + Cell reaction 6 H 2 O → 2 H 2(g) + O 2(g) + 4H + 4 OH 4 H 2O
The overall reaction is 2 H 2 O → 2H 2(g) + O 2(g)
19
The Electrolysis of Aqueous Potassium- pole Sulfate of battery + pole of battery Cell diagram e- flow
Battery, a source of direct current
- electrode
e- flow
+ electrode O2 gas
H2 gas aqueous K2SO4 2 H2O + 2e- → H2 (g) + 2 OHcathode reaction
2H2O → O2 (g) + 4H+ + 4eanode reaction
20
Electrolytic Cells
In all electrolytic cells the most easily reduced species is reduced and the most easily oxidized species is oxidized.
21
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.
1 faraday of electricity ≡ 6.022 ×10 e 23
-
22
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
1 faraday ≡ 6.022 × 10 e ≡ 96,487 coulombs 23
-
23
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.
1 equivalent of oxidizing agent ≡ gain of 6.022 × 10 23 e 1 equivalent of reducing agent ≡ loss of 6.022 × 10 23 e -
24
Counting Electrons: Coulometry and Faraday’s Law of Example 21-1: Calculate the mass of palladium produced by Electrolysis 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. 2+2 + -0 Cathode: Pd + 2e → Pd Cathode : Pd + 2e → 11mol mol 22mol mol 11mol mol
106 106gg 2(96,500) 2(96,500)106 106g g 3.20 amp = 3.20
C
s
60 s 3.20 C 106 g Pd ? g = 30.0 min × × × = 316 . g Pd min s 2( 96,500 C) 25
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. Anode Anode::22 H H22O O→ →OO22( g( g) )++4H 4H+ +++4e4e- 22 mol mol ? LSTP
11mol mol 44mol mol 44mol mol 22.4L 44((96,500 22.4 LSTP 96,500CC) ) STP
60 s 3.20 C 22.4 LSTP O 2 O 2 = 30.0 min × × × min s 4( 96,500 C ) = 0.334 LSTP O 2 or 334 mLSTP O 2
26
Commercial Applications of Electrolytic Cells Electrolytic 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.
27
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. 0 2+ − Anode impure Cu( s) → Cu( aq ) + 2e
Cathode very pure Net rxn.
2+ Cu( aq )
−
+ 2e →
0 Cu( s)
No net rxn. 28
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 0 2+ − metal. Zn → Zn + 2e 2+
−
Fe → Fe + 2e And so forth for other 0
active metals 29
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) 30
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:
Auto batteries
Flashlight batteries Computer and calculator batteries 31
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. 32
The Construction of Simple Voltaic Cells
33
The Zinc-Copper Cell Cell components for the Zn-Cu cell are:
A metallic Cu strip immersed in 1.0 M copper (II) sulfate. A metallic Zn strip immersed in 1.0 M zinc (II) sulfate. A wire and a salt bridge to complete circuit
The cell’s initial voltage is 1.10 volts
34
The Zinc-Copper Cell Anode reaction
Zn 0 → Zn 2+ + 2e −
Cathode reaction
Cu 2+ + 2e − → Cu 0
Overall cell reaction Zn 0 + Cu 2+ → Zn 2+ + Cu 0 This is a spontaneous reaction with E 0cell = +1.10V
In all voltaic cells, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode). 35
The Zinc-Copper Cell
There is a commonly used short hand notation for voltaic cells.
The Zn-Cu cell provides a good example.
species (and concentrations) in contact with electrode surfaces
Zn/Zn2+(1.0 M) || Cu2+(1.0 M)/Cu salt bridge electrode surfaces 36
The Copper - Silver Cell
Cell components:
A Cu strip immersed in 1.0 M copper (II) sulfate. A Ag strip immersed in 1.0 M silver (I) nitrate. A wire and a salt bridge to complete the circuit.
The initial cell voltage is 0.46 volts.
37
The Copper - Silver Cell Anode reaction
Cu 0 → Cu 2+ + 2e −
Cathode reaction
2 Ag + + e - → Ag 0
(
)
Overall cell reaction Cu 0 + 2 Ag + → Cu 2+ + 2 Ag 0 This is a spontaneous reaction with E 0cell = +0.46V
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. 38
The Copper - Silver Cell
These experimental facts demonstrate that Cu2+ is a stronger oxidizing agent than Zn2+.
Similarly, Ag+ is is a stronger oxidizing agent than Cu2+.
In other words Cu2+ oxidizes metallic Zn to Zn2+. Because Ag+ oxidizes metallic Cu to Cu 2+. 2+
2+
+
< Cuin order < Ag of increasing strengths, If we arrange theseZn species we see that: strength as oxidizing agent Zn 0 > Cu 0 > Ag 0 strength as reducing agent 39
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
40
The Zinc-SHE Cell For this cell the components are:
A Zn strip immersed in 1.0 M zinc (II) sulfate. The other electrode is the Standard Hydrogen Electrode. A wire and a salt bridge to complete the circuit.
The initial cell voltage is 0.763 volts.
41
The Zinc-SHE Cell E0 Anode reaction Zn 0 → Zn 2+ + 2 e -
0.763 V
Cathode reaction 2 H + + 2 e - → H 2( g )
0.000 V
Cell reaction Zn 0 + 2H + → Zn 2+ + H 2( g ) E 0cell = 0.763 V
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.
42
The Copper-SHE Cell The cell components are:
A Cu strip immersed in 1.0 M copper (II) sulfate. The other electrode is a Standard Hydrogen Electrode. A wire and a salt bridge to complete the circuit.
The initial cell voltage is 0.337 volts.
43
The Copper-SHE Cell E0 Anode reaction H 2( g ) → 2 H + + 2 e -
0.000 V
Cathode reaction Cu 2+ + 2e - → Cu 0
0.337 V
Cell reaction H 2( g ) + Cu 2+ → 2H + + Cu 0 E 0cell = +0.337 V
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.
44
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:
+
−
K +e →K
0
0
E = -2.925 V
The large negative value tells us that this reaction will occur only under extreme conditions. 45
Uses of Standard Electrode Potentials
Compare the potassium half-reaction to fluorine’s halfreaction:
0 F2
-
+ 2e → 2F
-
0
E = +2.87 V
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. 46
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. 47
Uses of Standard Electrode Potentials Choose the appropriate half-reactions from a table of
4 5
standard reduction potentials. Write the equation for the half-reaction with the more positive E0 value first, along with its E0 value. 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. Balance the electron transfer. 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. 48
Uses of Standard Electrode Potentials E0
Reduction
2(Ag + + e - → Ag 0 )
+ 0.7994 V
Oxidation
1(Zn → Zn 2+ + 2 e - )
- (-0.7628 V)
Cell reaction 2Ag + + Zn 0 → 2Ag 0 + Zn 2+ E 0cell = +1.5662 V
49
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.
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? 0 E
Reduction
1(MnO -4 + 8H + + 5e - → Mn 2+ + 4H 2 O)
Oxdation
5(Fe 2+ → Fe3+ + e - )
+ 1.51 V - (+0.771 V)
Cell reaction MnO -4 + 5Fe 2+ + 8H + → Mn 2+ + 4H 2 O + 5Fe3+ E 0cell = +0.74 V
Thus permanganate ions will oxidize iron (II) ions to iron (III) and are reduced to manganese (II) ions in acidic solution. 51
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!
E0
Reduction
2(NO3- + 4H + + 3e - → NO + 2H 2 O)
+ 0.96 V
Oxidation
3(H 3 AsO3 + H 2 O → H 3 AsO 4 + 2H + + 2e - )
- (+0.58 V)
Cell reaction 2NO 3- + 3H 3 AsO3 + 8H + + 3H 2 O → 2NO + 4H 2 O + 3H 3AsO4 + 6H + 2NO 3- + 3H 3 AsO3 + 2H + → 2NO + H 2 O + 3H 3 AsO 4 + 6H + E 0cell = +0.38 V
52
Corrosion
Metallic corrosion is the oxidation-reduction reactions of a metal with atmospheric components such as CO2, O2, and H2O.
4 Fe0 + 3 O 02 → 2 Fe 2 O 3 ( overall reaction ) The reaction occurs rapidly at exposed points.
53
Corrosion Protection 1
Some examples of corrosion protection. Plate a metal with a thin layer of a less active (less easily oxidized) metal.
"Tin plate" for steel. 54
Corrosion Protection 1.
Connect the metal to a sacrificial anode, a piece of a more active metal.
Soil pipes and ship hulls have Mg and Zn on the exterior as sacrificial anodes. 55
Corrosion Protection 1.
Allow a protective film to form naturally. 0 0 2 2 3
4 Al + 3 O → 2 Al O
Al2 O 3 forms a hard, transparent film on exterior of Al foil.
56
Corrosion Protection 4
Galvanizing, the coating of steel with zinc, provides a more active metal on the exterior.
The thin coat of Zn must be oxidized before Fe begins to rust. Zinc
Steel
57
Corrosion Protection 1.
Paint or coat with a polymeric material such as plastic or ceramic.
Steel bathtubs are coated with ceramic.
58
Effect of Concentrations (or Partial Pressures) on Nernst Equation ElectrodeThe Potentials
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.
59
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:
60
The Nernst Equation 2.303RT E=E log Q nF E = potential under condition of interest 0
E 0 = potential under standard conditions R = universal gas constant = 8.314 J mol K T = temperature in K n = number of electrons transferred F = the faraday = (96,487 C/mol e - ) × 1 J . CV = 96,487 J/V mol e Q = reaction quotient
(
) 61
The Nernst Equation
Substitution of the values of the constants into the Nernst equation at 25o C gives:
2.303 RT 2.303 × 8.314 J mol K × 298 K = = 0.0592 F 96,487 J/V mol e 0 0.0592 Thus E = E log Q n
62
The Nernst Equation
For this half-reaction:
Cu
2+
+ → + e ← Cu -
0
E = +0.153 V
The corresponding Nernst equation is:
[ [
+
] ]
Cu 0.0592 E=E log 2+ 1 Cu 0
63
The Nernst Equation
Substituting E0 into the above expression gives:
[ [
+
] ]
Cu 0.0592 E = 0.153 V log 2+ 1 Cu
If [Cu2+] and [Cu+] are both 1.0 M, i.e. at standard conditions, then E = E0 because the concentration term equals zero.
0.0592 1 E = 0.153 V log 1 1 64
The Nernst Equation
65
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.
Cu
2+
+ → + e ←Cu -
[ Cu ] 3 [Cu ] Q= = =3 [Cu ] [Cu ] +
2+
2+
2+
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. 0.0592 log Q 1 0.0592 E = 0.153 V log 3 1 E = E0 -
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. 0.0592 log Q 1 0.0592 E = 0.153 V log 3 1 E = 0.153V - 0.0592( 0.477 ) V E = ( 0.153 - 0.0282 ) V E = E0 -
E = 0.125 V 68
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.
Cu
2+
+ → + e ← Cu -
[ Cu ] [ Cu ] 3 Q= = = 0.333 [Cu ] [Cu ] +
2+
1
2+
2+
You do it! 69
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. 0.0592 E=E log Q 1 0.0592 E = 0.153 V log 1 3 1 0
70
The Nernst Equation 0 0.0592 E=E log Q 1 0.0592 E = 0.153 V log 1 3 1 E = 0.153V - 0.0592( - 0.477 ) V E = ( 0.153 + 0.0282 ) V E = 0.181 V 71
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.
→H 2H +2e ← 2 +
Q=
-
PH 2
=
( 0.50)
[H ] (1.0 ×10 ) + 2
Q = 5.0 ×10
−3 2
5
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.
(
0.0592 5 E=E log 5.0 ×10 2 0.0592 ( 5.699) E = 0− 2 E = −0.168 V 0
)
73
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.
74
The Nernst Equation
Calculate the E0 cell by the usual procedure.
Reduction Oxidation
( 1(Sn
2 Fe + e → Fe 3+
2+
2+
)
→ Sn 4+ + 2 e -
E0
)
0.771 V - ( 0.15) V
Cell reaction 2 Fe3+ + Sn 2+ → 2 Fe 2+ + Sn 4+ E 0cell = +0.62 V
75
The Nernst Equation
Substitute the ion concentrations into Q to calculate Ecell. E cell = E 0cell -
0.0592 log Q 2
= 0.62 V -
[ [
Fe
][ ] ] [ Sn ]
2+ 2
Sn 4+
2
2+
0.0592 log 2 Fe3+
0.0592 010 . ) ( 10 . ) ( = 0.62 V log −2 2 2 10 . × 10 ( 010 . ) 2
E cell
(
)
76
The Nernst Equation
( 0.0592 0.10 ) (1.0 ) = 0.62 V log 2 − 2 2 1.0 ×10 ( 0.10 ) 0.0592 ( 3.00) = 0.62 V 2 = 0.62 − 0.09 V 2
E cell E cell E cell
(
)
E cell = 0.53 V
77
Relationship of E0cell to ∆G0 and K
From previous chapters we know the relationship of ∆G0 and K for a reaction.
∆G = -RT lnK or 0
∆G = −2.303 RT log K 0
78
Relationship of E0cell to ∆G0 and K
The relationship between ∆G0 and E0cell is also a simple one.
∆G = -n F E 0
0 cell
where F = 96,487 J/V mol e n = number of e
-
-
79
Relationship of E0cell to ∆G0 and K
Combine these two relationships into a single relationship to relate E0cell to K.
-n FE
0 cell
= - RT lnK or 0 cell
n FE ln K = RT
80
Relationship of E0cell to ∆G0 and K
Example 21-10: Calculate the standard Gibbs free energy change, ∆G0 , at 250C for the following reaction. 2+ 2+
Pb + Cu
→ Pb + Cu
81
Relationship of E0cell to ∆G0 and K
Calculate E0cell using the appropriate halfreactions.
E 2+
−
Cu + 2 e → Cu 2+
Pb → Pb + 2 e 2+
0
0.337 V - ( - 0.126 V )
− 2+
0
Cu + Pb → Pb + Cu E 0
0 cell
= 0.463 V 82
Relationship of E0cell to ∆G0 and K
Now that we know E0cell , we can calculate ∆G0 .
∆∆GG ==−−nnFFEE 00
((
00 cell cell
)( )(
))
mole e 9696 ,500J J 0.463 ∆∆GG ==−−22mol ,500 V)V) - .(463 - (0 mol VV mol e e 0 4 ∆G = −8.94 ×10 J (per " mol of rxn" ) 00
−−
∆G 0 = -89.4 kJ 83
Relationship of E0cell to ∆G0 and K 21-11: Calculate the thermodynamic Example equilibrium constant for the reaction in example 21-10 at 250C.
∆G = − RT ln K ∴ 0
− 8.94 ×10 4 J
∆G ln K = = − RT - 8.314 J 0
(
ln K = 36.1∴ K = 5 ×10
15
mol ( 298 K )
) mol K
[ Pb ] = [Cu ] 2+
2+
84
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.
Zn + 2 Ag ( 0.30 M ) → 2Ag + Zn +
2+
( 0.50 M )
85
Relationship of E0cell to ∆G0 and K
Calculate the standard cell potential E0cell.
( ( Zn
+
2 Ag + e → Ag 0
-
2+
0
E
)
→ Zn + 2e
-
)
2 Ag + + Zn 0 → 2 Ag 0 + Zn 2+
0
0.799 V - ( - 0.763 V ) E 0cell = +1.562 V
86
Relationship of E0cell to ∆G0 and K
Use the Nernst equation to calculate Ecell for the given concentrations.
[ ( Zn ] 0.50 ) Q= = [Ag ] ( 0.30) 2+
+ 2
E cell
2
= 5.6
log 5.6 = 0.748 0.0592 0 = E cell − log Q 2 87
Relationship of E0cell to ∆G0 and K
[ Zn ] ( 0.50) Q= = [ Ag ] ( 0.30) 2+
+ 2
2
= 5 .6
log 5.6 = 0.748 0.0592 0 E cell = E cell − log Q 2 0.0592 E cell = 1.562 V log ( 5.6) V 2 88
Relationship of E0cell to ∆G0 and K [ Zn ] ( 0.50 ) Q= = = 5.6 2+
[ Ag ]
+ 2
( 0.30) 2
log 5.6 = 0.748 0.0592 0 E cell = E cell − log Q 2 0.0592 E cell = 1.562 V log ( 5.6) V 2 E cell = (1.562 - 0.022) V E cell = 1.540 V 89
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.
∆G = - n F E cell
(
)(
∆G = − 2 mol e - 96,500 J
) ( 1.540 V ) V mol e -
∆G = −2.97 ×105 J per mol of rxn. ∆G = -297 kJ/mol
The negative ∆G tells us that the reaction is spontaneous. 90
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. - n F E 0cell = − 2.303 RT log K - n F E 0cell log K = − 2.303 RT − ( 2 )( 96,500 )(1.562 ) log K = − ( 2.303)( 8.314 )( 298) log K = 52.8 K = 6 ×10
52
[ Zn ] = [Ag ] 2+
+ 2
91
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.
92
The Dry Cell 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:
ammonium chloride, NH4Cl
manganese (IV) oxide, MnO2
zinc chloride, ZnCl2
and a porous inactive solid. 93
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.
94
The Dry Cell
The anode reaction is:
0
Zn → Zn
2
2+
+2 e
-
The graphite rod is the positive electrode (cathode). Ammonium ions from the NH4Cl are reduced at the cathode.
+ NH 4
-
+ 2 e → 2 NH 3( g ) + H 2( g ) 95
The Dry Cell
The cell reaction is:
Anode reaction
Zn 0 → Zn 2+ + 2 e -
Cathode reaction 2 NH +4 + 2 e - → 2 NH 3( g ) + H 2( g ) Cell reaction
Zn 0 + 2 NH +4 → Zn 2+ + 2 NH 3( g ) + H 2( g )
96
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.
H 2( g ) + 2 MnO2( s) → 2 MnO(OH)( s)
At the anode, NH3 combines with Zn2+ to form a soluble complex and removing the Zn2+ ions from the reaction.
Zn
2+
+ 4 NH 3 → Zn(
)
2+ NH 3 4 97
The Dry Cell
98
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:
Anode reaction
Zn 0( s ) + 2 OH - → Zn( OH ) 2( s ) + 2 e −
Cathode reaction 2 MnO 2( s ) + 2 H 2O + 2 e - → 2 MnO(OH)( s ) + 2 OH Cell reaction
Zn 0( s ) + 2 MnO 2( s ) + 2 H 2O → Zn( OH ) 2( s ) + 2 MnO(OH)( s ) E 0cell = −1.5 V
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:
100
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.
101
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.
102
The Lead Storage Battery
Diagram of the lead storage battery.
103
The Lead Storage Battery
As the battery discharges, spongy lead is oxidized to lead ions and the plate becomes negatively charged. Pb 0( s ) → Pb 2+ + 2 e This is the anode reaction. The anode is the negative battery post during discharge.
The Pb2+ ions that are formed combine with SO42- from sulfuric acid to form solid lead sulfate on the Pb electrode. 2+ 2Pb + SO 4 →PbSO 4 ( s) 104
The Lead Storage Battery
The net reaction at the anode during discharge is:
0 Pb( s)
+
2SO4
→ PbSO4( s) + 2 e
-
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 SO42- to form PbSO4 and coat the PbO2 electrode. PbO2 electrode is the positive electrode (cathode). 105
The Lead Storage Battery
As the cell discharges, the cathode reaction is:
PbO2 ( s) + 4 H + + SO2+ 2 e → PbSO 4( s) + 2 H 2 O 4
The cell reaction for a discharging lead storage battery is:
Anode reaction
Pb ( s ) + SO 24- → PbSO 4( s ) + 2 e -
Cathode reaction PbO 2( s ) + 4 H + + SO 24- + 2 e - → PbSO 4( s ) + 2 H 2 O Cell reaction
Pb ( s ) + PbO 2( s ) + 4 H + + SO 24- → 2 PbSO 4( s ) + 2 H 2 O
106
The Lead Storage Battery
As the cell discharges, the cathode reaction is:
PbO2 ( s) + 4 H + + SO2+ 2 e → PbSO 4( s) + 2 H 2 O 4
The cell reaction for a discharging lead storage battery is:
107
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.
PbSO 4( s ) + 2 H 2 O → PbO 2 ( s ) + 4 H + + SO 24- + 2 e
The concentration of the H2SO4 decreases as the cell discharges.
Recharging the cell regenerates the H2SO4.
Pb ( s ) + PbO 2 ( s )
discharge → PbSO + 2H O + 4H + 2SO ← 4( s) 2 charge +
24
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.
PbSO 4( s ) + 2 H 2 O → PbO 2 ( s ) + 4 H + + SO 24- + 2 e
The concentration of the H2SO4 decreases as the cell discharges.
Recharging the cell regenerates the H2SO4.
109
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: Anode Cd ( s ) + 2 OH - → Cd(OH) 2 + 2 e Cathode NiO 2 ( s ) + 2 H 2 O + 2 e - → Ni(OH) 2 ( s ) + 2 OH Cell rxn Cd ( s ) + NiO 2( s ) + 2 H 2 O → Cd(OH) 2( s ) + Ni(OH) 2 ( s ) E 0 ≈ 1.4V 110
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.
111
The Hydrogen-Oxygen Fuel Cell Anode reaction 2( H ( ) + 2 OH → 2 H O + 2 e ) Cathode reaction ( O ( ) + 2 H O + 4 e → 4 OH ) (aq)
2 g
-
2
( )
-
2 g
Cell reaction
( )
2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( )
Notice that the overall reaction is the combination of hydrogen and oxygen to form water.
2
(aq)
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! 112
Synthesis Question
What are the explosive chemicals in the fuel cell that exploded aboard Apollo 13?
113
Synthesis Question
The Apollo 13 fuel cells contained hydrogen and oxygen. Both are explosive, especially when mixed. The oxygen tank aboard Apollo 13 exploded.
114
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?
115
End of Chapter 21
Electrochemistry is an important part of the electronics industry.
116
Electrochemistry
1
Chapter Goals 1. 2.
4. 5. 6. 7. 8.
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 Commercial Applications of Electrolytic Cells
2
Chapter Goals 2. 3. 4.
6. 7. 8. 9. 10.
Voltaic or Galvanic Cells The Construction of Simple Voltaic Cells The Zinc-Copper Cell The Copper-Silver Cell Standard Electrode Potentials The Standard Hydrogen Electrode The Zinc-SHE Cell The Copper-SHE Cell Standard Electrode Potentials Uses of Standard Electrode Potentials 3
Chapter Goals 1.
2. 3.
Standard Electrode Potentials for Other HalfReactions Corrosion Corrosion Protection Effect of Concentrations (or Partial Pressures) on Electrode Potentials The Nernst Equation Using Electrohemical Cells to Determine Concentrations The Relationship of Eocell to ∆Go and K
4
Chapter Goals Primary Voltaic Cells 2.
4. 5. 6.
Dry Cells Secondary Voltaic Cells The Lead Storage Battery The Nickel-Cadmium (Nicad) Cell The Hydrogen-Oxygen Fuel Cell
5
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. Electrochemical cells containing in nonspontaneous chemical reactions are called electrolytic cells. Electrochemical cells containing spontaneous chemical reactions are called voltaic or galvanic cells. 6
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. 7
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. 8
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.
9
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. 2.
The electroplating of jewelry and auto parts. The electrolysis of chemical compounds.
10
Electrolytic Cells Electrolytic cells consist of:
1. 2. 3.
A container for the reaction mixture. Two electrodes immersed in the reaction mixture. A source of direct current.
11
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 eoccurs at this electrode. Is this electrode the anode or cathode? 12
The Electrolysis of Molten Potassium Chloride Diagram of this electrolytic cell. eGenerator-source eof DC - electrode + electrode
molten KCl K+ + e- → K() cathode reaction
Porous barrier
2Cl- → Cl2 (g) + 2eanode reaction 13
The Electrolysis of Molten Potassium Chloride
The nonspontaneous redox reaction that occurs is:
Anode reaction
2 Cl → Cl 2(g) + 2 e -
(
+
Cathode reaction 2 K + e → K ( ) -
-
)
Cell reaction 2 Cl - + 2 K + → Cl 2 ( g ) + 2 K ( )
14
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).
15
The Electrolysis of Aqueous Potassium Chloride In this electrolytic cell, hydrogen gas is produced at Anode reaction 2 Cl → Cl 2(g) + 2 e one electrode.
The aqueous solution becomes basic near this electrode. Cathode reaction 2 H 2 O + 2 e → H 2( g ) + 2 OH What reaction is occurring at this electrode? You do it!
Cell reaction 2 Cl is+produced 2 H 2 O →atHthe + Cl 2electrode. Gaseous chlorine 2 ( g ) other ( g ) + 2 OH -
+
What reaction is occurring at this electrode? You do it!
-
K is a spectator ion. Note that wate r is electrolyz ed! These experimental facts lead us to the following nonspontaneous electrode reactions:
16
The Electrolysis of Aqueous Potassium Chloride Cell diagram e- flow
- pole of battery + pole of battery Battery, a source e- flow of direct current
- electrode H2 gas
+ electrode Cl2 gas
aqueous KCl 2 H2O + 2e- → H2 (g) + 2 OH2Cl- → Cl2 (g) + 2ecathode reaction anode reaction
17
The Electrolysis of Aqueous Potassium Sulfate
In this electrolysis, hydrogen gas is produced at one electrode.
Gaseous oxygen is produced at the other electrode
The solution becomes basic near this electrode. What reaction is occurring at this electrode? You do it! 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: 18
The Electrolysis of Aqueous Potassium Sulfate Anode reaction
2 H 2 O → O 2(g) + 4 H + + 4 e -
Cathode reaction 2(2 H 2 O + 2e- → H 2(g) + 2OH - ) + Cell reaction 6 H 2 O → 2 H 2(g) + O 2(g) + 4H + 4 OH 4 H 2O
The overall reaction is 2 H 2 O → 2H 2(g) + O 2(g)
19
The Electrolysis of Aqueous Potassium- pole Sulfate of battery + pole of battery Cell diagram e- flow
Battery, a source of direct current
- electrode
e- flow
+ electrode O2 gas
H2 gas aqueous K2SO4 2 H2O + 2e- → H2 (g) + 2 OHcathode reaction
2H2O → O2 (g) + 4H+ + 4eanode reaction
20
Electrolytic Cells
In all electrolytic cells the most easily reduced species is reduced and the most easily oxidized species is oxidized.
21
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.
1 faraday of electricity ≡ 6.022 ×10 e 23
-
22
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
1 faraday ≡ 6.022 × 10 e ≡ 96,487 coulombs 23
-
23
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.
1 equivalent of oxidizing agent ≡ gain of 6.022 × 10 23 e 1 equivalent of reducing agent ≡ loss of 6.022 × 10 23 e -
24
Counting Electrons: Coulometry and Faraday’s Law of Example 21-1: Calculate the mass of palladium produced by Electrolysis 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. 2+2 + -0 Cathode: Pd + 2e → Pd Cathode : Pd + 2e → 11mol mol 22mol mol 11mol mol
106 106gg 2(96,500) 2(96,500)106 106g g 3.20 amp = 3.20
C
s
60 s 3.20 C 106 g Pd ? g = 30.0 min × × × = 316 . g Pd min s 2( 96,500 C) 25
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. Anode Anode::22 H H22O O→ →OO22( g( g) )++4H 4H+ +++4e4e- 22 mol mol ? LSTP
11mol mol 44mol mol 44mol mol 22.4L 44((96,500 22.4 LSTP 96,500CC) ) STP
60 s 3.20 C 22.4 LSTP O 2 O 2 = 30.0 min × × × min s 4( 96,500 C ) = 0.334 LSTP O 2 or 334 mLSTP O 2
26
Commercial Applications of Electrolytic Cells Electrolytic 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.
27
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. 0 2+ − Anode impure Cu( s) → Cu( aq ) + 2e
Cathode very pure Net rxn.
2+ Cu( aq )
−
+ 2e →
0 Cu( s)
No net rxn. 28
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 0 2+ − metal. Zn → Zn + 2e 2+
−
Fe → Fe + 2e And so forth for other 0
active metals 29
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) 30
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:
Auto batteries
Flashlight batteries Computer and calculator batteries 31
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. 32
The Construction of Simple Voltaic Cells
33
The Zinc-Copper Cell Cell components for the Zn-Cu cell are:
A metallic Cu strip immersed in 1.0 M copper (II) sulfate. A metallic Zn strip immersed in 1.0 M zinc (II) sulfate. A wire and a salt bridge to complete circuit
The cell’s initial voltage is 1.10 volts
34
The Zinc-Copper Cell Anode reaction
Zn 0 → Zn 2+ + 2e −
Cathode reaction
Cu 2+ + 2e − → Cu 0
Overall cell reaction Zn 0 + Cu 2+ → Zn 2+ + Cu 0 This is a spontaneous reaction with E 0cell = +1.10V
In all voltaic cells, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode). 35
The Zinc-Copper Cell
There is a commonly used short hand notation for voltaic cells.
The Zn-Cu cell provides a good example.
species (and concentrations) in contact with electrode surfaces
Zn/Zn2+(1.0 M) || Cu2+(1.0 M)/Cu salt bridge electrode surfaces 36
The Copper - Silver Cell
Cell components:
A Cu strip immersed in 1.0 M copper (II) sulfate. A Ag strip immersed in 1.0 M silver (I) nitrate. A wire and a salt bridge to complete the circuit.
The initial cell voltage is 0.46 volts.
37
The Copper - Silver Cell Anode reaction
Cu 0 → Cu 2+ + 2e −
Cathode reaction
2 Ag + + e - → Ag 0
(
)
Overall cell reaction Cu 0 + 2 Ag + → Cu 2+ + 2 Ag 0 This is a spontaneous reaction with E 0cell = +0.46V
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. 38
The Copper - Silver Cell
These experimental facts demonstrate that Cu2+ is a stronger oxidizing agent than Zn2+.
Similarly, Ag+ is is a stronger oxidizing agent than Cu2+.
In other words Cu2+ oxidizes metallic Zn to Zn2+. Because Ag+ oxidizes metallic Cu to Cu 2+. 2+
2+
+
< Cuin order < Ag of increasing strengths, If we arrange theseZn species we see that: strength as oxidizing agent Zn 0 > Cu 0 > Ag 0 strength as reducing agent 39
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
40
The Zinc-SHE Cell For this cell the components are:
A Zn strip immersed in 1.0 M zinc (II) sulfate. The other electrode is the Standard Hydrogen Electrode. A wire and a salt bridge to complete the circuit.
The initial cell voltage is 0.763 volts.
41
The Zinc-SHE Cell E0 Anode reaction Zn 0 → Zn 2+ + 2 e -
0.763 V
Cathode reaction 2 H + + 2 e - → H 2( g )
0.000 V
Cell reaction Zn 0 + 2H + → Zn 2+ + H 2( g ) E 0cell = 0.763 V
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.
42
The Copper-SHE Cell The cell components are:
A Cu strip immersed in 1.0 M copper (II) sulfate. The other electrode is a Standard Hydrogen Electrode. A wire and a salt bridge to complete the circuit.
The initial cell voltage is 0.337 volts.
43
The Copper-SHE Cell E0 Anode reaction H 2( g ) → 2 H + + 2 e -
0.000 V
Cathode reaction Cu 2+ + 2e - → Cu 0
0.337 V
Cell reaction H 2( g ) + Cu 2+ → 2H + + Cu 0 E 0cell = +0.337 V
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.
44
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:
+
−
K +e →K
0
0
E = -2.925 V
The large negative value tells us that this reaction will occur only under extreme conditions. 45
Uses of Standard Electrode Potentials
Compare the potassium half-reaction to fluorine’s halfreaction:
0 F2
-
+ 2e → 2F
-
0
E = +2.87 V
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. 46
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. 47
Uses of Standard Electrode Potentials Choose the appropriate half-reactions from a table of
4 5
standard reduction potentials. Write the equation for the half-reaction with the more positive E0 value first, along with its E0 value. 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. Balance the electron transfer. 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. 48
Uses of Standard Electrode Potentials E0
Reduction
2(Ag + + e - → Ag 0 )
+ 0.7994 V
Oxidation
1(Zn → Zn 2+ + 2 e - )
- (-0.7628 V)
Cell reaction 2Ag + + Zn 0 → 2Ag 0 + Zn 2+ E 0cell = +1.5662 V
49
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.
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? 0 E
Reduction
1(MnO -4 + 8H + + 5e - → Mn 2+ + 4H 2 O)
Oxdation
5(Fe 2+ → Fe3+ + e - )
+ 1.51 V - (+0.771 V)
Cell reaction MnO -4 + 5Fe 2+ + 8H + → Mn 2+ + 4H 2 O + 5Fe3+ E 0cell = +0.74 V
Thus permanganate ions will oxidize iron (II) ions to iron (III) and are reduced to manganese (II) ions in acidic solution. 51
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!
E0
Reduction
2(NO3- + 4H + + 3e - → NO + 2H 2 O)
+ 0.96 V
Oxidation
3(H 3 AsO3 + H 2 O → H 3 AsO 4 + 2H + + 2e - )
- (+0.58 V)
Cell reaction 2NO 3- + 3H 3 AsO3 + 8H + + 3H 2 O → 2NO + 4H 2 O + 3H 3AsO4 + 6H + 2NO 3- + 3H 3 AsO3 + 2H + → 2NO + H 2 O + 3H 3 AsO 4 + 6H + E 0cell = +0.38 V
52
Corrosion
Metallic corrosion is the oxidation-reduction reactions of a metal with atmospheric components such as CO2, O2, and H2O.
4 Fe0 + 3 O 02 → 2 Fe 2 O 3 ( overall reaction ) The reaction occurs rapidly at exposed points.
53
Corrosion Protection 1
Some examples of corrosion protection. Plate a metal with a thin layer of a less active (less easily oxidized) metal.
"Tin plate" for steel. 54
Corrosion Protection 1.
Connect the metal to a sacrificial anode, a piece of a more active metal.
Soil pipes and ship hulls have Mg and Zn on the exterior as sacrificial anodes. 55
Corrosion Protection 1.
Allow a protective film to form naturally. 0 0 2 2 3
4 Al + 3 O → 2 Al O
Al2 O 3 forms a hard, transparent film on exterior of Al foil.
56
Corrosion Protection 4
Galvanizing, the coating of steel with zinc, provides a more active metal on the exterior.
The thin coat of Zn must be oxidized before Fe begins to rust. Zinc
Steel
57
Corrosion Protection 1.
Paint or coat with a polymeric material such as plastic or ceramic.
Steel bathtubs are coated with ceramic.
58
Effect of Concentrations (or Partial Pressures) on Nernst Equation ElectrodeThe Potentials
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.
59
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:
60
The Nernst Equation 2.303RT E=E log Q nF E = potential under condition of interest 0
E 0 = potential under standard conditions R = universal gas constant = 8.314 J mol K T = temperature in K n = number of electrons transferred F = the faraday = (96,487 C/mol e - ) × 1 J . CV = 96,487 J/V mol e Q = reaction quotient
(
) 61
The Nernst Equation
Substitution of the values of the constants into the Nernst equation at 25o C gives:
2.303 RT 2.303 × 8.314 J mol K × 298 K = = 0.0592 F 96,487 J/V mol e 0 0.0592 Thus E = E log Q n
62
The Nernst Equation
For this half-reaction:
Cu
2+
+ → + e ← Cu -
0
E = +0.153 V
The corresponding Nernst equation is:
[ [
+
] ]
Cu 0.0592 E=E log 2+ 1 Cu 0
63
The Nernst Equation
Substituting E0 into the above expression gives:
[ [
+
] ]
Cu 0.0592 E = 0.153 V log 2+ 1 Cu
If [Cu2+] and [Cu+] are both 1.0 M, i.e. at standard conditions, then E = E0 because the concentration term equals zero.
0.0592 1 E = 0.153 V log 1 1 64
The Nernst Equation
65
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.
Cu
2+
+ → + e ←Cu -
[ Cu ] 3 [Cu ] Q= = =3 [Cu ] [Cu ] +
2+
2+
2+
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. 0.0592 log Q 1 0.0592 E = 0.153 V log 3 1 E = E0 -
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. 0.0592 log Q 1 0.0592 E = 0.153 V log 3 1 E = 0.153V - 0.0592( 0.477 ) V E = ( 0.153 - 0.0282 ) V E = E0 -
E = 0.125 V 68
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.
Cu
2+
+ → + e ← Cu -
[ Cu ] [ Cu ] 3 Q= = = 0.333 [Cu ] [Cu ] +
2+
1
2+
2+
You do it! 69
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. 0.0592 E=E log Q 1 0.0592 E = 0.153 V log 1 3 1 0
70
The Nernst Equation 0 0.0592 E=E log Q 1 0.0592 E = 0.153 V log 1 3 1 E = 0.153V - 0.0592( - 0.477 ) V E = ( 0.153 + 0.0282 ) V E = 0.181 V 71
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.
→H 2H +2e ← 2 +
Q=
-
PH 2
=
( 0.50)
[H ] (1.0 ×10 ) + 2
Q = 5.0 ×10
−3 2
5
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.
(
0.0592 5 E=E log 5.0 ×10 2 0.0592 ( 5.699) E = 0− 2 E = −0.168 V 0
)
73
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.
74
The Nernst Equation
Calculate the E0 cell by the usual procedure.
Reduction Oxidation
( 1(Sn
2 Fe + e → Fe 3+
2+
2+
)
→ Sn 4+ + 2 e -
E0
)
0.771 V - ( 0.15) V
Cell reaction 2 Fe3+ + Sn 2+ → 2 Fe 2+ + Sn 4+ E 0cell = +0.62 V
75
The Nernst Equation
Substitute the ion concentrations into Q to calculate Ecell. E cell = E 0cell -
0.0592 log Q 2
= 0.62 V -
[ [
Fe
][ ] ] [ Sn ]
2+ 2
Sn 4+
2
2+
0.0592 log 2 Fe3+
0.0592 010 . ) ( 10 . ) ( = 0.62 V log −2 2 2 10 . × 10 ( 010 . ) 2
E cell
(
)
76
The Nernst Equation
( 0.0592 0.10 ) (1.0 ) = 0.62 V log 2 − 2 2 1.0 ×10 ( 0.10 ) 0.0592 ( 3.00) = 0.62 V 2 = 0.62 − 0.09 V 2
E cell E cell E cell
(
)
E cell = 0.53 V
77
Relationship of E0cell to ∆G0 and K
From previous chapters we know the relationship of ∆G0 and K for a reaction.
∆G = -RT lnK or 0
∆G = −2.303 RT log K 0
78
Relationship of E0cell to ∆G0 and K
The relationship between ∆G0 and E0cell is also a simple one.
∆G = -n F E 0
0 cell
where F = 96,487 J/V mol e n = number of e
-
-
79
Relationship of E0cell to ∆G0 and K
Combine these two relationships into a single relationship to relate E0cell to K.
-n FE
0 cell
= - RT lnK or 0 cell
n FE ln K = RT
80
Relationship of E0cell to ∆G0 and K
Example 21-10: Calculate the standard Gibbs free energy change, ∆G0 , at 250C for the following reaction. 2+ 2+
Pb + Cu
→ Pb + Cu
81
Relationship of E0cell to ∆G0 and K
Calculate E0cell using the appropriate halfreactions.
E 2+
−
Cu + 2 e → Cu 2+
Pb → Pb + 2 e 2+
0
0.337 V - ( - 0.126 V )
− 2+
0
Cu + Pb → Pb + Cu E 0
0 cell
= 0.463 V 82
Relationship of E0cell to ∆G0 and K
Now that we know E0cell , we can calculate ∆G0 .
∆∆GG ==−−nnFFEE 00
((
00 cell cell
)( )(
))
mole e 9696 ,500J J 0.463 ∆∆GG ==−−22mol ,500 V)V) - .(463 - (0 mol VV mol e e 0 4 ∆G = −8.94 ×10 J (per " mol of rxn" ) 00
−−
∆G 0 = -89.4 kJ 83
Relationship of E0cell to ∆G0 and K 21-11: Calculate the thermodynamic Example equilibrium constant for the reaction in example 21-10 at 250C.
∆G = − RT ln K ∴ 0
− 8.94 ×10 4 J
∆G ln K = = − RT - 8.314 J 0
(
ln K = 36.1∴ K = 5 ×10
15
mol ( 298 K )
) mol K
[ Pb ] = [Cu ] 2+
2+
84
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.
Zn + 2 Ag ( 0.30 M ) → 2Ag + Zn +
2+
( 0.50 M )
85
Relationship of E0cell to ∆G0 and K
Calculate the standard cell potential E0cell.
( ( Zn
+
2 Ag + e → Ag 0
-
2+
0
E
)
→ Zn + 2e
-
)
2 Ag + + Zn 0 → 2 Ag 0 + Zn 2+
0
0.799 V - ( - 0.763 V ) E 0cell = +1.562 V
86
Relationship of E0cell to ∆G0 and K
Use the Nernst equation to calculate Ecell for the given concentrations.
[ ( Zn ] 0.50 ) Q= = [Ag ] ( 0.30) 2+
+ 2
E cell
2
= 5.6
log 5.6 = 0.748 0.0592 0 = E cell − log Q 2 87
Relationship of E0cell to ∆G0 and K
[ Zn ] ( 0.50) Q= = [ Ag ] ( 0.30) 2+
+ 2
2
= 5 .6
log 5.6 = 0.748 0.0592 0 E cell = E cell − log Q 2 0.0592 E cell = 1.562 V log ( 5.6) V 2 88
Relationship of E0cell to ∆G0 and K [ Zn ] ( 0.50 ) Q= = = 5.6 2+
[ Ag ]
+ 2
( 0.30) 2
log 5.6 = 0.748 0.0592 0 E cell = E cell − log Q 2 0.0592 E cell = 1.562 V log ( 5.6) V 2 E cell = (1.562 - 0.022) V E cell = 1.540 V 89
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.
∆G = - n F E cell
(
)(
∆G = − 2 mol e - 96,500 J
) ( 1.540 V ) V mol e -
∆G = −2.97 ×105 J per mol of rxn. ∆G = -297 kJ/mol
The negative ∆G tells us that the reaction is spontaneous. 90
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. - n F E 0cell = − 2.303 RT log K - n F E 0cell log K = − 2.303 RT − ( 2 )( 96,500 )(1.562 ) log K = − ( 2.303)( 8.314 )( 298) log K = 52.8 K = 6 ×10
52
[ Zn ] = [Ag ] 2+
+ 2
91
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.
92
The Dry Cell 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:
ammonium chloride, NH4Cl
manganese (IV) oxide, MnO2
zinc chloride, ZnCl2
and a porous inactive solid. 93
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.
94
The Dry Cell
The anode reaction is:
0
Zn → Zn
2
2+
+2 e
-
The graphite rod is the positive electrode (cathode). Ammonium ions from the NH4Cl are reduced at the cathode.
+ NH 4
-
+ 2 e → 2 NH 3( g ) + H 2( g ) 95
The Dry Cell
The cell reaction is:
Anode reaction
Zn 0 → Zn 2+ + 2 e -
Cathode reaction 2 NH +4 + 2 e - → 2 NH 3( g ) + H 2( g ) Cell reaction
Zn 0 + 2 NH +4 → Zn 2+ + 2 NH 3( g ) + H 2( g )
96
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.
H 2( g ) + 2 MnO2( s) → 2 MnO(OH)( s)
At the anode, NH3 combines with Zn2+ to form a soluble complex and removing the Zn2+ ions from the reaction.
Zn
2+
+ 4 NH 3 → Zn(
)
2+ NH 3 4 97
The Dry Cell
98
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:
Anode reaction
Zn 0( s ) + 2 OH - → Zn( OH ) 2( s ) + 2 e −
Cathode reaction 2 MnO 2( s ) + 2 H 2O + 2 e - → 2 MnO(OH)( s ) + 2 OH Cell reaction
Zn 0( s ) + 2 MnO 2( s ) + 2 H 2O → Zn( OH ) 2( s ) + 2 MnO(OH)( s ) E 0cell = −1.5 V
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:
100
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.
101
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.
102
The Lead Storage Battery
Diagram of the lead storage battery.
103
The Lead Storage Battery
As the battery discharges, spongy lead is oxidized to lead ions and the plate becomes negatively charged. Pb 0( s ) → Pb 2+ + 2 e This is the anode reaction. The anode is the negative battery post during discharge.
The Pb2+ ions that are formed combine with SO42- from sulfuric acid to form solid lead sulfate on the Pb electrode. 2+ 2Pb + SO 4 →PbSO 4 ( s) 104
The Lead Storage Battery
The net reaction at the anode during discharge is:
0 Pb( s)
+
2SO4
→ PbSO4( s) + 2 e
-
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 SO42- to form PbSO4 and coat the PbO2 electrode. PbO2 electrode is the positive electrode (cathode). 105
The Lead Storage Battery
As the cell discharges, the cathode reaction is:
PbO2 ( s) + 4 H + + SO2+ 2 e → PbSO 4( s) + 2 H 2 O 4
The cell reaction for a discharging lead storage battery is:
Anode reaction
Pb ( s ) + SO 24- → PbSO 4( s ) + 2 e -
Cathode reaction PbO 2( s ) + 4 H + + SO 24- + 2 e - → PbSO 4( s ) + 2 H 2 O Cell reaction
Pb ( s ) + PbO 2( s ) + 4 H + + SO 24- → 2 PbSO 4( s ) + 2 H 2 O
106
The Lead Storage Battery
As the cell discharges, the cathode reaction is:
PbO2 ( s) + 4 H + + SO2+ 2 e → PbSO 4( s) + 2 H 2 O 4
The cell reaction for a discharging lead storage battery is:
107
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.
PbSO 4( s ) + 2 H 2 O → PbO 2 ( s ) + 4 H + + SO 24- + 2 e
The concentration of the H2SO4 decreases as the cell discharges.
Recharging the cell regenerates the H2SO4.
Pb ( s ) + PbO 2 ( s )
discharge → PbSO + 2H O + 4H + 2SO ← 4( s) 2 charge +
24
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.
PbSO 4( s ) + 2 H 2 O → PbO 2 ( s ) + 4 H + + SO 24- + 2 e
The concentration of the H2SO4 decreases as the cell discharges.
Recharging the cell regenerates the H2SO4.
109
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: Anode Cd ( s ) + 2 OH - → Cd(OH) 2 + 2 e Cathode NiO 2 ( s ) + 2 H 2 O + 2 e - → Ni(OH) 2 ( s ) + 2 OH Cell rxn Cd ( s ) + NiO 2( s ) + 2 H 2 O → Cd(OH) 2( s ) + Ni(OH) 2 ( s ) E 0 ≈ 1.4V 110
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.
111
The Hydrogen-Oxygen Fuel Cell Anode reaction 2( H ( ) + 2 OH → 2 H O + 2 e ) Cathode reaction ( O ( ) + 2 H O + 4 e → 4 OH ) (aq)
2 g
-
2
( )
-
2 g
Cell reaction
( )
2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( )
Notice that the overall reaction is the combination of hydrogen and oxygen to form water.
2
(aq)
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! 112
Synthesis Question
What are the explosive chemicals in the fuel cell that exploded aboard Apollo 13?
113
Synthesis Question
The Apollo 13 fuel cells contained hydrogen and oxygen. Both are explosive, especially when mixed. The oxygen tank aboard Apollo 13 exploded.
114
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?
115
End of Chapter 21
Electrochemistry is an important part of the electronics industry.
116
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