Chapter 21

1

Chapter Twenty One

Electrochemistry

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


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


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.

33

The Construction of Simple Voltaic Cells

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


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


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


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


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


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


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

65

The Nernst Equation

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


3 log 1

0.0592-V 0.153=E

Q log 1

0.0592-E=E 0

68

The Nernst Equation


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


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


and K Example 21-10: Calculate the standard Gibbs

free energy change, G0 , at 250C for the following reaction.

Cu PbCuPb 22

82


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


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


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


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


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


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


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


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


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

98

The Dry Cell

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.

103

The Lead Storage Battery Diagram of the lead storage battery.

104


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 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 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 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


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


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?

116

End of Chapter 21

Electrochemistry is an important part of the electronics industry.

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Chapter 21