31

Click here to load reader

Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

  • Upload
    vandang

  • View
    334

  • Download
    9

Embed Size (px)

Citation preview

Page 1: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

1

Chapter 20

Electrochemistry

Yonsei University

Lecture Presentation

© 2012 Pearson Education, Inc.

Electrochemistry

2

20.1 Oxidation States and Oxidation-Reduction Reactions

• Electrochemistry is the branch of chemistry that deals with relationships between electricity and chemical reactions

• Chemical reactions in which the oxidation state of one or more substances change are called oxidation-reduction reactions (redox reactions). Recall:

– Oxidation involves loss of electrons (OIL).

– Reduction involves gain of electrons (RIG).

Also:

– Oxidation involves an increase of an oxidation number.

– Reduction involves a decrease of an oxidation number.

© 2012 Pearson Education, Inc.

Page 2: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

3

Oxidation Numbers

In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

© 2012 Pearson Education, Inc.

Electrochemistry

4

Electronegativity

Page 3: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

5

Electrochemistry

6

Oxidation and Reduction

• A species is oxidized when it loses electrons.– Here, zinc loses two electrons to go from neutral

zinc metal to the Zn2+ ion.

• A species is reduced when it gains electrons.– Here, each of the H+ gains an electron, and they

combine to form H2.

© 2012 Pearson Education, Inc.

Page 4: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

7

Oxidation and Reduction

• What is reduced is the oxidizing agent.– H+ oxidizes Zn by taking electrons from it.

• What is oxidized is the reducing agent.– Zn reduces H+ by giving it electrons.

© 2012 Pearson Education, Inc.

Electrochemistry

8

Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents

Solution

The oxidation state of Cd increases from 0 to +2, and that of Ni decreases from +4 to +2. Thus, the Cd atom is oxidized (loses electrons) and is the reducing agent. The oxidation state of Ni decreases as NiO2 is converted into Ni(OH)2. Thus, NiO2 is reduced (gains electrons) and is the oxidizing agent.

The nickel-cadmium (nicad) battery uses the following redox reaction to generate electricity:Cd(s) + NiO2(s) + 2 H2O(l) Cd(OH)2(s) + Ni(OH)2(s)

Identify the substances that are oxidized and reduced, and indicate which is the oxidizing agent and which is the reducing agent.

Practice ExerciseIdentify the oxidizing and reducing agents in the reaction

2 H2O(l) + Al(s) + MnO4(aq) Al(OH)4

(aq) + MnO2(s)

Answer: Al(s) is the reducing agent; MnO4 is the oxidizing agent.

Page 5: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

9

20.2 Balancing Redox Equations

• Recall the law of conservation of mass: The amount of each element present at the beginning of the reaction must be present at the end.

• Conservation of charge: Electrons are not lost in a chemical reaction.

• Some redox equations may be easily balanced by inspection.– However, for many redox reactions we need to look

carefully at the transfer of electrons.

Electrochemistry

10

Half-Reactions

Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)

• The oxidation half-reaction is:

Sn2+(aq) Sn4+(aq) +2e–

• The reduction half-reaction is:

2Fe3+(aq) + 2e– 2Fe2+(aq)

© 2012 Pearson Education, Inc.

Page 6: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

11

The Half-Reaction Method

1. Assign oxidation numbers to determine what is oxidized and what is reduced.

2. Write the oxidation and reduction half-reactions.

3. Balance each half-reaction.a. Balance elements other than H and O.

b. Balance O by adding H2O.

c. Balance H by adding H+.

d. Balance charge by adding electrons.

4. Multiply the half-reactions by integers so that the electrons gained and lost are the same.

5. Add the half-reactions, subtracting things that appear on both sides.

6. Make sure the equation is balanced according to mass.7. Make sure the equation is balanced according to charge.

© 2012 Pearson Education, Inc.

Electrochemistry

12

The Half-Reaction Method

Consider the reaction between MnO4 and C2O4

2:

MnO4(aq) + C2O4

2(aq) Mn2+(aq) + CO2(aq)

© 2012 Pearson Education, Inc.

Page 7: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

13

MnO4(aq) + C2O4

2(aq) Mn2+(aq) + CO2(aq)

16H+(aq) + 2MnO4– (aq) + 5C2O4

2– (aq) 2Mn2+(aq) + 8H2O(l) + 10CO2(g)

Electrochemistry

14

Sample Exercise 20.2 Balancing Redox Equations in Acidic Solution

Solution

14 H+(aq) + Cr2O72(aq) + 6 Cl(aq) 2 Cr3+(aq) + 7 H2O(l) + 3 Cl2(g)

Complete and balance this equation by the method of half-reactions:Cr2O7

2(aq) + Cl(aq) Cr3+(aq) + Cl2(g) (acidic solution)

Practice ExerciseComplete and balance the following equations using the method of half-reactions. Both reactions occur in acidic solution.

(a) Cu(s) + NO3(aq) Cu2+(aq) + NO2(g)

(b) Mn2+(aq) + NaBiO3(s) Bi3+(aq) + MnO4(aq)

Answers: (a) Cu(s) + 4 H+(aq) + 2 NO3(aq) Cu2+(aq) + 2 NO2(g) + 2 H2O(l)

(b) 2 Mn2+(aq) + 5 NaBiO3(s) + 14 H+(aq) 2 MnO4(aq) + 5 Bi3+(aq) + 5 Na+(aq) + 7 H2O(l)

Page 8: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

15

Balancing in Basic Solution

• If a reaction occurs in a basic solution, one can balance it as if it occurred in acid.

• Once the equation is balanced, add OH

to each side to “neutralize” the H+ in the equation and create water in its place.

• If this produces water on both sides, you might have to subtract water from each side.

© 2012 Pearson Education, Inc.

Electrochemistry

16

Sample Exercise 20.3 Balancing Redox Equations in Basic Solution

Solution

3 CN(aq) + H2O(l) + 2 MnO4(aq) 3 CNO(aq) + 2 MnO2(s) + 2 OH(aq)

Complete and balance this equation for a redox reaction that takes place in basic solution:CN(aq) + MnO4

(aq) CNO(aq) + MnO2(s) (basic solution)

Practice ExerciseComplete and balance the following equations for oxidation-reduction reactions that occur in basic solution:

(a) NO2(aq) + Al(s) NH3(aq) + Al(OH)4

(aq)

(b) Cr(OH)3(s) + ClO(aq) CrO42(aq) + Cl2(g)

Answers: (a) NO2(aq) + 2 Al(s) + 5 H2O(l) + OH(aq) NH3(aq) + 2 Al(OH)4

(aq)

(b) 2 Cr(OH)3(s) + 6 ClO(aq) 2 CrO42(aq) + 3 Cl2(g) + 2 OH(aq) + 2 H2O(l)

Page 9: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

17

20.3 Voltaic Cells

In spontaneousoxidation-reduction (redox) reactions, electrons are transferred and energy is released.

© 2012 Pearson Education, Inc.

Electrochemistry

18

Voltaic Cells

• We can use that energy to do work if we make the electrons flow through an external device.

• We call such a setup a voltaic cell.

© 2012 Pearson Education, Inc.

Page 10: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

19

Voltaic Cells• The oxidation occurs at the anode.

• The reduction occurs at the cathode.

© 2012 Pearson Education, Inc.

Electrochemistry

20

Electrochemical Cells

Zn→Zn2++2e- Cu2++2e- →Cu

Page 11: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

21

Voltaic Cells

• Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

• Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced.– Cations move toward the

cathode.

– Anions move toward the anode.

© 2012 Pearson Education, Inc.

Electrochemistry

22

Electrochemical Cells (Anode)

Page 12: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

23

Electrochemical Cells (Cathode)

Electrochemistry

24

Voltaic Cells

• In the cell, then, electrons leave the anode and flow through the wire to the cathode.

• As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.

• As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode.

• The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

© 2012 Pearson Education, Inc.

Page 13: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

25

Cell Notation

• The Daniel Cell reaction is:

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

• Using cell notation to represent the cell:

Zn(s) Zn2+(aq) Cu2+(aq) Cu(s)

anode salt bridge cathode

Electrochemistry

26

20.4 Cell Potentials Under Standard Conditions

Electromotive Force (emf)

• Water only spontaneously flows one way in a waterfall.

• Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

© 2012 Pearson Education, Inc.

Page 14: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

27

Electromotive Force (emf)

• The potential difference between the anode and cathode in a cell is called the electromotive force (emf).

• It is also called the cell potential and is designated Ecell.

© 2012 Pearson Education, Inc.

• Cell potential is measured in volts (V).

1 V = 1 JC

Useful Units : 1C = 1A·s 1J = 1V·C

Electrochemistry

28

Standard Reduction Potentials

Standard reduction potentials, Eºred for many electrodes have been measured and tabulated.

Page 15: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

29

Standard Hydrogen Electrode

• Their values are referenced to a standard hydrogen electrode (SHE).

• By definition, the reduction potential for hydrogen is 0 V:

2 H+(aq, 1M) + 2e H2(g, 1 atm)

© 2012 Pearson Education, Inc.

Electrochemistry

30

Standard Cell Potentials

The cell potential at standard conditions can be found through this equation:

Ecell = Ered (cathode) Ered (anode)

Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

© 2012 Pearson Education, Inc.

Page 16: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

31

Cell Potentials

• For the oxidation in this cell,

• For the reduction,

Ered = 0.76 V

Ered = +0.34 V

© 2012 Pearson Education, Inc.

Ecell = Ered (cathode) Ered

(anode)

= +0.34 V (0.76 V)

= +1.10 V

- +

Electrochemistry

32

Cell Potentials

The greater the difference between the two, the greater the voltage of the cell.

© 2012 Pearson Education, Inc.

Page 17: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

33

Solution

For the Zn-Cu2+ voltaic cell shown in Figure 20.5, we have

Given that the standard reduction potential of Zn2+ to Zn(s) is 0.76 V, calculate the for the reduction of Cu2+ to Cu:

Cu2+(aq, 1 M) + 2 e Cu(s)

Sample Exercise 20.5 Calculating from

Electrochemistry

34

Oxidizing and Reducing Agents

• The strongest oxidizers have the most positive reduction potentials.

• The strongest reducers have the most negativereduction potentials.

© 2012 Pearson Education, Inc.

Page 18: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

35

20.5 Free Energy and Redox Reactions

G for a redox reaction can be found by using the equation

G = nFE

where n is the number of moles of electrons transferred, and F is a constant, the Faraday:

© 2012 Pearson Education, Inc.

)emol)(V(485,96

emol485,961

JCF

Electrochemistry

36

Free Energy

Under standard conditions,

G = nFE

© 2012 Pearson Education, Inc.

00 ln

lnG RT K RT

E KnF nF nF

Since ∆G˚ is related to the equilibrium constant, K, we can relate E° to K:

Page 19: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

37

Sample Exercise 20.9 Determining Spontaneity

Solution

Use Table 20.1 to determine whether the following reactions are spontaneous under standard conditions.(a) Cu(s) + 2 H+(aq) Cu2+(aq) + H2(g)(b) Cl2(g) + 2 I(aq) 2 Cl(aq) + I2(s)

(a) Cu2+(aq) + H2(g) Cu(s) + 2 H+(aq) E = +0.34 V, spontaneous

(b)

E = (1.36 V) (0.54 V) = +0.82 V

spontaneous

Electrochemistry

38

20.6 Cell Potentials Under Nonstandard Conditions

• A voltaic cell is functional until E = 0 at which point equilibrium has been reached.– The cell is then “dead.”

• The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction.

© 2012 Pearson Education, Inc.

Page 20: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

39

Nernst Equation

Dividing both sides by nF, we get the Nernst equation:

E = E RTnF

ln Q

or, using base-10 logarithms,

E = E 2.303RT

nFlog Q

© 2012 Pearson Education, Inc.

G = G + RT ln Q; nFE = nFE + RT ln Q

Electrochemistry

40

Nernst Equation

At room temperature (298 K),

Thus, the equation becomes

E = E 0.0592

nlog Q

2.303RTF

= 0.0592 V

© 2012 Pearson Education, Inc.

Page 21: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

41

Concentration Cells

• Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.

• For such a cell, would be 0, but Q would not.Ecell

• Therefore, as long as the concentrations are different, E will not be 0.

© 2012 Pearson Education, Inc.

Electrochemistry

42

Sample Exercise 20.11 Cell Potential under Nonstandard Conditions

Solution

Calculate the emf at 298 K generated by a voltaic cell in which the reaction isCr2O7

2(aq) + 14 H+(aq) + 6 I(aq) 2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)when [Cr2O7

2] = 2.0 M, [H+] = 1.0 M, [I] = 1.0 M, and [Cr3+] = 1.0 105 M.

Page 22: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

43

Sample Exercise 20.12 Calculating Concentrations in a Voltaic Cell

Solution

If the potential of a Zn-H+ cell (like that in Figure 20.9) is 0.45 V at 25 C when [Zn2+] = 1.0 M andPH2

= 1.0 atm, what is the H+ concentration?

Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g)

n = 2

Electrochemistry

44

20.7 Batteries and Fuel Cells

A battery is a portable, self-contained electrochemical power source consisting of one or more voltaic cells.

• Primary cells: cannot be recharged.

• Secondary cells: can be recharged from an external power source after its voltage has dropped.

© 2012 Pearson Education, Inc.

Page 23: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

45

Lead-Acid Battery

© 2012 Pearson Education, Inc.

Anode: Pb(s) + SO42–(aq) PbSO4(s) + 2e– E=0.127V

Cathode: PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– PbSO4(s) + 2H2O

E=1.687VPb(s)+ PbO2(s)+ 4H+(aq)+ 2SO4

2–(aq) 2PbSO4(s)+ 2H2O

Electrochemistry

46

Leclanche cell

© 2012 Pearson Education, Inc.

산화전극: Zn(s) Zn2+(aq) + 2e–

환원전극: MnO2(s)+ 2NH4+(aq) + 2e– Mn2O3(s)+ 2NH3 + H2O

Zn(s)+ MnO2(s)+ 2NH4+(aq) Zn2+(aq) + Mn2O3(s)+ 2NH3(g) + H2O

Page 24: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

47

Alkaline Batteries

© 2012 Pearson Education, Inc.

Anode : Zn(s) + 2OH–(aq) Zn(OH)2(aq) + 2e–

Cathode : 2MnO2(s) + 2H2O(l) + 2e– 2MnO(OH)(s) + 2OH– (aq)

V= 1.55 V

Electrochemistry

48

Nickel-Cadmium Batteries

© 2012 Pearson Education, Inc.

Cathode : 2NiO(OH)(s) + 2H2O(l) + 2e– 2Ni(OH)2(s) +2OH-(aq)Anode : Cd(s) + 2OH– (aq) Cd(OH)2(s) + 2e–

Working Cd(s)+ 2NiO (s)+ 2H2O Cd(OH)2(s) + Ni(OH)2(s) E= 1.30 V

Recharge Cd(OH)2(s) + Ni(OH)2(s) Cd(s)+ 2NiO2 (s)+ 2H2O E=-1.3 V

• Cadmium is a toxic heavy metal.

• Other rechargeable batteries have been developed.NiMH batteries (nickel-metal-hydride).Li-ion batteries (lithium-ion batteries).

Page 25: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

49

Hydrogen Fuel Cells

2H2(g) + 4OH– (aq) 4H2O(l) + 4e–

2H2O(l) + O2(g) + 4e– 4OH– (aq)

Electrochemistry

50

Hydrogen Economy

Page 26: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

51

20.8 Corrosion

© 2012 Pearson Education, Inc.

Electrochemistry

52

…Corrosion Prevention

© 2012 Pearson Education, Inc.

Zn2+(aq) +2e– Zn(s) E°red = -0.76 VFe2+(aq) + 2e– Fe(s) E°red = -0.44 V

Galvanized iron

Mg2+(aq) +2e– Mg(s) E°red = -2.37 VFe2+(aq) + 2e– Fe(s) E°red = -0.44 V

cathodic protection :the sacrificial anode is destroyed

Page 27: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

53

Electrical Work

• Free energy is a measure of the maximum amount of useful work that can be obtained from a system.

– We know: G = wmax and G = -nFE

thus: wmax = -nFE

– If Ecell is positive, wmax will be negative.

• Work is done by the system on the surroundings.

© 2012 Pearson Education, Inc.

Electrochemistry

54

Electrical Work

• The emf can be thought of as being a measure of the driving force for a redox process.– In an electrolytic cell an external source of energy is

required to force the reaction to proceed. : w = nFEexternal

– To drive the nonspontaneous reaction,

the external emf must be greater than Ecell.

– From physics we know that work is measured in units of watts: 1 W = 1 J/s

– Electric utilities use units of kilowatt-hours: kWh = 3.6 106 J.

© 2012 Pearson Education, Inc.

Page 28: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

55

20.9 Electrolysis

© 2012 Pearson Education, Inc.

• Electrolysis reactions: nonspontaneous reactions - Need an external current to force the reaction to proceed

: electrolytic cells.

• In both cells (voltaic and electrolytic)- reduction: cathode, oxidation: anode.- In electrolytic cells, electrons forced to flow from anode to cathode.thus, anode (positive terminal) cathode (negative).

- In voltaic cells, anode (negative ), cathode (positive).

Electrochemistry

56

Decomposition of molten NaCl

Cathode: 2Na+(l) + 2e– 2Na(l) E°red= -2.714 V2H2O + 2e– H2(g) + 2 OH–(aq) E°red = - 0.828 V

Anode: 2Cl– (l) Cl2(g) + 2e– : E°ox= -1.360 V

Page 29: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

57

Electrolytic Reactions

Electrowinning

2Al2O3 (l) + 3C(s)→4Al(l)+3CO2(g) E°≈-2.1V•Overvoltage

cryolite

Electrochemistry

58

Electroplating

Active electrodesnickel strip

steel strip

Page 30: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

59

Quantitative Aspects of Electrolysis

• Faraday observed that the amount of current applied to a cell is directly proportional to the amount of metal deposited.

- -

( ) ( ) :

e

I A t s I t M mol of metaln g

F F mol of e

A = ampere s = seconds

M = molar mass F = 96,485 C/mol of e-

Useful Units : 1C = 1A·s 1J = 1V·C

MeM 22For the case of eeM nmolmol2

1

2

1

Electrochemistry

60

Sample Exercise 20.14 Relating Electrical Charge and Quantity of Electrolysis

Solution

Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if theelectrical current is 10.0 A.

Al3+ + 3 e Al

Page 31: Chapter 20 Electrochemistry - Yonsei Universitychem.yonsei.ac.kr/.../20_Electrochemistry.pdf · Chapter 20 Electrochemistry Yonsei University ... Electrochemistry 5 Electrochemistry

Electrochemistry

61

Problems

• 8, 16, 28, 36, 40, 48, 56, 68, 94