Prentice-Hall © 2007 General Chemistry: Chapter 20 Slide 1 of 54 CHEMISTRY Ninth Edition GENERAL...

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Prentice-Hall © 2007 General Chemistry: Chapter 20 Slide 1 of 54 CHEMISTRY Ninth Editio n GENERAL Principles and Modern Applications Petrucci • Harwood • Herring • Madura Chapter 20: Electrochemistry

Transcript of Prentice-Hall © 2007 General Chemistry: Chapter 20 Slide 1 of 54 CHEMISTRY Ninth Edition GENERAL...

Page 1: Prentice-Hall © 2007 General Chemistry: Chapter 20 Slide 1 of 54 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci Harwood Herring.

Prentice-Hall © 2007General Chemistry: Chapter 20Slide 1 of 54

CHEMISTRYNinth

Edition GENERAL

Principles and Modern Applications

Petrucci • Harwood • Herring • Madura

Chapter 20: Electrochemistry

Page 2: Prentice-Hall © 2007 General Chemistry: Chapter 20 Slide 1 of 54 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci Harwood Herring.

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Contents

20-1 Electrode Potentials and Their Measurement

20-2 Standard Electrode Potentials

20-3 Ecell, ΔG, and Keq

20-4 Ecell as a Function of Concentration

20-5 Batteries: Producing Electricity Through Chemical Reactions

20-6 Corrosion: Unwanted Voltaic Cells

20-7 Electrolysis: Causing Non-spontaneous Reactions to Occur

20-8 Industrial Electrolysis Processes

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Prentice-Hall © 2007General Chemistry: Chapter 20Slide 3 of 54

20-1 Electrode Potentials and Their Measurement

Cu(s) + 2Ag+(aq)

Cu2+(aq) + 2 Ag(s)

Cu(s) + Zn2+(aq)

No reaction

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An Electrochemical Half Cell

Anode

Cathode

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An Electrochemical Cell

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Terminology

Electromotive force, Ecell.

The cell voltage or cell potential.

Cell diagram. Shows the components of the cell in a symbolic way. Anode (where oxidation occurs) on the left. Cathode (where reduction occurs) on the right.

◦ Boundary between phases shown by |.

◦ Boundary between half cells (usually a salt bridge) shown by ||.

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Terminology

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

Ecell = 1.103 V

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Terminology

Galvanic cells. Produce electricity as a result of spontaneous reactions.

Electrolytic cells. Non-spontaneous chemical change driven by electricity.

Couple, M|Mn+

A pair of species related by a change in number of e-.

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20-2 Standard Electrode Potentials

Cell voltages, the potential differences between electrodes, are among the most precise scientific measurements.

The potential of an individual electrode is difficult to establish.

Arbitrary zero is chosen.

The Standard Hydrogen Electrode (SHE)

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Standard Hydrogen Electrode2 H+(a = 1) + 2 e- H2(g, 1 bar) E° = 0 V

Pt|H2(g, 1 bar)|H+(a = 1)

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Standard Electrode Potential, E°

E° defined by international agreement. The tendency for a reduction process to occur at

an electrode. All ionic species present at a=1 (approximately 1 M). All gases are at 1 bar (approximately 1 atm). Where no metallic substance is indicated, the potential

is established on an inert metallic electrode (ex. Pt).

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Reduction Couples

Cu2+(1M) + 2 e- → Cu(s) E°Cu2+/Cu = ?

Pt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) E°cell = 0.340 V

Standard cell potential: the potential difference of a cell formed from two standard electrodes.

E°cell = E°cathode - E°anode

cathodeanode

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Standard Cell Potential

Pt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) E°cell = 0.340 V

E°cell = E°cathode - E°anode

E°cell = E°Cu2+/Cu - E°H+/H2

0.340 V = E°Cu2+/Cu - 0 V

E°Cu2+/Cu = +0.340 V

H2(g, 1 atm) + Cu2+(1 M) → H+(1 M) + Cu(s) E°cell = 0.340 V

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Measuring Standard Reduction Potential

anodeanode cathode cathode

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Standard Reduction Potentials

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20-3 Ecell, ΔG, and Keq

Cells do electrical work. Moving electric charge.

Faraday constant, F = 96,485 C mol-1

elec = -nFE

ΔG = -nFE

ΔG° = -nFE°

Michael Faraday 1791-1867

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Combining Half Reactions

Fe3+(aq) + 3e- → Fe(s) E°Fe3+/Fe = ?

Fe2+(aq) + 2e- → Fe(s) E°Fe2+/Fe = -0.440 V

Fe3+(aq) + 3e- → Fe2+(aq) E°Fe3+/Fe2+ = 0.771 V

Fe3+(aq) + 3e- → Fe(s)

ΔG° = +0.880 J

ΔG° = -0.771 J

ΔG° = +0.109 VE°Fe3+/Fe = +0.331 V

ΔG° = +0.109 V = -nFE°

E°Fe3+/Fe = +0.109 V /(-3F) = -0.0363 V

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Spontaneous Change

ΔG < 0 for spontaneous change. Therefore E°cell > 0 because ΔGcell = -nFE°cell

E°cell > 0 Reaction proceeds spontaneously as written.

E°cell = 0 Reaction is at equilibrium.

E°cell < 0 Reaction proceeds in the reverse direction spontaneously.

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The Behavior or Metals Toward Acids

M(s) → M2+(aq) + 2 e- E° = -E°M2+/M

2 H+(aq) + 2 e- → H2(g) E°H+/H2 = 0 V

2 H+(aq) + M(s) → H2(g) + M2+(aq)

E°cell = E°H+/H2 - E°M2+/M = -E°M2+/M

When E°M2+/M < 0, E°cell > 0. Therefore ΔG° < 0.

Metals with negative reduction potentials react with acids.

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Relationship Between E°cell and Keq

ΔG° = -RT ln Keq = -nFE°cell

E°cell = nF

RTln Keq

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Prentice-Hall © 2007General Chemistry: Chapter 20Slide 23 of 54

Summary of Thermodynamic, Equilibrium and Electrochemical Relationships.

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20-4 Ecell as a Function of Concentration

ΔG = ΔG° -RT ln Q

-nFEcell = -nFEcell° -RT ln Q

Ecell = Ecell° - ln QnF

RT

Convert to log10 and calculate constants.

Ecell = Ecell° - log Qn

0.0592 VThe Nernst Equation:

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Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)

Applying the Nernst Equation for Determining Ecell. What is the value of Ecell for the voltaic cell pictured below and diagrammed as follows?

EXAMPLE 20-8

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Ecell = Ecell° - log Qn

0.0592 V

Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)

Ecell = Ecell° - logn

0.0592 V [Fe3+][Fe2+] [Ag+]

Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag (s)

Ecell = 0.029 V – 0.018 V = 0.011 V

EXAMPLE 20-8

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Concentration Cells

Two half cells with identical electrodes but different ion concentrations.

2 H+(1 M) → 2 H+(x M)

Pt|H2 (1 atm)|H+(x M)||H+(1.0 M)|H2(1 atm)|Pt(s)

2 H+(1 M) + 2 e- → H2(g, 1 atm)

H2(g, 1 atm) → 2 H+(x M) + 2 e-

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Concentration Cells

Ecell = Ecell° - logn

0.0592 V x2

12

Ecell = 0 - log2

0.0592 V x2

1

Ecell = - 0.0592 V log x

Ecell = (0.0592 V) pH

2 H+(1 M) → 2 H+(x M)Ecell = Ecell° - log Qn

0.0592 V

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Measurement of Ksp

Ag+(0.100 M) → Ag+(sat’d M)

Ag|Ag+(sat’d AgI)||Ag+(0.10 M)|Ag(s)

Ag+(0.100 M) + e- → Ag(s)

Ag(s) → Ag+(sat’d) + e-

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Using a Voltaic Cell to Determine Ksp of a Slightly Soluble Solute. With the date given for the reaction on the previous slide, calculate Ksp for AgI.

AgI(s) → Ag+(aq) + I-(aq)

Let [Ag+] in a saturated Ag+ solution be x:

Ag+(0.100 M) → Ag+(sat’d M)

Ecell = Ecell° - log Q = n

0.0592 VEcell° - log

n

0.0592 V

[Ag+]0.10 M soln

[Ag+]sat’d AgI

EXAMPLE 20-10

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Ecell = Ecell° - log n

0.0592 V

[Ag+]0.10 M soln

[Ag+]sat’d AgI

Ecell = Ecell° - log n

0.0592 V

0.100

x

0.417 =0 - (log x – log 0.100) 1

0.0592 V

0.417log 0.100 -

0.0592log x = = -1 – 7.04 = -8.04

x = 10-8.04 = 9.110-9 Ksp = x2 = 8.310-17

EXAMPLE 20-10

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20-5 Batteries: Producing Electricity Through Chemical Reactions

Primary Cells (or batteries). Cell reaction is not reversible.

Secondary Cells. Cell reaction can be reversed by passing electricity

through the cell (charging).

Flow Batteries and Fuel Cells. Materials pass through the battery which converts

chemical energy to electric energy.

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The Leclanché (Dry) Cell

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Dry Cell

Zn(s) → Zn2+(aq) + 2 e-Oxidation:

2 MnO2(s) + H2O(l) + 2 e- → Mn2O3(s) + 2 OH-Reduction:

NH4+ + OH- → NH3(g) + H2O(l) Acid-base reaction:

NH3 + Zn2+(aq) + Cl- → [Zn(NH3)2]Cl2(s)Precipitation reaction:

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Alkaline Dry Cell

Zn2+(aq) + 2 OH- → Zn (OH)2(s)

Zn(s) → Zn2+(aq) + 2 e-

Oxidation reaction can be thought of in two steps:

2 MnO2(s) + H2O(l) + 2 e- → Mn2O3(s) + 2 OH-Reduction:

Zn (s) + 2 OH- → Zn (OH)2(s) + 2 e-

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Lead-Acid (Storage) Battery

The most common secondary battery.

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Lead-Acid Battery

PbO2(s) + 3 H+(aq) + HSO4-(aq) + 2 e- → PbSO4(s) + 2 H2O(l)

Oxidation:

Reduction:

Pb (s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2 e-

PbO2(s) + Pb(s) + 2 H+(aq) + HSO4-(aq) → 2 PbSO4(s) + 2 H2O(l)

E°cell = E°PbO2/PbSO4 - E°PbSO4/Pb = 1.74 V – (-0.28 V) = 2.02 V

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The Silver-Zinc Cell: A Button Battery

Zn(s),ZnO(s)|KOH(sat’d)|Ag2O(s),Ag(s)

Zn(s) + Ag2O(s) → ZnO(s) + 2 Ag(s) Ecell = 1.8 V

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The Nickel-Cadmium Cell

Cd(s) + 2 NiO(OH)(s) + 2 H2O(L) → 2 Ni(OH)2(s) + Cd(OH)2(s)

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Fuel Cells

O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq)

2{H2(g) + 2 OH-(aq) → 2 H2O(l) + 2 e-}

2H2(g) + O2(g) → 2 H2O(l)

E°cell = E°O2/OH- - E°H2O/H2

= 0.401 V – (-0.828 V) = 1.229 V

= ΔG°/ ΔH° = 0.83

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Air Batteries

4 Al(s) + 3 O2(g) + 6 H2O(l) + 4 OH- → 4 [Al(OH)4](aq)

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20-6 Corrosion: Unwanted Voltaic Cells

O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq)

2 Fe(s) → 2 Fe2+(aq) + 4 e-

2 Fe(s) + O2(g) + 2 H2O(l) → 2 Fe2+(aq) + 4 OH-(aq)

Ecell = 0.841 V

EO2/OH- = 0.401 V

EFe/Fe2+ = -0.440 V

In neutral solution:

In acidic solution:

O2(g) + 4 H+(aq) + 4 e- → 4 H2O (aq) EO2/OH- = 1.229 V

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Corrosion

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Corrosion Protection

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Corrosion Protection

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20-7 Electrolysis: Causing Non-spontaneous Reactions to Occur

Galvanic Cell:

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) EO2/OH- = 1.103 V

Electolytic Cell:

Zn2+(aq) + Cu(s) → Zn(s) + Cu2+(aq) EO2/OH- = -1.103 V

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Predicting Electrolysis Reaction

An Electrolytic Cell e- is the reverse of the

voltaic cell. Battery must have a

voltage in excess of 1.103 V in order to force the non-spontaneous reaction.

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Complications in Electrolytic Cells

Overpotential. Competing reactions. Non-standard states. Nature of electrodes.

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Quantitative Aspects of Electrolysis

1 mol e- = 96485 C

Charge (C) = current (C/s) time (s)

ne- = I t

F

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20-8 Industrial Electrolysis Processes

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Chlor-Alkali Process