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Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring.

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Presentation on theme: "Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring."— Presentation transcript:

1 Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition Chapter 20: Electrochemistry

2 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 2 of 52 Contents 20-1Electrode Potentials and Their Measurement 20-2Standard Electrode Potentials 20-3E cell, ΔG, and K eq 20-4 E cell as a Function of Concentration 20-5Batteries: Producing Electricity Through Chemical Reactions. 20-6Corrosion: Unwanted Voltaic Cells 20-7Electrolysis: Causing Non-spontaneous Reactions to Occur 20-8Industrial Electolysis Processes Focus On Membrane Potentials

3 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 3 of Electrode Potentials and Their Measurement Cu(s) + 2Ag + (aq) Cu 2+ (aq) + 2 Ag(s) Cu(s) + Zn 2+ (aq) No reaction

4 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 4 of 52 An Electrochemical Half Cell Anode Cathode

5 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 5 of 52 An Electrochemical Cell

6 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 6 of 52 Terminology Electromotive force, E cell. –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 ||.

7 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 7 of 52 Terminology Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) E cell = V

8 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 8 of 52 Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) Ecell = V

9 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 9 of 52 Terminology Galvanic cells (Voltaic, or electrochemical). –Produce electricity as a result of spontaneous reactions. Electrolytic cells. –Non-spontaneous chemical change driven by electricity. Couple, M|M n+ –A pair of species related by a change in number of e -.

10 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 10 of 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)

11 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 11 of 52 Standard Hydrogen Electrode 2 H + (a = 1) + 2 e - ↔ H 2 (g, 1 bar) E° = 0 V Pt|H 2 (g, 1 bar)|H + (aq, a = 1)

12 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 12 of 52 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).

13 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 13 of 52 Reduction Couples Cu 2+ (1M) + 2 e - → Cu(s)E° Cu/Cu 2+ = ? Pt|H 2 (g, 1 bar)|H + (a = 1) || Cu 2+ (1 M)|Cu(s) E° cell = V Standard cell potential: the potential difference of a cell formed from two standard electrodes. E° cell = E° cathode - E° anode cathodeanode

14 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 14 of 52 Standard Cell Potential Pt|H 2 (g, 1 bar)|H + (a = 1) || Cu 2+ (1 M)|Cu(s) E° cell = V E° cell = E° cathode - E° anode E° cell = E° Cu 2+ /Cu - E° H + /H V = E° Cu 2+ /Cu - 0 V E° Cu 2+ /Cu = V H 2 (g, 1 atm) + Cu 2+ (1 M) → H + (1 M) + Cu(s) E° cell = V

15 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 15 of 52 Measuring Standard Reduction Potential cathode anode

16 Table 19: Standard reduction potentials SystemE 0 (V) Li(s)/Li K(s)/K Ba(s)/Ba Sr(s)/Sr Ca(s)/Ca Na(s)/Na General Chemistry: Chapter 21Slide 16 of 52 A temperature of K (25 °C). An activity of unity for each pure solid, pure liquid, or for water (solvent). Legend: (s) – solid; (l) – liquid; (g) – gas; (aq) – aqueous (default for all charged species). For example Li(s)/Li + denotes the half reaction Li + + e - → Li(s), where Li + (aqueous)

17 SystemE 0 (V) H 2(g) /2H Sn(s)/Sn Sn 2+ /Sn Cu + /Cu OH - /O 2 (g)+2H 2 O(l)0.410 Fe 2+ /Fe Ag(s)/Ag Cl - /Cl 2 (g)1.360 Pb 2+ /Pb F - /F 2 (g)2.850 General Chemistry: Chapter 21Slide 17 of 52

18 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 18 of E cell, ΔG, and K eq Cells do electrical work. –Moving electric charge. Faraday constant, F = 96,485 C mol -1 w maxusef = w elec = -nFE ΔG = -nFE ΔG° = -nFE°

19 ΔG° = ·F J -3·F·E° = ΔG° = ·F J Prentice-Hall © 2002General Chemistry: Chapter 21Slide 19 of 52 Combining Half Reactions Fe 3+ (aq) + 3e - → Fe(s) E° Fe 3+ /Fe = ? Fe 2+ (aq) + 2e - → Fe(s) E° Fe 2+ /Fe = V Fe 3+ (aq) + 1e - → Fe 2+ (aq) E° Fe 3+ /Fe 2+ = V Fe 3+ (aq) + 3e - → Fe(s) ΔG° = ·F J ΔG° = ·F J E° Fe 3+ /Fe = V E° Fe 3+ /Fe = ·F J /(-3·F C) = V = (2 E° Fe 2+ /Fe + E° Fe 3+ /Fe 2+ )/3 = V

20 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 20 of 52 Spontaneous Change ΔG < 0 for spontaneous change. Therefore E° cell > 0 because ΔG° cell = -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.

21 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 21 of 52 The Behavior of Metals Toward Acids M(s) → M 2+ (aq) + 2 e - E° = -E° M 2+ /M 2 H + (aq) + 2 e - → H 2 (g) E° H + /H 2 = 0 V 2 H + (aq) + M(s) → H 2 (g) + M 2+ (aq) E° cell = E° H + /H 2 - E° M 2+ /M = -E° M 2+ /M When E° M 2+ /M 0. Therefore ΔG° < 0. Metals with negative reduction potentials react with acids

22 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 22 of 52 Relationship Between E° cell and K eq ΔG° = -RT ln K eq = -nFE° cell E° cell = nF RT ln K eq

23 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 23 of 52 Summary of Thermodynamic, Equilibrium and Electrochemical Relationships.

24 General Chemistry: Chapter 21Slide 24 of E cell as a Function of Concentration ΔG = ΔG° +RT ln Q -nFE cell = -nFE cell ° +RT ln Q E cell = E cell ° - ln Q nF RT Convert to log 10 and calculate constants E cell = E cell ° - log Q n V The Nernst Equation: log QE

25 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 25 of 52 Example 20-8 Pt|Fe 2+ (0.10 M),Fe 3+ (0.20 M)||Ag + (1.0 M)|Ag(s) Applying the Nernst Equation for Determining E cell. What is the value of E cell for the voltaic cell pictured below and diagrammed as follows?

26 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 26 of 52 Example 20-8 E cell = E cell ° - log Q n V Pt|Fe 2+ (0.10 M),Fe 3+ (0.20 M)||Ag + (1.0 M)|Ag(s) E cell = E cell ° - log n V [Fe 3+ ] [Fe 2+ ] [Ag + ] Fe 2+ (aq) + Ag + (aq) → Fe 3+ (aq) + Ag (s) E cell = V – V = V

27 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 27 of 52 Concentration Cells Two half cells with identical electrodes but different ion concentrations. 2 H + (1 M) → 2 H + (x M) Pt|H 2 (1 atm)|H + (x M)||H + (1.0 M)|H 2 (1 atm)|Pt(s) 2 H + (1 M) + 2 e - → H 2 (g, 1 atm) H 2 (g, 1 atm) → 2 H + (x M) + 2 e -

28 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 28 of 52 Concentration Cells E cell = E cell ° - log n V x2x E cell = 0 - log V x2x2 1 E cell = V log x E cell = ( V) pH 2 H + (1 M) → 2 H + (x M) E cell = E cell ° - log Q n V

29 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 29 of 52 Measurement of K sp 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 -

30 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 30 of 52 Example Using a Voltaic Cell to Determine K sp of a Slightly Soluble Solute. With the data given for the reaction on the previous slide, calculate K sp 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) E cell = E cell ° - log Q = n V E cell ° - log n V [Ag + ] 0.10 M soln [Ag + ] sat’d AgI

31 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 31 of 52 Example E cell = E cell ° - log n V [Ag + ] 0.10 M soln [Ag + ] sat’d AgI E cell = E cell ° - log n V x = 0 - (log x – log 0.100) V log log x == -1 – 7.04 = x = = 9.1·10 -9 K sp = x 2 = 8.3·10 -17

32 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 32 of 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.

33 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 33 of 52 The Leclanché (Dry) Cell

34 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 34 of 52 Dry Cell Zn(s) → Zn 2+ (aq) + 2 e - Oxidation: 2 MnO 2 (s) + H 2 O(l) + 2 e - → Mn 2 O 3 (s) + 2 OH - Reduction: NH OH - → NH 3 (g) + H 2 O(l)Acid-base reaction: NH 3 + Zn 2+ (aq) + Cl - → [Zn(NH 3 ) 2 ]Cl 2 (s)Precipitation reaction:

35 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 35 of 52 Alkaline Dry Cell Zn 2+ (aq) + 2 OH - → Zn (OH) 2 (s) Zn(s) → Zn 2+ (aq) + 2 e - Oxidation reaction can be thought of in two steps: 2 MnO 2 (s) + H 2 O(l) + 2 e - → Mn 2 O 3 (s) + 2 OH - Reduction: Zn (s) + 2 OH - → Zn (OH) 2 (s) + 2 e -

36 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 36 of 52 Lead-Acid (Storage) Battery The most common secondary battery

37 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 37 of 52 Lead-Acid Battery PbO 2(s) + 3 H + (aq) + HSO 4 - (aq) + 2 e - → PbSO 4(s) + 2 H 2 O (l) Oxidation: Reduction: Pb (s) + HSO 4 - (aq) → PbSO 4(s) + H + (aq) + 2 e - PbO 2(s) + Pb (s) + 2 H + (aq) + 2HSO 4 - (aq) → 2 PbSO 4(s) + 2 H 2 O (l) E° cell = E° PbO 2 /PbSO 4 - E° PbSO 4 /Pb = 1.74 V – (-0.28 V) = 2.02 V

38 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 38 of 52 The Silver-Zinc Cell: A Button Battery Zn(s),ZnO(s)|KOH(sat’d)|Ag 2 O(s),Ag(s) Zn(s) + Ag 2 O(s) → ZnO(s) + 2 Ag(s) E cell = 1.8 V

39 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 39 of 52 The Nickel-Cadmium Cell Cd(s) + 2 NiO(OH)(s) + 2 H 2 O(l) → 2 Ni(OH) 2 (s) + Cd(OH) 2 (s)

40 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 40 of 52 Fuel Cells O 2 (g) + 2 H 2 O(l) + 4 e - → 4 OH - (aq) 2{H 2 (g) + 2 OH - (aq) → 2 H 2 O(l) + 2 e - } 2H 2 (g) + O 2 (g) → 2 H 2 O(l) E° cell = E° O 2 /OH - - E° H 2 O/H 2 = V – ( V) = V  = ΔG°/ ΔH° = 0.83

41 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 41 of 52 Air Batteries 4 Al(s) + 3 O 2 (g) + 6 H 2 O(l) + 4 OH - → 4 [Al(OH) 4 ] - (aq)

42 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 42 of Corrosion: Unwanted Voltaic Cells O 2 (g) + 2 H 2 O(l) + 4 e - → 4 OH - (aq) 2 Fe 2+ (aq) + 4 e - → 2 Fe(s) 2 Fe(s) + O 2 (g) + 2 H 2 O(l) → 2 Fe 2+ (aq) + 4 OH - (aq) E cell = V E O 2 /OH - = V E Fe/Fe 2+ = V In neutral solution: In acidic solution: O 2 (g) + 4 H + (aq) + 4 e - → 4 H 2 O (aq) E O 2 /OH - = V

43 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 43 of 52 Stability of Water pH2H + /H 2 O 2(aq) /4OH

44 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 44 of 52 Corrosion

45 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 45 of 52 Corrosion Protection

46 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 46 of 52 Corrosion Protection

47 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 47 of Electrolysis: Causing Non-spontaneous Reactions to Occur Galvanic Cell: Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) E = V Electolytic Cell: Zn 2+ (aq) + Cu(s) → Zn(s) + Cu 2+ (aq) E = V

48 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 48 of 52 Complications in Electrolytic Cells Overpotential. Competing reactions. Non-standard states. Nature of electrodes.

49 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 49 of 52 Quantitative Aspects of Electrolysis 1 mol e - = C = F Charge (C) = current (C/s) · time (s) n e - = I · t F

50 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 50 of Industrial Electrolysis Processes

51 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 51 of 52 Electroplating

52 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 52 of 52 Chlor-Alkali Process

53 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 53 of 52 Focus On Membrane Potentials

54 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 54 of 52 Nernst Potential, Δ 

55 Ion transports General Chemistry: Chapter 21Slide 55 of 52

56 Calculate the reversal potential, Δ  Calculate the reversal (Nernst) potential (ΔΦ) for Na + for a cell in assuming the following ion activities: a int [Na + ]=10.18×10 −3, a ext [Na + ]=149.8×10 −3, where a int [Na + ] is the intracellular activity (≈ concentration in M) of sodium, and a ext [Na + ] is the extracellular activity (≈ concentration in M) of sodium. What is the ΔΦ reversal potential (mV) at 37 ºC? The Boltzmann constant: k = 1.381×10 −23 J/K. The electron's charge: e = 1.602×10 −19 C. General Chemistry: Chapter 21Slide 56 of 52

57 Prentice-Hall © 2002General Chemistry: Chapter 21Slide 57 of 52 Chapter 21 Questions Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding. Choose a variety of problems from the text as examples. Practice good techniques and get coaching from people who have been here before.


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