Presentation is loading. Please wait.

Presentation is loading. Please wait.

Electric power conversion in electrochemistry Chemical Reactions Electric Power Electrolysis / Power consumption Electrochemical battery / Power generation.

Similar presentations


Presentation on theme: "Electric power conversion in electrochemistry Chemical Reactions Electric Power Electrolysis / Power consumption Electrochemical battery / Power generation."— Presentation transcript:

1 Electric power conversion in electrochemistry Chemical Reactions Electric Power Electrolysis / Power consumption Electrochemical battery / Power generation

2 battery +- inert electrodes power source vessel e-e- e-e- conductive medium Cell Construction Sign or polarity of electrodes (-)(+)

3 What chemical species would be present in a vessel of molten sodium chloride, NaCl (l)? Na + Cl - Let’s examine the electrolytic cell for molten NaCl.

4 +- battery Na (l) electrode half-cell Molten NaCl Na + Cl - Na + Na + + e -  Na2Cl -  Cl 2 + 2e - Cl 2 (g) escapes Observe the reactions at the electrodes NaCl (l) (-) Cl - (+)

5 +- battery e-e- e-e- NaCl (l) (-)(+) cathode anode Molten NaCl Na + Cl - Na + Na + + e -  Na 2Cl -  Cl 2 + 2e - cations migrate toward (-) electrode anions migrate toward (+) electrode At the microscopic level

6 Molten NaCl Electrolytic Cell cathode half-cell (-) REDUCTION Na + + e -  Na anode half-cell (+) OXIDATION2Cl -  Cl 2 + 2e - overall cell reaction 2Na + + 2Cl -  2Na + Cl 2 X 2 Non-spontaneous reaction!

7 Definitions: CATHODE REDUCTION occurs at this electrode ANODE OXIDATION occurs at this electrode

8 What chemical species would be present in a vessel of aqueous sodium chloride, NaCl (aq)? Na + Cl - H2OH2O Will the half-cell reactions be the same or different?

9 Water Complications in Electrolysis In an electrolysis, the most easily oxidized and most easily reduced reaction occurs. When water is present in an electrolysis reaction, then water (H 2 O) can be oxidized or reduced according to the reaction shown. ElectrodeIons...Anode RxnCathode Rxn E° Pt (inert)H 2 O H 2 O (l) + 2e-  H 2(g) + 2OH - (aq) V H 2 O 2 H 2 O (l)  4e - + 4H + (g) + O 2(g) V Net Rxn Occurring: 2 H 2 O  2 H 2(g) + O 2 (g) E° = V

10 battery +- power source e-e- e-e- NaCl (aq) (-)(+) cathode different half-cell Aqueous NaCl anode 2Cl -  Cl 2 + 2e - Na + Cl - H2OH2O What could be reduced at the cathode?

11 Aqueous NaCl Electrolytic Cell possible cathode half-cells (-) REDUCTION Na + + e -  Na 2H 2 O + 2e -  H 2 + 2OH - possible anode half-cells (+) OXIDATION2Cl -  Cl 2 + 2e - 2H 2 O  O 2 + 4H + + 4e - overall cell reaction 2Cl - + 2H 2 O  H 2 + Cl 2 + 2OH -

12 e-e- Ag + Ag For every electron, an atom of silver is plated on the electrode. Ag + + e -  Ag Electrical current is expressed in terms of the ampere, which is defined as that strength of current which, when passed thru a solution of AgNO 3 (aq) under standard conditions, will deposit silver at the rate of g Ag/sec 1 amp = g Ag/sec

13 Faraday’s Law The mass deposited or eroded from an electrode depends on the quantity of electricity. Quantity of electricity – coulomb (Q) Q is the product of current in amps times time in seconds Q = It coulomb current in amperes (amp) time in seconds 1 coulomb = 1 amp-sec = g Ag

14 Ag + + e -  Ag 1.00 mole e - = 1.00 mole Ag = g Ag g Ag/mole e g Ag/coul = 96,485 coul/mole e - 1 Faraday ( F ) mole e - = Q/ F mass = mole metal x MM mole metal depends on the half-cell reaction

15 Examples using Faraday’s Law How many grams of Cu will be deposited in 3.00 hours by a current of 4.00 amps? Cu e -  Cu The charge on a single electron is x coulomb. Calculate Avogadro’s number from the fact that 1 F = 96,487 coulombs/mole e -.

16

17

18 21-8 Industrial Electrolysis Processes Prentice-Hall © 2002General Chemistry: Chapter 21 Slide 18 of 52

19 A series of solutions have 50,000 coulombs passed thru them, if the solutions were Au +3, Zn +2, and Ag +, and Au, Zn, and Ag were plated out respectively, calculate the amount of metal deposited at each anode. battery M Au M Zn M Ag + Au e -  AuZn e -  ZnAg + + e -  Ag e-e- e-e- e-e- e-e-

20 Volta’s battery (1800) Alessandro Volta Paper moisturized with NaCl solution Cu Zn

21 Lesson 9 NEEP Galvanic cells and electrodes  To sustain the cell reaction, the charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells.  Relative amounts of charge can be carried by negative or positive ions (depends on their relative mobilities) through the solution.  Salt bridge, consists of an intermediate compartment filled with saturated salt solution and fitted with porous barriers at each end, is used for precise measurements. The purpose of salt bridge is to minimize the natural potential difference (junction potential).

22 with Galvanic Cells 19.2 spontaneous redox reaction anode oxidation cathode reduction

23 Electrodes are passive (not involved in the reaction) Olmsted Williams

24 Cu 1.0 M CuSO 4 Zn 1.0 M ZnSO 4 Salt bridge – KCl in agar Provides conduction between half-cells Cell Construction Observe the electrodes to see what is occurring.

25 Cu 1.0 M CuSO 4 Zn 1.0 M ZnSO 4 Cu plates out or deposits on electrode Zn electrode erodes or dissolves cathode half-cell Cu e -  Cu anode half-cell Zn  Zn e What about half-cell reactions? What about the sign of the electrodes? What happened at each electrode? Why?

26 Cu 1.0 M CuSO 4 Zn 1.0 M ZnSO 4 cathode half-cell Cu e -  Cu anode half-cell Zn  Zn e Now replace the light bulb with a volt meter. 1.1 volts

27 H 2 input 1.00 atm inert metal We need a standard electrode to make measurements against! The Standard Hydrogen Electrode (SHE) Pt 1.00 M H + 25 o C 1.00 M H atm H 2 Half-cell 2H + + 2e -  H 2 E o SHE = 0.0 volts

28 How do we calculate Standard Redox Potentials? We must compare the half reactions to a standard What is that standard? 2 H 3 O+(aq) + 2e-  H 2 (g) + 2 H 2 O(l) E°= 0.00 V This is called the standard hydrogen electrode or SHE Now that we have a standard, we can calculate standard redox potential by using the table of standard redox potentials

29 19.3 E 0 is for the reaction as written The more positive E 0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0

30 Copyright 1999, PRENTICE HALLChapter Cell EMF Oxidizing and Reducing Agents

31 H atm Pt 1.0 M H + Cu 1.0 M CuSO v cathode half-cell Cu e -  Cu anode half-cell H 2  2H + + 2e - KCl in agar + Now let’s combine the copper half-cell with the SHE E o = v

32 H atm Pt 1.0 M H M ZnSO v cathode half-cell 2H + + 2e -  H 2 anode half-cell Zn  Zn e - KCl in agar Zn - Now let’s combine the zinc half-cell with the SHE E o = v

33 Assigning the E o Al e -  AlE o = v Zn e -  ZnE o = v 2H + + 2e -  H 2 E o = 0.00 v Cu e -  CuE o = Ag + + e -  AgE o = v Write a reduction half-cell, assign the voltage measured, and the sign of the electrode to the voltage. Increasing activity

34 Measuring Standard Reduction Potential Prentice-Hall © 2002General Chemistry: Chapter 21 Slide 34 of 52 cathode anode

35 What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e - Cd (s) E 0 = V Cr 3+ (aq) + 3e - Cr (s) E 0 = V Cd is the stronger oxidizer Cd will oxidize Cr 2e - + Cd 2+ (1 M) Cd (s) Cr (s) Cr 3+ (1 M) + 3e - Anode (oxidation): Cathode (reduction): 2Cr (s) + 3Cd 2+ (1 M)  3Cd (s) + 2Cr 3+ (1 M) x 2 x 3 E 0 = E cathode - E anode cell 00 E 0 = – (-0.74) cell E 0 = 0.34 V cell 19.3

36 Calculating the cell potential, E o cell, at standard conditions Fe e -  Fe E o = v O 2 (g) + 2H 2 O + 4e -  4 OH - E o = v This is corrosion or the oxidation of a metal. Consider a drop of oxygenated water on an iron object Fe H 2 O with O 2 Fe  Fe e - -E o = v2x 2Fe + O 2 (g) + 2H 2 O  2Fe(OH) 2 (s) E o cell = v reverse

37 E =? -Concentration -Related half reaction kinetic e - +2Fe 3+  Fe E0 = e- +2H+  H2 E0=0.000 Fe +H +  Fe 3+ !!!!!!!!!!!!!!!!!!!!!!!! =================================================================================================== e- +Fe 3+  Fe 2+ E0= e +Fe2+  Fe E0=-0.44 Fe  Fe  Fe ================================================= Cu  Cu  Cu

38 Whereas potential and free energy are related, and free energy and equilibrium are related, equilibrium and potential must be related to one another.  G° = –nFE° cell and  G° = –RT ln K eq therefore –RT ln K eq = –nFE o cell Equilibrium Constants in Redox Reactions RT ln K eq RT E° cell = ––––––––– = –––– ln K eq nF nF V E° cell = –––––––– ln K eq n R and F are constant, therefore at 298 K:

39 1) 2e +Fe2+  Fe E0= ) e +Fe3+  Fe2+ E0= e +Fe3+  Fe E0=??? ===============

40 1)  G0=-2(-0.440)F=+0.880F 2)  G0=-1(+0.771)F=-0.771F F  G0 =-nfE0=+0.109F=3FE0 E)=-0.036

41 41 Effect of Concentration on Cell EMF A voltaic cell is functional until E = 0 at which point equilibrium has been reached. The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction. The Nernst Equation The Nernst equation relates emf to concentration using and noting that

42 and the previous relationship:  G o = -n F E o cell from thermodynamics:  G o = RT log K -n F E o cell = RT log K at 25 o C: E o cell = log K n where n is the number of electrons for the balanced reaction

43 What happens to the electrode potential if conditions are not at standard conditions? The Nernst equation adjusts for non-standard conditions For a reduction potential: ox + ne  red at 25 o C: E = E o log (red) n (ox) Calculate the E for the hydrogen electrode where 0.50 M H + and 0.95 atm H 2. in general: E = E o – RT ln (red) n F (ox)

44 1) An example: Ni (s) | Ni 2+ (0.600M) || Sn 2+ (0.300M) | Sn (s) According to the reduction potentials: 2 e - + Ni 2+  Ni (s) V 2 e - + Sn 2+  Sn (s) V One of these needs to be reversed to get a positive voltage, and a spontaneous reaction.

45 Ni (s)  2 e - + Ni V 2 e - + Sn 2+  Sn (s) V Ni (s) + Sn 2+  Ni 2+ + Sn (s) V = E o

46 E) Calculation of the equilibrium constant 1) at equilibrium E = 0 ; Q = ____ From the Nernst Equation: For the cell: Ni (s) | Ni 2+ (0.600M) || Sn 2+ (0.300M) | Sn (s)

47

48 2) An example: Sn (s) | Sn 2+ (1.0M) || Pb 2+ (0.0010M) | Pb (s) According to the reduction potentials: 2 e - + Pb 2+  Pb (s) E0= V 2 e - + Sn 2+  Sn (s) E0=-0.136V One of these needs to be reversed to get a positive voltage, and a spontaneous reaction.!!!!!!!!!!!!!!!!!!!!!!!!!

49 E=E /2log[Sn2+]/[Pb2+] E=-0.079

50  G o = -n F E o cell Free Energy and the Cell Potential Cu  Cu e - -E o = Ag + + e -  Ag E o = v 2x Cu + 2Ag +  Cu Ag E o cell = v where n is the number of electrons for the balanced reaction What is the free energy for the cell? 1 F = 96,500 J/v

51 Electrolysis of Copper A net reaction of zero, yet a process does take place. A concentration cell based on the Cu/Cu2+ half-reaction. A, Even though the half-reactions involve the same components, the cell operates because the half-cell concentrations are different. B, The cell operates spontaneously until the half-cell concentrations are equal. Note the change in electrodes (exaggerated here for clarity) and the equal color of solutions.

52 Cu+Cu2+(1.0 M)  Cu2+(0.1M)+Cu E=E /2Log(0.1/1.0) =

53 Concentration Cells Prentice-Hall © 2002General Chemistry: Chapter 21 Slide 53 of 52 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 -

54 Concentration Cells Prentice-Hall © 2002General Chemistry: Chapter 21 Slide 54 of 52 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

55 F)The pH meter is a special case of the Nernst Equation 1/2 H 2  H e - Q = [H + ]

56 pH = - log [H + ]

57 E = E o pH The scale on the pH meter is marked off so that a change of 1 pH unit equals volts or 59.2 millivolts.

58 The pH Meter In practice, a special pH electrode is much more convenient than using platinum electrodes and a tank of hydrogen gas! A stable reference electrode and a glass-membrane electrode are contained within a combination pH electrode. The electrode is merely dipped into a solution, and the potential difference between the electrodes is displayed as pH.

59 galvanicelectrolytic need power source two electrodes produces electrical current anode (-) cathode (+) anode (+) cathode (-) salt bridgevessel conductive medium Comparison of Electrochemical Cells  G < 0  G > 0

60 Copyright 1999, PRENTICE HALLChapter Corrosion Corrosion of Iron Since E  red (Fe 2+ ) < E  red (O 2 ) iron can be oxidized by oxygen. Cathode: O 2 (g) + 4H + (aq) + 4e -  2H 2 O(l). Anode: Fe(s)  Fe 2+ (aq) + 2e -. Dissolved oxygen in water usually causes the oxidation of iron. Fe 2+ initially formed can be further oxidized to Fe 3+ which forms rust, Fe 2 O 3.xH 2 O(s). Oxidation occurs at the site with the greatest concentration of O 2.

61 21-6 Corrosion: Unwanted Voltaic Cells Prentice-Hall © 2002General Chemistry: Chapter 21 Slide 61 of 52 O 2 (g) + 2 H 2 O(l) + 4 e - → 4 OH - (aq) 2 Fe(s) → 2 Fe 2+ (aq) + 4 e - 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

62 Copyright 1999, PRENTICE HALLChapter Corrosion Corrosion of Iron

63

64 Rust formation: 4Fe 2+ (aq) + O 2 (g) + 4H + (aq)  4Fe 3+ (aq) + 2H 2 O(l) 2Fe 3+ (aq) + 4H 2 O(l)  Fe 2 O 3  H 2 O(s) + 6H + (aq)

65 Prevention of Corrosion Cover the Fe surface with a protective coating Paint Tin Zn Galvanized iron

66 Copyright 1999, PRENTICE HALLChapter Corrosion Preventing the Corrosion of Iron Corrosion can be prevented by coating the iron with paint or another metal. Galvanized iron is coated with a thin layer of zinc. Zinc protects the iron since Zn is the anode and Fe the cathode: Zn 2+ (aq) +2e -  Zn(s), E  red = V Fe 2+ (aq) + 2e -  Fe(s), E  red = V With the above standard reduction potentials, Zn is easier to oxidize than Fe.

67 Corrosion Protection Prentice-Hall © 2002General Chemistry: Chapter 21 Slide 67 of 52

68 Cathodic Protection galvanized steel (Fe)

69 Cathodic Protection (cathode) (electrolyte) (anode)

70 Cathodic Protection In cathodic protection, an iron object to be protected is connected to a chunk of an active metal. The iron serves as the reduction electrode and remains metallic. The active metal is oxidized. Water heaters often employ a magnesium anode for cathodic protection.


Download ppt "Electric power conversion in electrochemistry Chemical Reactions Electric Power Electrolysis / Power consumption Electrochemical battery / Power generation."

Similar presentations


Ads by Google