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Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Electrochemistry The study of the interchange of chemical and electrical energy.

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Presentation on theme: "Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Electrochemistry The study of the interchange of chemical and electrical energy."— Presentation transcript:

1 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Electrochemistry The study of the interchange of chemical and electrical energy.

2 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 2 Review of Terms oxidation-reduction (redox) reaction: involves a transfer of electrons from the reducing agent to the oxidizing agent. oxidation: loss of electrons reduction: gain of electrons

3 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 3 2Mg (s) + O 2 (g) 2MgO (s) 2Mg 2Mg 2+ + 4e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) 19.1 Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur 002+2-

4 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 4 Half-Reactions The overall reaction is split into two half-reactions, one involving oxidation and one reduction. 8H + + MnO 4  + 5Fe 2+  Mn 2+ + 5Fe 3+ + 4H 2 O Reduction: 8H + + MnO 4  + 5e   Mn 2+ + 4H 2 O Oxidation: 5Fe 2+  5Fe 3+ + 5e 

5 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 5 Galvanic Cell A device in which chemical energy is changed to electrical energy.

6 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 Anode and Cathode OXIDATION occurs at the ANODE. REDUCTION occurs at the CATHODE.

7 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7 Electrochemical Cells 19.2 spontaneous redox reaction anode oxidation cathode reduction

8 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 Electrochemical Cells 19.2 The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Cell Diagram Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) [Cu 2+ ] = 1 M & [Zn 2+ ] = 1 M Zn (s) | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu (s) anodecathode

9 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 Standard Electrode Potentials 19.3 Standard reduction potential (E 0 ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. E 0 = 0 V Standard hydrogen electrode (SHE) 2e - + 2H + (1 M) H 2 (1 atm) Reduction Reaction

10 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10 Standard Electrode Potentials 19.3 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s) 2e - + 2H + (1 M) H 2 (1 atm) Zn (s) Zn 2+ (1 M) + 2e - Anode (oxidation): Cathode (reduction): Zn (s) + 2H + (1 M) Zn 2+ + H 2 (1 atm)

11 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 19.3 E 0 = 0.76 V cell Standard emf (E 0 ) cell 0.76 V = 0 - E Zn /Zn 0 2+ E Zn /Zn = -0.76 V 0 2+ Zn 2+ (1 M) + 2e - Zn E 0 = -0.76 V E 0 = E H /H - E Zn /Zn cell 00 + 2+ 2 Standard Electrode Potentials E 0 = E cathode - E anode cell 00 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s)

12 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 12 Standard Electrode Potentials 19.3 Pt (s) | H 2 (1 atm) | H + (1 M) || Cu 2+ (1 M) | Cu (s) 2e - + Cu 2+ (1 M) Cu (s) H 2 (1 atm) 2H + (1 M) + 2e - Anode (oxidation): Cathode (reduction): H 2 (1 atm) + Cu 2+ (1 M) Cu (s) + 2H + (1 M) E 0 = E cathode - E anode cell 00 E 0 = 0.34 V cell E cell = E Cu /Cu – E H /H 2++ 2 000 0.34 = E Cu /Cu - 0 0 2+ E Cu /Cu = 0.34 V 2+ 0

13 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 13 Standard Reduction Potentials The E  values corresponding to reduction half-reactions with all solutes at 1M and all gases at 1 atm. Cu 2+ + 2e   Cu E  = 0.34 V vs. SHE SO 4 2  + 4H + + 2e   H 2 SO 3 + H 2 O E  = 0.20 V vs. SHE

14 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 14 Cell Potential Cell Potential or Electromotive Force (emf): The “pull” or driving force on the electrons.

15 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 15 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

16 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 16 emf and Work

17 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 17 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 = -0.40 V Cr 3+ (aq) + 3e - Cr (s) E 0 = -0.74 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 = -0.40 – (-0.74) cell E 0 = 0.34 V cell 19.3

18 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 18 Free Energy and Cell Potential  G  =  nFE  n = number of moles of electrons F = Faraday = 96,485 coulombs per mole of electrons

19 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 19 Concentration Cell...a cell in which both compartments have the same components but at different concentrations.

20 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 20 The Nernst Equation We can calculate the potential of a cell in which some or all of the components are not in their standard states.

21 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 21 Calculation of Equilibrium Constants for Redox Reactions At equilibrium, E cell = 0 and Q = K.

22 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 22 19.4 Spontaneity of Redox Reactions  G = -nFE cell  G 0 = -nFE cell 0 n = number of moles of electrons in reaction F = 96,500 J V mol = 96,500 C/mol  G 0 = -RT ln K = -nFE cell 0 E cell 0 = RT nF ln K (8.314 J/K mol)(298 K) n (96,500 J/V mol) ln K = = 0.0257 V n ln K E cell 0 = 0.0592 V n log K E cell 0

23 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 23 Spontaneity of Redox Reactions 19.4

24 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 24 2e - + Fe 2+ Fe 2Ag 2Ag + + 2e - Oxidation: Reduction: What is the equilibrium constant for the following reaction at 25 0 C? Fe 2+ (aq) + 2Ag (s) Fe (s) + 2Ag + (aq) = 0.0257 V n ln K E cell 0 19.4 E 0 = -0.44 – (0.80) E 0 = -1.24 V 0.0257 V x nE0E0 cell exp K = n = 2 0.0257 V x 2x 2-1.24 V = exp K = 1.23 x 10 -42 E 0 = E Fe /Fe – E Ag /Ag 00 2+ +

25 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 25 Batteries A battery is a galvanic cell or, more commonly, a group of galvanic cells connected in series.

26 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 26 Batteries 19.6 Leclanché cell Dry cell Zn (s) Zn 2+ (aq) + 2e - Anode: Cathode: 2NH 4 (aq) + 2MnO 2 (s) + 2e - Mn 2 O 3 (s) + 2NH 3 (aq) + H 2 O (l) + Zn (s) + 2NH 4 (aq) + 2MnO 2 (s) Zn 2+ (aq) + 2NH 3 (aq) + H 2 O (l) + Mn 2 O 3 (s)

27 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 27 Batteries Zn(Hg) + 2OH - (aq) ZnO (s) + H 2 O (l) + 2e - Anode: Cathode: HgO (s) + H 2 O (l) + 2e - Hg (l) + 2OH - (aq) Zn(Hg) + HgO (s) ZnO (s) + Hg (l) Mercury Battery 19.6

28 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 28 Batteries 19.6 Anode: Cathode: Lead storage battery PbO 2 (s) + 4H + (aq) + SO 2- (aq) + 2e - PbSO 4 (s) + 2H 2 O (l) 4 Pb (s) + SO 2- (aq) PbSO 4 (s) + 2e - 4 Pb (s) + PbO 2 (s) + 4H + (aq) + 2SO 2- (aq) 2PbSO 4 (s) + 2H 2 O (l) 4

29 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 29 Fuel Cells...galvanic cells for which the reactants are continuously supplied. 2H 2 (g) + O 2 (g)  2H 2 O(l) anode: 2H 2 + 4OH   4H 2 O + 4e  cathode: 4e  + O 2 + 2H 2 O  4OH 

30 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 30 Batteries 19.6 A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: Cathode: O 2 (g) + 2H 2 O (l) + 4e - 4OH - (aq) 2H 2 (g) + 4OH - (aq) 4H 2 O (l) + 4e - 2H 2 (g) + O 2 (g) 2H 2 O (l)

31 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 31 Corrosion Some metals, such as copper, gold, silver and platinum, are relatively difficult to oxidize. These are often called noble metals.

32 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 32 Corrosion 19.7

33 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 33 Cathodic Protection of an Iron Storage Tank 19.7

34 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 34 Electrolysis...forcing a current through a cell to produce a chemical change for which the cell potential is negative.

35 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 35 19.8 Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.

36 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 36 Electrolysis of Water 19.8

37 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 37 Stoichiometry of Electrolysis 4 How much chemical change occurs with the flow of a given current for a specified time? current and time  quantity of charge  moles of electrons  moles of analyte  grams of analyte

38 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 38 Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mole e - = 96,500 C 19.8

39 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 39 How much Ca will be produced in an electrolytic cell of molten CaCl 2 if a current of 0.452 A is passed through the cell for 1.5 hours? Anode: Cathode: Ca 2+ (l) + 2e - Ca (s) 2Cl - (l) Cl 2 (g) + 2e - Ca 2+ (l) + 2Cl - (l) Ca (s) + Cl 2 (g) 2 mole e - = 1 mole Ca mol Ca = 0.452 C s x 1.5 hr x 3600 s hr96,500 C 1 mol e - x 2 mol e - 1 mol Ca x = 0.0126 mol Ca = 0.50 g Ca 19.8


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