1. 1 Chapter Eighteen Electrochemistry

2. 2 Electrochemical reactions are oxidation-reduction reactions. The two parts of the reaction are physically separated. –oxidation occurs at one cell –reduction occurs in the other cell There are two kinds electrochemical cells. Electrolytic cells - nonspontaneous chemical reactions Voltaic or galvanic cells - spontaneous chemical reactions

3. 3 Electrical Conduction Ionic or electrolytic conduction Ionic motion transports the electrons Positively charged ions, cations, move toward the negative electrode. Negatively charged ions, anions, move toward the positive electrode.

4. 4 Electrodes Conventions for electrodes: Cathode - electrode at which reduction occurs Anode - electrode at which oxidation occurs Inert electrodes do not react with the liquids or products of the electrochemical reaction. Graphite and Platinum are common inert electrodes.

5. 5 Voltaic or Galvanic Cells Electrochemical cells in which a spontaneous chemical reaction produces electrical energy. Cell halves are physically separated so that electrons (from redox reaction) are forced to travel through wires and creating a potential difference. Examples include:

6. 6 The Construction of Simple Voltaic Cells Half-cell contains the oxidized and reduced forms of an element (or other chemical species) in contact with each other. Simple cells consist of: –two pieces of metal immersed in solutions of their ions –wire to connect the two half-cells –salt bridge to complete circuit maintain neutrality prevent solution mixing

7. 7 The Zinc-Copper Cell Cell components: Cu strip immersed in 1.0 M copper (II) sulfate Zn strip immersed in 1.0 M zinc (II) sulfate wire and a salt bridge to complete circuit Initial voltage is 1.10 volts

8. 8 The Zinc-Copper Cell

9. 9 In all voltaic cells, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode).

10. 10 The Zinc-Copper Cell Short hand notation for voltaic cells –Zn-Cu cell example

11. 11 The Copper - Silver Cell Cell components: Cu strip immersed in 1.0 M copper (II) sulfate Ag strip immersed in 1.0 M silver (I) nitrate wire and a salt bridge to complete circuit Initial voltage is 0.46 volts

12. 12 The Copper - Silver Cell

13. 13 The Copper - Silver Cell Compare the Zn-Cu cell to the Cu-Ag cell Cu electrode is cathode in Zn-Cu cell Cu electrode is anode in Cu-Ag cell Whether a particular electrode behaves as an anode or as a cathode depends on what the other electrode of the cell is.

14. 14 The Copper - Silver Cell Demonstrates that Cu 2+ is a stronger oxidizing agent than Zn 2+ Cu 2+ oxidizes metallic Zn to Zn 2+ Ag + is is a stronger oxidizing agent than Cu 2+ Ag + oxidizes metallic Cu to Cu 2+ Arrange these species in order of increasing strengths

15. 15 Standard Electrode Potential Establish an arbitrary standard to measure potentials of a variety of electrodes Standard Hydrogen Electrode (SHE) –assigned an arbitrary voltage of 0.000000… V

16. 16 Standard Electrode Potential

17. 17 The Zinc-SHE Cell Cell components: Zn strip immersed in 1.0 M zinc (II) sulfate other electrode is a SHE wire and a salt bridge to complete circuit Initial voltage is 0.763 volts

18. 18 The Zinc-SHE Cell

19. 19 The Zinc-SHE Cell SHE is the cathode Zn reduces H + to H 2 Zn is the anode

20. 20 The Copper-SHE Cell Cell components: Cu strip immersed in 1.0 M copper (II) sulfate other electrode is a SHE wire and a salt bridge to complete circuit Initial voltage is 0.337 volts

21. 21 The Copper-SHE Cell

22. 22 The Copper-SHE Cell SHE is the anode Cu 2+ ions oxidize hydrogen to H + Cu is the cathode

23. 23 The Electromotive (Activity) Series of the Elements Electrodes that force the SHE to act as an anode are assigned positive standard reduction potentials. Electrodes that force the SHE to act as the cathode are assigned negative standard reduction potentials. Table 18.1 in Text lists the Activity of some of the elements

24. 24 Uses of the Electromotive Series Standard electrode (reduction) potentials tell us the tendencies of half-reactions to occur as written. For example, the half-reaction for the standard potassium electrode is: The large negative value tells us that this reaction will occur only under extreme conditions.

25. 25 Uses of the Electromotive Series Compare the potassium half-reaction to fluorine’s half-reaction: The large positive value denotes that this reaction occurs readily as written. Positive E 0 values tell us that the reaction tends to occur to the right –larger the value, greater tendency to occur to the right Opposite for negative values

26. 26 Uses of the Electromotive Series Use standard electrode potentials to predict whether an electrochemical reaction at standard state conditions will occur spontaneously. Steps for obtaining the equation for the spontaneous reaction.

27. 27 Uses of the Electromotive Series 1Choose the appropriate half-reactions from a table of standard reduction potentials.  Write the equation for the half-reaction with the more positive E 0 value first, along with its E 0 value. 3Write the equation for the other half-reaction as an oxidation with its oxidation potential, i.e. reverse the tabulated reduction half-reaction and change the sign of the tabulated E 0.

28. 28 Uses of the Electromotive Series 4Balance the electron transfer. 5Add the reduction and oxidation half-reactions and their potentials. This produces the equation for the reaction for which E 0 cell is positive, which indicates that the forward reaction is spontaneous.

29. 29 Electrode Potentials for Other Half-Reactions Example: Will permanganate ions, MnO 4 -, oxidize iron (II) ions to iron (III) ions, or will iron (III) ions oxidize manganese ions to permanganate ions in acidic solution? Follow the steps outlined in the previous slides. E 0 values are not multiplied by any stoichiometric relationships in this procedure.

30. 30 Electrode Potentials for Other Half-Reactions Example: Will permanganate ions, MnO 4 -, oxidize iron (II) ions to iron (III) ions, or will iron (III) ions oxidize manganese ions to permanganate ions in acidic solution?

31. 31 Electrode Potentials for Other Half-Reactions Example: Will permanganate ions, MnO 4 -, oxidize iron (II) ions to iron (III) ions, or will iron (III) ions oxidize manganese ions to permanganate ions in acidic solution?

32. 32 Electrode Potentials for Other Half-Reactions Example: Will permanganate ions, MnO 4 -, oxidize iron (II) ions to iron (III) ions, or will iron (III) ions oxidize manganese ions to permanganate ions in acidic solution?

33. 33 Electrode Potentials for Other Half-Reactions Example: Will permanganate ions, MnO 4 -, oxidize iron (II) ions to iron (III) ions, or will iron (III) ions oxidize manganese ions to permanganate ions in acidic solution? Thus permanganate ions will oxidize iron (II) ions to iron (III) and are reduced to manganese (II) ions in acidic solution.

34. 34 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation Standard electrode potentials are determined at thermodynamic standard conditions. 1 M solutions 1 atm of pressure for gases liquids and solids in their standard states temperature of 25 0 C

35. 35 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation Potentials change if conditions are nonstandard. Nernst equation describes the electrode potentials at nonstandard conditions.

36. 36 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation

37. 37 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation Substitution of the values of the constants into the Nernst equation at 25 0 C gives:

38. 38 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation For a typical half-reaction: The corresponding Nernst equation is

39. 39 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation Substituting E 0 into the above expression gives

40. 40 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation If [Cu 2+ ] and [Cu + ] are both 1.0 M, standard conditions, then E = E 0 because concentration term is zero. Since log 1= 0, we have

41. 41 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation Example: Calculate the potential for the Cu 2+ / Cu + electrode at 25 0 C when the concentration of Cu + ions is three times that of Cu 2+ ions.

42. 42 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation Example: Calculate the potential for the Cu 2+ / Cu + electrode at 25 0 C when the concentration of Cu + ions is three times that of Cu 2+ ions.

43. 43 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation The Nernst equation can also be used to calculate the potential for a cell that consists of two nonstandard electrodes. Example: Calculate the initial potential of a cell that consists of an Fe 3+ /Fe 2+ electrode in which [Fe 3+ ]=1.0 x 10 -2 M and [Fe 2+ ]=0.1 M connected to a Sn 4+ /Sn 2+ electrode in which [Sn 4+ ]=1.0 M and [Sn 2+ ]=0.10 M. A wire and salt bridge complete the circuit.

44. 44 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation Calculate the E 0 cell by the usual procedure.

45. 45 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation Substitute the ion concentrations into Q to calculate E cell.

46. 46 Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation Substitute the ion concentrations into Q to calculate E cell.

47. 47 Relationship of E 0 cell to  G 0 and K From previous chapters we know the relationship of  G 0 and K for a reaction.

48. 48 Relationship of E 0 cell to  G 0 and K The relationship between  G 0 and E 0 cell is also a simple one.

49. 49 Relationship of E 0 cell to  G 0 and K Combine these two relationships into a single relationship to relate E 0 cell to K.

50. 50 Relationship of E 0 cell to  G 0 and K Example: Calculate the standard Gibbs free energy change,  G 0, at 25 0 C for the following reaction.

51. 51 Relationship of E 0 cell to  G 0 and K Calculate E 0 cell using the appropriate half-reactions.

52. 52 Relationship of E 0 cell to  G 0 and K Now that we know E 0 cell, we can calculate  G 0. The negative value tells us that the reaction is spontaneous as written.