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Chapter 18

Chapter 18. Electrochemistry: electrical charge can be produced from a spontaneous chemical reaction (chemical energy can be converted to electrical energy). Chapter 18.

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Chapter 18

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  1. Chapter 18 • Electrochemistry: electrical charge can be produced from a spontaneous chemical reaction (chemical energy can be converted to electrical energy).

  2. Chapter 18 • Reactions that do not occur spontaneously can be forced to take place by supplying energy with an external current (electrolytic reactions).

  3. Chapter 18 • Background: Faraday’s Law - one Faraday is one mole of electrons (1F = 1 mol of e- ~ 96,500 coulombs).

  4. Chapter 18 • One coulomb is the amount of charge that moves past any given point in a circuit when a current of 1 ampere is supplied for 1 second.

  5. Chapter 18 • Charge = 1 C = 1 A*s = 1 J/V • Current = 1 A = 1 C/s • Potential = 1 V = 1 J/C • Power = 1 W = 1 J/s • Energy = 1 J = 1 V*C

  6. Chapter 18 • Faraday’s Law: during electrolysis, the passage of 1F through the circuit brings about the ox. of one equivalent weight at one electrode and red. of one eq. wt. at the other electrode.

  7. Chapter 18 • Equivalent weight = the stoichiometric mass required to transfer one mol of e-.

  8. Chapter 18 • In all cells, oxidation occurs at the anode and reduction at the cathode.

  9. Galvanic Cells • (Aka Voltaic cells) chemical energy is converted into electrical energy. The anode is negative. The cathode is positive.

  10. Galvanic Cells • The force with which the electrons flow from the negative to the positive electrode through an external wire is called the electromotive force, emf, and is measured in volts.

  11. Galvanic Cells • 1V = 1 J/C • The greater the tendency of the two half reactions to occur spontaneously, the greater the emf of the cell.

  12. Galvanic Cells • Standard cell potential, Eo, can be related to Go, and Keq, etc.

  13. Voltaic Cells • Salt bridge cells: chemical cells must be designed to make electron transfers occur indirectly. • Ex. Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s)

  14. Voltaic Cells • Anode: Zn(s) --> Zn2+(aq) + 2e- • Cathode: Cu2+ (aq) + 2e- ---> Cu

  15. Voltaic Cells • Salt bridge allows current to flow but prevents contact between zinc and copper(II) ions, which would short circuit the cell.

  16. Voltaic Cells • Usually the metals used in the red-ox reaction serve as the electrodes. • For the cell: Zn(s) + H+(aq) --> Zn2+(aq) + H2(g); use an inert electrode, like Pt, for the hydrogen half cell.

  17. Cell Notation • Zn-Cu2+ cell: Zn|Zn2+||Cu2+|Cu. • Zn- H+ cell: Zn|Zn2+||H+|H2|Pt • Which is the anode? Which is the cathode?

  18. Standard Voltages • Eo = cell voltage when all species are at standard concentrations (1 atm of gases, 1 M for solutions in water).

  19. Standard Voltages • Zn(s) + 2 H+(aq, 1M) --> Zn2+(aq, 1M) + H2(g, 1atm) • Eo = + 0.762 V = EooxZn + EoredH+

  20. Standard Voltages • See Table 18.1. EoredH+ is arbitrarily set at 0.000 V; EooxZn = + 0.762 V. (Half reactions can not be measured directly).

  21. Standard Voltages • Table 18.1 lists values of Eored; you can get Eoox by changing the sign. • Ex. Cu(s) --> Cu2+(aq) + 2e-; EooxCu = -EoredCu = -0.339V

  22. Standard Voltages • Relative strengths of oxidizing and reducing agents: • 1.) Oxidizing agents (left column, Table 18.1). The larger (more positive) the value of Eored, the stronger the oxidizing agent.

  23. Standard Voltages • Relative strengths of oxidizing and reducing agents: • 2.) Reducing agents (right column, Table 18.1). The larger (more positive) the value of Eoox, the stronger the reducing agent.

  24. Standard Voltages • 3.) Oxidizing agents become stronger moving down the left column. Reducing agents become weaker moving down the right column.

  25. Calculation of Eo • Eo = Eoox + Eored. • Cl2(g) + 2Br-(aq) --> 2 Cl-(aq) + Br2(l) • Eo = EoredCl2 + EooxBr- = 1.360V -1.007V = +0.283V

  26. Calculation of Eo • A positive value for Eo indicates that the reaction can occur in a voltaic cell.

  27. Redox Spontaneity • Determination of whether a redox reaction will occur: • ex-What, if anything, happens when bromine is added to a solution of tin(II) chloride?

  28. Redox Spontaneity • Possible oxidations: • Sn2+(aq) --> Sn4+(aq) + 2e- Eoox = -0.154 V • 2Cl-(aq) --> Cl2(g) + 2e- Eoox = -1.360 V

  29. Redox Spontaneity • Possible reductions: • Sn2+(aq) + 2e- --> Sn(s) Eored = -0.141 V • Br2(l) + 2e- --> 2Br-(aq) Eored = 1.077 V

  30. Redox Spontaneity • Reaction that occurs: • Sn2+(aq) + Br2(l) --> Sn4+(aq) + 2Br-(aq) • Eo = 0.923 V

  31. Eo, Go, and K • Go = -nFEo • Free energy is a measurement of the amount of useful work that can be obtained by a reaction at constant T and P

  32. Eo, Go, and K • So, Go = wmax = -Q’E • wmax is the useful work done on the system. • Q’E is the electrical work done by the system.

  33. Eo, Go, and K • Q’E = product of the charge times the voltage. • Q’ = (9.648 x 104 C/mol)*n (n = mols of e- exchanged in the reaction).

  34. Eo, Go, and K • Go = -nFEo • Faraday’s constant = 9.648 x 104 J/mol*V • Note: when spontaneous, Go is negative while Eo is positive.

  35. Eo, Go, and K • Table 18.2 - K vs. Eo : if Eo is greater than +0.10V then K is very large, if less than -0.10V K is very small.

  36. Eo, Go, and K • lnK = nEo/0.0257 • Note: when spontaneous, Go is negative, Eo is positive, lnK is positive, K > 1.

  37. Eo, Go, and K • From the last lecture: Cl2(g) + 2Br-(aq) --> 2 Cl-(aq) + Br2(l) • Eo = 1.360V -1.077V = +0.283V • What isGo and K for this reaction?

  38. Eo, Go, and K • Cl2(g) + 2Br-(aq) --> 2 Cl-(aq) + Br2(l); What isGo and K for this reaction? Go = -2(96.5)(0.283)kJ = -54.6kJ. lnK = 2(0.283)/0.0257 = 22.0; K = 4 x 109

  39. Conc. and voltage • Effect of conc. on voltage: Nernst Equation: E = Eo - [(RT)/(nF)] ln Q = Eo - (0.0257/n) ln Q • In expressions of Q gases are in partial pressures, solutions in molarities.

  40. Conc. and voltage • Cl2(g) + 2Br-(aq) --> 2 Cl-(aq) + Br2(l); Calculate the voltage when [Br-] = 1 M, P Cl2 = 1 atm, [Cl-] = 0.01 M? (R = 8.31 J/mol*K, T = 298K)  V

  41. Conc. and voltage • Technique: Use theNernst Equation to determine the concentration of ions in solution.

  42. Conc. and voltage • Given: Zn(s) + 2H+(aq) --> Zn2+(aq) + H2(g). • E = 0.762 V - 0.0257/2 ln Q • Q = ?

  43. Conc. and voltage • Suppose [Zn2+] = 1 M, P H2 = 1 atm • E = ? • You can then measure the voltage and calculate [H+]

  44. Conc. and voltage • Suppose E = 0.200 V. What is [H+]?

  45. Conc. and voltage • Suppose E = 0.200 V. What is [H+]? • ln [H+] = (-0.562)/(0.0257) = -21.9; [H+] = 3 x 10-10 M

  46. Electrolytic Cells • Occur when electrical energy is supplied to bring about a nonspontaneous redox reaction.

  47. Electrolytic Cells • See figure 18.10. P. 494 • The storage battery acts as a source of direct electrical current.

  48. Electrolytic Cells • From the terminals of the battery, two wires lead to the cell. The anode and cathode are dipped into a solution containing M+ and X-.

  49. Electrolytic Cells • The battery pumps electrons into the cathode and removes them from the anode.

  50. Electrolytic Cells • The process that allows the consumption of e- at C and the liberation of them at A is called electrolysis (a nonspontaneous redox reaction).

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