1. ELECTROCHEMISTRY Chap 20

2. Electrochemistry Sample Exercise 20.6 Calculating E° cell from E° red Using the standard reduction potentials listed in Table 20.1 (p 857), calculate the standard emf for the voltaic cell, which is based on the reaction: E cell  = E red (cathode) − E red (anode)  Cathode: Cr 2 O 7 2− (aq) + 14 H + (aq) + 6 e − → 2 Cr 3+ (aq) + 7 H 2 O(l) E o = +1.33 V Anode:3 I 2 (s) + 6 e − → 6 I − (aq) E o = +0.54 V E cell  = E red (cathode) − E red (anode) = 1.33 V − 0.54 V = 0.79 V 

3. Table 20.1 Standard Reduction Potentials in Water at 25 o C Very poor oxidizing agent (very good reducing agent) Very good oxidizing agent (poor reducing agent)

4. Electrochemistry Fig 20.12 Standard cell potential of a voltaic cell E cell  = E red (cathode) − E red (anode) 

5. Electrochemistry FIG 20.13 H ALF - CELL POTENTIALS The greater the difference between the two, the greater the voltage of the cell E cell  = E red (cathode) − E red (anode) 

6. © 2009, Prentice-Hall, Inc.

7. Electrochemistry Fig 20.14 Relative strengths of oxidizing and reducing agents Strongest oxidizers have the most positive reduction potentials Strongest reducers have the most negative reduction potentials

8. Effect of Concentration on Electrode Potentials E is measure of the extent to which the existing concentrations in a half-cell differ from their equilibrium values e.g., Ag + + e − ⇌ Ag (s) If [Ag + ] increases then E will also increase (larger); ΔG = -n F E will become more negative For the half-rxn: aA + ne − ⇌ bB Nernst Equation

9. Electrochemistry Free Energy and Redox Reactions For a redox reaction:  G = −nFE where n ≡ number of moles of electrons transferred F = 96,485 C/mol; the Faraday constant Under standard conditions:  G  = −nFE 

10. Electrochemistry Free Energy and Redox Reactions Recall that:  G =  G  + RT ln Q by substitution: −nFE = −nFE  + RT ln Q Dividing both sides by −nF,

11. Nernst Equation E° = standard reduction potential when [A] = [B] = 1 M R = gas constant = 8.314 J/mol∙K = 8.314 V∙C/mol∙K n = number of electrons in the half-rxn F = Faraday’s constant = 9.6485 x 10 4 C/mol Reaction quotient, Q

12. Nernst Equation At T= 298.15 K and converting ln to log: Most useful form of Nernst equation

13. Write the Nernst expression for: (a) Fe 2+ + 2e − ⇌ Fe (s) (b) CrO 7 2- + 14 H + + 6e − ⇌ 2 Cr 3+ + 7 H 2 O  Multiplying half-rxn by some number does not change E° or E e.g., 2 Fe 2+ + 4e ⇌ 2 Fe (s)

14. Electrochemistry Fig 20.16 Concentration Cells based on the Ni 2+ / Ni cell reaction The Nernst equation implies that a cell could be created that has the same species at both electrodes For such a cell, = 0, but Q ≠ 0E cell  Therefore, as long as the concentrations are different, E ≠ 0

15. Electrochemistry Nernst Equation for a Complete Reaction E cell = E + − E − = E cathode − E anode Given: Cd (s) │ Cd(NO 3 ) 2 (aq); M A ║ FeCl 2 (aq); M C │ Fe (s) (a) [Cd 2+ ] = 0.50 M; [Fe 2+ ] = 0.10 M Ans = −0.058 V (b) [Cd 2+ ] = 0.010 M; [Fe 2+ ] = 1.0 M; Ans = +0.021 V Only valid when both ½-rxns written as reductions

16. Calculate the cell potential for: Cu │ CuCl 2 (0.0200 M) ║ AgNO 3 (0.0200 M) │ Ag In Class Exercise: Calculate the cell potential for: Pt │ UO 2 2+ (0.0150 M); U 4+ (0.200 M); H + (0.0300 M ║ Fe 2+ (0.0100 M); Fe 3+ (0.0250 M) │ Pt Ans = +0.410 V Ans = +0.638 V UO 2 2+ + 4H + + 2e − → U 4+ + 2H 2 O; E o = +0.334 V