1. CHM 103 Lesson Plan 6/1/2004

2. Electrochemistry Introduction Voltaic Cells Standard Voltage: E° Relations Between K, ΔG°, & E° Effect of Concentration on Voltage Electrolytic Cells Commercial Cells (for independent reading)

3. Introduction Voltaic Cell Spontaneous redox reaction Generates useful electrical energy Electrolytic Cell Non-spontaneous Redox Reaction Driven by application of electrical energy

4. Introduction (2) Cathode Reduction Consumes electrons Attracts cations Anode Oxidation Produces electrons Attracts anions

5. Introduction (3) Voltage A measure of spontaneity

6. Voltaic Cells Overall Reaction: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Component Half-Reactions Zn(s) Zn 2+ (aq) + 2e - (oxidation) Cu 2+ (aq) + 2e - Cu(s) (reduction)

7. Voltaic Cells (2) Runs spontaneously Generates heat No useable electrical energy

8. Voltaic Cells (3) Spontaneous Electric current flows Cu plates onto cathode Zn dissolves from anode

9. Standard Voltages (1) Run at Standard Conditions: P H 2 = 1 atm [H + ] = 1 M [Zn 2+ ] = 1 M Get Standard Voltage E° = 0.762 V Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g)

10. Standard Voltages (2) Define: E° red : Standard reduction voltage E° ox : Standard oxidation voltage E° = E° red + E° ox Example ( Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) ): E° = 0.762 V = E° red (H + H 2 ) + E° ox (Zn Zn 2+ )

11. Standard Voltages (3) Problem: Can’t measure E° red or E° ox directly. Solution: Arbitrarily set E° red (H + H 2 ) = 0.000 V

12. Standard Voltages (4) Now finish: E° = 0.762 V = E° red (H + H 2 ) + E° ox (Zn Zn 2+ ) = 0.00 + E° ox (Zn Zn 2+ ) E° ox (Zn Zn 2+ ) = 0.000 V + 0.762 V E° ox (Zn Zn 2+ ) = 0.762 V

13. Standard Voltages (5) Determine E° red (Cu +2 Cu) Measure: E° = +1.101 V for Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq) E° = E° red (Cu +2 Cu) + E° ox (Zn Zn 2+ ) E° red (Cu +2 Cu) = E° - E° ox (Zn Zn 2+ ) E° red (Cu +2 Cu) = +1.101 V – 0.762 V E° red (Cu +2 Cu) = + 0.339 V

14. Standard Voltages (6) Determine E° red (Zn +2 Zn) from Zn(s) Zn 2+ (aq) + 2e - E° ox = 0.762 V For the reverse reaction, change the sign, thus Zn 2+ (aq) + 2e - Zn(s) E° = - 0.762 V But, this E° is the E° red we want, so: E° red (Zn +2 Zn) = -0.762 V

15. Standard Voltages (7) Measure E° for lots of voltaic cells Build a list of E° ox and E° red Convert the E° ox values to E° red using E° red = - E° ox Obtain a table like 18.1

16. Standard Voltages (8)

17. Standard Voltages (9) Calculate E° For a Reaction From E° ox & E° red Split into oxidation and reduction half-reactions Look up E° red for the reduction Find E° red for the reverse of the oxidation. Apply E° ox = -E° red Then: E° = E° ox + E° red

18. Standard Voltages (10) Example: 2Ag + (aq) + Cd(s) 2Ag(s) + Cd 2+ (aq) Half-reactions: 2Ag + (aq) + 2e - 2Ag(s) E° red = 0.799 V (Don’t double this E° red.) Cd(s) Cd 2+ (aq) + 2e - E° ox = -(-0.402) V Result: E° = E° red + E° ox = 0.799 + 0.402 V = +1.201 V

19. Standard Voltages (11) E° and Spontaneity If E° > 0, then the reaction is spontaneous. If E° < 0, then the reaction is non- spontaneous. E° is always positive for reaction in a voltaic cell.

20. K, ΔG°, & E° (1) ΔG° = -nFE° ΔG° is the standard free energy change (in J) E° is the standard voltage (in V) n is the number of moles of electrons F is Faraday’s constant value: 9.648 x 10 4 J/mol· V

21. K, ΔG°, & E° (2) E° and K Recall that: ΔG° = -RTlnK Substitute: -nFE° = ΔG° This gives: nFE° = RTlnK Solve for E°: E° = (RT/nF)lnK Or solve for K: nFE°/RT = lnK K = e (nFE°/RT)

22. K, ΔG°, & E° (3) Getting to Eq. 18.3 Evaluate (RT/F) R = 8.31 J/mol· K T = 298 K F = 9.648 x 10 -4 J/mol· V Result: (RT/F) = 0.0257 V So: E° = (0.0257/n)lnK V And: K = e (nE°/0.0257)

23. Concentration vs. Voltage (1) E > E° if: a reactant concentration > 1 M a product concentration < 1 M E < E° if: a reactant concentration < 1 M a product concentration > 1 M

24. Concentration vs. Voltage (2) When A Voltaic Cell Runs Reactant concentrations diminish Product concentrations build up E decreases Eventually the Cell “Dies” Reaction stops The cell is at equilibrium E has become 0 (no more voltage)

25. Concentration vs. Voltage (3) The Nernst Equation Recall: ΔG = ΔG° + RTlnQ Substitute: ΔG = -nFE So: -nFE = -nFE° + RTlnQ Result: E = E° - (RT/nF)lnQ E = E° - (0.0257/n)lnQ (in V)

26. Concentration vs. Voltage (4) E vs. E° and Q E = E° - (RT/nF)lnQ Q > 1 Means Product concentrations are relatively high vs. reactants lnQ > 0 E < E°

27. Concentration vs. Voltage (5) E vs. E° and Q E = E° - (RT/nF)lnQ Q < 1 Means Reactant concentrations are relatively high vs. products lnQ < 0 E > E°

28. Concentration vs. Voltage (6) Example (18.6): Find E if: O 2 (g) + 4H + (aq) + 4Br – (aq) 2H 2 O(l) + 2Br 2 (l) P O 2 = 1.0 atm [H + ] = [Br – ] = 0.10 M Strategy: Use Nernst Equation 1.Determine Q 2.Determine E° 3.Solve for E (work at board)

29. Concentration vs. Voltage (7) Example (18.7): Find [H + ] if: Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) P H 2 = 1.0 atm [Zn 2+ ] = 1.0 M E = +0.560 V Strategy: Use Nernst Equation 1.Determine Q (as function of [H + ]) 2.Solve for [H + ] (work at board)

30. Electrolytic Cells (1) Example Zn anode Cu cathode Zn(NO 3 ) 2 electrolyte No spontaneous reaction When current flows Zn 2+ (aq) + 2e - Zn(s) at cathode Zn(s) Zn 2+ (aq) + 2e - at anode No net reaction

31. Electrolytic Cells (2) Electricity vs. Deposit on Cathode Ag + (aq) + e - Ag(s) 1 mol e - 1mol (107.9 g) Ag(s) Cu 2+ (aq) + 2e - Cu(s) 2 mol e - 1mol (63.55 g) Cu(s) Au 3+ (aq) + 3e - Au(s) 3 mol e - 1mol (197.0 g) Au(s)

32. Electrolytic Cells (3) How Much Is 1 mol of Electrons? 1 mol electrons = 9.648x10 4 coulombs Conversion Factors (Table 18.3) Chargecoulomb (C)1C = 1A· s = 1J/V Currentampere (A)1A = 1C/s Potentialvolt (V)1V = 1J/C Powerwatt (W)1W = 1J/s Energyjoule (J)1J = 1V/s

33. Electrolytic Cells (4) Examples of Problems How many C of charge to plate x g of metal? How long to plate x g of metal with a current of y A? How much electric energy to plate x g of metal at y V potential? These are simple stoichiometry/ conversion problems

34. Electrolytic Cells (5) Cell Reactions (Aqueous Solutions) At the Cathode Reduction of cation to metal Ag + (aq) + e - Ag(s) E° = +0.799 V Reduction of water to hydrogen gas for a difficult-to-reduce cation like Na + or K + 2H 2 O + 2e - H 2 (g) + 2OH - (aq) E° = -0.828 V

35. Electrolytic Cells (6) Cell Reactions (Aqueous Solutions) At the Anode Oxidation of anion to non-metal 2I - (aq) I 2 (s) + 2e - E° = -0.534 V Oxidation of water to oxygen gas for an anion like NO 3 - or SO 3 2- that cannot be oxidized 2H 2 O O 2 (g) + 4H + (aq) E° = -0.828 V -1.299 V

36. Commercial Cells Electrolysis of Aqueous NaCl Primary (non-rechargeable) Voltaic Cells Storage (rechargeable) Voltaic Cells Fuel Cells (Read for yourselves: pp. 493-497.)