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CHM 103

CHM 103. Lesson Plan 6/1/2004. Electrochemistry. Introduction Voltaic Cells Standard Voltage: E° Relations Between K, ΔG°, & E° Effect of Concentration on Voltage Electrolytic Cells Commercial Cells (for independent reading). Introduction. Voltaic Cell Spontaneous redox reaction

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CHM 103

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

  5. Introduction (3) • Voltage • A measure of spontaneity

  6. Voltaic Cells • Overall Reaction: • Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) • Component Half-Reactions • Zn(s) Zn2+(aq) + 2e- (oxidation) • Cu2+(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: • PH2= 1 atm • [H+] = 1 M • [Zn2+] = 1 M • Get Standard Voltage • E° = 0.762 V Zn(s) + 2H+(aq) Zn2+(aq) + H2(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) Zn2+(aq) + H2(g)): E° = 0.762 V = E°red (H+H2) + E°ox (Zn Zn2+)

  11. Standard Voltages (3) • Problem: Can’t measure E°red or E°ox directly. • Solution: Arbitrarily set E°red (H+H2) = 0.000 V

  12. Standard Voltages (4) • Now finish: • E° = 0.762 V = E°red (H+H2) + E°ox (Zn Zn2+) = 0.00 + E°ox (Zn Zn2+) • E°ox (Zn Zn2+) = 0.000 V + 0.762 V • E°ox (Zn Zn2+) = 0.762 V

  13. Standard Voltages (5) • Determine E°red (Cu+2 Cu) • Measure: E° = +1.101 V for Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) • E° = E°red (Cu+2 Cu) + E°ox (Zn Zn2+) • E°red (Cu+2 Cu) = E° - E°ox (Zn Zn2+) • 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) Zn2+(aq) + 2e- E°ox = 0.762 V • For the reverse reaction, change the sign, thus • Zn2+(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) + Cd2+(aq) • Half-reactions: • 2Ag+(aq) + 2e- 2Ag(s) E°red = 0.799 V (Don’t double this E°red.) • Cd(s) Cd2+(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 104 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: • O2(g) + 4H+(aq) + 4Br–(aq) 2H2O(l) + 2Br2(l) • PO2 = 1.0 atm • [H+] = [Br–] = 0.10 M • Strategy: Use Nernst Equation • Determine Q • Determine E° • Solve for E (work at board)

  29. Concentration vs. Voltage (7) • Example (18.7): Find [H+] if: • Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) • PH2 = 1.0 atm • [Zn2+] = 1.0 M • E = +0.560 V • Strategy: Use Nernst Equation • Determine Q (as function of [H+]) • Solve for [H+] (work at board)

  30. Electrolytic Cells (1) • Example • Zn anode • Cu cathode • Zn(NO3)2 electrolyte • No spontaneous reaction • When current flows • Zn2+(aq)+ 2e- Zn(s) at cathode • Zn(s) Zn2+(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) • Cu2+(aq) + 2e- Cu(s) 2 mol e- 1mol (63.55 g) Cu(s) • Au3+(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 = 9.648x104 coulombs • Conversion Factors (Table 18.3)

  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+ 2H2O + 2e- H2(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) I2(s) + 2e- E° = -0.534 V • Oxidation of water to oxygen gas for an anion like NO3- or SO32- that cannot be oxidized 2H2O O2(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.)

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