1. Brain Warmup Half-Reaction ℰ ° (V) Ag + + e -  Ag 0.80 Cu 2+ + 2e -  Cu 0.34 Zn 2+ + 2e -  Zn-0.76 Al 3+ + 3e -  Al-1.66 What is ℰ ° for each of the following reactions? Which reaction(s) are spontaneous? 3 Ag + (aq) + Al (s)  3 Ag (s) + Al 3+ (aq) Cu 2+ (aq) + Zn (s)  Cu (s) + Zn 2+ (aq) 2 Al 3+ (aq) + 3 Zn (s)  2 Al (s) + 3 Zn 2+ (aq) ℰ°ℰ° 2.46 V 1.10 V -0.90 V Spontaneous? Y Y N Zn can reduce Cu 2+, but not Al 3+

2. 3 Ag + (aq) + Al (s)  3 Ag (s) + Al 3+ (aq) Polishing Silver with Aluminum Foil ℰ ° = 2.46 V ΔG° = -nF ℰ ° = -712 kJ 3 Ag 2 S + 2 Al + 3 H 2 O  6 Ag + Al 2 O 3 + 3 H 2 S  Stinky!! (Add NaHCO 3 to neutralize H 2 S)

3. Redox Reaction Equilibrium Constants ΔG° = -nF ℰ ° ΔG° = -RT lnK nF ℰ ° = RT lnK lnK = nF ℰ °/RT What is K at 25°C for the aluminum/silver polishing reaction? 3 mol (96485 C/mol)(2.46 J/C) (8.314 J/mol K)(298 K) 3 Ag + (aq) + Al (s)  3 Ag (s) + Al 3+ (aq) ℰ ° = 2.46 V lnK == 287 K = e 287 = 4.39  10 124

4. Concentration Cells A voltage is generated just by a difference in concentration between the two half-cells Cu (s)  Cu 2+ (aq) (0.5 M)  Cu 2+ (aq) (2.5 M)  Cu (s) ℰ = ℰ ° - Cu 2+ (aq) + 2e -  Cu (s) Cu (s)  Cu 2+ (aq) + 2e - Oxidation Reduction ℰ ° = 0 ℰ = ℰ ° - = 0.02 V = 0 -

5. Concentration Cells in Nature Living cells contain low-pH vesicles surrounded by neutral-pH cytoplasm –The [H + ] concentration gradient creates a voltage pH = 3 pH = 7 ℰ cell = ℰ ° cell - 4 H + + O 2 + 4e -  2 H 2 O ℰ ° = 1.23 V (10 -7 ) 4 (10 -3 ) 4 Q = ([H + ] outside ) 4 ([H + ] inside ) 4 = ℰ cell = -(0.059/4) log(10 -16 ) = 0.24 V ℰ cell = ℰ ° cell -

6. Batteries Car Battery (Lead storage battery) –Anode: Pb + HSO 4 -  PbSO 4 + H + + 2e - –Cathode: PbO 2 + HSO 4 - + 3H + + 2e -  PbSO 4 + 2H 2 O –Cell: Pb + PbO 2 + 2H + + 2 HSO 4 -  2 PbSO 4 + 2H 2 O –2 volts per cell, 6 cells to a battery  12 volt battery Alkaline Battery –Anode: Zn  Zn 2+ + 2e - –Cathode: 2 MnO 2 + H 2 O + 2e -  Mn 2 O 3 + 2OH - –1.5 volts Lemon Battery –Anode: Zn  Zn 2+ + 2e - –Cathode: Cu 2+ + 2e -  Cu –Cell reaction: Zn + Cu 2+  Zn 2+ + Cu –1.1 volts (if all goes well) Want more volts? Link cells in series...

7. Electrolytic Cells Galvanic cell (battery): spontaneous reaction, ℰ ° > 0 Electrolytic cell: non-spontaneous reaction, ℰ ° < 0 –An external power source is used to force the reaction to occur Used for: –Charging (rechargeable) batteries –Producing or purifying metals (aluminum, copper) –Electroplating Charles Hall Discovered how to produce aluminum by electrolysis

8. Electrolysis of Water Spontaneous reaction: formation of water –2 H 2(g) + O 2(g)  2 H 2 O (l) ΔG° = -474 kJ 0 0 +1 -2 –As a galvanic cell: Cathode: O 2  H 2 O(O 2(g) + 4 H + (aq) + 4 e -  2 H 2 O (l) ) Anode: H 2  H 2 O(2 H 2(g) + 4 OH - (aq)  4 H 2 O (l) + 4 e - ) –ℰ ° = 2.06 V (pure water, ℰ = 1.23 V) Non-spontaneous reaction: electrolysis of water –2 H 2 O (l)  2 H 2(g) + O 2(g) ΔG° = +474 kJ –As an electrolytic cell: Anode (oxidation):2 H 2 O (l)  O 2(g) + 4 H + (aq) + 4 e - Cathode (reduction):4 H 2 O (l) + 4 e -  2 H 2(g) + 4 OH - (aq) – ℰ ° = -2.06 V (pure water, ℰ = -1.23 V)

9. Electrolysis of Water 2 H 2 O (l)  2 H 2(g) + O 2(g) ℰ ° = -2.06 V Cathode (reduction): 4 H 2 O + 4 e -  2 H 2 + 4 OH - Produces: 2 mol gas Base Anode (oxidation): 2 H 2 O  O 2 + 4 H + + 4 e - Produces: 1 mol gas Acid Battery > 2.06 V + − Pt electrodes e- →e- → ← e -

10. Electrolysis Calculations How many grams of O 2(g) will a 3.0 amp power source produce in 5 minutes? –Amperes (A) = electric current = coulombs/second (C/s) André-Marie Ampère Oxygen half-reaction: 2 H 2 O  O 2 + 4 H + + 4 e - Current and time ChargeMoles of e - Moles of product Grams of product 3.0 C/s  300 s = 900 C 900 C  1 mol e - /96485 C = 0.0093 mol e - 0.0093 mol e -  1 mol O 2 /4 mol e - = 0.0023 mol O 2 0.0023 mol O 2  32 g O 2 /1 mol O 2 = 0.075 g O 2 CsCs  s  1 mol e - 96485 C mol product mol e - g product mol product =g product

11. Electroplating Cathode Cu 2+ (aq) + 2e -  Cu (s) Anode Cu (s)  Cu 2+ (aq) + 2e - Battery + − Cu  Cu 2+ Cu 2+  e- →e- → ← e - How long will it take a 15 amp power source to deposit 5.9 g of copper? 15 C s  t  1 mol e - 96485 C 1 mol Cu 2 mol e - 63.55 g Cu 1 mol Cu =5.9 g Cu t = 1200 s = 20 minutes