Lecture 284/3/06 Seminar today. Secondary Batteries (rechargeable) Lead Acid battery E° = 2.04 V Anode:Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + + 2e -

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Lecture 284/3/06 Seminar today

Secondary Batteries (rechargeable) Lead Acid battery E° = 2.04 V Anode:Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + + 2e - Cathode:PbO 2 (s) + HSO 4 - (aq) + 3H + + 2e -  PbSO 4 (s) + 2H 2 O Combine 6 cells to get 12 V

Electrolysis Need to put in energy 2 kinds Molten Aqueous Overpotential (or overvoltage) Extra voltage need to overcome electron transfer limitations between electrode and solution

Electrolysis of molten salts Cathode:2Na + (l) + 2e -  2Na(l)E° = V Anode:2Cl - (l)  Cl 2 (g) + 2e - E° = 1.36 V Net:2NaCl (l)  Cl 2 (g) + 2Na(l)E° =

Electrolysis of aqueous solutions What would happen if you put electricity into a solution of KI? Possible oxidation reactions? Possible reduction reactions? Actual Reaction?

General rules in aqueous solutions Reduction 6H 2 O + 2e -  H 2 (g) + 2OH - E° = V Oxidation 6H 2 O  O 2 (g) + 4H 3 O + + 4e - E° = V

Counting electrons

1.50 amps flow through a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited?

The anode reaction in a lead storage battery is: Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e - If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?