1. FARADAY’S LAW

2. NaCl (s) → Na + (l) + Cl – (l) E° R = -2.71 V E° O = -1.36 V 2 Cl - (l) → Cl 2(g) + 2e - 2 Na + (l) + 2e - → 2 Na (s) Electrolytic cell Using electricity to force a nonspontaneous rxn. Electrodes are oppositely charged versus voltaic cells.

3. I is the current in amperes (A) Q is the charge transferred in coulombs (C) t is the time in seconds (s) 1 coulomb is BILLIONS of electrons worth of charge. Current (I) – defined as 1 coulomb of charge flowing past a point in one second. I = Q t

4. Faraday (1791-1867) Coined terms "electrode", "anode", "cathode", "electrolyte", "ion", "anion" and "cation". Amount of product formed at each electrode is directly proportional to the amount of current passed through the cell. Faraday (F) is the amount of charge carried by one mole of electrons.

5. 1 e - has a charge of 1.60 x 10 –19 C Therefore: 1 mole e - = 96 500 C Faraday’s Constant: 1 F = 96 500 C/mol e – Coefficients mean moles or particles: Na + + 1e – → Na 1 F will reduce a mole of sodium. Al 3+ + 3e – → Al 1 F will reduce a third of a mole of aluminum.

6. Q · 1 mol e – 96 500 C Q = I · t I = Q t Rearrange for Charge Multiply by Faraday’s Constant = mol e - Solve for Charge

7. Determine the number of moles of electrons supplied by a battery with a current of 0.100 A for 50.0 min. Q =I · t = 0.00311 mol e - 300 C · 1 mol e – 96 500 C 0.100 · (50min x 60 sec/min) Q = 300 C

8. Calculate a mass of aluminum produced by 7.50 A passing through molten aluminum oxide for 6 hours, 20 minutes and 10 seconds. Al 3+ + 3e – → Al Q =I · t = 1.77 mol e - 1.71 x 10 5 C· 1 mol e – 96 500 C 7.50 A · (22810 s) Q = 1.71 x 10 5 C

9. 1 mol Al 27.0 g Al 1.77 mol e - 3 mol e - 1 mol Al = 15.9 g Al produced Al 3+ + 3e – → Al Every other question type is a variation of this procedure – some backwards, some with a different missing variables but the same basic steps and overall approach.