1. Electrolysis
non-spontaneous reaction is caused by the passage of an electric current through a solution

3. Electrolysis Electrolysis of Sodium chloride (chlor-alkali process)
molten reactants => liquid sodium and chlorine gas
aqueous reactants => caustic soda (sodium hydroxide) and chlorine gas

4. Electrolysis
Preparation of Aluminum (Hall process)

5. Electrolytic Refining of Copper
Cu(s) + Cu+2(aq) --> Cu+2(aq) + Cu(s)
impure pure
anode cathode
impurities: anode mud; Ag, Au, Pb

7. Quantitative Aspects of Electrolysis
1 coulomb = 1 amp sec
1 mole e- = 96,500 coulombs

8. Electroplating
EXAMPLE: How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?

9. Electroplating
EXAMPLE: How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
(45 min)
#g Cr =

10. Electroplating
EXAMPLE: How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
definition of minute
(45 min)(60 sec)
#g Cr =
(1 min)

11. Electroplating
EXAMPLE: How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
(45) (60 sec) (25 amp)
#g Cr =
(1)

12. Electroplating
EXAMPLE: How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
definition of a coulomb
(45)(60 sec)(25 amp)(1 C)
#g Cr =
(1) (1 amp sec)

13. Electroplating
EXAMPLE: How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
Faraday’s constant
(45)(25)(60)(1 C)(1 mol e-)
#g Cr =
(1)(1)(96,500 C)

14. Electroplating
EXAMPLE: How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
atomic weight
(45)(60)(25)(1)(1 mol e-)(52 g Cr)
#g Cr =
(1)(1)(96,500) (6 mol e-)

15. Electroplating
EXAMPLE: How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
(45)(60)(25)(1)(1 mol e-)(52 g Cr)
#g Cr =
(1)(1)(96,500)(6 mol e-)
= 58 g Cr

17. O2(g) + 4 H+(aq) + 4 e- -----> 2 H2O(l)
Corrosion
O2(g) H+(aq) e > 2 H2O(l)
Eo = 1.23 V
Rusting
Fe(s) > Fe+2(aq) e- Eo = 0.44 V
O2(g) H+(aq) e > 2 H2O(l) Eo = 1.23 V
2 Fe(s) + O2(g) H+(aq) >
2 H2O(l) + Fe+2(aq) Eo = 1.67 V

18. Preventing Corrosion
painting
galvanizing
sacrificial anode