Electrolysis nonspontaneous reaction is caused by the passage
Electrolysis • non-spontaneous reaction is caused by the passage of an electric current through a solution
Electrolysis of Sodium chloride (chlor-alkali process) molten reactants => liquid sodium and chlorine gas aqueous reactants => caustic soda (sodium hydroxide) and chlorine gas
Electrolysis Preparation of Aluminum (Hall process)
Electrolytic Refining of Copper Cu(s) + Cu+2(aq) --> Cu+2(aq) + Cu(s) impure anode cathode impurities: anode mud; Ag, Au, Pb
Quantitative Aspects of Electrolysis • 1 coulomb = 1 amp sec • 1 mole e- = 96, 500 coulombs
Electroplating EXAMPLE: How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
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 = ------
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)
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)
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)
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)
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-)
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
Corrosion O 2(g) + 4 H+(aq) + 4 e- -----> 2 H 2 O(l) Eo = 1. 23 V Rusting Fe(s) -----> Fe+2(aq) + 2 e- Eo = 0. 44 V O 2(g) + 4 H+(aq) + 4 e- -----> 2 H 2 O(l) Eo = 1. 23 V ---------------------- 2 Fe(s) + O 2(g) + 4 H+(aq) -----> 2 H 2 O(l) + Fe+2(aq) Eo = 1. 67 V
Preventing Corrosion painting galvanizing sacrificial anode
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