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Title: What time in minutes is required to plate 2.08 grams of copper at a constant cureent flow of 1.26 Amps? Post by: lemonsplk on Mar 1, 2013 Cu2+ + 2e- == Cu
Cu = 63.9 grams/mol 1amp= 1 coulomb/sec 1 faraday = 96,485 coulombs Title: What time in minutes is required to plate 2.08 grams of copper at a constant cureent flow of 1.26 Amps? Post by: micheleann on Mar 1, 2013 Reduction equation:
Cu^2+(aq) + 2e- -------> Cu(s) Mole of Cu to be plated: 2.08 g / 63.9 g/mol = 0.0325 mol To reduce 1 mol Cu^2+ , 2 moles of electron are rquired. To reduce 0.0325 mol Cu^2+, 2x0.0325 = 0.065 mol electron is required. 1 mol e- carries a charge of 1 faraday, or 96485 coulombs. 0.065 mol e- carries 0.065 x 96485 = 6271.5 coulombs Ampere = Coulomb / sec Second = Coulomb / Ampere = 6271.5 / 1.26 = 4977.4 sec 4977.4 sec / 60 sec/min = 82.95 or 83 min |