1. Current of 0.5 A passed through cell for 2 h with CuSO4
a. anode reaction : Cu(s) ---> Cu2+(aq) + 2e-
b. Coulombs = I x t = 0.5 x 2 x 60 x 60 = 3600 C
c. Faraday generated = 3600/96485 = 0.0373 Faradays
d. moles of electrons = 0.0373 x 2 = 0.0746 mol e-
e. change in mass at cathode = 0.0746 x 63.546 = 4.740 g
A current of 0.50 amperes flowed through a cell for two hours in which CuSO_4 is...
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