Question

A current of 156A is applied to an electrolytic cell containing molten CuCl2 for 45 minutes. What substance and how many grams of it will form on the cathode?

Answer 1

Cu^2+ + 2e ==> Cu...........This is gain of electrons; therefore, reduction.
2Cl^- ==> Cl2 + 2e............This is a loss of electrons; therefore, oxidation.
Oxidation occurs at the anode; reduction occurs at the cathod. Therefore, we want the Cu half reaction.
How many coulombs do we generate? That's
Coulombs = amperes x seconds = 146 A x 45 min x (60 sec/min) = 394,200.
1 Faraday or 96,485 coulombs will deposit 63.54 g/2 = 31.77 g Cu metal.
So you can deposit 31.77 g Cu x (394,200/96,485) = ? g Cu