Question

How many seconds are required to deposit 0.271 grams of aluminum metal from a solution that contains al3 ions, if a current of 0.828 a is applied.

Answer 1

Mass of Al atom = 26.982 u

1 u = 1.660E-27 kg

Mass of 3 Al atoms = 80.946 u = 1.344E-25 kg

2.71E-4 / 1.344E-25 = 2.016E21 ions deposited

Since charge on 1 ion is 3 one needs

3 * 2.016E21 = 6.048E21 electrons

6.048E21 * 1.602E-19 = 968.9 Coulombs of charge required

968.9 C / (.828 C / s) = 1170 sec

1170 sec = 19.5 min