Question

For the above current at 4.24 V, how much electrical energy (in kw-hr) is consumed in the production of 1 kg of aluminum? (The electrical energy or work =qV and 1 kW-hr is 3.6 x106 J.)

Answer 1

mass of aluminium = 1 kg = 1000 g

moles of Al = 1000 / 27 = 37.04

1 mol Al gives -------> 3 mol of e-

37.03 mol Al   --------------> ??

moles of e- = 37.04 x 3 = 111.1 mol

1 mol e- carries -----------> 96500 C charge

111.1 mol e- -----------> ??

total charge carried = 111.1 x 96500

                                = 1.07 x 10^7 C

Energy required = 1.07 x 10^7 x 4.24

                          = 4.55 x 10^7 J

1 Kw-hr = 3.6 x 10^6 J

Energy required = 12.6 kw-hr