Chemistry help please?

Question

Calculate the number of kilowatt hours of electricity required to produce 1.0•10^3kg (1metric ton) of aluminum by electrolysis of AL^3+ if the applied voltage is 4.60 V and the process is 48% efficient.

Answer 1

This is a fun example of a Faraday's calculation with lots of little tricky bits in there! There is an equation that CAN help with these calculations, but with all the potential variations, I find it easier to use dimensional analysis.

First, determine the energy required to complete the reaction given starting with the mass of the Aluminum plated.

convert kg->g g-> mol mol->mol e- faraday const. voltage (V= J/c)

1.0E3 kg x 1000 g x 1 mol Al x 3 mole e- x 96485 C x 4.60 J = 4.935E10 J

1 kg 26.98g 1 mol Al 1 mol e-

Then, convert J to KwH

1 J = 1 Watt second

4.935E10 Watt second x 1 kW x 1 min x 1 hour = 13708 J (if it was 100% efficient)

1000 W 60 sec 60 min

Finally, since it is only 48% efficient, calculate the KwH required to complete the reaction

13708 kWH = X * 0.48

X = 28559 KwH --> 29000 KwH with significant figures

This could all be done in one big problem, but I thought breaking it down might help!