Question

I could not find the info to update so here is different question:

Aluminum is produced by electrolysis of molten Al2O3-Na3AlF6 mixture. In a production process of aluminum, electrolytic cells operating at 4.5 V and 5.0 x 105 A were used. (a) How many kilograms of Al can be produced in 1.00 hour of electrolysis? (b) How much energy (in kWh) is consumed during this operation? (1 kWh = 3.6 x 106 J)

Answer 1

Coulombs of electric charge passed in electrolytic cell = ampheres*seconds

= 5.0*10⁵ A * 60*60(secs /min)

= 1.8*10⁹ C

Number of faradays of electrical charge passing into cell = Moles of electron

= 1.8*10⁹ C * (1mole^- / 96,485 )

= 1.87*10⁴ mol e^-

Half reaction of reduction of Al 3+

Al³⁺ + 3e^- -----> Al

Thus 3 F of electrical charge is required for the formation of Al

Numb of moles = 1.87*10⁴ mol e^- (1mol Al / 3 mol e^-)

= 6233.3 mol Al

Converting moles to grams = 6233.3 * (26.98 g Al / 1mol Al )

where 26.98 is the std atomic weight of Al

= 1.68 * 10⁵ g Al

=1.68*10² kg Al

The reduction uses cell with voltage 4.5 V , the energy thus consumed = 4.5V * 1.8*10⁹C

= 8.1*10⁹  CV = 8100*10⁶ / 3.6*10⁶

Energy = 2250kWh