Calculate the minimum number of kilowatt-hours of electricity required to produce 100 kg of Al by electrolysis of `Al^(3+)` if the required emf is 4.50 V.

Explanation

The no. of faraday required for electrolysis
= number of eq. of Al deposited
`= (10^(6))/(27//3)F = (10^(6))/(9)F`.
Charge `= (10^(6))/(9) xx 96500` coulombs.
`= 1.07 xx 10^(10)C`.
`therefore` electric energy `= 1.07 xx 10^(10) xx 4.50 J = 4.815 xx 10^(10)J " "(because J = C xx V)`
kilowatt-hours `= (4.815 xx 10^(10))/(3.6 xx 10^(6))kWh " "(because 1 kWh = 3.6 xx 10^(6)J)`
`= 1.34 xx 10^(4) kWh`.