Aluminium oxide may be electrolysed at `1000^(@)`C to furnish aluminium metal (Atomic mass = 27 amu, 1 Faraday = 96500 Coulomb). The cathode reaction is `Al^(3+) + 3e^(-) rightarrow Al`. To prepare 5.12 kg of aluminium metal by this method would require:

Options

A. `5.49xx10^(1)C` of electricity
B. `5.49xx10^(4)C` of electricity
C. `1.83xx10^(7)C` o electricity
D. `5.49xx10^(7)C` of electricity

Answer

D

Explanation

`Al^(3+)+3e^(-)toAl`
w=zQ
where w=amount of metal
=5.12 kg
`=5.12xx10^(3)g`
z=electrochemical equivalent
`=("equivalent weight")/(96500)`
`=("atomic mass")/("electrons"xx96500)`
`=(27)/(3xx96500)`
`5.12xx10^(3)=(27)/(3xx96500)xxQ`
`Q=(5.12xx10^(3)xx3xx96500)/(27)C`
`=5.49xx10^(7)C`