`Al_2 O_3 ` is reduced by electrolysis at ow potentials and high currents, If `4.0 xx 10^(4)` amperes of current is passed through molten `Al_2O_3` fro `6` hours , what mass of alumininum is produced ? (Assume ` 100%` current efficiency gt At mass of ` Al = 27 g ` "mol"^(-) )`.

Options

A. `8.1xx10^(4)g`
B. `2.4xx10^(5)g`
C. `1.3xx10^(4)g`
D. `9.0xx10^(3)g`

Answer

A

Explanation

`Al^(3+)+3e^(-)toAl`
Quantity of electricity passed
`=(4.0xx10^(4)A)xx(6xx60xx60s)`
`=864xx10^(6)`coulombs
3F, i.e., `3xx96500C` produce `Al=1` mole, i.e., 27g
`therefore864xx10^(6)C` will produce Al ltbRgt `=(27)/(3xx96500)xx864xx10^(6)g`
`=8.058xx10^(4)g=8.1xx10^(4)g`.