What is the amount of `Al` deposited on the electrolysis of molten `Al_(2)O_(3)` when a current of `9.65 A` is passed for `10.0 s` .

Explanation

Molten `Al_(2)O_(3)overset(El e trolysis)rarr 2Al^(3)+3O^(2-)`
Number of Faradays `=(Ixxts)/(96500C)`
`=(9.65Axx10s)/(96500C)=10^(-3)F`
First method
Reduction of `Al^(3+) ` at cathode
`Al^(3+)+3e^(-) rarr Al`
`3e^(-)=3F=1 mol Al=27g Al`
`:. 10^(-3)F=(27)/(3)xx10^(-3)g`
`=0.009g Al ` deposited
Second method
`1F=1` equivalent of `Al=(1 mol)/(Charg e)=(27g)/(3)=9g`
`:. 10^(-3)F=9 xx 10^(-3)g Al` deposited