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` .

What is the amount of `Al` deposited on the electrolysis of molten `Al

1.	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` .
Answer» 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

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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` .

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