1.

To deposit 1 mol of aluminium from molten `Al_(2)O_(3)`. What is the amount of electricity (in coulombs) required ?

Answer» Correct Answer - `2.895xx10^(5) C`
`Al_(2)O_(3)(l) to 2Al(s)+3//2O_(2)(g)`
The ionic equation for the reaction is :
`Al^(3+)(aq)+underset(3xx96500C)(3e^(-)) to underset(1" mol")Al(s)`
`=3xx96500C=2.895xx10^(65)" C"`.


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