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