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Calculate the number of coulombs required to deposit 40.5 g of Al when the electrode reaction is , `Al^(3+)+3e^(-)rarr Al` |
Answer» Correct Answer - `4.342xx10^(5) C` `Al^(3+)(aq)+underset(3xx96500C)(3e^(-)) to underset(27 g)(Al(s))` To deposit 27 g of Al(s), the amount of charge required `=3xx96500" C"` To deposit, 40.5 g of Al (s), the amount of charge required`=((3xx96500C))/((27 g))xx(40.5 g)=4.342xx10^(5)C`. |
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