calculate the number of coulombs required to deposit 40.5g of Al when the electrode reaction is: Al3+ +3e- -> Al

Asked by shivangi sachan | 1st Jul, 2010, 12:00: AM

Expert Answer:

The reaction is:

Al3+             +            3e- -------------->  Al

1mole                3 mole                 1 mole

1 mole             3 Faradays             1 mole

 

1 mole of Al =  27 g is deposited by 3 faraday of electricity

40.5 g of Al is deposited by = (3/27) *40.5 F

            = 3/27 *40.5 *96500 Coulombs

     =  434250 Coulombs

Answered by  | 17th Jul, 2010, 10:57: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.