Calculate the no. Of coulpmb required to deposit 40.5g AL when the electrode rxn is  AL3+ +3e--->AL

 

Asked by Harshfarwaha | 23rd Jul, 2020, 03:27: PM

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 space straight g space of space Al space is space deposited space by space equals space 3 over 27 cross times 40.5 space straight F space space space space space space space space space
equals space space 3 over 27 cross times 40.5 cross times 96500 space Coulombs space space space space space space space
equals space space 434250 space equals space 4.342 space cross times 10 to the power of 5 space Coulombs

Answered by Ramandeep | 23rd Jul, 2020, 06:08: PM