Question

Calculate the number of electrons required to deposit 40.5g of Al

Answer 1

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

3.4

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Comments Report

jatinpanchal098 Helping Hand

Moles of al= 40.5/27= 1.5 (approx.)

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

 

1mole            3F                    1mole

Now, 3×1.5=4.5mol

Answer 2

Answer:

Number of electrodes required to deposit 40.5 of Al are 4.5 mol