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How to calculate the no. of electrons required to deposit 40.5 g of Al

How to calculate the no. of electrons required to deposit 40.5 g of Al

Grade:12

1 Answers

varun
304 Points
4 years ago
HELLO NADIA,
 The reaction is,
Al3+        +            3e-   ----------------------->   Al
1mole                    3mole                            1mole
1mole                     3faradays                      1 mole
 
1 mole of Al = 27gm is deposited by 3 faraday of electricity
therefore,
40.5gm of Al is deposited by = 3/27 X 40.5F
 
                                             = 3/27 X 40.5* 96500
                                             = 434250 Coulombs

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