Grade 12Inorganic Chemistry
How to calculate the no. of electrons required to deposit 40.5 g of Al
1 Answer
varun
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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