Question

The number of coulombs required to deposit
5.4g of Al when the electrode reaction is-
A13+ + 3e- → Al
(1) 1.83 x 10$C (2) 57900C
(3) 5.86 x 105C (4) None of the above

Answer 1

Answer:

There are three electrons required to reduce 1mole of Al3+ to Al,

Thus Coulombs required will be 3*F = 3*96500