Question

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

Answer 1

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

Answer 2

Answer:

The electrons required to deposit $40.5$g of Al is  $434250$ Coulombs

Explanation:

• For getting how many electrons are required for $40.5$ g of Aluminium we need to see the reaction for depositing 1 mole of aluminium.
  $Al^3^+$  +  $3e^-$   --->   $Al$

• So for 1mole of aluminium 3moles of electrons are required which means for depositing 1mole of aluminium 3 faraday charge of electron is required.
• Now as we know 1 mole aluminium has 27 gm of weight so for 40.5 gm we require = $3 * \frac{40.5}{27} * 96500$ coulombs ( as 1 faraday =  $96500$coulombs )
• Hence we require $434250$ coulombs of electrons