Question

Calculate the number of coulombs required to electroplate 4.75 g of aluminium when electrode reaction is \[\ce{Al^{3+} + 3e- -> Al}\].

(Given: Atomic weight of Al = 27 g mol⁻¹, 1 Faraday = 96,500 coulombs)

Answer 1

Given: Atomic weight of Al = 27 g/mol

1 Faraday = 96,500 coulombs

Electrode reaction: \[\ce{Al^{3+} + 3e- -> Al}\]

This means 3 Faradays of charge are required to deposit 1 mole (27 g) of Al.

Now, we will calculate the moles of Al to be deposited:

Moles of Al = `"Mass of Al deposited"/"Atomic weight of Al"`

= `4.75/27`

= 0.1759 moles

Since 3 Faradays are required to deposit 1 mole of Al, the charge required for 0.1759 moles is:

Charge required = 0.1759 × 96500

= 50930.56 C

Thus, the number of coulombs required is approximately 50,930.56 C.