Question

What mass of aluminum will be deposited on the cathode if an electric current of 0.15 a is run for 4 5s when a solution of al(no3 )3 is subjected to electrolysis?

Answer 1

Mass of Al, m= ?

Molar mass of copper Al, M = 27.0 g/mol

time, t = 45 s

current , c = 0.15 A

valency of Al in Al(NO₃)₃ , z = +3

From Faraday's 1st law,

m = Mct/zF

where F = Faraday constant = 96500 C/mol

Substitute all the values,

m = [(27.0 g/mol) (0.15 A) (45 s)]/ [3] [96500 C/mol]

= 6.3 x 10⁻⁴ g

Mass of aluminium = 6.3 x 10⁻⁴ g

Therefore, mass of aluminium plated on electrode =  6.3 x 10⁻⁴ g