Question

A.) How many seconds are required to deposit 0.247 grams of magnesium metal from a solution that contains Mg²⁺ ions, if a current of 0.840 A is applied.

B.) How many amperes are required to deposit 0.120 grams of aluminum metal in 732 seconds, from a solution that contains Al³⁺ ions .

Answer 1

(A) According to Faraday’s law ,W = (ECt) / 96500

Where W = mass of metal deposited = 0.247 g

           E = Equivalent weight of Mg = molar mass / valence

              = 24.3 / 2

              = 12.15

           C = current = 0.840 A

           t = time taken = ?

Plug the values we get t = 96500 W / EC

                                   = 2335 s

(B) According to Faraday’s law ,W = (ECt) / 96500

Where W = mass of metal deposited = 0.120 g

           E = Equivalent weight of Al = Molar mass / valence

              = 27.0 / 3

              = 9.0

           C = current = ?

           t = time taken = 732 s

Plug the values we get C = (96500xW ) / Et

                                     = ( 96500x0.120) / (9.0x732)

                                     = 1.758 A