During the electrolysis of a salt of a metal M, a current of 0.5 A flows from for 32 minutes 10 seconds and deposits 0.325 g of M, What is the charge of the metal ion? (M = 65, If = 96,500 C per mole of electron)
A. 1 B. 2 C. 3 D. 4
Correct Answer: B
Explanation
Ampere = (Coulomb) x (Sec) x 1930 secs = 965 coulombs (unit of current) But quantity of electricity = current x time = (Coulomb)/(Sec) x time If 0.5 A for 32 mins 10 secs; Quantity of electricity = (0.5 coulomb)/(Sec x 1930) secs = 965 coulombs Since 965 coulombs give 0.325 g M i.e1 coulombs = (0.325)/(965) g ∴96500 coulombs = (0.30)/(965) x 96500 =32.5 g ∴(65 g)/(32.5) = 2 inference: it takes 2 Faraday of electricity to deposit 1 mole of M. Therefore, the change on M ion = 2.
