Question

How many grams of cadmium are deposited from an aqueous solution of cadmium sulfate, CdSO₄, when an electric current of 1.64 A flows through the solution for 240 min?

Answer 1

Charge, Q = It

I = current = 1.64 A

t = 240 min = 240 min x 60 s / 1 min = 14400 s

Thus, total charge transported by the current = 1.64 A x 14400 s = 23616 C

Charge of 1 mol of electron = 1 Faraday = 96500 C

Now,

Cd²⁺(aq) + 2e^- Cd(s)

Thus, 2 mol of electrons or 2 x 96500 C of electron is required to deposit 1 mol of Cd or 112.411 g of Cd

Therefore, 23616 C will deposit = (112.411 g x 23616 C)/(2 x 96500 C) = 13.8 g of Cd

Therefore, the grams of cadmium are deposited from an aqueous solution of cadmium sulfate = 13.8 g