The electrolysis of BiO plus produces pure bismuth How long would it take to produce 10.0 g of Bi by the electrolysis of a BiO plus solution using a current of 25.0 amps?

Answer 1

Farady's Equation for his Law of Electrolysis applies. The law states "The mass of a substance altered at an electrode during electrolysis is directly proportional to the quantity of electricity transferred at that electrode. Quantity of electricity refers to the quantity of electrical charge, typically measured in coulomb."

The equation is m = (Q/F)*(M/z)

where m is the mass of material, Q is the total charge passed through the electrode, F is the Faraday constant (equal to the charge of 1 mole of electrons), M is the molar mass of the substance and z is the charge on the ion to be reduced.

From elementary circuit theory, we know the Q = I*t. That is to say the current the for a given current Q passed through a point for a given time t, the total charge is know.

Substitute I*t for Q and solve for t:

t = mFz/IM

The Faraday constant F is 96485 Coulombs/mol

The Molar Mass of Bismuth, M, is 208.9804 grams/mol

m is given as 10g

I is given as 25amps (25 Coulombs/mol)

We still need z. The oxidation state of oxygen is almost always O2-. So in BiO+, Bi will be Bi3+. Therefor, z = 3.

t = (10g * 96485C/mol * 3) / (25C/sec * 208.9804 g/mol)

t = 554.0328 sec or 550 sec

This answer assumes 100% efficiency in your apparatus and 100% yield, which will never, ever happen.