Question

A 1.0L solution of BeF2 was electrolyzed for 18.8 hours to give 20.01g of beryllium. Assuming the minimum voltage needed was available, what amperage would be needed to compare the electrolysis in the given time?

Answer 1

Be²⁺ + 2e^- Be

Molar mass of Be = 9.01 g/mol

Mass of Be produced = 20.01 g

Moles of Be produced = 20.01/9.01 = 2.22 mol

Producing 1 mol Be requires 2 mol of electrons

Therefore, producing 2.22 mol Be requires = 2 x 2.22 = 4.44 mol of electrons

Charge of 1 mol of electron = 96485 coulomb

Therefore, the charge of 4.44 mol of electron = 96485 x 4.44 = 428393.4 coulomb

Now,

current x time = charge

time = 18.8 hours = 18.8 x 3600 s = 67680 s

Thus,

current = 428393.4/67680 = 6.330 ampere

Therefore the amperage would be needed = 6.330 ampere