Question

What mass of copper could be deposited from a copper(ll) sulfate solution using a current of 0.50 A over 100 seconds?

Answer 1

Current: I = 0.50 A

Time: t = 10O seconds

F = 96,500 C mol^-1

But quantity of electricity is Q = I x t

  I = 0.50 A

  t = 10 seconds

Q = 0.50 × 100= 50 C

Moles of electrons given = Q/F

Moles = 50/96,500 C mol^-1

Moles = 5.18 x 10^-4 mol

Cu²⁺ + 2e^- → Cu(s)

but I mol of Cu deposited 2 moles of electrons

therefore, 1/2 x 5.18 x 10^-4 mol = 2.59 x 10^-4 mol

molar mass (Cu) = 63.55 g mol^-1 and moles of Cu is 2.59 x 10^-4 mol

and mass = moles x molar mass

63.55 g mol^-1 x 2.59 x 10^-4 mol

mass of Cu = 1.65 x 10^-3 g or 1.65 mg