Question

If an electric current of 6.00 A flows for 3.75 hours through an electrolytic cell containing copper-sulfate (CuSO4) solution, then how much copper is deposited on the cathode (the negative electrode) of the cell? (Copper ions carry two units of positive elementary charge, and the atomic mass of copper is 63.5 g/mol.)

Answer 1

Improved version

Total charge (Q) passed = 6.00 X 3.75 X 3600 Coulombs

For deposition of one atom, the no; of electrons to pass through the solution is equal to the number of charges on the cation in our case Cu atom has +2 charge so it needs 2 electrons for the deposition of one copper atom.

For deposition of one mole of the copper atoms we nee 2x6.022×10²³ electrons (two for charge and 6.022x10²³ for 1mole that is 6.022x10²³ atoms )

The charge required to pass to deposit one mole of copper atoms is

2x(charge on one electron)x(6.022x10²³) or

2x1.60217662x10^-19x6.02214179x10²³ = 2x96485 C where 96485 is the amount of electric charge on 1 mole of electrons.

So the number of moles deposited is when a charge of 81000C passes is

81000/(2x96485) = 0.91975moles

mass of copper deposited is given by

(number of moles deposited)x(mass of one mole of the copper atom) = 0.91975x63.5 g = 26.6544g