Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic c...

Answer Details

M = $\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}$

= $\frac{MmIT}{96500n}$

Copper II Chloride = CuCl2

CuCl2 → Cu2+ + 2Cl2

Mass of compound deposited = $\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}$

Q = IT

I = 0.5A

T = 45 × 60

T = 2700s

Q = 0.5 × 2700

= 1350c

Molarmass = 64gmol-1

no of charge = + 2

Mass = $\frac{64\times 1350}{96500\times 2}$

Mass = 0.448g