Question

Copper is being electrodeposited from a CuSO\(_4\) bath onto a stainless steel cathode of total surface area of 2 m\(^2\) in an electrolytic cell operated at a current density of 200 A m\(^{-2}\) with a current efficiency of 90%. The mass of copper deposited in 24 h is _________ kg (rounded off to two decimal places). Given: Faraday's constant = 96500 C mol\(^{-1}\), Atomic mass of copper = 63.5 g mol\(^{-1}\).

Hint:

When calculating the mass of an element deposited by electrolysis, use Faraday's law: \( {mass} = \frac{Q}{nF} \times M \), where \( M \) is the molar mass and \( n \) is the number of electrons involved in the reaction. Don't forget to apply the current efficiency to get the actual mass deposited.

Answer 1

First, we need to calculate the total charge passed through the electrolyte: \[ {Current density} = 200 \, {A/m}^2, \quad {Area} = 2 \, {m}^2 \] The total current \( I \) is: \[ I = 200 \times 2 = 400 \, {A} \] Next, we calculate the total charge passed in 24 hours: \[ Q = I \times t = 400 \times (24 \times 3600) = 34,560,000 \, {C} \] Now, using Faraday's law, the amount of substance deposited can be calculated. For copper, the number of moles of copper deposited is given by: \[ {moles of Cu} = \frac{Q}{nF} \] where \( n = 2 \) (since copper undergoes a 2-electron transfer), and \( F = 96500 \, {C/mol} \). So: \[ {moles of Cu} = \frac{34,560,000}{2 \times 96500} = 179.8 \, {mol} \] The mass of copper deposited is: \[ {mass of Cu} = 179.8 \times 63.5 = 11,419.3 \, {g} = 11.42 \, {kg} \] Since the current efficiency is 90%, we adjust the mass accordingly: \[ {mass deposited} = 11.42 \times 0.90 = 10.28 \, {kg} \] Thus, the mass of copper deposited is approximately 10.20 kg to 10.30 kg.