How long does it take to pass a current through a solution of copper sulfate so that 2.4 g

How long does it take to pass a current through a solution of copper sulfate so that 2.4 g of copper is released at the cathode with a current strength of 2A?

To determine the duration of the current passing through the solution of the taken copper sulfate, we use the first Faraday law: m = k * I * t, whence we express: t = m / (k * I).

Constants and variables: m is the mass of copper precipitated at the cathode (m = 2.4 g = 2.4 * 10 ^ -3 kg); k is the electrochemical equivalent of copper (k = 32.9 * 10 ^ -8 kg / C); I is the strength of the transmitted current (I = 2 A).

Calculation: t = m / (k * I) = 2.4 * 10 ^ -3 / (32.9 * 10 ^ -8 * 2) ≈ 3647 s ≈ 1 h.

Answer: The current must be passed for 1 hour.