Industrially, copper is purified by electrolysis. The impure copper acts as the anode, and the cathode is made of pure copper. The electrodes are immersed in a \(\mathrm{CuSO}_{4}\) solution. During electrolysis, copper at the anode enters the solution as \(\mathrm{Cu}^{2+}\) while \(\mathrm{Cu}^{2+}\) ions are reduced at the cathode. (a) Write half-cell reactions and the overall reaction for the electrolytic process. (b) Suppose the anode was contaminated with \(\mathrm{Zn}\) and \(\mathrm{Ag} .\) Explain what happens to these impurities during electrolysis. (c) How many hours will it take to obtain \(1.00 \mathrm{~kg}\) of \(\mathrm{Cu}\) at a current of \(18.9 \mathrm{~A} ?\)

The half-cell reactions for electrolysis of copper are: Anode: \( \mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2\mathrm{e}^{-} \) Cathode: \( \mathrm{Cu}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cu} \) The overall reaction can be gotten by adding the half-cell reactions: \( \mathrm{Cu} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Cu}^{2+} + \mathrm{Cu} \) This simplifies to: \( \mathrm{Cu} \rightarrow \mathrm{Cu} \)

During electrolysis, the anodic impurities (\(\mathrm{Zn}\) and \(\mathrm{Ag}\)) remain in the solution and do not form ions. This is because the standard electrode potentials of \(\mathrm{Zn}\) and \(\mathrm{Ag}\) are both less than \(\mathrm{Cu}\), which means that they are less likely to dissolve into the solution as ions. Hence, they will be left at the bottom of the electrolytic cell.

The amount of substance deposited on the cathode during electrolysis is directly proportional to the quantity of electricity passed (Faraday's law). We know that current = charge/time. Given: - Mass of copper (\(\mathrm{Cu}\)) = 1 kg = 1000 grams - Atomic weight of (\(\mathrm{Cu}\)) = 63.5 grams/mole - Quantity of charge required to deposit one mole of \(\mathrm{Cu}\) = 2 F (Faraday's constant = 96485 Coulombs/mole of electrons) 1. We will first calculate the number of moles in 1 kg of \(\mathrm{Cu}\). This is given by \(\frac{mass}{atomic weight} = \frac{1000}{63.5}\). 2. Then, calculate the total charge \(\= \) moles \(\times\) Faraday constant. 3. Then divide the result by the current to get the time. The final result will be in seconds, convert it to hours by dividing by 3600.