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 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 halfcell 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 \(18.9 \mathrm{~A} ?\)

Short Answer

Expert verified
\[\begin{align*}& \text{(a)} \quad &\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)} \& &\mathrm{Cu(s)\rightarrow Cu^{2+}(aq)+2e^{-}} \& &\mathrm{Cu(s) + Cu^{2+}(aq) \rightarrow Cu^{2+}(aq) + Cu(s)} \& \text{(b)} \quad &\text{Silver will not react and fall to the bottom of the cell, while zinc will dissolve.} \& \text{(c)} \quad &\text{Use Faraday's formula to find the time, applying the given values.}\end{align*}\]

Step by step solution

01

Writing Halfcell and Overall Reactions

In the CuSO4 electrolyte solution, the Cu^2+ ions are reduced at the cathode, giving us a half-reaction of: \[\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}\]At the anode (impure copper), copper atoms are oxidized to copper ions, and we can write:\[\mathrm{Cu(s)\rightarrow Cu^{2+}(aq)+2e^{-}}\]Adding these two half-cell reactions yields the overall reaction for the electrolytic process, which is:\[\mathrm{Cu(s) + Cu^{2+}(aq) \rightarrow Cu^{2+}(aq) + Cu(s)}\]
02

Behavior of Zn and Ag during Electrolysis

During electrolysis, 'nobler' metals like silver (Ag) would remain unreacted and would fall to the bottom of the electrochemical cell. In contrast, metals more 'reactive' than copper, such as zinc (Zn), would dissolve into the solution along with copper.
03

Calculation of Electrolysis Time

We use Faraday’s laws of electrolysis, in which the mass of substance deposited is directly proportional to the quantity of electricity. The molar mass of copper, \(M_{\mathrm{Cu}}\), is 63.55 g/mol. The Faraday constant, \(F\), is 96485 C/mol. Therefore, to deposit 1kg (or 1000 g) of copper, the total charge needed (\(Q\)) is:\[Q=\frac{1000 \, \mathrm{g}}{63.55 \, \mathrm{g/mol}} \times 2 \, \mathrm{F}\]Then, the time (\(t\)) required (at a current (\(I\)) of 18.9 A), can be calculated using:\[t=\frac{Q}{I}\]

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