Industrially, copper metal can be purified electrolytically according to the following arrangement. The anode is made of the impure Cu electrode and the cathode is the pure Cu electrode. The electrodes are immersed in a \(\mathrm{CuSO}_{4}\) solution. (a) Write the half-cell reactions at the electrodes. (b) Calculate the mass (in grams) of Cu purified after passing a current of 20 A for \(10 \mathrm{~h}\). (c) Explain why impurities such as \(\mathrm{Zn}, \mathrm{Fe}, \mathrm{Au},\) and \(\mathrm{Ag}\) are not deposited at the electrodes.

In the electrolytic purification of copper, half-cell reactions play a crucial role. During this process, the anode is composed of impure copper which undergoes oxidation. Oxidation is the loss of electrons, and in this case, copper atoms lose two electrons to form copper ions (\text{Cu}^{2+}) that dissolve into the solution. This can be written as:
\[\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-\].
At the cathode, reduction takes place, which means gaining electrons. The Cu2+ ions in the solution are attracted to the cathode and gain electrons to become pure copper metal. This is expressed as:
\[\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\].
Having a clear understanding of these reactions helps one to grasp why and how the purification occurs.

Faraday's laws of electrolysis quantify the relationship between the amount of electrical charge passed through an electrolyte and the quantity of substance that undergoes oxidation or reduction at each electrode. The first Faraday's law states that the amount of chemical change is proportional to the electric charge passed through the electrolyte.
To calculate the mass of copper purified, one must first calculate the total charge (\text{Q}) that has passed through the system using the equation \text{Q} = \text{current} (\text{i}) \times \text{time} (\text{t}). Subsequently, the number of moles of electrons can be determined by dividing the total charge by the charge of one mole of electrons, known as a Faraday (\text{F} = 96485 \text{C}):\[\text{moles of e}^- = \frac{\text{Q}}{\text{F}}\].
Each copper ion requires two electrons (2 \text{e}^-) to be reduced to a copper atom. Thus, the amount of copper deposited can be deduced by halving the moles of electrons. This demonstrates how Faraday's laws are applied to obtain a quantitative measure of electrolytic purification.

The electrochemical series is a ranking of elements based on their standard electrode potentials. In the context of electrolyting purification, this ranking is critical in determining which metal will be reduced at the cathode. Copper is higher in the electrochemical series than elements such as zinc (Zn), iron (Fe), gold (Au), and silver (Ag), meaning it has a greater tendency to be reduced from its ionic state to the metallic state.
In an electrolyte solution containing various metal ions, those with higher electrode potentials are reduced preferentially. Since copper ions have a higher electrode potential compared to the impurities present, copper is the metal that gets plated onto the cathode. Understanding the electrochemical series explains why, when purifying copper, the desired metal is deposited at the cathode, leaving impurities behind in the solution.

During electrolysis, impurities are not deposited at the electrodes mainly due to their position in the electrochemical series. Elements such as Zn, Fe, Au, and Ag have different electrode potentials when compared to Cu. Copper, having a relatively mid-level electrode potential, will be reduced from Cu2+ to Cu before the other impurities are reduced. This implies that impurity metals with lower electrode potentials than copper will remain in the solution as ions.
For instance, zinc and iron have lower electrode potentials and require a higher reduction potential, which is not achieved under the conditions set for copper purification. Conversely, noble metals like gold (Au) and silver (Ag), although having higher electrode potentials, do not deposit because their ions are present in very low concentrations and the electrolyte conditions are not optimized for their deposition.
This selective deposition ensures that the copper obtained at the cathode is of high purity, with impurities remaining dissolved or settling as anode sludge.