QUESTION 9 (Start on a new page.) The simplified diagram below represents an electrolytic cell used to purify copper.

Electrolysis is the process of using electrical energy to drive a non-spontaneous chemical reaction.

Electrode T consists of pure copper.
Reason: In electrolytic refining, the pure metal is deposited at the cathode, which is connected to the negative terminal of the power supply (electrode T).

Electrode R is the anode (connected to the positive terminal). At the anode, oxidation occurs. Since R is the impure copper, copper metal from the impure electrode will be oxidized.

The concentration of the electrolyte $\text{CuSO}_4(aq)$ will $\boxed{\text{REMAIN THE SAME}}$.
Reason: At the anode (R), copper metal oxidizes to form $\text{Cu}^{2+}$ ions, increasing the concentration of $\text{Cu}^{2+}$ in the solution. At the cathode (T), $\text{Cu}^{2+}$ ions from the solution are reduced and deposited as copper metal, decreasing the concentration of $\text{Cu}^{2+}$. In an ideal electrolytic refining process, the rate of $\text{Cu}^{2+}$ formation at the anode is equal to the rate of $\text{Cu}^{2+}$ consumption at the cathode, thus maintaining a constant concentration of $\text{Cu}^{2+}$ in the electrolyte.

The presence of zinc ions will $\boxed{\text{NO}}$T influence the quality of the refined copper.
Reason: At the anode, both $\text{Cu}$ and $\text{Zn}$ (impurity) will oxidize to form $\text{Cu}^{2+}$ and $\text{Zn}^{2+}$ ions in the solution. At the cathode, both $\text{Cu}^{2+}$ and $\text{Zn}^{2+}$ ions are present. However, $\text{Cu}^{2+}$ ions are stronger oxidizing agents ($E^\circ = +0.34 \text{ V}$) than $\text{Zn}^{2+}$ ions ($E^\circ = -0.76 \text{ V}$). Therefore, $\text{Cu}^{2+}$ ions will be preferentially reduced and deposited as pure copper on the cathode, while $\text{Zn}^{2+}$ ions will remain in the solution.