This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Here are the solutions to Question 9: 9.1 Electrolysis is the process of using electrical energy to drive a non-spontaneous chemical reaction. 9.2 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). 9.3 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. Cu(s) Cu^2+(aq) + 2e^- 9.4 The concentration of the electrolyte CuSO_4(aq) will REMAIN THE SAME. Reason: At the anode (R), copper metal oxidizes to form Cu^2+ ions, increasing the concentration of Cu^2+ in the solution. At the cathode (T), Cu^2+ ions from the solution are reduced and deposited as copper metal, decreasing the concentration of Cu^2+. In an ideal electrolytic refining process, the rate of Cu^2+ formation at the anode is equal to the rate of Cu^2+ consumption at the cathode, thus maintaining a constant concentration of Cu^2+ in the electrolyte. 9.5 The presence of zinc ions will NOT influence the quality of the refined copper. Reason: At the anode, both Cu and Zn (impurity) will oxidize to form Cu^2+ and Zn^2+ ions in the solution. At the cathode, both Cu^2+ and Zn^2+ ions are present. However, Cu^2+ ions are stronger oxidizing agents (E^ = +0.34 V) than Zn^2+ ions (E^ = -0.76 V). Therefore, Cu^2+ ions will be preferentially reduced and deposited as pure copper on the cathode, while Zn^2+ ions will remain in the solution. 9.6 Step 1: Convert time to seconds. t = 30 minutes × 60 s/minute = 1800 s Step 2: Calculate the moles of copper deposited. The molar mass of copper (Cu) is 63.55 g/mol. n_Cu = massmolar mass = 1.72 g63.55 g/mol ≈ 0.027065 mol Step 3: Determine the moles of electrons transferred. The half-reaction at the cathode is Cu^2+(aq) + 2e^- Cu(s). From the stoichiometry, 2 mol of electrons are required for every 1 mol of Cu deposited. n_e^- = n_Cu × 2 = 0.027065 mol × 2 = 0.05413 mol e^- Step 4: Calculate the total charge (Q) transferred. Using Faraday's constant (F = 96485 C/mol e^-): Q = n_e^- × F = 0.05413 mol e^- × 96485 C/mol e^- ≈ 5222.8 C Step 5: Calculate the current (I) using the formula Q = I × t. I = (Q)/(t) = 5222.8 C1800 s ≈ 2.9016 A The reading on the ammeter is 2.90 A. 3 done, 2 left today. You're making progress.