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.
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Here are the answers for question 38: a) Step 1: Identify the species present in the aqueous copper(II) sulphate solution. The solution contains Cu^2+(aq), SO_4^2-(aq), H^+(aq), and OH^-(aq) (from the dissociation of water). The electrodes are inert carbon. Step 2: Determine the reactions at the cathode (negative electrode). At the cathode, reduction occurs. Possible species to be reduced are Cu^2+ and H_2O. Cu^2+(aq) + 2e^- → Cu(s) (E^ = +0.34 V) 2H_2O(l) + 2e^- → H_2(g) + 2OH^-(aq) (E^ = -0.83 V at pH 7) Since Cu^2+ has a more positive standard reduction potential, copper metal will be deposited. Observation 1: A reddish-brown solid is deposited on the cathode. Step 3: Determine the reactions at the anode (positive electrode). At the anode, oxidation occurs. Possible species to be oxidized are SO_4^2- and H_2O (or OH^-). 2SO_4^2-(aq) → S_2O_8^2-(aq) + 2e^- (requires very high potential) 2H_2O(l) → O_2(g) + 4H^+(aq) + 4e^- (E^ = +1.23 V) Water is more easily oxidized than sulphate ions. Observation 2: Colorless gas bubbles are produced at the anode. Step 4: Consider changes to the solution. As Cu^2+ ions are removed from the solution, the characteristic blue color of copper(II) sulphate will fade. Observation 3: The blue color of the solution fades. Any two of these observations are acceptable. The two observations are: 1. A reddish-brown solid is deposited on the cathode. 2. Colorless gas bubbles are produced at the anode. b) Step 1: Write the half-reactions occurring at each electrode. At the cathode: Cu^2+(aq) + 2e^- → Cu(s) At the anode: 2H_2O(l) → O_2(g) + 4H^+(aq) + 4e^- Step 2: Analyze the effect of the reactions on the solution's composition. At the anode, water is oxidized, producing oxygen gas and hydrogen ions (H^+). The H^+ ions are released into the solution. Step 3: Explain the change in pH. The accumulation of H^+ ions in the solution increases the concentration of hydrogen ions. An increase in H^+ concentration makes the solution more acidic. Therefore, the pH of the resultant solution will decrease. Explanation: The pH of the resultant solution will decrease. This is because water is oxidized at the anode, producing hydrogen ions (H^+). The increase in H^+ concentration makes the solution more acidic, leading to a lower pH. That's 2 down. 3 left today — send the next one.