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
You're on a roll — d) i) An acid, according to the Brønsted-Lowry concept, is a proton donor. It donates a hydrogen ion (H^+) to another substance. ii) Step 1: Write the dissociation equation for sulfuric acid (H_2SO_4). Sulfuric acid is a strong acid and dissociates completely in water. H_2SO_4(aq) 2H^+(aq) + SO_4^2-(aq) Step 2: Determine the concentration of H^+ ions. From the dissociation equation, 1 mole of H_2SO_4 produces 2 moles of H^+ ions. Given concentration of H_2SO_4 = 0.001 M. Concentration of H^+ = 2 × 0.001 M = 0.002 M. Step 3: Calculate the pH using the formula pH = -[H^+]. pH = -(0.002) pH = -(2 × 10^-3) pH = -( 2 + 10^-3) pH = -(0.301 - 3) pH = -(-2.699) pH = 2.699 The pH of the solution is approximately 2.70. iii) A heterogeneous catalyst is a catalyst that is in a different phase from the reactants. For example, a solid catalyst used in a reaction involving gaseous or liquid reactants. e) i) To identify the strongest oxidizing and reducing agents, we look at the standard electrode potentials (E^). • The strongest oxidizing agent is the species on the left side of the half-reaction with the most positive (or least negative) E^ value. This species is most easily reduced. • The strongest reducing agent is the species on the right side of the half-reaction with the most negative (or least positive) E^ value. This species is most easily oxidized. Given electrode potentials: A: Pb^2+(aq) + 2e^- Pb(s), E^ = -0.13 V B: I_2(aq) + 2e^- 2I^-(aq), E^ = +0.54 V C: MnO_4^-(aq) + 8H^+(aq) + 5e^- Mn^2+(aq) + 4H_2O(l), E^ = +1.51 V Comparing the E^ values: +1.51 V is the most positive. The strongest oxidizing agent is MnO_4^-(aq). Comparing the E^ values: -0.13 V is the most negative. The strongest reducing agent is Pb(s). Strongest oxidizing agent: MnO_4^-(aq) Strongest reducing agent: Pb(s) ii) Step 1: Determine the anode and cathode when cells A and B are coupled. For a spontaneous reaction, the species with the more positive E^ will be reduced (cathode), and the species with the more negative E^ will be oxidized (anode). Cell A: Pb^2+(aq) + 2e^- Pb(s), E^ = -0.13 V Cell B: I_2(aq) + 2e^- 2I^-(aq), E^ = +0.54 V Since E^_B (+0.54 V) is more positive than E^_A (-0.13 V), reaction B will occur at the cathode (reduction) and reaction A will occur at the anode (oxidation, reversed). Anode (oxidation): Pb(s) Pb^2+(aq) + 2e^- Cathode (reduction): I_2(aq) + 2e^- 2I^-(aq) Step 2: Write the cell diagram. The cell diagram notation is: Anode | Anode ion || Cathode ion | Cathode. Pb(s) | Pb^2+(aq) || I_2(aq) | I^-(aq) Step 3: Calculate the electromotive force (emf) of the cell. E^_cell = E^_cathode - E^_anode E^_cell = E^_B - E^_A E^_cell = (+0.54 V) - (-0.13 V) E^_cell = 0.54 V + 0.13 V E^_cell = 0.67 V The emf of the cell is 0.67 V. iii) Electrons flow from the anode to the cathode in the external circuit. In this cell, Pb(s) is oxidized at the anode, and I_2(aq) is reduced at the cathode. Therefore, electrons will flow from the lead electrode (Pb) to the iodine electrode (I_2). What's next?