Solve: 1. You are titrating a 50.0 mL solution of 0.100 M iodide ions (I−) with 0.100 M lead(II) nitrate (Pb(NO₃)₂) using a lead electrode. The standard electrode potential for the lead electrode is E∘=−0.126 V. Calculate the potential of the solution at the following points: 4.5 marks a) Before adding any lead(II) nitrate. b) After adding 25.0 mL of lead(II) nitrate. c) At the equivalence point. d) After adding excess lead(II) nitrate (e.g., 75.0 mL). 2. You are titrating a 50.0 mL solution of 0.050 M hydrogen peroxide (H2O2) with 0.100 M potassium permanganate (KMnO4) using a platinum electrode. The standard electrode potential for the platinum electrode is E∘=0 V. Calculate the potential of the solution at the following points: 4.5 marks a) Before adding any potassium permanganate. b) After adding 25.0 mL of potassium permanganate. c) At the equivalence point. d) After adding excess potassium permanganate (e.g., 75.0 mL).

a) Before adding any lead(II) nitrate, the solution consists only of iodide ions. Therefore, the potential of the solution is equal to the standard electrode potential of the lead electrode, which is -0.126 V.

b) After adding 25.0 mL of lead(II) nitrate, the reaction is not at equilibrium, so we need to use the Nernst equation to calculate the potential. The Nernst equation is E = E° - (RT/nF) * ln(Q), where E is the potential, E° is the standard electrode potential, R is the gas constant, T is the temperature, n is the number of electrons transferred in the reaction, F is Faraday's constant, and Q is the reaction quotient. Assuming the temperature is 25°C (298K), R is 8.314 J/(molK), F is 96485 C/mol, and n is 2 (since Pb2+ + 2I- -> PbI2), we can calculate Q as [Pb2+]/[I-]^2 = (0.025 L * 0.100 M) / (0.050 L * 0.100 M)^2 = 1. Then, E = -0.126 V - (8.314 J/(molK) * 298 K / (2 * 96485 C/mol)) * ln(1) = -0.126 V.

c) At the equivalence point, all the iodide ions have reacted with the lead(II) nitrate, so the potential of the solution is again equal to the standard electrode potential of the lead electrode, which is -0.126 V.

d) After adding excess lead(II) nitrate, the solution contains excess Pb2+ ions. The potential of the solution will be greater than -0.126 V, but we cannot calculate the exact value without knowing the concentration of Pb2+ ions in the excess solution.

a) Before adding any potassium permanganate, the solution consists only of hydrogen peroxide. Therefore, the potential of the solution is equal to the standard electrode potential of the platinum electrode, which is 0 V.

b) After adding 25.0 mL of potassium permanganate, the reaction is not at equilibrium, so we need to use the Nernst equation to calculate the potential. Assuming the temperature is 25°C (298K), R is 8.314 J/(molK), F is 96485 C/mol, and n is 2 (since 2MnO4- + 5H2O2 -> 2Mn2+ + 5O2 + 4H2O), we can calculate Q as [Mn2+]^2/[H2O2]^5 = (0.025 L * 0.100 M)^2 / (0.050 L * 0.050 M)^5 = 0.16. Then, E = 0 V - (8.314 J/(molK) * 298 K / (2 * 96485 C/mol)) * ln(0.16) = 0.014 V.

c) At the equivalence point, all the hydrogen peroxide has reacted with the potassium permanganate, so the potential of the solution is again equal to the standard electrode potential of the platinum electrode, which is 0 V.

d) After adding excess potassium permanganate, the solution contains excess MnO4- ions. The potential of the solution will be less than 0 V, but we cannot calculate the exact value without knowing the concentration of MnO4- ions in the excess solution.