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's the solution for part (b): b) i) The student was surprised that no calcium was produced because in an aqueous solution of calcium chloride, there are Ca^2+ ions and water molecules (H_2O) present. At the cathode (negative electrode), water is preferentially reduced over calcium ions because water has a less negative (more positive) standard reduction potential (E^(H_2O) = -0.83 V) compared to calcium ions (E^(Ca^2+) = -2.87 V). Therefore, hydrogen gas is produced instead of calcium metal. ii) The equation for the reaction that occurred at the anode if the solution was concentrated is: 2Cl^-(aq) Cl_2(g) + 2e^- In a concentrated chloride solution, chloride ions are preferentially oxidized over water due to their high concentration and overpotential effects, producing chlorine gas. iii) Step 1: Convert time to seconds. t = 1 hour × 60 min/hour × 60 s/min = 3600 s Step 2: Calculate the total charge (Q) passed. Q = I × t Q = 4 A × 3600 s Q = 14400 C Step 3: Determine the number of moles of electrons (n_e) transferred. n_e = (Q)/(F) n_e = 14400 C96500 C/mol e⁻ n_e ≈ 0.14922 mol e⁻ Step 4: Use the stoichiometry from the anode reaction (2Cl^-(aq) Cl_2(g) + 2e^-) to find moles of Cl_2. From the reaction, 2 moles of electrons produce 1 mole of Cl_2. n_Cl_2 = (n_e)/(2) n_Cl_2 = 0.14922 mol e⁻2 mol e⁻/mol Cl_2 n_Cl_2 ≈ 0.07461 mol Cl_2 Step 5: Calculate the mass of Cl_2 produced. The molar mass of Cl_2 = 2 × 35.5 g/mol = 71.0 g/mol. m_Cl_2 = n_Cl_2 × M_Cl_2 m_Cl_2 = 0.07461 mol × 71.0 g/mol m_Cl_2 ≈ 5.297 g The mass of the product at the anode is 5.30 g. That's 2 down. 3 left today — send the next one.