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 $\text{Ca}^{2+}$ ions and water molecules ($\text{H}_2\text{O}$) present. At the cathode (negative electrode), water is preferentially reduced over calcium ions because water has a less negative (more positive) standard reduction potential ($E^\circ(\text{H}_2\text{O}) = -0.83 \text{ V}$) compared to calcium ions ($E^\circ(\text{Ca}^{2+}) = -2.87 \text{ V}$). 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: $$2\text{Cl}^-(aq) \to \text{Cl}_2(g) + 2\text{e}^-$$ 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. $$t = 1 \text{ hour} \times 60 \text{ min/hour} \times 60 \text{ s/min} = 3600 \text{ s}$$ Step 2: Calculate the total charge ($Q$) passed. $$Q = I \times t$$ $$Q = 4 \text{ A} \times 3600 \text{ s}$$ $$Q = 14400 \text{ C}$$ Step 3: Determine the number of moles of electrons ($n_e$) transferred. $$n_e = \frac{Q}{F}$$ $$n_e = \frac{14400 \text{ C}}{96500 \text{ C/mol e⁻}}$$ $$n_e \approx 0.14922 \text{ mol e⁻}$$ Step 4: Use the stoichiometry from the anode reaction ($2\text{Cl}^-(aq) \to \text{Cl}_2(g) + 2\text{e}^-$) to find moles of $\text{Cl}_2$. From the reaction, $2$ moles of electrons produce $1$ mole of $\text{Cl}_2$. $$n_{\text{Cl}_2} = \frac{n_e}{2}$$ $$n_{\text{Cl}_2} = \frac{0.14922 \text{ mol e⁻}}{2 \text{ mol e⁻/mol Cl}_2}$$ $$n_{\text{Cl}_2} \approx 0.07461 \text{ mol Cl}_2$$ Step 5: Calculate the mass of $\text{Cl}_2$ produced. The molar mass of $\text{Cl}_2 = 2 \times 35.5 \text{ g/mol} = 71.0 \text{ g/mol}$. $$m_{\text{Cl}_2} = n_{\text{Cl}_2} \times M_{\text{Cl}_2}$$ $$m_{\text{Cl}_2} = 0.07461 \text{ mol} \times 71.0 \text{ g/mol}$$ $$m_{\text{Cl}_2} \approx 5.297 \text{ g}$$ The mass of the product at the anode is $\boxed{\text{5.30 g}}$. That's 2 down. 3 left today — send the next one.