Step 1: Write the half-reaction for the production of magnesium. Magnesium ions ($Mg^{2+}$) gain two electrons to form solid magnesium ($Mg$). $$Mg^{2+} + 2e^- \to Mg$$ Step 2: Calculate the total charge ($Q$) passed through the cell. The current ($I$) is 500 amperes and the time ($t$) is 24 hours. First, convert time to seconds: $$t = 24 \text{ hours} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 86400 \text{ s}$$ Now, calculate the charge: $$Q = I \times t$$ $$Q = 500 \text{ A} \times 86400 \text{ s} = 43200000 \text{ C}$$ Step 3: Calculate the moles of electrons ($n_e$) transferred. Using Faraday's constant ($F = 96500 \text{ C/mol}$): $$n_e = \frac{Q}{F}$$ $$n_e = \frac{43200000 \text{ C}}{96500 \text{ C/mol}} \approx 447.668 \text{ mol}$$ Step 4: Determine the moles of magnesium ($n_{Mg}$) produced. From the half-reaction, 2 moles of electrons are required to produce 1 mole of magnesium. $$n_{Mg} = \frac{n_e}{2}$$ $$n_{Mg} = \frac{447.668 \text{ mol}}{2} \approx 223.834 \text{ mol}$$ Step 5: Calculate the mass of magnesium ($m_{Mg}$) produced. The molar mass of magnesium ($Mg$) is $24 \text{ g/mol}$. $$m_{Mg} = n_{Mg} \times \text{Molar Mass of Mg}$$ $$m_{Mg} = 223.834 \text{ mol} \times 24 \text{ g/mol} = 5372.016 \text{ g}$$ Step 6: Convert the mass from grams to kilograms. $$m_{Mg} = \frac{5372.016 \text{ g}}{1000 \text{ g/kg}} \approx 5.372 \text{ kg}$$ The mass of magnesium produced is $\boxed{5.372 \text{ kg}}$. That's 2 down. 3 left today — send the next one.