Calculate the mass, in kilograms, of magnesium produced by the electrolysis of magnesium (II) chloride in a cell operating for 24 hours at 500 amperes. [1 Faraday = 96500c mol-1, Mg = 24]

Magnesium ions ($Mg^{2+}$) gain two electrons to form solid magnesium ($Mg$).

$$Mg^{2+} + 2e^- \to Mg$$

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:

$$Q = I \times t$$ $$Q = 500 \text{ A} \times 86400 \text{ s} = 43200000 \text{ C}$$

Using Faraday's constant ($F = 96500 \text{ C/mol}$):

$$n_e = \frac{Q}{F}$$ $$n_e = \frac{43200000 \text{ C}}{96500 \text{ C/mol}} \approx 447.668 \text{ mol}$$

From the half-reaction, 2 moles of electrons are required to produce 1 mole of magnesium.

$$n_{Mg} = \frac{n_e}{2}$$ $$n_{Mg} = \frac{447.668 \text{ mol}}{2} \approx 223.834 \text{ mol}$$

The molar mass of magnesium ($Mg$) is $24 \text{ g/mol}$.

$$m_{Mg} = n_{Mg} \times \text{Molar Mass of Mg}$$ $$m_{Mg} = 223.834 \text{ mol} \times 24 \text{ g/mol} = 5372.016 \text{ g}$$

$$m_{Mg} = \frac{5372.016 \text{ g}}{1000 \text{ g/kg}} \approx 5.372 \text{ kg}$$

The mass of magnesium produced is $5.372 \text{ kg}$.