An electrolytic cell produces aluminum from \(\mathrm{Al}_{2} \mathrm{O}_{3}\) at the rate of ten kilograms a day. Assuming a yield of \(100 \%\), (a) how many moles of electrons must pass through the cell in one day? (b) how many amperes are passing through the cell? (c) how many moles of oxygen \(\left(\mathrm{O}_{2}\right)\) are being produced simultaneously?

To convert the mass of aluminum produced into moles, we will use the molar mass of aluminum. The molar mass of aluminum (\(\mathrm{Al}\)) is approximately 26.98 g/mol. So, we have: \(10\,\text{kg} \cdot \frac{1000\,\text{g}}{1\,\text{kg}} \cdot \frac{1\,\text{mol}}{26.98\,\text{g}} \approx 370.64\,\text{moles}\) of \(\mathrm{Al}\) So, the cell produces approximately 370.64 moles of aluminum per day.

Next, we need to recall the balanced redox reaction for aluminum production. In this case, we have: \(\mathrm{Al}^{3+} + 3\,\text{e}^- \rightarrow \mathrm{Al}\) Thus, three moles of electrons are required for each mole of aluminum produced. So, the moles of electrons required for this process are: \(370.64\,\text{moles}\,\mathrm{Al} \cdot \frac{3\,\text{moles\,electrons}}{1\,\text{mole\,Al}} = 1111.93\,\text{moles\,electrons}\) Therefore, 1111.93 moles of electrons must pass through the cell in one day.

Now, we can calculate the current passing through the cell. We know that current is measured in amperes (A), and one ampere is equal to one coulomb of charge per second: \(1\,\text{A} = 1\,\frac{\text{C}}{\text{s}}\) We know that the charge of one mole of electrons is equal to Avogadro's number times the elementary charge. So we have: \(1\,\text{mole}\,\text{electrons=}6.022 \times 10^{23}\,\text{electrons} \cdot 1.6 \times 10^{-19}\,\text{C} \approx 96485\,\text{C}\) Thus, to find the current, we need to convert the moles of electrons per day to coulombs per second: \(\frac{1111.93\,\text{moles\,electrons}}{1\,\text{day}} \cdot \frac{96485\,\text{C}}{\text{mole}\,\text{electrons}} \cdot \frac{1\,\text{day}}{86400\,\text{s}} \approx 1272.74\,\text{A}\) So, approximately 1272.74 amperes are passing through the cell.

Lastly, we will find the moles of oxygen (\(\mathrm{O}_{2}\)) produced simultaneously. The balanced redox reaction for oxygen production is: \(2\,\mathrm{O}^{2-} \rightarrow \mathrm{O}_{2} + 4\,e^-\) We can see from this reaction that two moles of oxygen ions (\(\mathrm{O}^{2-}\)) produce one mole of oxygen gas (\(\mathrm{O}_{2}\)). Since aluminum oxide contains two moles of oxygen ions for each mole of aluminum ions, we can use the moles of aluminum produced to find the moles of oxygen produced: \(370.64\,\text{moles}\,\mathrm{Al} \cdot \frac{1\,\text{mole}\,\mathrm{O}_{2}}{1\,\text{mole}\,\mathrm{Al}} = 370.64\,\text{moles}\,\mathrm{O}_{2}\) Thus, 370.64 moles of oxygen gas are produced simultaneously by the electrolytic cell.