Question

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?

Short answer

How many moles of oxygen are produced simultaneously? Answer: (a) 1111.92 moles of electrons pass through the cell in one day. (b) 1240.93 amperes are passing through the cell. (c) 555.96 moles of oxygen are being produced simultaneously.

Step-by-step solution

Convert the mass of aluminum into moles

Given, the mass of aluminum produced = 10 kg. To convert this mass into moles, we can use the molar mass of aluminum which is \(26.98\, \mathrm{g/mol}\). Note that 1 kg = 1000 g, so our mass in grams is 10000 g. Moles of aluminum = \(\frac{\text{mass of aluminum}}{\text{molar mass of aluminum}} = \frac{10000\,\text{g}}{26.98\, \mathrm{g/mol}} = 370.64\,\text{moles}\)

Calculate the moles of electrons required to produce the given moles of aluminum

The reaction to produce aluminum from \(\mathrm{Al}_{2}\mathrm{O}_{3}\) can be given as: \(\mathrm{Al}_{2}\mathrm{O}_{3} + 6 e^{-} \rightarrow 2 \mathrm{Al} + 3 \mathrm{O}_{2}\) From the above reaction, 6 moles of electrons are required to produce 2 moles of aluminum. Moles of electrons required = \(\frac{6\,\text{moles of electrons}}{2\,\text{moles of aluminum}} \times 370.64\,\text{moles of aluminum} = 1111.92\,\text{moles of electrons}\)

Calculate the current in amperes using the moles of electrons

We can use Faraday's law of electrolysis to calculate the current in amperes. The total charge transferred (Q) in a day can be calculated using the formula: \(Q = n \times F\) where n = moles of electrons and F = Faraday's constant (\(96485\, \mathrm{C/mol}\)) \(Q = 1111.92 \, \text{moles} \times 96485\,\mathrm{C/mol} = 107242724\,\mathrm{C}\) Given that one day has \(24 \times 3600 = 86,400\) seconds, current (I) can be calculated as: \(I = \frac{Q}{t} = \frac{107242724\, C}{86400\, s} = 1240.93\, A\)

Calculate the moles of oxygen produced simultaneously

From the balanced equation, \(\mathrm{Al}_{2}\mathrm{O}_{3} + 6 e^{-} \rightarrow 2 \mathrm{Al} + 3 \mathrm{O}_{2}\) 3 moles of \(\mathrm{O}_{2}\) are produced for every 2 moles of aluminum. Moles of \(\mathrm{O}_{2}\) produced = \(\frac{3\,\text{moles of}\,\mathrm{O}_{2}}{2\,\text{moles of aluminum}} \times 370.64\,\text{moles of aluminum} = 555.96\,\text{moles of}\,\mathrm{O}_{2}\) To summarize: (a) 1111.92 moles of electrons must pass through the cell in one day. (b) 1240.93 amperes are passing through the cell. (c) 555.96 moles of oxygen are being produced simultaneously.