A roll of Reynolds Wrap aluminum foil contains 1.57 kg of aluminum.For how long would a current of 1000.0 amp need to be passed through a molten AlF₃ electrolysis cell to produce this amount of aluminum?

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Quizplus · Accepted Answer

The production of aluminum through electrolysis can be understood through Faraday's laws of electrolysis. The amount of substance produced at an electrode is directly proportional to the quantity of electricity that passes through the electrolyte. The relevant equation is:$$ m = \frac{ItM}{nF} $$where:- $m$ is the mass of the substance produced (in grams),- $I$ is the current (in amperes),- $t$ is the time (in seconds),- $M$ is the molar mass of the substance (for aluminum, $M = 26.98 \, \text{g/mol}$),- $n$ is the number of electrons involved in the electrolysis reaction (for aluminum, $n = 3$),- $F$ is Faraday's constant ($96485 \, \text{C/mol}$).Given:- $m = 1.57 \, \text{kg} = 1570 \, \text{g}$,- $I = 1000.0 \, \text{A}$,- $M = 26.98 \, \text{g/mol}$,- $n = 3$,- $F = 96485 \, \text{C/mol}$.Rearranging the formula to solve for $t$:$$ t = \frac{m \cdot n \cdot F}{I \cdot M} $$Plugging in the values:$$ t = \frac{1570 \cdot 3 \cdot 96485}{1000.0 \cdot 26.98} $$$$ t = 16807.6 \, \text{seconds} $$To convert seconds to hours:$$ t = \frac{16807.6}{3600} \approx 4.67 \, \text{hours} $$Rounding to two decimal places, the closest answer is 4.68 hours.

A Roll of Reynolds Wrap Aluminum Foil Contains 1