Calculate the number of grams of aluminum produced in the electrolytic process when 2.5 × 10⁵ coulombs are provided to the reaction.Assume 100% efficiency.

Question

Quizplus · Accepted Answer

The amount of aluminum produced in an electrolytic process can be calculated using Faraday's laws of electrolysis. The mass of a substance produced at an electrode is directly proportional to the quantity of electricity that passes through the electrolyte. The formula to calculate the mass is given by:$$ m = \frac{ItM}{nF} $$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 process (for aluminum, $ n = 3 $ since it forms $ Al^{3+} $ ions),- $ F $ is Faraday's constant ($ 96485 \, \text{C/mol} $).Given that $ 2.5 \times 10^{15} $ coulombs are provided, this value is essentially the product of $ I $ and $ t $ since $ Q = It $. Therefore, we can directly substitute $ Q $ for $ It $ in the formula:$$ m = \frac{2.5 \times 10^{15} \times 26.98}{3 \times 96485} $$$$ m \approx 23 \, \text{g} $$Thus, the correct answer is 23 g, which corresponds to choice A.

Calculate the Number of Grams of Aluminum Produced in the Electrolytic