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Ch.19 - Electrochemistry
Chapter 19, Problem 102

What mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt using a current of 25 A?

Verified step by step guidance
1
Identify the relevant electrochemical reaction: The reduction of aluminum ions to aluminum metal is represented by the half-reaction: \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \).
Determine the number of moles of electrons required: From the half-reaction, 3 moles of electrons are needed to produce 1 mole of aluminum.
Use Faraday's Law of Electrolysis: The amount of substance produced at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The formula is \( n = \frac{I \cdot t}{n_e \cdot F} \), where \( n \) is the number of moles of aluminum, \( I \) is the current in amperes, \( t \) is the time in seconds, \( n_e \) is the number of moles of electrons, and \( F \) is Faraday's constant (approximately 96485 C/mol).
Calculate the time in seconds for one hour: Since the problem asks for the mass produced per hour, convert 1 hour into seconds (1 hour = 3600 seconds).
Calculate the mass of aluminum produced: Use the molar mass of aluminum (approximately 26.98 g/mol) to convert the moles of aluminum to grams.