In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,0,) dissolved in molten cryolite (Na, AlF), resulting in the reduction of the Al,0, to pure aluminum. Suppose a current of 940. A is passed through a Hall-Heroult cell for 71.0 seconds. Calculate the mass of pure aluminum produced.
In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,0,) dissolved in molten cryolite (Na, AlF), resulting in the reduction of the Al,0, to pure aluminum. Suppose a current of 940. A is passed through a Hall-Heroult cell for 71.0 seconds. Calculate the mass of pure aluminum produced.
Principles of Modern Chemistry
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ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter17: Electrochemistry
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Transcribed Image Text:In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,0,) dissolved in molten cryolite (Na, AlF), resulting
in the reduction of the Al,0, to pure aluminum.
Suppose a current of 940. A is passed through a Hall-Heroult cell for 71.0 seconds. Calculate the mass of pure aluminum produced.
Be sure your answer has a unit symbol and the correct number of significant digits.
Expert Solution
Step 1
we will apply the combined Faraday's law
W = (Q*M )/F*n
W = weight of mass in grams
Q = quantity of electricity passed
M = molar mass of metal
t = time of electricity passed
F = 1 Faraday = 96500C
n = valency of metal
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