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14) In an electrochemical cell made of Al and Cu, Al loses 3 electrons per atom while Cu gains 2 electrons per atom. What are the two half-reactions (oxidation and reduction) combined together and balanced? (A total of 6 electrons should be lost and gained.) a) Al³+ + Cu(s) → Al(s) + Cu²+ b) 2Al³+ + 3Cu(s) → 2AI (s) + 3Cu²+ c) Al (s) + Cu²+ → Al³+ + Cu(s) d) 2Al(s) + 3Cu²+ → 2Al³+ + 3Cu(s)

14) In an electrochemical cell made of Al and Cu, Al loses 3 electrons per atom while Cu gains 2 electrons per atom. What are the two half-reactions (oxidation and reduction) combined together and balanced? (A total of 6 electrons should be lost and gained.) a) Al³+ + Cu(s) → Al(s) + Cu²+ b) 2Al³+ + 3Cu(s) → 2AI (s) + 3Cu²+ c) Al (s) + Cu²+ → Al³+ + Cu(s) d) 2Al(s) + 3Cu²+ → 2Al³+ + 3Cu(s)

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter19: Electrochemistry
Section: Chapter Questions
Problem 19.118QP
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14) In an electrochemical cell made of Al and Cu, Al loses 3 electrons per atom while Cu gains 2 electrons per atom. What are the two half-reactions (oxidation and reduction) combined together and balanced? (A total of 6 electrons should be lost and gained.) a) Al³+ + Cu(s) → Al (s) + Cu²+ b) 2Al³+ + 3Cu(s) ➜ 2AI (s) + 3Cu²+ c) Al (s) + Cu²+ → Al³+ + Cu(s) d) 2Al(s) + 3Cu²+ ➜ 2Al³+ + 3Cu(s)
Transcribed Image Text:14) In an electrochemical cell made of Al and Cu, Al loses 3 electrons per atom while Cu gains 2 electrons per atom. What are the two half-reactions (oxidation and reduction) combined together and balanced? (A total of 6 electrons should be lost and gained.) a) Al³+ + Cu(s) → Al (s) + Cu²+ b) 2Al³+ + 3Cu(s) ➜ 2AI (s) + 3Cu²+ c) Al (s) + Cu²+ → Al³+ + Cu(s) d) 2Al(s) + 3Cu²+ ➜ 2Al³+ + 3Cu(s)
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