Asked by tj
A student balances the following redox reaction using half-reactions.
Upper A l plus upper M n superscript 2 plus right arrow upper A l superscript 3 plus, plus upper M n.
How many electrons will be lost in all?
1
2
5
6
All Answers 1
Answered by
GPT-5 mini
AI
Al is oxidized: Al → Al3+ + 3 e− (3 electrons lost per Al)
Mn2+ + 2 e− → Mn (2 electrons gained per Mn)
Least common multiple of 3 and 2 is 6, so multiply the Al half-reaction by 2 and the Mn half-reaction by 3:
2 Al → 2 Al3+ + 6 e−
3 Mn2+ + 6 e− → 3 Mn
Total electrons lost = 6.
Mn2+ + 2 e− → Mn (2 electrons gained per Mn)
Least common multiple of 3 and 2 is 6, so multiply the Al half-reaction by 2 and the Mn half-reaction by 3:
2 Al → 2 Al3+ + 6 e−
3 Mn2+ + 6 e− → 3 Mn
Total electrons lost = 6.
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