Asked by tt
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Cu2+ reduces to Cu, and Al oxidizes to AI3+.
Cu2+ oxidizes to Cu, and Al reduces to AI3+.
Cu2+ is the reducing agent, and Al is the oxidizing agent.
Cu is the reducing agent, and AI3+ is the oxidizing agent.
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Answered by
GPT-5 mini
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Only the first statement is correct.
Why: Al(s) is oxidized and Cu2+(aq) is reduced.
Half-reactions:
- Oxidation: Al → Al3+ + 3 e−
- Reduction: Cu2+ + 2 e− → Cu
Thus Al is the reducing agent (it loses electrons) and Cu2+ is the oxidizing agent (it gains electrons). The fully balanced redox equation (lowest whole-number coefficients) is:
2 Al + 3 Cu2+ → 2 Al3+ + 3 Cu.
Why: Al(s) is oxidized and Cu2+(aq) is reduced.
Half-reactions:
- Oxidation: Al → Al3+ + 3 e−
- Reduction: Cu2+ + 2 e− → Cu
Thus Al is the reducing agent (it loses electrons) and Cu2+ is the oxidizing agent (it gains electrons). The fully balanced redox equation (lowest whole-number coefficients) is:
2 Al + 3 Cu2+ → 2 Al3+ + 3 Cu.
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