The number of electrons transferred when the following unbalanced redox equation is balanced using simplest whole number ratio is Cu+(aq) + Fe(s) → Cu(s) + Fe3+(aq)

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The number of electrons transferred when the following unbalanced redox equation is balanced using simplest whole number ratio is Cu+(aq) + Fe(s) →  Cu(s) + Fe3+(aq)

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First, restating the problem in my own words: we have the unbalanced redox equation Cu+(aq) + Fe(s) → Cu(s) + Fe3+(aq), and we want the number of electrons transferred when it is balanced in the simplest whole-number ratio.

Option a: 2 electrons. If 2 electrons were transferred, we would need the oxidation and reduction half-reactions to balance electrons one-for-one with a total of 2 e− exchanged. The oxidation half-reaction for iron would be Fe(s) → Fe3+(aq) + 3 e−, which releases 3 electrons. The reduction half-reaction for copper would be Cu+(aq) + e− → Cu(s) if it were to gain electrons, but Cu+ to Cu(s) requires 1 electron gained. To match 2 electrons, you would need a combination that yields exactly 2 e− total, which is not consistent with the actual oxidation state change (Fe going from 0 to +3 releases 3 e−, not 2, and Cu+ gaining 1 e− per copper atom cannot pair cleanly to yield 2). Thus, 2 e− is not the proper electron count for this redox couple.

Option b: 1 electron. Achieving a net transfer of only 1 electron would require half-reactions that sum to 1 e−. However, the iron oxidation step inherently transfers 3 e− (Fe → Fe3+ + 3 e−). Balancing the reaction with 1 e− would leave unbalanced charge and mass, and you cannot reconcile a 3 e− donation with a single 1 e− acceptance in this pair. Therefore, 1 e− is not feasible here.

Option c: 6 electrons. If 6 electrons were transferred, you would need to scale the half-reactions such that the electrons gained and lost sum to 6. Since the Fe oxidation produces 3 e− per Fe atom and Cu+ reduction consumes 1 e− per Cu+, you could attempt to multiply Cu+ reduction by 6 and Fe oxidation by 6 as well, but that would imply altering the actual stoichiometry inconsistently with the given species. The simplest balanced transfer for this specific pair is not 6 e−; the electron count must correspond to the smallest common multiple of the elementary transfer values (which are 3 e− from Fe and 1 e− for Cu+), leading to 3 e− as the appropriate total transfer.

Option d: 3 electrons. This matches the intuitive and standard balancing approach: the oxidation half-reaction is Fe(s) → Fe3+(aq) + 3 e−, releasing 3 electrons. The reduction half-reaction is Cu+(aq) + 1 e− → Cu(s), consuming 1 electron per copper ion. To balance the electrons, the Cu+ reduction must occur with three Cu+ species for every Fe atom oxidized, giving 3 × 1 e− = 3 e− consumed. When you multiply the copper half-reaction by 3, you obtain 3 Cu+(aq) + 3 e− → 3 Cu(s). The iron half-reaction remains Fe(s) → Fe3+(aq) + 3 e−. The electrons cancel and the overall balanced equation becomes Fe(s) + 3 Cu+(aq) → Fe3+(aq) + 3 Cu(s). Therefore, the number of electrons transferred in this balanced process is 3.

Summarizing the reasoning for each option: while 2 e− and 1 e− fail to align with the actual electron bookkeeping of Fe oxidation and Cu+ reduction, and 6 e− would overcount relative to the smallest integer balancing, the correct balance requires 3 electrons to be transferred overall, which corresponds to the stoichiometry Fe(s) + 3 Cu+(aq) → Fe3+(aq) + 3 Cu(s).

MUF0042 Chemistry Unit 2 - Semester 2, 2025 10.13 Quiz: Balancing Redox Reactions

The number of electrons transferred when the following unbalanced redox equation is balanced using simplest whole number ratio is Cu+(aq) + Fe(s) → Cu(s) + Fe3+(aq)

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