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)

Question

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)

BlackTom AI · Accepted Answer

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)

View Explanation

Log in for full answers

Similar Questions

Which of the below is the correctly balanced redox reaction of ammonium ion (NH4+) to the nitrite ion (NO2−).

During redox reactions:

The number of electrons transferred when the following unbalanced redox equation is balanced using simplest whole number ratio is Fe + H+ → Fe3+ + H2

Consider the following unbalanced redox equation: H+ + BrO3– + Fe2+ → Fe3+ + Br– + H2O When this equation is completely balanced using simplest whole number ratio, the number of electrons transferred will be

Consider the following unbalanced redox equation: H+ + Cr2O72– + CH3CHO → CH3COOH + Cr3+ + H2O When this equation is completely balanced using simplest whole number ratio, the number of protons required will be

How many protons will be produced when the following redox reaction is balanced with simplest whole number ratio? H2O2 + Ce4+ → O2 + Ce3+

Question at position 28 In the reaction below, what happens to sodium (Na)? Na + Cl --> Na+ + Cl-Sodium is a catalystSodium is reducedSodium is oxidized

More Practical Tools for International Students

Homework AI Solver

Stylized Writing Assistant

Plagiarism Checker Assistant

Citation Assistant

In-Class Translation Assistant

Class Notes Assistant

Quiz Search Assistant

Past Exam Questions by School

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler