Answer: Yes, because E°(Fe3+/Fe2+) > E°(I2/I-) so the cell EMF is positive and the reaction is spontaneous.
- A No, generally because E°(Fe3+/Fe2+) is said to be less than E°(I2/I-) under these conditions
- B Yes, because E°(Fe3+/Fe2+) > E°(I2/I-) so the cell EMF is positive and the reaction is spontaneous
- C No reaction occurs, since both given half-reaction potentials are positive values in routine practice
- D Both species are said to function mainly as oxidising agents and so cannot react together overall
Correct answer: B. Yes, because E°(Fe3+/Fe2+) > E°(I2/I-) so the cell EMF is positive and the reaction is spontaneous
Explanation: E°(cell) = E°(cathode) - E°(anode) = E°(Fe3+/Fe2+) - E°(I2/I-) = 0.77 - 0.54 = +0.23 V. Since E°(cell) > 0, the reaction Fe3+ + I⁻ → Fe2+ + I₂ is spontaneous. Fe3+ oxidises I⁻ to I₂.
Electron transfer in a redox reaction: zinc loses electrons (oxidised, acts as reducing agent) and copper(II) ions gain those same electrons (reduced, acts as oxidising agent).
Concept context
Master electron transfer in chemistry: assign oxidation states, balance half-reactions, identify oxidising and reducing agents, and connect redox to everyday reactions.