Number of electrons exchanged during the following reaction are/is
\( \ce{2Fe^3+(aq) + 2I^-(aq) -> 2Fe^2+(aq) + I2(s)} \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
The number of electrons exchanged (n) in a redox reaction is found by splitting it into its oxidation and reduction half-reactions and balancing the electrons.
\( \displaystyle \text{Reduction:}\quad \ce{2Fe^3+ + 2e^- -> 2Fe^2+}\quad (+3 \rightarrow +2,\ \text{one electron per Fe}) \)
\( \displaystyle \text{Oxidation:}\quad \ce{2I^- -> I2 + 2e^-}\quad (-1 \rightarrow 0,\ \text{one electron per I}) \)
\( \displaystyle \text{Net reaction:}\quad \ce{2Fe^3+(aq) + 2I^-(aq) -> 2Fe^2+(aq) + I2(s)} \)
Two electrons pass from the two iodide ions to the two \( \ce{Fe^{3+}} \) ions, so n = 2 for the reaction as written.