During the electrolysis of molten NaI, which reaction occurs at the anode?

• A. I_2 + 2 e^- → 2 I^-
• B. Na^+ + e^- → Na
• C. 2 I^- → I_2 + 2 e^-
• D. 2 Na → 2 Na^+ + 2 e^-

Answer: (C)

Verified step by step guidance

1. Identify the type of electrolysis: Since the problem involves molten NaI, the compound is in a molten (liquid) state, meaning it is composed of Na⁺ and I⁻ ions free to move.
2. Recall that during electrolysis, oxidation occurs at the anode and reduction occurs at the cathode. Oxidation means loss of electrons, and reduction means gain of electrons.
3. Look at the ions present: Na⁺ (cation) and I⁻ (anion). At the cathode (negative electrode), cations gain electrons (reduction), so Na⁺ will be reduced to Na metal: \(\mathrm{Na^+ + e^- \rightarrow Na}\).
4. At the anode (positive electrode), anions lose electrons (oxidation). The iodide ion (I⁻) will be oxidized to iodine (I₂) by losing electrons: \(\mathrm{2 I^- \rightarrow I_2 + 2 e^-}\).
5. Therefore, the reaction occurring at the anode is the oxidation of iodide ions to iodine, which matches the equation \(\mathrm{2 I^- \rightarrow I_2 + 2 e^-}\).