2018 H2 Chem P1 Q10 — Redox Solution

The answer is B (+5).
Iodine atoms: 0.006 mol \(\text{I}_2\) contains 0.012 mol I atoms. The 0.01 mol \(\text{I}^-\) supplies 0.010 mol of these, so the oxide supplies \(0.012 - 0.010 = 0.002\) mol I atoms. 0.001 mol oxide gives 0.002 mol I, so each formula unit contains 2 I atoms: \(\text{I}_2\text{O}_n\).
Electron balance: each \(\text{I}^-\) is oxidised from \(-1\) to 0, giving \(0.01 \times 1 = 0.01\) mol electrons. The 0.002 mol I atoms in the oxide, of oxidation number \(+x\), are reduced to 0 and take up these electrons:
\(0.002 \times x = 0.01 \Rightarrow x = +5\)
So the oxide is \(\text{I}_2\text{O}_5\): \(\text{I}_2\text{O}_5 + 10\text{I}^- + 10\text{H}^+ \rightarrow 6\text{I}_2 + 5\text{H}_2\text{O}\), which matches the 1 : 10 : 6 mole ratio given.
• A is wrong: iodine at +1 in 0.002 mol I atoms would accept only 0.002 mol electrons, enough to oxidise only 0.002 mol \(\text{I}^-\), not the 0.01 mol that reacted.
• C is wrong: iodine at +6 would accept \(0.002 \times 6 = 0.012\) mol electrons, needing 0.012 mol \(\text{I}^-\), but only 0.010 mol \(\text{I}^-\) reacted.
• D is wrong: +10 is the value obtained if the oxide is assumed to contain only one I atom (\(0.001 \times x = 0.01\)); the iodine atom count shows there are two. Iodine has only 7 valence electrons, so +10 is impossible.