Iodide ion reacts with manganese(IV) oxide according to: MnO2(s) + 2 I– (aq) + 4 H+(aq) → I2(aq) + Mn2+(aq) + 2 H2O(l) Which statement correctly describes this reaction?

Question

Iodide ion reacts with manganese(IV) oxide according to: MnO2(s) + 2 I– (aq) + 4 H+(aq) →  I2(aq) + Mn2+(aq) + 2 H2O(l) Which statement correctly describes this reaction?

BlackTom AI · Accepted Answer

Mn in MnO2 has oxidation state +4, while iodide (I–) is –1; these change during the reaction, so assign oxidation numbers and track electron transfer.

1) Relevant knowledge points and concepts
- Oxidation is loss of electrons (increase in oxidation number); reduction is gain of electrons (decrease in oxidation number).
- An oxidizing agent is reduced (it accepts electrons). A reducing agent is oxidized (it donates electrons).
- Write half-reactions to see electron transfer.

2) Step-by-step reasoning with half-reactions
- Determine oxidation states: in MnO2, O is –2, so Mn + 2(–2) = 0 → Mn = +4. In Mn2+ the oxidation state is +2. I– is –1, and in I2 each I is 0.
- Reduction half-reaction (Mn): MnO2 + 4 H+ + 2 e– → Mn2+ + 2 H2O. Mn goes +4 → +2 (gain of 2 e–).
- Oxidation half-reaction (I): 2 I– → I2 + 2 e–. Each I– goes –1 → 0 (loss of 1 e– per I, total 2 e–).
- Combine half-reactions: electrons cancel, giving the overall balanced equation as provided.

3) Intermediate calculation steps shown
- Oxidation number change for Mn: +4 → +2 (change = gain 2 e–).
- Oxidation number change for each I: –1 → 0 (each loses 1 e–; two iodides give 2 e– total).
- Electrons transferred: 2 e– from the iodide oxidation are accepted by MnO2 reduction.

4) Why the correct answer is correct
- Because Mn in MnO2 is reduced from +4 to +2, MnO2 is the oxidizing agent. I– is oxidized to I2, so I– is the reducing agent. Statement (a) correctly states “Mn in MnO2 is reduced to Mn2+ and I– is oxidised to I2.”

5) Why the other options are incorrect (brief)
- Any choice claiming I– is reduced is wrong: I– is oxidized to I2. 
- Any choice reversing the agents (saying MnO2 is reducing agent or I– is oxidizing agent) is incorrect because MnO2 accepts electrons (is reduced), so it is the oxidizing agent.
- Statements that claim H+ is oxidized or is the main oxidizing agent are wrong: H+ participates to form water but does not undergo oxidation (it does not lose electrons).

6) Conclusion
- The reasoning and half-reactions confirm option (a): Mn in MnO2 is reduced to Mn2+ and I– is oxidised to I2, matching the provided answer.

Iodide ion reacts with manganese(IV) oxide according to: MnO2(s) + 2 I– (aq) + 4 H+(aq) → I2(aq) + Mn2+(aq) + 2 H2O(l) Which statement correctly describes this reaction?单项选择题

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