A ship that has sunk to the bottom of the ocean may still exhibit corrosion. One difference in the c — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Recall the normal iron corrosion cell.** In normal atmospheric corrosion, the corrosion cell consists of: - **Anode:** $\ce{Fe(s) -> Fe^{2+}(aq) + 2e-}$ (iron dissolves) - **Cathode (reduction):** $\ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)}, \quad E^\circ = +1.229\ \text{V}$ The oxygen reduction at the cathode is central to the normal corrosion mechanism. **Step 2: Analyse conditions at the ocean floor.** At the bottom of the ocean, water pressure is enormous and dissolved oxygen concentration is **extremely low** (oxygen has limited solubility at high pressures and is consumed by deep-sea organisms). **Step 3: Determine what changes in the corrosion mechanism.** Without dissolved $\text{O}_2$, the standard cathode reaction ($\text{O}_2$ reduction) **cannot proceed**. The reduction half-reaction must involve another species (e.g., water, $\text{H}^+$, or other dissolved species). The **oxidation reaction** of the iron hull remains the same. Therefore: **the reduction reaction does NOT include $\text{O}_2$ as a reactant** in a sunken ship. **Step 4: Evaluate options.** - A: Wrong — oxidation reaction still involves iron metal; H₂O may or may not be a reactant there - B: Wrong — the oxidation reaction doesn't specifically need O₂ as a reactant regardless (O₂ is consumed at cathode in normal corrosion) - C: Wrong — it is the reduction reaction, not oxidation, that changes - D: ✓ Correct — reduction reaction cannot proceed via O₂ (absent at ocean floor) $$\boxed{\text{Answer: D — The reduction reaction does not include O}_2\text{ as a reactant}}$$