Which of the following materials could not serve as a sacrificial anode for lead? — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Recall the principle of sacrificial anodes.** A sacrificial anode is a metal that is **more easily oxidised** (has a **lower standard reduction potential**) than the metal it protects. It corrodes preferentially, protecting the protected metal. **Step 2: Find the standard reduction potential of lead.** $$E^\circ(\text{Pb}^{2+}/\text{Pb}) = -0.126\ \text{V}$$ For a metal to act as a sacrificial anode for lead, it must have: $$E^\circ_{\text{metal}} < -0.126\ \text{V}$$ **Step 3: Compare the reduction potentials of the options.** | Metal | $E^\circ$ (V) | Can protect Pb? | |-------|--------------|----------------| | Copper | $+0.34\ \text{V}$ | **No** — $E^\circ$ is greater than Pb (less easily oxidised) | | Iron | $-0.44\ \text{V}$ | Yes — more easily oxidised | | Magnesium | $-2.37\ \text{V}$ | Yes — more easily oxidised | | Manganese | $-1.18\ \text{V}$ | Yes — more easily oxidised | **Step 4: Identify the answer.** Copper has a **higher** reduction potential than lead, meaning it is **less easily oxidised**. If copper is attached to lead, lead would actually corrode preferentially. Copper **cannot** serve as a sacrificial anode for lead. $$\boxed{\text{Answer: A — Copper cannot serve as a sacrificial anode for lead}}$$