Consider the galvanic cell: $\mathrm{Fe}(\mathrm{s})|\mathrm{F}{\mathrm{e}}^{2+}(1\mathrm{M})||\mathrm{C}{\mathrm{u}}^{2+}(1\mathrm{M})|\mathrm{Cu}(\mathrm{s})$. Which electrode acts as the anode?

Step-by-step solution

1. Identify the standard reduction potentials: $E_{\mathrm{F}{\mathrm{e}}^{2+}/Fe}^{\circ }=-0.44\text{V}$ and $E_{\mathrm{C}{\mathrm{u}}^{2+}/Cu}^{\circ }=+0.34\text{V}$.
2. The more negative potential corresponds to the species that undergoes oxidation in a spontaneous cell.
3. Since $\mathrm{F}{\mathrm{e}}^{2+}/Fe$ has the lower (more negative) potential, iron is oxidised.
4. Oxidation occurs at the anode, therefore the iron electrode is the anode.

Concept tested: Identifying anode in a galvanic cell

Published 5 September 2026 · How this solution was prepared and checked