In the galvanic cell Cu(s) | $C{u}^{2+}$(0.5 M) || $F{e}^{2+}$(1 M) | Fe(s), which electrode acts as the anode at 25 °C? Standard potentials: $E_{C{u}^{2+}/Cu}^{\circ }$=+0.34 V, $E_{F{e}^{2+}/Fe}^{\circ }$=−0.44 V.

Step-by-step solution

1. Step 1: Find the reduction potential of the Cu electrode using the Nernst equation: $E_{Cu}=0.34-\frac{0.0592}{2}log(0.5)=0.34+0.009=0.349$ V.
2. Step 2: The Fe electrode has $E_{Fe}=-0.44$ V (concentration 1 M, so Nernst term is zero).
3. Step 3: The electrode with the higher (more positive) reduction potential will act as the cathode; thus Cu is cathode.
4. Step 4: The other electrode, Fe, must be the anode.

Concept tested: Identifying anode/cathode from electrode potentials

Published 13 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked