Calculate the emf of the galvanic cell Fe(s) | $F{e}^{3+}$(0.02 M) || $C{u}^{2+}$(1 M) | Cu(s) at 25 °C. Standard potentials: $E_{F{e}^{3+}/Fe}^{\circ }$=+0.04 V and $E_{C{u}^{2+}/Cu}^{\circ }$=+0.34 V.

Step-by-step solution

1. Step 1: Identify the reduction half‑cell that will act as the cathode – $C{u}^{2+}$/Cu has a higher standard potential, so it is the cathode.
2. Step 2: Write the reduction potential for the Cu electrode, which is independent of concentration: $E_{Cu}=+0.34\text{V}$ (n=2).
3. Step 3: Write the Nernst equation for the Fe electrode (oxidation): $E_{Fe}=E{°}_{F{e}^{3+}/Fe}-\frac{0.0592}{n}log\frac{[F{e}^{3+}]}{[Fe]}$ with $n=3$ and $[Fe]=1$.
4. Step 4: Substitute the values: $E_{Fe}=0.04-\frac{0.0592}{3}log(0.02)=0.04+\frac{0.0592}{3}\times 3.912=0.04+0.077=0.073\text{V}$.
5. Step 5: Compute the cell emf: $E_{cell}=E_{cathode}-E_{anode}=0.34-0.073=0.267\text{V}$.
6. Step 6: Round to two decimal places: $E_{cell}\approx 0.27\text{V}$.

Concept tested: Nernst equation for non‑standard concentrations

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