Calculate the emf of the galvanic cell: Fe(s) | Fe³⁺(0.02 M) || Cu²⁺(1 M) | Cu(s) at 25 °C. Standard potentials: $E_{F e^{3 +} / F e}^{\circ}$ =+0.04 V and $E_{C u^{2 +} / C u}^{\circ}$ =+0.34 V.

easy 3 marks Class 12 · Chemistry · Electrochemistry

Answer

0.33 V

The correct answer is 0.33 V — Compute cell emf: Ecell = Ecathode – Eanode = 0.34 V – 0.0065 V = 0.3335 V ≈ 0.33 V.

Step-by-step solution

Write the two half‑reactions for reduction.
Use the Nernst equation for Fe³⁺/Fe: E = E° – (0.0592/3) log(1/[Fe³⁺]) = +0.04 V – 0.01973 log(50) = +0.0065 V.
Use the Nernst equation for Cu²⁺/Cu: E = +0.34 V – (0.0592/2) log(1/1) = +0.34 V.
Identify Fe as the anode (oxidation) and Cu as the cathode.
Compute cell emf: Ecell = Ecathode – Eanode = 0.34 V – 0.0065 V = 0.3335 V ≈ 0.33 V.

Common mistake: Using the wrong sign when subtracting anode potential

Concept tested: Cell emf using Nernst equation

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked