Q. Given: (i) Cu²⁺ + 2e⁻ → Cu, E° = 0.337 V (ii) Cu²⁺ + e⁻ → Cu⁺, E° = 0.153 V Electrode potential, E° for the reaction, Cu⁺ + e⁻ → Cu, will be:
A) 0.90 V
B) 0.30 V
C) 0.38 V
D) 0.52 V
Explanation
[Solution from source] Combine free energies, not potentials: ΔG° for Cu⁺ + e⁻ → Cu = ΔG°(i) − ΔG°(ii) = (−2 × 0.337F) − (−1 × 0.153F) = −0.521F. With n = 1, E° = +0.52 V. Simply subtracting the potentials gives 0.18 V and adding them gives 0.49 V, both wrong. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.