Q. The electrode potentials for Cu²⁺(aq) + e⁻ → Cu⁺(aq) and Cu⁺(aq) + e⁻ → Cu(s) are +0.15 V and +0.50 V respectively. The value of E°(Cu²⁺/Cu) will be
A) 0.150 V
B) 0.500 V
C) 0.325 V
D) 0.650 V
Explanation
[Solution from source] Potentials are not additive but free energies are: ΔG° = −nFE°. −2F·E° = −1F(0.15) − 1F(0.50), so E°(Cu²⁺/Cu) = 0.65/2 = 0.325 V. Simply adding the potentials gives 0.65 V, the trap. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.