For the reaction: Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V. The reduction potential at [Cu²⁺] = 0.1 M is
Q1. For the reaction: Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V. The reduction potential at [Cu²⁺] = 0.1 M is
Answer: +0.31 V
Explanation: E = E° - (0.059/n) * log (1 / [Cu²⁺]), E = 0.34 - (0.059/2) * log (1 / 0.1) = 0.34 - 0.0295 = 0.31 V.