What is the reduction potential of a half-cell consisting of a Pt electrode dipped in 2.2 M Fe ²⁺ and 0.04 M Fe ³⁺ solution where the reaction taking place is conversion of Fe ³⁺ ions to Fe ²⁺ ? E ^0 ( Fe ³⁺ / Fe ²⁺ )=0.771 ~V

Step-by-step solution

The reduction half-reaction is given by Fe ³⁺ + e⁻ Fe ²⁺ . Using the Nernst equation for the electrode potential at 298 K : E = E⁰ - 0.0591 n [ Fe ²⁺] [ Fe ³⁺] Given E⁰ = 0.771 V , n = 1 , [ Fe ²⁺] = 2.2 M , and [ Fe ³⁺] = 0.04 M . Substituting the values: E = 0.771 - 0.0591 1 ( 2.2 0.04 ) E = 0.771 - 0.0591 (55) Since (55) 1.74036 : E = 0.771 - 0.0591 1.74036 E = 0.771 - 0.10285 E 0.66815 V Answer: 0.668 V