Calculate the emf of the following galvanic cell at 298 K : Cu(s) + 2 Ag ⁺( aq ) Cu ²⁺( aq ) + 2 Ag(s) Given: E^ _ cell = 0.46 V , [ Cu ²⁺] = 0.1 M , [ Ag ⁺] = 0.01 M , and 2.303 RT F = 0.059 V at 298 K .

Step-by-step solution

The given cell reaction is: Cu(s) + 2 Ag ⁺( aq ) Cu ²⁺( aq ) + 2 Ag(s) The number of electrons transferred, n = 2 . According to the Nernst equation: E_ cell = E^ _ cell - 0.059 n [ Cu ²⁺] [ Ag ⁺]² Substituting the given values: E_ cell = 0.46 - 0.059 2 0.1 (0.01)² E_ cell = 0.46 - 0.0295 0.1 10⁻⁴ E_ cell = 0.46 - 0.0295 (10³) E_ cell = 0.46 - 0.0295 3 E_ cell = 0.46 - 0.0885 = 0.3715 V If [ Ag ⁺] is not squared in the reaction quotient, it leads to the incorrect value of 0.4305 V . Answer: 0.3715 V