If E ^ ( Mg ⁺² _ ( aq ) Mg _ ( s ) )=-2.37 ~V . What is potential for Mg _ ( s ) Mg ⁺²(0.1 M )+2 e at 298 K ?

Step-by-step solution

The standard oxidation potential E^ _ ox is determined from the given reduction potential E^ ( Mg ⁺²_ ( aq ) Mg _ ( s ) ) = -2.37 , V : E^ _ ox = -E^ _ red = +2.37 , V For the reaction Mg _ ( s ) Mg ⁺²(0.1 , M ) + 2e^- , the number of electrons transferred is n = 2 and the reaction quotient is Q = [ Mg ⁺²] = 0.1 . The Nernst equation gives the oxidation potential as: E = E^ _ ox - 0.0592 n Q = 2.37 - 0.0592 2 (0.1) Since (0.1) = -1 , the calculation becomes: E = 2.37 - 0.0296 (-1) = 2.37 + 0.0296 = 2.3996 , V Final