Consider the disproportionation reaction of Gold(I) ions in an aqueous solution: 3 Au ⁺_ ( aq ) Au ³⁺_ ( aq ) + 2 Au _ ( s ) If the standard cell potential for this disproportionation reaction is E₁ , and the standard reduction potential for the half-reaction Au ³⁺ + 3 e ⁻ Au is E₂ , what is the standard reduction potential of the half-reaction Au ⁺ + e ⁻ Au ?

Step-by-step solution

Let the standard reduction potential for Au ⁺ + e ⁻ Au be x . The disproportionation reaction is: 3 Au ⁺ Au ³⁺ + 2 Au This can be split into two half-reactions: Oxidation (Anode): Au ⁺ Au ³⁺ + 2 e ⁻ Reduction (Cathode): Au ⁺ + e ⁻ Au The standard cell potential is given by: E₁ = E^ _ cathode - E^ _ anode = x - E^ _ Au ³⁺/ Au ⁺ To find E^ _ Au ³⁺/ Au ⁺ , we use the standard Gibbs free energy change ( G^ = -nFE^ ): 1) Au ³⁺ + 3 e ⁻ Au G^ ₁ = -3FE₂ 2) Au ⁺ + e ⁻ Au G^ ₂ = -Fx Subtracting equation (2) from (1) gives th