Calculate the equilibrium constant K of the reaction?

Au⁺(aq) + e⁻ → Au(s) …… E° = +1.69 V
Au(s) + 2CN⁻(aq) → [Au(CN)₂]⁻(aq) …… E° = -0.60 V

Add the above two equations, and cancel e⁻ and Au(s) on the both sides.
Au⁺(aq) 2CN⁻(aq) → [Au(CN)₂]⁻(aq) …… E°(cell) = (+1.69) + (-0.60) V = +1.09 V

E°(cell) = [RT/(nF)] ln(K)
+1.09 = [8.314 × 298 / (1 × 96500)] ln(K)
K = e^[1.09 × 96500 / (8.314 × 298)]
Equilibrium constant, K = 2.74 × 10¹⁸

K = Kf ÷ Kb = products ÷ reactants
Delta G° = -2.303 RT log K