Use standard reduction potentials to calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}^{-}(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned}\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ} &=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ} &=-1.26 \mathrm{~V} \end{aligned}$$

Since we are given the standard reduction potentials of both half-reactions, we can calculate the standard cell potential by subtracting the reduction potential of the half-reaction that needs to be reversed (oxidation) from the reduction potential of the reaction that will remain as it is (reduction). $$\mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{red} - \mathscr{E}^{\circ}_{ox}$$ We need to reverse the second half-reaction because Zn needs to be oxidized, not reduced. Therefore, we will keep the first half-reaction's standard reduction potential as is and reverse the second half-reaction's standard reduction potential. $$\mathscr{E}^{\circ}_{cell} = (-0.60\,\text{V}) - (-1.26\,\text{V}) = 0.66\,\text{V}$$

To calculate the standard Gibbs free energy change, we can use the following formula: $$\Delta G^{\circ}=-nF\mathscr{E}^{\circ}$$ where \(n\) is the total number of electrons transferred in the balanced redox reaction \(F\) is the Faraday constant, equal to \(96485\frac{\text{C}}{\text{mol}}\) \(\mathscr{E}^{\circ}\) is the standard cell potential For our reaction, \(n = 2\) for transferring two electrons. $$\Delta G^{\circ} = - (2\,\text{mol} \cdot 96485\,\frac{\text{C}}{\text{mol}} \cdot 0.66\,\text{V}) = -127599\,\text{J/mol}$$ Converting to kJ/mol for convenience: $$\Delta G^{\circ} = -127.6\,\text{kJ/mol}$$

We can relate the equilibrium constant to the standard Gibbs free energy change using the following equation: $$\Delta G^{\circ} = -RT\ln{K}$$ where \(R\) is the universal gas constant, equal to \(8.314\frac{\text{J}}{\text{mol}\cdot\text{K}}\) \(T\) is the absolute temperature in Kelvin \(K\) is the equilibrium constant We are given the temperature as 298 K. Rearranging the equation, we have: $$K = e^{\frac{-\Delta G^{\circ}}{RT}}$$ Plugging in the known values: $$K = e^{\frac{-(-127599\,\text{J/mol})}{(8.314\,\frac{\text{J}}{\text{mol}\cdot\text{K}})(298\,\text{K})}}$$ Calculating for \(K\): $$K \approx 1.49\times10^{17}$$ In conclusion, we have found that the standard cell potential for the gold production reaction is 0.66 V, the standard Gibbs free energy change is -127.6 kJ/mol, and the equilibrium constant is \(1.49\times10^{17}\).