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 $$ \begin{aligned} \mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{E}^{\circ}=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ}=-1.26 \mathrm{~V} \end{aligned} $$

To calculate the standard cell potential for the given redox reaction, we will use the standard reduction potentials of the half-reactions. Remember that the standard cell potential is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode. Here, we should identify which half-reaction is the reduction and which half-reaction is the oxidation. The given reaction is: $$ 2 \mathrm{Au} (\mathrm{CN})_{2}^{-}(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q) $$ The first half-reaction is: $$ \mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} \qquad \mathscr{E}^{\circ}=-0.60 \mathrm{~V} $$ The second half-reaction is: $$ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} \quad \mathscr{E}^{\circ}=-1.26 \mathrm{~V} $$ In the overall redox reaction, Zn is oxidized to Zn(CN)\(_4^{2-}\) and Au(CN)\(_2^-\) is reduced to Au. Therefore, the first half-reaction is the reduction half-reaction (cathode) and the second half-reaction is the oxidation half-reaction (anode). Now, let's calculate the standard cell potential (\(\mathscr{E}^{\circ}\)): $$ \mathscr{E}^{\circ} = \mathscr{E}_{cathode}^{\circ} - \mathscr{E}_{anode}^{\circ} = (-0.60) -(-1.26) = 0.66 \mathrm{~V} $$

To calculate the standard Gibbs free energy change, we will use the following formula: $$ \Delta G^{\circ} = -nFE^{\circ} $$ Here, \(n\) is the number of electrons transferred, \(F\) is Faraday's constant (\(96485 \mathrm{~C/mol}\)), and \(\mathscr{E}^{\circ}\) is the standard cell potential. In this case, 2 electrons are transferred in the overall redox reaction (both half-reactions show the transfer of 2 electrons and are already balanced). Now, plug in the values and calculate \(\Delta G^{\circ}\): $$ \Delta G^{\circ} = -(2)(96485 \mathrm{~C/mol})(0.66 \mathrm{~V}) = -127680.6 \mathrm{~J/mol} = -127.68 \mathrm{~kJ/mol} $$

To calculate the equilibrium constant, we will use the following formula: $$ \Delta G^{\circ} = -RT\ln {K} $$ Where \(R\) is the gas constant (\(8.314 \mathrm{~J/(mol \cdot K)}\)), \(T\) is the temperature in Kelvin (given as \(298 \mathrm{~K}\)), and \(K\) is the equilibrium constant. To find \(K\), we can rearrange this formula: $$ K = e^{\frac{-\Delta G^{\circ}}{RT}} $$ Now, plug in the values and calculate \(K\): $$ K = e^{\frac{-( -127680.6 \mathrm{~J/mol})}{(8.314 \mathrm{~J/(mol \cdot K)})(298 \mathrm{~K})}} = 4.39\times10^{21} $$ The equilibrium constant (\(K\)) for the reaction is approximately \(4.39\times10^{21}\).