Use standard reduction potentials to calculate $\mathscr{C}^{\circ}, \Delta G^{\circ},\( and \)K$ (at 298 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 $$\begin{aligned} \operatorname{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-0.60 \mathrm{V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-1.26 \mathrm{V} \end{aligned}$$

We are provided with the balanced redox reaction already: $$2\;\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{Zn}(s)\longrightarrow 2\;\mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}$$

First, let's identify the two half-reactions involved. The given reduction half-reactions are: $$\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-}\longrightarrow \mathrm{Au}+2\;\mathrm{CN}^{-}\;\;\;\;\mathscr{C}^{\circ}=-0.60\;\mathrm{V}$$ $$\mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2\;\mathrm{e}^{-}\longrightarrow \mathrm{Zn}+4\;\mathrm{CN}^{-}\;\;\;\;\mathscr{C}^{\circ}=-1.26\;\mathrm{V}$$ For the second equation to match the overall reaction, it must be reversed and turned into an oxidation half-reaction: $$\mathrm{Zn}+4\;\mathrm{CN}^{-} \longrightarrow\mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2\;\mathrm{e}^{-}\;\;\;\;\mathscr{C}^{\circ}=1.26\;\mathrm{V}$$ Now, we can find the standard cell potential by adding the standard reduction potentials of the two half-reactions: $$\mathscr{C}^{\circ}_{\text{cell}}= \mathscr{C}^{\circ}_{\text{red}}+\mathscr{C}^{\circ}_{\text{ox}}= -0.60\;\mathrm{V}+1.26\;\mathrm{V}=0.66\;\mathrm{V}$$

We can calculate the standard free energy change using the standard cell potential: $$\Delta G^{\circ}=-nFE_{\text{cell}}^{\circ}$$ F is the Faraday constant \((96,\!485\;\mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{V}^{-1})\), n is the number of transferred electrons (2 in this case), and \(E_{\text{cell}}^{\circ}\, (= \mathscr{C}_{\text{cell}}^{\circ})\) is the standard cell potential. $$\Delta G^{\circ}=-(2\;\text{mol}\;\mathrm{e}^{-})(96,\!485\;\mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{V}^{-1})(0.66\;\mathrm{V})=-127,\!640\;\mathrm{J}\cdot\mathrm{mol}^{-1}$$

Finally, we can calculate the equilibrium constant using the following relationship: $$\Delta G^{\circ}=-RT\ln K$$ R is the gas constant \((8.314\;\mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1})\), and T is the temperature (298 K). $$K = \exp{\left(-\frac{\Delta G^{\circ}}{RT}\right)}= \exp{\left(-\frac{-127,\!640\;\mathrm{J}\cdot\mathrm{mol}^{-1}}{(8.314\;\mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1})(298\;\mathrm{K})}\right)}\approx 5.90\times10^{20}$$ In conclusion, for this redox reaction, the standard cell potential is 0.66 V, the standard free energy change is -127,640 J/mol, and the equilibrium constant is approximately \(5.90\times10^{20}\).