The standard Gibbs energy change ($\Delta_r G^{\circ}$) is related to the standard cell potential ($E^{\circ}_{cell}$) by the equation: $\Delta_r G^{\circ} = -nFE^{\circ}_{cell}$ For the given cell reaction, $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$, the number of electrons transferred ($n$) is $2$. Given: $E^{\circ}_{cell} = 1.1\text{ V}$ $F = 96487\text{ C mol}^{-1}$ Substitute the values into the equation: $\Delta_r G^{\circ} = -2 \times 96487\text{ C mol}^{-1} \times 1.1\text{ V}$ $\Delta_r G^{\circ} = -212271.4\text{ J mol}^{-1}$ $\Delta_r G^{\circ} = -212.2714\text{ kJ mol}^{-1}$ Therefore, the standard Gibbs energy change for the cell reaction is $-212.27\text{ kJ mol}^{-1}$.