Answer
$2.47\times10^{54}$
Work Step by Step
$E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=E^{\circ}_{Zn^{2+}/Zn}-E^{\circ}_{Mg^{2+}/Mg}$
$=-0.76\,V-(-2.37\,V)=1.61\,V$
$E^{\circ}_{cell}=\frac{0.0592\,V}{n}\log K$
where $n$ is the number of electrons transferred and $K$ is the equilibrium constant.
$n=2$
$\log K=\frac{nE^{\circ}_{cell}}{0.0592\,V}=\frac{2\times1.61\,V}{0.0592\,V}=54.392$
$K=10^{54.392}=2.47\times10^{54}$
