The standard potentials at ${25}^{o}C$ for the following half reactions are given against them 

$\begin{array}{l}{\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Zn},{\mathrm{E}}^{\mathrm{o}}=-0.762\mathrm{V}\\ {\mathrm{Mg}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mg},{\mathrm{E}}^{\mathrm{o}}=-2.37\mathrm{V}\end{array}$

When zinc dust is added to a solution of magnesium chloride 

$\begin{array}{l}\mathrm{Zn}+{\mathrm{MgCl}}_2\to {\mathrm{ZnCl}}_2+\mathrm{Mg}\\ \therefore {\mathrm{E}}_{\text{cell}}^{\mathrm{o}}={\mathrm{E}}_{{\mathrm{Zn}}^{+2}/\mathrm{Zn}}^{\mathrm{o}}-{\mathrm{E}}_{{\mathrm{Mg}}^{+2}/\mathrm{Mg}}^{\mathrm{o}}=-0.762-\left(-2.37\right)\\ =+1.608\mathrm{V}\end{array}$

Here, ${\mathrm{E}}_{\text{cell}}^{\mathrm{o}}$ is  positive so  ${\mathrm{Zn}}^{+2}$ is reduced by $\mathrm{Mg}$ but $\mathrm{Zn}$ can't reduces ${\mathrm{Mg}}^{+2}$ ions