$$\begin{gathered} {\text{E}}^{\circ } for {\mathrm{Fe}}^{2+}+2\mathrm{e}⟶\mathrm{Fe} is -0.44\mathrm{V} \\ {\mathrm{E}}^{\circ } \mathrm{for} {\mathrm{Zn}}^{2+}+2\mathrm{e}⟶\mathrm{Zn} \mathrm{is} -0.76\mathrm{V}.\mathrm{Then} \end{gathered}$$

The E^o values for the reduction process are given;

$$\begin{gathered} {\mathrm{E}}_{\mathrm{red}}^{\mathrm{o}} \mathrm{for} {\mathrm{Zn}}^{2+}/\mathrm{Zn} = -0.76\mathrm{V} \\ {\mathrm{E}}_{\mathrm{red} }^{\mathrm{o}}\mathrm{for} {\mathrm{Fe}}^{2+}/\mathrm{Fe} = -0.44\mathrm{V} \end{gathered}$$

Metal with smaller reduction potential value has  more tendency to oxidize(loose lectrons) 

Hence Zn is more electropositive than Fe.