The standard reduction potential E0 for the half reactions are as

$\begin{array}{r}\mathrm{Zn}={\mathrm{Zn}}^{2+}+2e^{-};E^{\circ }=+0.76\mathrm{V}\\ \mathrm{Fe}={\mathrm{Fe}}^{2+}+2e^{-};E^{\circ }=+0.41\mathrm{V}\end{array}$

The $EMF$ for cell reaction  ${\mathrm{Fe}}^{2+}+\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+\mathrm{Fe}$ is 

In this reaction, reduction is taken place at cathode and oxidation takes place at anode. Fe forms at cathode whereas Zn forms at anode.

$$\begin{gathered} \mathrm{EMF}={\mathrm{E}}_{\text{cathode }}-{\mathrm{E}}_{\text{anode }} \\ =-0.41 \mathrm{V}-(-0.76 \mathrm{V})=+0.35\mathrm{V} \end{gathered}$$