The standard potentials $\left(E^0\right)$ for the half reactions are as

$\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}; {\mathrm{E}}^{{ }^0}=+0.76\mathrm{V}$

$\mathrm{Fe}\to {\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}; {\mathrm{E}}^0=+0.41\mathrm{V}$

Calculate the EMF for the given cell reaction

${\mathrm{Fe}}^{2+}+\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+\mathrm{Fe}$

Standard reduction potential of Zn is -0.70 V and for Fe, -0.41 V.

Standard Reduction potential of Zn is less than that of Fe, so Zinc at anode and Fe at cathode.

Cell reaction is:

$\mathrm{Zn}+{\mathrm{Fe}}^{+2}\to {\mathrm{Zn}}^{+2}+\mathrm{Fe}$

$$\begin{gathered} \mathrm{EMF}={\mathrm{E}}_{\mathrm{cathode}}-{\mathrm{E}}_{\mathrm{anode}} \\ \mathrm{EMF}={\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\circ }-{\mathrm{E}}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}^{\circ } \\ \mathrm{EMF}=-0.41 \mathrm{V}-\left(-0.76 \mathrm{V}\right)=+0.35 \mathrm{V} \end{gathered}$$