The standard potentials (E⁰) for the half reactions are as 
$\begin{array}{r}\mathrm{Zn}⟶{\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-},{\mathrm{E}}^0=+0.76\mathrm{V}\\ \mathrm{Fe}⟶{\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-},{\mathrm{E}}^0=+0.41\mathrm{V}\end{array}$
the emf for the cell reaction ${\mathrm{Fe}}^{2+}+\mathrm{Zn}⟶{\mathrm{Zn}}^{2+}+\mathrm{Fe}\text{ is }$

$$\begin{gathered} {\mathrm{Zn}}^{2+\hspace{0.17em}}+\hspace{0.17em}2 {\mathrm{e}}^{-}\to \mathrm{Zn}; {\mathrm{E}}^{\mathrm{o}}= +0.76\mathrm{V} \\ \mathrm{Fe}\to {\mathrm{Fe}}^{2+}+\hspace{0.17em}2{\mathrm{e}}^{-} ; {\mathrm{E}}^{\mathrm{o} }= +0.41\mathrm{V}\hspace{0.17em} \end{gathered}$$

Given are oxidation potentials.
Reduction potentials are equal and opposite. 
Fe forms cathode and Zn forms anode.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}= {\mathrm{E}}_{\mathrm{reduction} }^{\mathrm{o}}+\hspace{0.17em}{\mathrm{E}}_{\mathrm{Oxidation} }^{\mathrm{o}} \\ = (-0.41)\hspace{0.17em}+\hspace{0.17em}(0.76)\hspace{0.17em} \\ =+ 0.35\mathrm{V}\hspace{0.17em} \end{gathered}$$