The standard electrode potential for the half cell reactions are: 

$\begin{array}{l}{\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}⟶\mathrm{Zn};\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} {\mathrm{E}}^{\circ }=-0.76\mathrm{V}\\ {\mathrm{Fe}}^{++}+2{\mathrm{e}}^{-}⟶\mathrm{Fe};\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} {\mathrm{E}}^{\circ }=-0.44\mathrm{V}\end{array}$

The e.m.f of the cell reaction, ${\mathrm{Fe}}^{2+}+\mathrm{Zn}⟶{\mathrm{Zn}}^{++}+\mathrm{Fe}$ is:

• In the cell reaction, ${\mathrm{Fe}}^{2+}+\mathrm{Zn}⟶{\mathrm{Zn}}^{2+}+\mathrm{Fe},$
• Zn is oxidized to ${\mathrm{Zn}}^{2+}$ and ${\mathrm{Fe}}^{2+}$ is reduced to Fe. 
• So, the cell can be represented as:
• $\mathrm{Zn}/{\mathrm{Zn}}^{2+}\left|\right|{\mathrm{Fe}}^{2+}/\mathrm{Fe}$

$\begin{array}{r}\therefore {\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{right }}-{\mathrm{E}}_{\text{left }}={\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}-{\mathrm{E}}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}\\ =(-0.44\mathrm{V})-(-0.76\mathrm{V})\\ =0.76\mathrm{V}-0.44\mathrm{V}=0.32\mathrm{V}\end{array}$