What is the EMF of the cell for the reaction $F{e}^{2+}+Zn\to Z{n}^{2+}+Fe.$Given that $E_{Zn/Z{n}^{+2}}^0=0.76V;E_{Fe/F{e}^{+2}}^0=0.44V$ 

Among Zn and Fe electrodes, Zn is more electropositive as its SRP value(-0.76V) is less than that of Fe(-0.44V), hence Zn acts as anode and Fe acts as cathode. $E_{cell}^0=E_{Fe}^0-E_{Zn}^0=-0.44V-\left(-0.76V\right)=+0.32V$ 
Note: Standard potentials mentioned in the question represent oxidation potential, hence changes to get reduction potentials