What is the EMF of the cell for the reaction ${\mathrm{Fe}}^{2+}+\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+\mathrm{Fe}$.  Given that ${\mathrm{E}}^0\mathrm{Zn}/{\mathrm{Zn}}^{2+}\left(1.0\mathrm{M}\right)=0.76\mathrm{V};{\mathrm{E}}^0\mathrm{Fe}/{\mathrm{Fe}}^{2+}\left(1.0\mathrm{M}\right)=0.41\mathrm{V}$

Among Zn and Fe electrodes, Zn is more  electropositive as its SRP value(-0.76V) is  less than that of Fe (- 0.41V ), hence Zn  acts as anode and Fe acts as cathode.
E⁰cell = E⁰Fe – E⁰Zn = -0.41V – (-0.76V) = +0.35V
Note: Standard potentials mentioned in th  question represent oxidation potential,  hence change sign to get reduction  potentials