Calculate the EMF of the cell,
$\mathrm{Na}/{\mathrm{Na}}^{+}\left(1\mathrm{M}\right)//{\mathrm{Fe}}^{+2}\left(1\mathrm{M}\right)/\mathrm{Fe}$
Given, ${\mathrm{E}}_{{\mathrm{Na}}^{+}/\mathrm{Na}}^0=-2.71\mathrm{V}\text{ and }{\mathrm{E}}_{{\mathrm{Fe}}^{+2}/\mathrm{Fe}}^0=-0.44\mathrm{V}$

${\mathrm{E}}_{\mathrm{Na}+/\mathrm{Na}}^0=-2.71\mathrm{V}\text{ and }{\mathrm{E}}_{{\mathrm{Fe}}^{+2}/\mathrm{Fe}}^0=-0.44\mathrm{V}$
Both are S.R.P values ; 
From the cell notation ,
Sodium electrode( L.H.E ) is the Anode, 
Iron electrode ( R.H.E ) is the Cathode,
${\mathrm{E}}_{\text{cell }}^0={\mathrm{E}}_{\text{Cathode }}^0-{\mathrm{E}}_{\text{Anode }}^0=-0.44\mathrm{V}-(-2.71)\mathrm{V}=+2.27\mathrm{V}$