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  ${\mathrm{E}}_{{\mathrm{Na}}^{+}/\mathrm{Na}}^0=-2.71 \mathrm{V} \mathrm{and} {\mathrm{E}}_{{\mathrm{Fe}}^{+2}/\mathrm{Fe}}^0=-0.44 \mathrm{V} ;$

$$\begin{gathered} {\mathrm{E}}_{{\mathrm{Na}}^{+}/\mathrm{Na}}^0=-2.71 \mathrm{V} \mathrm{and} {\mathrm{E}}_{{\mathrm{Fe}}^{+2}/\mathrm{Fe}}^0=-0.44 \mathrm{V}.; \\ \mathrm{Both} \mathrm{are} \mathrm{S}.\mathrm{R}.\mathrm{P} \mathrm{values}.; \end{gathered}$$

From the cell notation ,

Sodium electrode( L.H.E ) is the Anode,

Iron electrode ( R.H.E ) is the Cathode,

${\mathrm{E}}_{\mathrm{cell}}^{°} ={\mathrm{E}}_{\mathrm{Cathode}}^{°}-{\mathrm{E}}_{\mathrm{Anode}}^{°}=-0.44 \mathrm{V}-(-2.71) \mathrm{V} =+2.27 \mathrm{V}$