So, $Fe \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-}, \mathrm{E}^{\circ}=+0.441 \mathrm{V} \ldots .$ (i)

and $E_{F e^{3}+/ F_{e}^{2}}^{0}=0.771 \mathrm{V}$

So, $Fe ^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}, \mathrm{E}^{\circ}=0.771 \mathrm{V} \ldots \ldots$ (ii)

Cell reaction (i) $Fe \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-}, \quad \mathrm{E}^{\circ}=0.441 \mathrm{V}$

(ii) $2 \mathrm{Fe}^{3+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Fe}^{2,+} \quad \mathrm{E}^{\circ}=+0.771 \mathrm{V}$

$\mathrm{Fe}^{2+}+2 \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+}, \quad E_{\mathrm{cell}}^{0}=1.212 \mathrm{V}$

So, on the basis of cell reaction following half-cell reactions are written

At anode:

$\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \quad$ (oxidation)

At cathode:

$2 \mathrm{Fe}^{3+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Fe}^{2+}(\text { reduction })$

$\mathrm{So}$

$\mathrm{E}_{\mathrm{cell}}^{0}=\mathrm{E}_{\text {cathode }}^{0}-\mathrm{E}_{\text {anode }}^{0}$

$=E_{F e^{3+} / / F e^{2 +}}^{0}-E_{F e^{2+}{ / F e}}^{0}$

$(0+.771)-(-0.441)=+1.212 \mathrm{V}$