$\mathrm{E}_{\mathrm{Fe} 2+/ \mathrm{Fe}}^{\mathrm{o}}=-0.441 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Fe}^{3+}…
$\mathrm{E}_{\mathrm{Fe} 2+/ \mathrm{Fe}}^{\mathrm{o}}=-0.441 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Fe}^{3+} \mathrm{Fe}^{2+}}^{\mathrm{O}}=0.771$
$\mathrm{V}$, the standard EMF of the reaction $\mathrm{Fe}$ $+2 \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+}$ will be:
$0.111 \mathrm{~V}$
$0.330 \mathrm{~V}$
$1.653 \mathrm{~V}$
$1.212 \mathrm{~V}$
Solution
For the cell reaction
$\begin{aligned}
& \mathrm{Fe}+2 \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+} \\
& \text { Anode reaction is } \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \\
& \text {Cathode reaction in } 2 \mathrm{Fe}^{3+}+2 \mathrm{e}^{-} \rightarrow \\
& 2 \mathrm{Fe}^{2+} \\
& \mathrm{E}_{\mathrm{Cell}}^{\circ}=\mathrm{E}_{\text {Cathode }}^{\circ}-\mathrm{E}_{\text {Anode }}^{\circ} \\
& =0.771-(-0.441) \\
& \mathrm{E}_{\mathrm{Cell}}^{\circ}=1.212 \mathrm{~V} \\
&
\end{aligned}$
Related Theory
If the cell potential is negative, the reaction is reversed. In this case, the electrode of the galvanic cell should be written in a reversed order.
Caution
The positive potential indicates a spontaneous reaction