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A solution contains $\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}$ and $\mathrm{I}^{-}$ions. This solution was treated with iodine at $35^{\circ} \mathrm{C}$. $E^{\circ}$ for $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ is $+0.77 \mathrm{~V}$ and $\mathrm{E}^{\circ}$ for $\mathrm{I}_2 / 2 \mathrm{I}^{-}=0.536 \mathrm{~V}$. The favourable redox reaction is
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Verified Answer
The correct answer is:
$\mathrm{I}^{-}$will be oxidised to $\mathrm{I}_2$
$2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_2+2 e^{-}$(Oxidation half-reaction)
$$
E_{\text {oxi. }}^{\circ}=-0.536 \mathrm{~V} \text {. }
$$
$\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+}$ (Reduction half-reaction)
$$
\begin{aligned}
E^{\circ} & =E_{\text {oxi }}^{\circ}+E_{\text {red }}^{\circ} \\
& =+\mathrm{ve}
\end{aligned}
$$
So, reaction will take place.
$$
E_{\text {oxi. }}^{\circ}=-0.536 \mathrm{~V} \text {. }
$$
$\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+}$ (Reduction half-reaction)
$$
\begin{aligned}
E^{\circ} & =E_{\text {oxi }}^{\circ}+E_{\text {red }}^{\circ} \\
& =+\mathrm{ve}
\end{aligned}
$$
So, reaction will take place.
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