A solution contains $\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}$ and $\mathrm{I}^{-}$ ions. This solution was…

A solution contains $\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}$ and $\mathrm{I}^{-}$ ions. This solution was treated with iodine at $35^{\circ} \mathrm{C} . \mathrm{E}^{\circ}$ 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 :
  1. $\mathrm{I}_{2}$ will be reduced to $\mathrm{I}^{-}$
  2. There will be no redox reaction
  3. $\mathrm{I}^{-}$ will be oxidised to $\mathrm{I}_{2}$
  4. $\mathrm{Fe}^{2+}$ will be oxidised to $\mathrm{Fe}^{3+}$

Solution

Given $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=+0.77 \mathrm{~V}$
and $\mathrm{I}_{2} / 2 \mathrm{I}^{-}=0.536 \mathrm{~V}$
$2\left(\mathrm{e}^{-}+\mathrm{Fe}^{3+} \longrightarrow \mathrm{Fe}^{2+}ight) E^{\circ}=0.77 \mathrm{~V}$
$2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_{2}+2 \mathrm{e}^{-}$
$E^{\circ}=-0.536 \mathrm{~V}$
$2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_{2}$
$E^{\circ}=E_{{ }_{\mathrm{ox}}}^{\circ}+E^{\circ}{ }_{\mathrm{red}}$
$=0.77-0.536=0.164 \mathrm{~V}$
So, reaction will take place. *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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