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

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
  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

$2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_2+2 e^{-}$(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.

Asked in: NEET 2011 (Mains)

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