Question

A solution contains ${Fe^{2+}, Fe^{3+}}$ and $I^-$ ions. This solution was treated with iodine at ${ 35^{\circ} C. E^{\circ}}$ for ${Fe^{3+} /Fe^{2+}}$ is $+0.77 \, V$ and $E^{\circ}$ for $I_2/2I^- = 0.536 \, V.$ The favourable redox reaction is-

• A. ${Fe^{2+}}$ will be oxidized to $ {Fe^{3+}}$
• B. $I_2$ will be the reduced to $I^-$
• C. There will be no redox reaction
• D. $I^-$ will be oxidized to $I_2$

Answer 1

The Correct Option is D

To determine the favourable redox reaction in the given solution, we need to compare the standard electrode potentials (E^{\circ}) of the reactions involved. The reactions provided are:

1. The oxidation of Fe^{2+} to Fe^{3+}:

   The standard electrode potential for this half-reaction is given as E^{\circ}_{Fe^{3+}/Fe^{2+}} = +0.77 \, V.

2. The reduction of I_2 to I^-\:

   The standard electrode potential for this half-reaction is given as E^{\circ}_{I_2/2I^-} = +0.536 \, V.

The tendency of the reaction to occur is determined by the reduction potential. Higher reduction potential means a higher tendency for reduction to occur. However, since both reactions given are in standard reduction potential form, we will consider them as is:

• Fe^{3+} + e^- \rightarrow Fe^{2+} has E^{\circ} = +0.77 \, V.
• I_2 + 2e^- \rightarrow 2I^-\ has E^{\circ} = +0.536 \, V.

For both an oxidation and reduction reaction to occur simultaneously (redox), one must act as an oxidizing agent and the other as a reducing agent. The reaction with the higher E^{\circ} value acts as the cathode (reduction), and the one with the lower value acts as the anode (oxidation).

Since E^{\circ}_{Fe^{3+}/Fe^{2+}} = +0.77 \, V is higher, Fe^{3+} would potentially be reduced, but since the solution also contains I^-\, it acts as a reducing agent.

The correct favourable reaction, considering E^{\circ} values, is:

• I^-\ is oxidized to I_2.

This is because Fe^{3+} has a higher tendency to be reduced more than I_2 at 35^{\circ} C.