Question

A solution contains $F{{e}^{2+}},F{{e}^{3+}}$ and ${{I}^{-}}$ ions. The solution was treated with iodine at $35{}^\circ C$. $E{}^\circ $for $F{{e}^{3+}}/F{{e}^{2+}}$ is +0.77V and $E{}^\circ $ for ${{I}_{2}}/2{{I}^{-}}$= 0.536 V. The favorable redox reaction is:
 A. ${{I}_{2}}$ will be reduced to${{I}^{-}}$.
B. There will be no redox reaction.
C. ${{I}^{-}}$will be oxidized to ${{I}_{2}}$.
D. $F{{e}^{2+}}$will be oxidized to $F{{e}^{3+}}$

Answer 1

Hint: Redox reaction can be defined as that chemical reaction in which electrons are transferred between the reactants participating in the chemical reaction. The transfer of electrons involves the changes in the oxidation state of the reacting species.
Step by step explanation: In the given case oxidation is possible which favors forward reaction in this case oxidation of iodine will oxidize with iodide ion i.e. ${{I}^{-}}$to ${{I}_{2}}$. Reactions of both can be shown as:
\[2{{e}^{-}}+F{{e}^{3+}}\to F{{e}^{2+}}\]and the value of $E{}^\circ $is given i.e. 0.77 V similarly we can write the equation for iodine and this can be shown as:
\[2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}\]and the value of $E{}^\circ $is 0.536 V which is given already. Now we have to add both reactions to balance the overall reaction. Hence the overall reaction will be shown as follows:
\[2F{{e}^{3+}}+2{{I}^{-}}\to 2F{{e}^{2+}}+{{I}_{2}}\]
Now we know that \[E={{E}_{ox}}+{{E}_{red}}\], here \[{{E}_{ox}}\]corresponds to oxidation and \[{{E}_{red}}\]corresponds to reduction and here we can see from the reactions that iron will treat as oxidizing agent and iodine as reducing agent.
Therefore \[E=0.77-0.536=0.134V\]
Value of cell potential is positive so we can say that a reaction will take place.

Thus option C is the correct answer.

Note:
Oxidation is defined as the loss of electrons or we can say that increase in the oxidation state of an atom, ion or molecule while reduction can be defined as the gain of electrons or decreases in the oxidation state of any given atom, ion or molecule.