question_answer

A solution contains \[F{{e}^{2+}},F{{e}^{3+}}\]and \[{{\text{I}}^{-}}\] ions. This solution was treated with iodine at\[35{}^\circ C\]. \[E{}^\circ \] for \[F{{e}^{3+}}/F{{e}^{2+}}\]is + 0.77 V and \[E{}^\circ \] for \[{{\text{I}}_{\text{2}}}\text{/2}{{\text{I}}^{-}}=\text{0}\text{.536}\,\text{V}\text{.}\]The favorable redox reaction is               

A) \[{{\text{I}}_{\text{2}}}\] will be reduced to \[{{\text{I}}^{-}}\]

B)  There will be no redox reaction

C) \[{{\text{I}}^{-}}\] will be oxidised to \[{{\text{I}}_{\text{2}}}\]

D)  \[\text{F}{{\text{e}}^{\text{2+}}}\]will be oxidised to \[\text{F}{{\text{e}}^{\text{3+}}}\]

Correct Answer: C

Solution :

\[2{{I}^{-}}\xrightarrow[{}]{{}}{{I}_{2}}+2{{e}^{-}}\](Oxidation half-reaction) \[E_{oxi.}^{o}=-0.536\,V.\] \[F{{e}^{3+}}+{{e}^{-}}\xrightarrow[{}]{{}}F{{e}^{2+}}\](Reduction half-reaction) \[{{E}^{o}}={{E}^{o}}_{oxi}+{{E}^{o}}_{red}\] \[=+ve\] So, reaction will take place.