question_answer

A solution contains \[F{{e}^{2+}},F{{e}^{3+}}\] and \[{{I}^{-}}\] 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 \[{{I}_{2}}/2{{I}^{-}}=0.536\,V\]. The favorable redox reaction is                       [AIPMT (M) 2011]

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

B) There will be no redox reaction

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

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

Correct Answer: C

Solution :

[c] \[2{{I}^{-}}\xrightarrow[{}]{{}}{{I}_{2}}+2{{e}^{-}}\](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.