Redox Reactions And Electrochemistry Question 576
Question: 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 $ +0.77V $ and $ {{E}^{{}^\circ }} $ for $ I_2/2{{I}^{-}}=0.536V. $ The favourable redox reaction is:
Options:
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+}} $
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Answer:
Correct Answer: C
Solution:
[c] Given $ F{{e}^{3+}}/F{{e}^{2+}}=+0.77V $ and $ I_2/2{{I}^{-}}=0.536V $
$ \begin{aligned} & 2(e^{-}+Fe^{3+} \xrightarrow{} Fe^{2+})E^{\circ} =0.77V \\ & \underline{2I^{-}}\xrightarrow{} I_2+2e^{-},E^{\circ} =-0.536V \\ & 2Fe^{3+}+2I^{-} \xrightarrow{} 2Fe^{2+}+I_2 \\ \end{aligned} $
$ E{}^\circ =E^{\circ } _{ox}+E^{\circ } _{red} $ $ =0.77-0.536 $ =0.164V
$ \therefore $ Since value of $ E{}^\circ $ is + ve reaction will take place.
BETA
