Redox Reactions And Electrochemistry Ques 558
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 + 0.77 V and $ E{}^\circ $ for $ I_2/2{I^{-}}=0.536V. $ he 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+}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given $ F{e^{3+}}/F{e^{2+}}=+0.77V $ and $ I_2/2{I^{-}}=0.536V $
$ 2({e^{-}}+F{e^{3+}}\xrightarrow{{}}F{e^{2+}}),E{}^\circ =0.77V $
$ 2{I^{-}}\xrightarrow{{}},I_2+2{e^{-}}E{}^\circ =-0.536V $
$ 2F{e^{3+}}+2{I^{-}}\xrightarrow{{}}2F{e^{2+}}+I_2 $
$ E{}^\circ =E{{{}^\circ } _{ox}}+E{{{}^\circ } _{red}} $ $ =0.77-0.536=0.164V $ So, reaction will take place.
