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] Given \[F{{e}^{3+}}/F{{e}^{2+}}=+0.77V\] and \[{{I}_{2}}/2{{I}^{-}}=0.536V\] \[\begin{align} & 2({{e}^{-}}+F{{e}^{3+}}\xrightarrow{{}}F{{e}^{2+}})\,\,\,\,E{}^\circ =0.77V \\ & \underline{2{{I}^{-}}\xrightarrow{{}}{{I}_{2}}+2{{e}^{-}}~\,\,\,\,\,\,\,\,\,\,\,\,\,E{}^\circ =-\,0.536V} \\ & 2F{{e}^{3+}}+2{{I}^{-}}~\xrightarrow{{}}2F{{e}^{2+}}+{{I}_{2}} \\ \end{align}\] \[E{}^\circ =E{{{}^\circ }_{ox}}+E{{{}^\circ }_{red}}\] \[=0.77-0.536\] =0.164V \[\therefore \] Since value of \[E{}^\circ \]is + ve reaction will take place.
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