question_answer 44)

  Using the standard electrode potential, find out the pair between which redox reaction is not feasible. \[{{E}^{O-}}values:\,\,\,F{{e}^{3+}}/F{{e}^{2+}}=+0.77;{{I}_{2}}/{{I}^{-}}=+0.54;\]                \[C{{u}^{2+}}/Cu=+0.34;Ag/Ag=+0.80V\] (a) \[F{{e}^{3+}}\]and\[{{I}^{-}}\]                             (b) \[A{{g}^{+}}\] and Cu                              (c) \[F{{e}^{3+}}\]and\[\text{Cu}\]                                (d) \[\text{Ag}\]and \[F{{e}^{3+}}\]

Answer:

  (d)\[Ag+F{{e}^{3+}}\to A{{g}^{+}}+F{{e}^{2+}}\] \[E_{\operatorname{Re}dox\,process}^{{}^\circ }=E_{\operatorname{Re}duced\,species}^{{}^\circ }-E_{OXidised\,species}^{{}^\circ }\] \[=E_{F{{e}^{3+}}/F{{e}^{2+}}}^{{}^\circ }-E_{A{{g}^{+}}/Ag}^{{}^\circ }\] \[=+0.77-0.80=-0.03V\] Negative value shows that the redox process is not feasible.