• # 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 $F{{e}^{3+}}$

(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.