Using the standard electrode potential, find out the pair between which redox reaction is not feasib

Question

Using the standard electrode potential, find out the pair between which redox reaction is not feasible.\ $E^{\circ}$ values : $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=+0.77 ; \mathrm{I}_2 / \mathrm{I}^{-}=+0.54$;\ $\mathrm{Cu}^{2+} / \mathrm{Cu}=+0.34 ; \mathrm{Ag}^{+} / \mathrm{Ag}=+0.80 \mathrm{~V}$

Professor PC Thomas Classes · Answer

(d): For the reaction,\ $2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} ightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2$ \ $ E_{\mathrm{cell}}^{\circ}=E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}-E_{\mathrm{I}_2 / \mathrm{I}^{-}}^{\circ}=0.77-(0.54) \ =+0.23 \mathrm{~V} $\ Here, $E_{	ext {cell }}^{\circ}$ is + ve so, reaction is feasible. \ For the reaction, $\mathrm{Cu}+2 \mathrm{Ag}^{+} ightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag}$ $ \ E_{\mathrm{cell}}^{\circ}=E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}-E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ} \ =0.80-(0.34)=+0.46 \mathrm{~V} $ \ Here, $E_{	ext {cell }}^{\circ}$ is + ve so, the reaction is feasible. \ For the reaction, \ $2 \mathrm{Fe}^{3+}+\mathrm{Cu} ightarrow 2 \mathrm{Fe}^{2+}+\mathrm{Cu}^{2+}$ $ \  E_{\mathrm{cell}}^{\circ} =E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}-E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ} \ =0.77-(0.34)=+0.43 \mathrm{~V} $ \ Here, $E_{	ext {cell }}^{\circ}$ is + ve so, the reaction is feasible. \ For the reaction, $\mathrm{Ag}+\mathrm{Fe}^{3+} ightarrow \mathrm{Ag}^{+}+\mathrm{Fe}^{2+}$ \ $ E_{\mathrm{cell}}^{\circ}=E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}-E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=0.77-(0.80)=-0.03 \mathrm{~V} $ \ Here, $E_{	ext {cell }}^{\circ}$ is negative so, \ the reaction is not feasible.