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Chemistry Question on Redox reactions

Using the standard electrode potential, find out the pair between which redox reaction is not feasible. EE^{\circ} values : Fe3+/Fe2+=+0.77;I2(s)/I=+0.54Fe^{3+}/Fe^{2+} = + 0.77; I_{2\left(s\right)}/I^- = + 0.54; Cu2+/Cu=+0.34Cu^{2+}/Cu = + 0.34; Ag+/Ag=+0.80VAg^{+}/Ag = + 0.80\, V

A

Fe3+Fe^{3+} and II^-

B

Ag+Ag^{+} and CuCu

C

Fe3+Fe^{3+} and CuCu

D

AgAg and Fe3+Fe^{3+}

Answer

AgAg and Fe3+Fe^{3+}

Explanation

Solution

For the reaction, 2Fe3++2I2Fe2++I22Fe^{3+} + 2I^{-} \to 2Fe^{2+} + I_{2} Ecell=EFe3+/Fe2+EI2/I=0.77(0.54)=+0.23VE^{\circ}_{cell} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{I_2/I^{-}} = 0.77 - \left(0.54\right) = + 0.23 \,V Here, EcellE^{\circ }_{cell} is +ve+ve so, reaction is feasible. For the reaction, Cu+2Ag+Cu2++2AgCu + 2Ag^{+} \to Cu^{2+} + 2Ag Ecell=EAg3+/AgECu2+/Cu=0.80(0.34)=+0.46VE^{\circ }_{cell} = E^{\circ }_{Ag^{3+}/Ag} - E^{\circ }_{Cu^{2+}/Cu}= 0.80 - \left(0.34\right) = + 0.46 \,V Here, EcellE^{\circ }_{cell} is +ve+ve so, the reaction is feasible. For the reaction, 2Fe3++Cu2Fe2++Cu2+2Fe^{3+ }+ Cu \to 2Fe^{2+} + Cu^{2+} Ecell=EFe3+/Fe2+ECu2+/Cu=0.77(0.34)=+0.43VE^{\circ }_{cell} = E^{\circ }_{Fe^{3+}/Fe^{2+}} - E^{\circ }_{Cu^{2+}/Cu} = 0.77 - \left(0.34\right) = + 0.43 \,V Here, EcellE^{\circ }_{cell} is +ve+ve so, the reaction is feasible. For the reaction, Ag+Fe3+Ag++Fe2+Ag + Fe^{3+} \to Ag^{+} + Fe^{2+} Ecell=EFe3+/Fe2+EAg3+/Ag=0.77(0.80)=0.03VE^{\circ }_{cell} = E^{\circ }_{Fe^{3+}/Fe^{2+}} -E^{\circ }_{Ag^{3+}/Ag} = 0.77 - \left(0.80\right) = - 0.03\, V Here, EcellE^{\circ }_{cell} is negative so, the reaction is not feasible.