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Question

Chemistry Question on Nernst Equation

Given, standard electrode potentials of following reactions, Fe2++2eFe;E=0.440VFe ^{2+}+2 e^{-} \longrightarrow Fe ; E^{\circ}=-0.440\, V Fe3++3eFe;E=0.036VFe ^{3+}+3 e^{-} \longrightarrow Fe ; E^{\circ}=-0.036\, V The standard electrode potential (E)\left(E^{\circ}\right) for Fe3++eFe2+Fe ^{3+}+e^{-} \longrightarrow Fe ^{2+}, is

A

-0.476 V

B

-0.404 V

C

+ 0.404 V

D

+ 0.772 V

Answer

+ 0.772 V

Explanation

Solution

ΔG=nFE\Delta G^{\circ}=-n F E^{\circ}
Fe2++2eFeFe ^{2+}+2 e^{-} \longrightarrow Fe ... (i)
ΔG=2×F×(0.440V)\Delta G^{\circ} =-2 \times F \times(-0.440 \,V )
=0.880F=0.880\, F
Fe3++3eFeFe ^{3+}+3 e^{-} \longrightarrow Fe ... (ii)
ΔG=3×F×(0.036)=0.108F\Delta G^{\circ}=-3 \times F \times(-0.036)=0.108\, F
On subtracting E (i) from E (ii)
Fe3++eFe2+Fe ^{3+}+e^{-} \longrightarrow Fe ^{2+}
ΔG=0.108F0.880F=0.772F\Delta G^{\circ}=0.108\, F -0.880 \,F =-0.772 \,F
Ecell =ΔGnFE_{\text {cell }}^{\circ}=\frac{-\Delta G^{\circ}}{n F}
=(0.772F)1×F=+0.772V=-\frac{(-0.772 F)}{1 \times F}=+0.772 \,V