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Question

Chemistry Question on Gibbs Free Energy

Given standard electrode potentials Fe2:+2eFeE=0440VFe ^{2}:+2 e ^{-} \longrightarrow Fe E ^{\circ}=-0 \cdot 440 \,V Fe3++3eFeE=0036VFe ^{3+}+3 e ^{-} \longrightarrow Fe E ^{\circ}=-0 \cdot 036\, V The standard electrode potential (E)\left(E^{\circ}\right) for Fe3++eFe2+Fe ^{3+}+ e ^{-} \longrightarrow Fe ^{2+} is :

A

+ 0.772 V

B

- 0.772 V

C

+ 0.417 V

D

-0.414 V

Answer

+ 0.772 V

Explanation

Solution

We know, G=nFE\triangle G ^{\circ}=- nFE ^{\circ}
Fe2++2eFeFe ^{2+}+2 e ^{-} \rightarrow Fe
ΔG=2×F×(0.440V)=0.880F...\Delta G ^{\circ}=-2 \times F \times(-0.440 V )=0.880 \,F\,\,\,\,... (1)
Fe3++3eFeFe ^{3+}+3 e ^{-} \rightarrow Fe
ΔG=3×F×(0.036)=0.108F...\Delta G ^{\circ}=-3 \times F \times(-0.036)=0.108 \,F\,\,\,\,... -(2)
On subtracting Eqs. (1) from (2),
Fe3++eFe2+Fe ^{3+}+ e ^{-} \rightarrow Fe ^{2+}
ΔG=0.108F0.880F=0.772F\Delta G ^{\circ}=0.108 F -0.880 F =-0.772 F
ΔG=nFE\Delta G ^{\circ}=- nFE ^{\circ}
E=ΔGnF=0.772F1×F=+0.772VE ^{\circ}=\frac{\Delta G ^{\circ}}{- nF }=\frac{-0.772 F }{-1 \times F }=+0.772\, V