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Chemistry Question on Gibbs Free Energy

The EoE^{o} values of the following reduction reactions are given: Fe3+(aq)+e−→Fe2+(aq),Eo=0.771 VF e^{3+}(a q)+e^{-} \rightarrow F e^{2+}(a q), E^{o}=0.771\, V Fe2+(aq)+2e−→Fe(s),Eo=−0.447 VF e^{2+}(a q)+2 e^{-} \rightarrow F e(s), E^{o}=-0.447\, V What will be the free energy change for the reaction? Fe3+(aq)+3e−→Fe(s)(F=96485 C mol−1)Fe ^{3+}( aq )+3 e^{-} \rightarrow Fe ( s )\left( F =96485\, C\, mol ^{-1}\right)

A

+18.51 kJ mol−1+18.51\, kJ\, mol ^{-1}

B

+11.87 kJ mol−1+11.87\, kJ\, mol ^{-1}

C

−8.10 kJ mol−1-8.10\, kJ\, mol ^{-1}

D

−10.41 kJ mol−1-10.41\, kJ\, mol ^{-1}

Answer

+11.87 kJ mol−1+11.87\, kJ\, mol ^{-1}

Explanation

Solution

Fe2++2e−→Fe;n1=2,E1o=−0.447 VF e^{2+}+2 e^{-} \rightarrow F e ; n_{1}=2, E_{1}^{o}=-0.447\, V
Fe3++e−→Fe2+;n2=1,E2o=0.771 VF e^{3+}+e^{-} \rightarrow F e^{2+} ; n_{2}=1, E_{2}^{o}=0.771\, V
Fe3++3e−→Fe;n3=3,E3o=?F e^{3+}+3 e^{-} \rightarrow F e ; n_{3}=3, E_{3}^{o}=?
ΔG3=ΔG1+ΔG23E3o=−2E1o+E2o\Delta G_{3}=\Delta G_{1}+\Delta G_{2} 3 E_{3}^{o}=-2 E_{1}^{o}+E_{2}^{o}
E3o=(−0.477×2)+0.7713=−0.041VΔGo=−nFEoE_{3}^{o}=\frac{(-0.477 \times 2)+0.771}{3}=-0.041 V \Delta G^{o}=-n F E^{o}
ΔGo=−3×96485×(−0.041 V)\Delta G^{o}=-3 \times 96485 \times(-0.041\, V )
ΔGo=+11867.65 mol−1≃+11.87 kJ mol−1\Delta G^{o}=+11867.65\, mol ^{-1} \simeq+11.87\, kJ\, mol ^{-1}