Question

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

• A. $+18.51\, kJ\, mol ^{-1}$
• B. $+11.87\, kJ\, mol ^{-1}$
• C. $-8.10\, kJ\, mol ^{-1}$
• D. $-10.41\, kJ\, mol ^{-1}$

Answer 1

Answer: (B)

$F e^{2+}+2 e^{-} \rightarrow F e ; n_{1}=2, E_{1}^{o}=-0.447\, V$
$F e^{3+}+e^{-} \rightarrow F e^{2+} ; n_{2}=1, E_{2}^{o}=0.771\, V$
$F e^{3+}+3 e^{-} \rightarrow F e ; n_{3}=3, E_{3}^{o}=?$
$\Delta G_{3}=\Delta G_{1}+\Delta G_{2} 3 E_{3}^{o}=-2 E_{1}^{o}+E_{2}^{o}$
$E_{3}^{o}=\frac{(-0.477 \times 2)+0.771}{3}=-0.041 V \Delta G^{o}=-n F E^{o}$
$\Delta G^{o}=-3 \times 96485 \times(-0.041\, V )$
$\Delta G^{o}=+11867.65\, mol ^{-1} \simeq+11.87\, kJ\, mol ^{-1}$