Question

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

• A. $+18.51kJmo{l}^{-1}$
• B. $+11.87kJmo{l}^{-1}$
• C. $-8.10kJmo{l}^{-1}$
• D. $-10.41kJmo{l}^{-1}$

Answer 1

Answer: (B)

$F{e}^{2+}+2e^{-}\to Fe;n_1=2,E_1^o=-0.447V$
$F{e}^{3+}+e^{-}\to F{e}^{2+};n_2=1,E_2^o=0.771V$
$F{e}^{3+}+3e^{-}\to Fe;n_3=3,E_3^o=?$
$\Delta G_3=\Delta G_1+\Delta G_{2}3E_3^o=-2E_1^o+E_2^o$
$E_3^o=\frac{(-0.477\times 2)+0.771}{3}=-0.041V\Delta G^o=-nF{E}^o$
$\Delta G^o=-3\times 96485\times (-0.041V)$
$\Delta G^o=+11867.65mo{l}^{-1}\simeq +11.87kJmo{l}^{-1}$