Question

For the cell reaction
$2Fe^{3+} (aq) + 2I^- (aq) \to 2Fe^{2+} (aq) + I_2$ (aq)
$E^{\circ}_{cell} = 0.24 V$ at $298\, K$
The standard Gibbs energy ($\Delta_{r}G^{\circ}$) of the cell reaction is
[Given that Faraday constant $F\,=\, 96500 \,C\, mol^{-1}$]

• A. $46.32\, kJ\, mol^{-1}$
• B. $23.16\, kJ\, mol^{-1}$
• C. $- 46.32\, kJ \,mol^{-1}$
• D. $- 23.16 \,kJ\, mol^{-1}$

Answer 1

Answer: (C)

$\Delta G^{\ominus} = - n F \; E^{\ominus}_{cell}$
$ = - 2 \times 96500 \times 0.24 \; J \; mol^{-1} $
= - 46320 J $mol^{-1}$
= - 46.32 kJ $mol^{-1}$