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For the cell reaction $2Fe^{3+}(aq)+2I^-(aq)\rightarrow2Fe^{2+}(aq)+I_2(aq)$, $E^\ominus_{cell}=0.24\ V$ at 298 K. The standard Gibbs energy ($\Delta_rG^\ominus$) 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
$23.16\ kJ\,mol^{-1}$
Detailed Solution
$\Delta_rG^\ominus=-nFE^\ominus_{cell}$, with $n=2$
$=-2\times96500\times0.24\ J\,mol^{-1}$
$=-46320\ J\,mol^{-1}=-46.32\ kJ\,mol^{-1}$