The standard Gibbs' energy change for Daniel cell reaction is $$ \begin{aligned} &…
The standard Gibbs' energy change for Daniel cell reaction is
$$
\begin{aligned}
& \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s) \\
& E_{\text {cell }}^{\circ}=1.1 \mathrm{~V} \text {. }
\end{aligned}
$$
$-212.3 \mathrm{~kJ}$
$106.15 \mathrm{~kJ}$
$+212.3 \mathrm{~kJ}$
$100 \mathrm{~kJ}$
Solution
Given,
$
\begin{aligned}
n, E_{\text {cell }}^{\circ} & =1.1 \mathrm{~V} \\
n & =2 \\
F & =96500 \mathrm{C} / \mathrm{mol} \\
\Delta G & =-n F E^{\circ} \\
& =-2 \times 96500 \times 1.1 \\
& =-212300 \mathrm{~J} \\
& =-212.3 \mathrm{~kJ}
\end{aligned}
$