Calculate E^o_ cell and _rG^o for the following reaction at 25^oC… - Vidyadip

Question

Calculate $E^o_{cell}$ and $\Delta_rG^o$ for the following reaction at $25^oC$:
$\text{A}^{2+}+\text{B}^{+}\xrightarrow{\text{ }\ \ \ \ \ \ \ \ \ \ \ \ }\text{A}^{3+}+\text{B}$
Given: $K_c =10^{10},1F =96500C mol^{-1}$

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Answer

$\text{A}^{2+}+\text{B}^{+}\xrightarrow{\text{ }\ \ \ \ \ \ \ \ \ \ \ \ }\text{A}^{3+}+\text{B }\text{ (n=1)}$
$K_c= 10^{10}F = 96500C/mol T = 25^0C = 298K$
$\Delta\text{G}^{o}=\ ?$ $\text{E}^{o}= \ ?$ R = 8.314J/K/mol
$\Delta\text{G}^{o}=-2.303\text{ RT log Kc}$
$\Delta\text{G}^{o}\text{= -2.303}\times\text{8.314J/K/mol}\times\text{298K}\times\log{10}^{10}$
$\therefore\Delta\text{G}^{o}=\text{-57058.4J/mol or -57.0584 kJ/mol}$
$\Delta\text{G}^{o}=\text{-57058.4J/mol = -nFE}^{o}=\text{-1}\times\text{96500}\times\text{E}^{o}$
$\therefore\text{E}^{o}=\frac{\text{-57084.4}}{\text{-96500}}=\text{0.591V}$

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