For the disproportionation reaction

Question:

For the disproportionation reaction

$2 \mathrm{Cu}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq})$ at $298 \mathrm{~K}$,

$\ln K$ (where $K$ is the equilibrium constant) is ________$\times 10^{-1}$.

Given

$\left(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}}^{0}=0.16 \mathrm{~V}\right.$

$\mathrm{E}_{\mathrm{Cu}^{+} / \mathrm{Cu}}^{0}=0.52 \mathrm{~V}$

$\left.\frac{\mathrm{RT}}{\mathrm{F}}=0.025\right)$

 

Solution:

$\mathrm{E}_{\mathrm{cell}}^{0}=\mathrm{E}_{\mathrm{Cu}^{+} / \mathrm{Cu}}^{0}-\mathrm{E}_{\mathrm{Cu}^{2} / \mathrm{Cu}^{+}}^{0}$

$=0.52-0.16$

$=0.36 \mathrm{~V}$

At equilibrium $\rightarrow \mathrm{E}_{\text {cell }}=0$

$\mathrm{E}_{\text {cell }}^{o}=\frac{\mathrm{RT}}{\mathrm{nF}} \ln \mathrm{K}$

$\ln \mathrm{K}=\frac{\mathrm{E}_{\text {cell }}^{\mathrm{o}} \times \mathrm{nF}}{\mathrm{RT}}$

$\ln \mathrm{K}=\frac{0.36 \times 1}{0.025}$

$=14.4=144 \times 10^{-1}$

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