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 $298 \mathrm{~K}, \ln \mathrm{K}$ (where $\mathrm{K}$ is the equilibrium constant) is $\times 10^{-1}$. Given $\left(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}}^{0}=0.16 \mathrm{~V} ; \mathrm{E}_{\mathrm{Cu}^{+} / \mathrm{Cu}}^{0}=0.52 \mathrm{~V} ; \frac{\mathrm{RT}}{\mathrm{F}}=0.025\right)$

Solution:

(144)

$2 \mathrm{Cu}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq})$

$\mathrm{E}_{\text {cell }}^{\mathrm{o}}=\mathrm{E}_{\mathrm{Cu}^{+} / \mathrm{Cu}}^{\mathrm{o}}-\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}}^{\mathrm{o}}=0.52-0.16=0.36 \mathrm{~V}$

$E_{\text {cell }}^{\mathrm{o}}=\frac{\mathrm{RT}}{\mathrm{nF}} \ln \mathrm{K}_{\mathrm{eq}} \Rightarrow 0.36=\frac{0.025}{1} \ln \mathrm{K}$

$\Rightarrow \ln \mathrm{K}=14.4=144 \times 10^{-1}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.