Electrochemistry Result Question 37
37. Find the equilibrium constant for the reaction
$ \mathrm{Cu}^{2+}+\mathrm{In}^{2+} \rightleftharpoons \mathrm{Cu}^{+}+\mathrm{In}^{3+} $
Given that
$ \begin{aligned} E_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}}^{\circ} & =0.15 \mathrm{V}, \\ E_{\mathrm{In}^{2+} / \mathrm{In}^{+}}^{\mathrm{O}} & =-0.4 \mathrm{V}, \\ E_{\mathrm{In}}^{\circ} / \mathrm{In}^{3+}{ }^{+} & =-0.42 \mathrm{V} \end{aligned} $
$ (2004, 4M) $
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Answer:
Correct Answer: 37. ( $ 10^ {10} $ )
Solution:
- Given, $\mathrm{In}^{2+}+e^{-} \longrightarrow \mathrm{In}^{+} \quad E^{\circ}=-0.40 $ …..(i)
$\Rightarrow \Delta G^{\circ}=0.40 \mathrm{F} $
$\mathrm{In}^{3+}+2 e^{-} \longrightarrow \mathrm{In}^{+} \quad E^{\circ}=-0.42 $ ……(ii)
$\Rightarrow \Delta G^{\circ}=0.84 \mathrm{F}$
Subtracting (i) from (ii)
$ \begin{aligned} & \mathrm{In}^{3+}+e \longrightarrow \operatorname{In}^{2+} ; \quad \Delta G^{\circ}=0.44 \mathrm{F}=-E^{\circ} F \\ & \Rightarrow \quad E^{\circ}=-0.44 \mathrm{V} \\ \end{aligned} $
Now, for: $\mathrm{Cu}^{2+}+\mathrm{In}^{2+} \longrightarrow \mathrm{Cu}^{+}+\mathrm{In}^{3+}$
$ \begin{aligned} E^{\circ} & =E^{\circ}\left(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\right)-E^{\circ}\left(\mathrm{In}^{3+} / \mathrm{In}^{2+}\right) \\ & =0.15-(-0.44)=0.59 \mathrm{V} \end{aligned} $
Also $E^{\circ} =0.0590 \log K $
$\Rightarrow \log K =\frac{E^{\circ}}{0.059}=10 \Rightarrow K=10^{10}$
BETA
