Electrochemistry 2 Question 38
37. Find the equilibrium constant for the reaction
$$ \mathrm{Cu}^{2+}+\mathrm{In}^{2+} \rightleftharpoons \mathrm{Cu}^{+}+\mathrm{In}^{3+} $$
Given that $E_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}}^{\circ}=0.15 \mathrm{~V}$,
$$ E_{\operatorname{In}^{2+} / \mathrm{In}^{+}}^{0}=-0.4 \mathrm{~V}, $$
$$ E_{\mathrm{In}^{3+} / \mathrm{In}^{+}}^{\circ}=-0.42 \mathrm{~V} $$
(2004, 4M)
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Solution:
- Given,
$$ \begin{array}{ll} & \operatorname{In}^{2+}+e^{-} \longrightarrow \operatorname{In}^{+} \quad E^{\circ}=-0.40 \ \Rightarrow \Delta G^{\circ}=0.40 \mathrm{~F} \ & \operatorname{In}^{3+}+2 e^{-} \longrightarrow \operatorname{In}^{+} \quad E^{\circ}=-0.42 \ \Rightarrow \Delta G^{\circ}=0.84 \mathrm{~F} \end{array} $$
Subtracting (i) from (ii)
$$ \begin{array}{rlrl} & \operatorname{In}^{3+}+e \longrightarrow \operatorname{In}^{2+} ; & \Delta G^{\circ} & =0.44 \mathrm{~F}=-E^{\circ} F \ & & E^{\circ}=-0.44 \mathrm{~V} \end{array} $$
Now, for: $\mathrm{Cu}^{2+}+\mathrm{In}^{2+} \longrightarrow \mathrm{Cu}^{+}+\mathrm{In}^{3+}$
$$ \begin{aligned} E^{\circ} & =E^{\circ}\left(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\right)-E^{\circ}\left(\mathrm{In}^{3+} / \mathrm{In}^{2+}\right) \ & =0.15-(-0.44)=0.59 \mathrm{~V} \end{aligned} $$
$$ \begin{array}{ll} \text { Also } & E^{\circ}=0.0590 \log K \ \Rightarrow & \log K=\frac{E^{\circ}}{0.059}=10 \Rightarrow K=10^{10} \end{array} $$
