Cu^{+}_{(aq)} is unstable in solution and und | vedclass

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Question

(A) The given reaction is a disproportionation reaction: $2 Cu^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cu_{(s)}$.This can be split into two half-react

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$+0.38 \ V$

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(A) The given reaction is a disproportionation reaction: $2 Cu^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cu_{(s)}$.This can be split into two half-reactions:$1$. Oxidation: $Cu^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + e^{-}$,where $E^{\circ}_{ox} = -E^{\circ}_{Cu^{2+}/Cu^{+}} = -0.15 \ V$.$2$. Reduction: $Cu^{+}_{(aq)} + e^{-} \rightarrow Cu_{(s)}$,where $E^{\circ}_{red} = E^{\circ}_{Cu^{+}/Cu}$.To find $E^{\circ}_{Cu^{+}/Cu}$,we use the relation: $\Delta G^{\circ}_{Cu^{2+}/Cu} = \Delta G^{\circ}_{Cu^{2+}/Cu^{+}} + \Delta G^{\circ}_{Cu^{+}/Cu}$.$-(2)F(E^{\circ}_{Cu^{2+}/Cu}) = -(1)F(E^{\circ}_{Cu^{2+}/Cu^{+}}) + -(1)F(E^{\circ}_{Cu^{+}/Cu})$.$2(0.34) = 0.15 + E^{\circ}_{Cu^{+}/Cu} \implies E^{\circ}_{Cu^{+}/Cu} = 0.68 - 0.15 = 0.53 \ V$.Now,$E^{\circ}_{cell} = E^{\circ}_{ox} + E^{\circ}_{red} = -0.15 \ V + 0.53 \ V = +0.38 \ V$.

List – $I$	List – $II$
$(a)$ Henry’s law	$(i)$ $p_1 = x_1 p_1^o$
$(b)$ Raoult’s law	(ii) $p = K_H x$
$(c)$ First law of thermodynamics	(iii) $\Lambda_m^o = \nu_+ \lambda_+^o + \nu_- \lambda_-^o$
$(d)$ Kohlrausch’s law	(iv) $\Delta U = q + w$

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