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

Vedclass · Accepted Answer

$+0.38 \ V$

Vedclass · Answer

(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,$E^{\circ}_{cell} = E^{\circ}_{ox} + E^{\circ}_{red} = -0.15 \ V + 0.53 \ V = +0.38 \ V$.

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The Gibbs energy change (in $J$) for the given reaction at $[Cu^{2+}] = [Sn^{2+}] = 1 \, M$ and $298 \, K$ is:
$Cu_{(s)} + Sn^{2+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Sn_{(s)}$
$(E^{\circ}_{Sn^{2+}|Sn} = -0.16 \, V, E^{\circ}_{Cu^{2+}|Cu} = 0.34 \, V, F = 96500 \, C \, mol^{-1})$

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