The Gibbs energy change (in J) for the given | vedclass

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Question

(A) The cell reaction is $Cu_{(s)} + Sn^{2+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Sn_{(s)}$.The standard cell potential is $E^{\circ}_{cell} = E^{\circ

Vedclass · Accepted Answer

$96500$

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(A) The cell reaction is $Cu_{(s)} + Sn^{2+}_{(aq)} ightarrow Cu^{2+}_{(aq)} + Sn_{(s)}$.The standard cell potential is $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Sn^{2+}|Sn} - E^{\circ}_{Cu^{2+}|Cu}$.$E^{\circ}_{cell} = -0.16 \, V - 0.34 \, V = -0.50 \, V$.The standard Gibbs energy change is $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.Here,$n = 2$ (number of electrons transferred).$\Delta G^{\circ} = -2 	imes 96500 \, C \, mol^{-1} 	imes (-0.50 \, V) = 96500 \, J \, mol^{-1}$.Since the concentrations are $[Cu^{2+}] = [Sn^{2+}] = 1 \, M$,the reaction quotient $Q = \frac{[Cu^{2+}]}{[Sn^{2+}]} = 1$.Using the Nernst equation,$\Delta G = \Delta G^{\circ} + RT \ln Q$.Since $\ln(1) = 0$,$\Delta G = \Delta G^{\circ} = 96500 \, J \, mol^{-1}$.

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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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})$