Consider an electrochemical cell: A_{(s)} | A | vedclass

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(B) The cell reaction is: $2A_{(s)} + B^{2n+}_{(aq)} \rightleftharpoons 2A^{n+}_{(aq)} + B_{(s)}$.At equilibrium,the emf is zero,so $\Delta G = 0$.The

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$-11.62$

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(B) The cell reaction is: $2A_{(s)} + B^{2n+}_{(aq)} ightleftharpoons 2A^{n+}_{(aq)} + B_{(s)}$.At equilibrium,the emf is zero,so $\Delta G = 0$.The standard Gibbs energy change is $\Delta G^{\ominus} = -RT \ln K_c$.$K_c = \frac{[A^{n+}]^2}{[B^{2n+}]} = \frac{(2)^2}{1} = 4$.Thus,$\Delta G^{\ominus} = -RT \ln(4) = -2RT \ln(2)$.Given $\Delta H^{\ominus} = 2 \Delta G^{\ominus}$.From the relation $\Delta G^{\ominus} = \Delta H^{\ominus} - T \Delta S^{\ominus}$,we substitute $\Delta H^{\ominus}$:$\Delta G^{\ominus} = 2 \Delta G^{\ominus} - T \Delta S^{\ominus} \implies \Delta S^{\ominus} = \frac{\Delta G^{\ominus}}{T}$.$\Delta S^{\ominus} = \frac{-2RT \ln(2)}{T} = -2R \ln(2)$.Substituting the values: $\Delta S^{\ominus} = -2 	imes 8.3 	imes 0.7 = -11.62 \ J \ K^{-1} \ mol^{-1}$.

Electrolyte	$\Lambda_m^\circ$ $(S\text{ cm}^2\text{ mol}^{-1})$
$BaCl_2$	$x_1$
$H_2SO_4$	$x_2$
$HCl$	$x_3$

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