The standard free energy change (Delta G^{cir | vedclass

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Question

(A) The cell reaction is: $3 Ca_{(s)} + 2 Au^{3+}(aq) \rightarrow 3 Ca^{2+}(aq) + 2 Au_{(s)}$$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anod

Vedclass · Accepted Answer

$-2.53 	imes 10^3$

Vedclass · Answer

(A) The cell reaction is: $3 Ca_{(s)} + 2 Au^{3+}(aq) ightarrow 3 Ca^{2+}(aq) + 2 Au_{(s)}$$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$E^{\circ}_{cell} = E^{\circ}_{Au^{3+}/Au} - E^{\circ}_{Ca^{2+}/Ca} = 1.50 \ V - (-2.87 \ V) = 4.37 \ V$The number of electrons transferred $(n)$ is $6$.Using the formula $\Delta G^{\circ} = -nFE^{\circ}_{cell}$:$\Delta G^{\circ} = -6 	imes 96500 \ C \ mol^{-1} 	imes 4.37 \ V$$\Delta G^{\circ} = -2530230 \ J \ mol^{-1} = -2530.23 \ kJ \ mol^{-1}$Rounding to three significant figures,$\Delta G^{\circ} = -2.53 	imes 10^3 \ kJ \ mol^{-1}$.

Electrolyte	$\Lambda_{m}^{o} \, (S \, cm^{2} \, mol^{-1})$
$NaCl$	$126.5$
$NaBr$	$128.2$
$KCl$	$149.5$

The standard free energy change $(\Delta G^{\circ})$ for the following reaction (in $kJ$) at $25^{\circ} C$ is $3 Ca_{(s)} + 2 Au^{3+}(aq, 1 M) \rightarrow 3 Ca^{2+}(aq, 1 M) + 2 Au_{(s)}$ (given: $E^{\circ}_{Au^{3+}/Au} = +1.50 \ V, E^{\circ}_{Ca^{2+}/Ca} = -2.87 \ V, 1 \ F = 96500 \ C \ mol^{-1}$)

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Electrolyte $\Lambda_{m}^{o} \, (S \, cm^{2} \, mol^{-1})$

$NaCl$ $126.5$

$NaBr$ $128.2$

$KCl$ $149.5$

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