The standard Gibbs free energy change (Delta | vedclass

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(C) The reaction in a $Daniell$ cell is: $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$.For $1 \, mole$ of $Zn$ oxidation,$n = 2$ e

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$-424.6$

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(C) The reaction in a $Daniell$ cell is: $Zn_{(s)} + Cu^{2+}_{(aq)} ightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$.For $1 \, mole$ of $Zn$ oxidation,$n = 2$ electrons are transferred.For $2 \, moles$ of $Zn$ oxidation,the number of electrons transferred is $n = 2 	imes 2 = 4$.Given $E^{\circ}_{cell} = 1.1 \, V$ and $F = 96500 \, C \, mol^{-1}$.The formula for standard Gibbs free energy change is $\Delta G^{\circ} = -n F E^{\circ}_{cell}$.Substituting the values: $\Delta G^{\circ} = -4 	imes 96500 \, C \, mol^{-1} 	imes 1.1 \, V$.$\Delta G^{\circ} = -424600 \, J \, mol^{-1} = -424.6 \, kJ \, mol^{-1}$.

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

The standard Gibbs free energy change ($\Delta G^{\circ}$ in $kJ \, mol^{-1}$),in a $Daniell$ cell $(E^{\circ}_{cell} = 1.1 \, V)$,when $2 \, moles$ of $Zn_{(s)}$ is oxidised at $298 \, K$,is closest to

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