At 300 K,the E_{cell}^{circ} of A_{(s)} + B^ | vedclass

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(A) Given: $\Delta_r S^{\circ} = 100 \ J \ K^{-1} \ mol^{-1}$,$E_{cell}^{\circ} = 1.0 \ V$,$n = 2$,$T = 300 \ K$,$F = 96500 \ C \ mol^{-1}$.Using the

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(A) Given: $\Delta_r S^{\circ} = 100 \ J \ K^{-1} \ mol^{-1}$,$E_{cell}^{\circ} = 1.0 \ V$,$n = 2$,$T = 300 \ K$,$F = 96500 \ C \ mol^{-1}$.Using the relation $\Delta G^{\circ} = -nFE_{cell}^{\circ}$:$\Delta G^{\circ} = -2 	imes 96500 	imes 1 = -193000 \ J \ mol^{-1} = -193 \ kJ \ mol^{-1}$.Using the Gibbs-Helmholtz equation: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.$\Delta H^{\circ} = \Delta G^{\circ} + T\Delta S^{\circ}$.$\Delta H^{\circ} = -193 \ kJ \ mol^{-1} + (300 \ K 	imes 100 \ J \ K^{-1} \ mol^{-1}) / 1000$.$\Delta H^{\circ} = -193 \ kJ \ mol^{-1} + 30 \ kJ \ mol^{-1} = -163 \ kJ \ mol^{-1}$.

List-$I$ (Conversion)	List-$II$ (Number of Faraday required)
$A$. $1 \ mol$ of $H_2O$ to $O_2$	$I$. $3 \ F$
$B$. $1 \ mol$ of $MnO_4^{-}$ to $Mn^{2+}$	$II$. $2 \ F$
$C$. $1.5 \ mol$ of $Ca$ from molten $CaCl_2$	$III$. $1 \ F$
$D$. $1 \ mol$ of $FeO$ to $Fe_2O_3$	$IV$. $5 \ F$

At $300 \ K$,the $E_{cell}^{\circ}$ of $A_{(s)} + B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + B_{(s)}$ is $1.0 \ V$. If $\Delta_r S^{\circ}$ of this reaction is $100 \ J \ K^{-1} \ mol^{-1}$,what is $\Delta_r H^{\circ}$ (in $kJ \ mol^{-1}$) of this reaction? $(F = 96500 \ C \ mol^{-1})$

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