At 298 K,the equilibrium constant is 2 times | vedclass

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(D) Given reaction: $Cu_{(s)} + 2 Ag^{+}_{(aq)} ightleftharpoons Cu^{2+}_{(aq)} + 2 Ag_{(s)}$ with $K_{eq} = 2 	imes 10^{15}$.The target reaction i

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$2$

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(D) Given reaction: $Cu_{(s)} + 2 Ag^{+}_{(aq)} ightleftharpoons Cu^{2+}_{(aq)} + 2 Ag_{(s)}$ with $K_{eq} = 2 	imes 10^{15}$.The target reaction is $\frac{1}{2} Cu^{2+}_{(aq)} + Ag_{(s)} ightleftharpoons \frac{1}{2} Cu_{(s)} + Ag^{+}_{(aq)}$.This target reaction is the reverse of the original reaction multiplied by a factor of $\frac{1}{2}$.Therefore,the new equilibrium constant $K_{eq}^{\prime} = (K_{eq})^{-1/2} = \frac{1}{\sqrt{K_{eq}}}$.$K_{eq}^{\prime} = \frac{1}{\sqrt{2 	imes 10^{15}}} = \frac{1}{\sqrt{20 	imes 10^{14}}} = \frac{1}{\sqrt{20} 	imes 10^7} = \frac{1}{4.472 	imes 10^7} \approx 0.2236 	imes 10^{-7} = 2.236 	imes 10^{-8}$.Comparing this with $x 	imes 10^{-8}$,we get $x = 2.236$.The nearest integer value of $x$ is $2$.

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 $Fe_2O_3$	$IV$. $5 \ F$

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