Match List-I with List-II. List-I (Conve | vedclass

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(A) The number of Faradays $(F)$ required is equal to the change in oxidation state (number of electrons transferred) per mole of substance.$A$. $2H_2

Vedclass · Accepted Answer

$A-II, B-IV, C-I, D-III$

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(A) The number of Faradays $(F)$ required is equal to the change in oxidation state (number of electrons transferred) per mole of substance.$A$. $2H_2O ightarrow O_2 + 4H^{+} + 4e^{-}$. For $2 \ mol$ of $H_2O$,$4 \ F$ is required. So,for $1 \ mol$ of $H_2O$,$2 \ F$ is required. $(A-II)$$B$. $MnO_4^{-} (Mn^{+7}) ightarrow Mn^{2+} (Mn^{+2})$. Change in oxidation state is $7 - 2 = 5$. So,$5 \ F$ is required. $(B-IV)$$C$. $Ca^{2+} + 2e^{-} ightarrow Ca$. For $1 \ mol$ of $Ca$,$2 \ F$ is required. For $1.5 \ mol$ of $Ca$,$1.5 	imes 2 = 3 \ F$ is required. $(C-I)$$D$. $FeO (Fe^{+2}) ightarrow Fe_2O_3 (Fe^{+3})$. Change in oxidation state is $3 - 2 = 1$. So,$1 \ F$ is required. $(D-III)$Therefore,the correct matching is $A-II, B-IV, C-I, D-III$.

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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