Match List I with List II. List I (Conversion | vedclass

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(A) . For $H_2O \rightarrow H_2 + \frac{1}{2}O_2$, the reaction is $H_2O \rightarrow \frac{1}{2}O_2 + 2H^+ + 2e^-$. Thus, $2F$ are required.$B$. For $

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

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(A) . For $H_2O ightarrow H_2 + \frac{1}{2}O_2$, the reaction is $H_2O ightarrow \frac{1}{2}O_2 + 2H^+ + 2e^-$. Thus, $2F$ are required.$B$. For $MnO_4^- ightarrow Mn^{2+}$, the change in oxidation state of $Mn$ is from $+7$ to $+2$. The reaction is $MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$. Thus, $5F$ are required.$C$. For $Ca^{2+} + 2e^- ightarrow Ca$, $1 	ext{ mol}$ of $Ca$ requires $2F$. So, $1.5 	ext{ mol}$ of $Ca$ requires $1.5 	imes 2 = 3F$.$D$. For $FeO ightarrow \frac{1}{2}Fe_2O_3$, the reaction is $Fe^{2+} ightarrow Fe^{3+} + e^-$. Thus, $1F$ is required.Matching: $A-II, B-IV, C-I, D-III$.

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

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