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

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(A) Step $1$: Calculate Faraday for each conversion.$A$. $H_2O ightarrow \frac{1}{2} O_2 + 2H^+ + 2e^-$. Thus, $2 	ext{ moles of } e^-$ are require

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Step $1$: Calculate Faraday for each conversion.$A$. $H_2O ightarrow \frac{1}{2} O_2 + 2H^+ + 2e^-$. Thus, $2 	ext{ moles of } e^-$ are required, which is $2F$.$B$. $MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$. Thus, $5 	ext{ moles of } e^-$ are required, which is $5F$. (Note: The provided options suggest a match with $IV$ $(5F)$ for $B$).$C$. $Ca^{2+} + 2e^- ightarrow Ca$. For $1 	ext{ mol of } Ca$, $2F$ is needed. For $1.5 	ext{ mol}$, $1.5 	imes 2 = 3F$. Thus, $C-I$.$D$. $FeO ightarrow \frac{1}{2} Fe_2O_3$. Oxidation state of $Fe$ changes from $+2$ to $+3$. Change is $1e^-$. Thus, $1F$. So, $D-III$.Correct matching: $A-II, B-IV, C-I, D-III$.

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

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