Match List-$I$ with List-$II$.

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

Choose the correct answer from the options given below:

An aqueous solution containing $6.5 \ g$ of $NaCl$ of $90 \%$ purity was subjected to electrolysis. After the complete electrolysis,the solution was evaporated to get solid $NaOH$. The volume of $1 \ M$ acetic acid required to neutralize $NaOH$ obtained above is (in $cm^{3}$)

Which of the following statements is not correct?

Based on the following information,arrange four metals,$A$,$B$,$C$,and $D$,in order of increasing ability to act as reducing agents:
$(I)$ Only $C$ reacts with $1 \ M \ HCl$ to give $H_{2(g)}$.
$(II)$ When $A$ is added to a solution of the other metal salts,metallic $D$ is formed,but not $B$ or $C$.

Copper reduces $NO_{3}^{-}$ into $NO$ and $NO_{2}$ depending upon the concentration of $HNO_{3}$ in solution. Assuming fixed $[Cu^{2+}]$ and $P_{NO} = P_{NO_{2}} = 1 \ bar$,the $HNO_{3}$ concentration at which the thermodynamic tendency for reduction of $NO_{3}^{-}$ into $NO$ and $NO_{2}$ by copper is the same is $10^{x} \ M$. The value of $2x$ is ...... .
$[Given: E_{Cu^{2+} / Cu}^{\circ} = 0.34 \ V, E_{NO_{3}^{-} / NO}^{\circ} = 0.96 \ V, E_{NO_{3}^{-} / NO_{2}}^{\circ} = 0.79 \ V$ and at $298 \ K, \frac{RT}{F}(2.303) = 0.059]$

Which of the following statements is wrong?