During electrolysis of aqueous $NaOH$,$4 \ g$ of $O_2$ gas is liberated at $NTP$ at the anode. The volume of $H_2$ gas liberated at the cathode in $litres$ is $..............$

Given below are the half-cell reactions:
$Mn^{2+} + 2e^{-} \rightarrow Mn; E^{o} = -1.18 \ V$
$2(Mn^{3+} + e^{-} \rightarrow Mn^{2+}); E^{o} = +1.51 \ V$
The $E^{o}$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ will be:

All the energy released from the reaction $X \rightarrow Y, \Delta_{r}G^0 = -193 \ kJ \ mol^{-1}$ is used for oxidizing $M^{+}$ as $M^{+} \rightarrow M^{3+} + 2e^-, E^0 = -0.25 \ V$. Under standard conditions,the number of moles of $M^{+}$ oxidized when one mole of $X$ is converted to $Y$ is $\left[F = 96500 \ C \ mol^{-1}\right]$.

Consider the following redox reaction :
$MnO_4^{-} + H^{+} + H_2C_2O_4 \rightleftharpoons Mn^{2+} + H_2O + CO_2$
The standard reduction potentials are given as below $(E_{red}^{\circ})$ :
$E_{MnO_4^{-} / Mn^{2+}}^{\circ} = +1.51 \ V$
$E_{CO_2 / H_2C_2O_4}^{\circ} = -0.49 \ V$
If the equilibrium constant of the above reaction is given as $K_{eq} = 10^x$,then the value of $x = $ . . . . . . (nearest integer).

During the electrolysis of carnallite,$MgCl_2$ is decomposed and not $KCl$. This is because of

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

List-$I$	List-$II$
$A$. $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$	$I$. Primary battery
$B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$	$II$. Discharging of secondary battery
$C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$	$III$. Fuel cell
$D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$	$IV$. Charging of secondary battery

Choose the correct answer from the options given below.