$100 \text{ mL}$ of $0.2 \text{ M}$ acetic acid is completely neutralized using a standard solution of $NaOH$. The volume of ethane obtained at $STP$ after complete electrolysis of the resulting solution is: (in $\text{ L}$)

Match the following columns:

Column-$I$	Column-$II$
$A$. Leclanche cell (Dry cell)	$P$. Converts energy of combustion into electrical energy
$B$. Lead storage battery	$Q$. Reaction at cathode: $O_{2(g)} + 4H^+_{(aq)} + 4e^- \rightarrow 2H_2O_{(\ell)}$
$C$. Fuel cell	$R$. Reaction at cathode: $MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3$
$D$. Rusting	$S$. Reaction at anode: $Pb_{\text{(s)}} + SO_4^{2-}{_{\text{(aq)}}} \rightarrow PbSO_{4\text{(s)}} + 2e^{-}$

For a spontaneous reaction,determine the values of $\Delta G^o$,equilibrium constant $K$,and $E^o_{cell}$ respectively.

For a saturated solution of $Ag_{2}CrO_{4}$ at infinite dilution,$\lambda_{m}^{\infty}(Ag^{+}) = 127 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ and $\lambda_{m}^{\infty}(CrO_{4}^{2-}) = 246 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$. If the specific conductance of the solution is $2 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$,calculate the solubility product $(K_{sp})$ of $Ag_{2}CrO_{4}$.

Based on the following information,arrange four metals $A, B, C$ and $D$ in order of decreasing ability to act as reducing agents:
$[I]$ Only $A, B$ and $C$ react with $1 \ M \ HCl$ to give $H_2 \ (g)$.
$[II]$ When $C$ is added to solutions of the other metal ions,metallic $B$ and $D$ are formed.
$[III]$ Metal $C$ does not reduce $A^{n+}$.

The electrochemical cell shown below is a concentration cell.
$M \mid M^{2+} (\text{saturated solution of a sparingly soluble salt, } MX_2) \mid M^{2+} (0.001 \ mol \ dm^{-3}) \mid M$
The emf of the cell depends on the difference in concentration of $M^{2+}$ ions at the two electrodes. The emf of the cell at $298 \ K$ is $0.059 \ V$.
$1.$ The solubility product $(K_{sp}; \ mol^3 \ dm^{-9})$ of $MX_2$ at $298 \ K$ based on the information available for the given concentration cell is (take $2.303 \times R \times 298 / F = 0.059 \ V$):
$(A) \ 1 \times 10^{-15} \quad (B) \ 4 \times 10^{-15}$
$(C) \ 1 \times 10^{-12} \quad (D) \ 4 \times 10^{-12}$
$2.$ The value of $\Delta G \ (kJ \ mol^{-1})$ for the given cell is (take $1 \ F = 96500 \ C \ mol^{-1}$):
$(A) \ -5.7 \quad (B) \ 5.7 \quad (C) \ 11.4 \quad (D) \ -11.4$
Give the answer for question $1$ and $2$.