Plotting $1 / \Lambda_{m}$ against $c \Lambda_{m}$ for aqueous solutions of a monobasic weak acid $(HX)$ resulted in a straight line with $y$-axis intercept of $P$ and slope of $S$. The ratio $P / S$ is
$[\Lambda_{m} =$ molar conductivity
$\Lambda_{m}^{\circ} =$ limiting molar conductivity
$c =$ molar concentration
$K_{a} =$ dissociation constant of $HX$ ]

Given ${E^o}_{Ag^{+}/Ag} = 0.80 \ V$,${E^o}_{Mg^{2+}/Mg} = -2.37 \ V$,${E^o}_{Cu^{2+}/Cu} = 0.34 \ V$,${E^o}_{Hg^{2+}/Hg} = 0.79 \ V$.
Which of the following statements is correct?

Match the column $I$ with column $II$ and mark the appropriate choice.

Column $I$	Column $II$
$A$. Kohlrausch law	$i$. $\Lambda _{m}^o = \nu _+ \lambda _+^o + \nu _- \lambda _-^o$
$B$. Molar Conductivity	$ii$. $\Lambda _m = \frac{\kappa \times 1000}{M}$
$C$. Degree of Dissociation	$iii$. $\alpha = \frac{\Lambda _m}{\Lambda _m^o}$
$D$. Dissociation Constant	$iv$. $K_a = \frac{C\alpha ^2}{1 - \alpha}$

Which one of the following statements is correct?

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 $2$.

Consider the following electrochemical cell at standard condition: $Au_{(s)} | QH_2, Q | NH_4X(0.01 \ M) || Ag^{+}(1 \ M) | Ag_{(s)}$. Given $E_{\text{cell}} = +0.4 \ V$. The couple $QH_2 / Q$ represents the quinhydrone electrode,and the half-cell reaction is given as: $Q + 2e^- + 2H^+ \rightarrow QH_2$ with $E^o_{Q/QH_2} = +0.7 \ V$. Given: $E^o_{Ag^+/Ag} = +0.8 \ V$ and $\frac{2.303 \ RT}{F} = 0.06 \ V$. The $pK_b$ value of the ammonium halide salt $(NH_4X)$ used here is $.........$ (nearest integer).