Consider the following data.

Electrolyte	$\Lambda_m^\circ$ $(S\text{ cm}^2\text{ mol}^{-1})$
$BaCl_2$	$x_1$
$H_2SO_4$	$x_2$
$HCl$	$x_3$

$BaSO_4$ is sparingly soluble in water. If the conductivity of the saturated $BaSO_4$ solution is $x\text{ S cm}^{-1}$, then the solubility product of $BaSO_4$ can be given as (Here $\Lambda_m = \Lambda_m^\circ$)

Match the following:

List-$I$	List-$II$
$(A)$ Potential of hydrogen electrode at $pH = 10$	$(I)$ $0.76 \ V$
$(B)$ $Cu^{2+}|Cu$	$(II)$ $0.059$
$(C)$ $Zn|Zn^{2+}$	$(III)$ $-0.591 \ V$
$(D)$ $\frac{2.303RT}{F}$	$(IV)$ $0.337 \ V$
	$(V)$ $-0.76 \ V$

$(a)$ $A-III, B-I, C-II, D-V$
$(b)$ $A-II, B-V, C-I, D-IV$
$(c)$ $A-III, B-IV, C-I, D-II$
$(d)$ $A-V, B-I, C-IV, D-II$

The molar conductivity of a solution of a weak acid $HX$ $(0.01 \ M)$ is $10$ times smaller than the molar conductivity of a solution of a weak acid $HY$ $(0.10 \ M)$. If $\lambda_{X^{-}}^0 \approx \lambda_{Y^{-}}^0$,the difference in their $pK_a$ values,$pK_a(HX) - pK_a(HY)$,is (consider degree of ionization of both acids to be $\ll 1$)

$A$ and $B$ are two metals. The standard reduction potentials of $A^{+}_{(aq)} / A_{(s)}$ and $B^{+}_{(aq)} / B_{(s)}$ are $-0.5 \ V$ and $+0.5 \ V$ respectively. What is the $\log K_C$ value for the following reaction at $298 \ K$?
$A_{(s)} + B^{+}_{(aq)} \rightleftharpoons A^{+}_{(aq)} + B_{(s)}$
(Given: $\frac{2.303 RT}{F} = 0.06 \ V$)

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 $2$.

What are the uses of standard half-cell potential?