Copper from copper sulphate solution can be displaced by .............. The standard reduction potentials of some electrodes are given below:
$E^o (Fe^{2+}, Fe) = -0.44 \ V$
$E^o (Zn^{2+}, Zn) = -0.76 \ V$
$E^o (Cu^{2+}, Cu) = +0.34 \ V$
$E^o (Cr^{2+}, Cr) = -0.74 \ V$
$E^o (H^{+}, 1/2H_2) = 0.00 \ V$

At $298 \ K$,the equilibrium constant is $2 \times 10^{15}$ for the reaction:
$Cu_{(s)} + 2 Ag^{+}_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} + 2 Ag_{(s)}$
The equilibrium constant for the reaction $\frac{1}{2} Cu^{2+}_{(aq)} + Ag_{(s)} \rightleftharpoons \frac{1}{2} Cu_{(s)} + Ag^{+}_{(aq)}$ is $x \times 10^{-8}$. The value of $x$ is (Nearest Integer).

On the basis of standard electrode potential values,suggest which of the following reactions would take place?

At $300 \ K$,the conductivity of $0.01 \ mol \ dm^{-3}$ aqueous solution of acetic acid is $19.5 \times 10^{-5} \ S \ cm^{-1}$ and the limiting molar conductivity of acetic acid at the same temperature is $390 \ S \ cm^2 \ mol^{-1}$. The degree of dissociation of acetic acid is:

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 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 $\Lambda_m = \Lambda_m^\circ$)