$A$ hydrogen electrode is immersed in a solution with $pH = 0$ $(HCl)$. By how much will the potential (reduction) change if an equivalent amount of $NaOH$ is added to the solution? (Take $pH_2 = 1 \ atm$,$T = 298 \ K$).

Using the standard electrode potentials given in the Table $8.1$,predict if the reaction between the following is feasible:
$(a)$ $Fe^{3+}_{(aq)}$ and $I^{-}_{(aq)}$
$(b)$ $Ag^{+}_{(aq)}$ and $Cu_{(s)}$
$(c)$ $Fe^{3+}_{(aq)}$ and $Cu_{(s)}$
$(d)$ $Ag_{(s)}$ and $Fe^{3+}_{(aq)}$
$(e)$ $Br_{2(aq)}$ and $Fe^{2+}_{(aq)}$

The number of correct statements from the following is :
$A.$ $E_{cell}$ is an intensive parameter.
$B.$ $A$ negative $E^{\Theta}$ means that the redox couple is a stronger reducing agent than the $H^{+}/H_2$ couple.
$C.$ The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction.
$D.$ The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

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 $2$.

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).

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	$iv$. $K_a = \frac{C\alpha ^2}{1 - \alpha}$