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).

For the reaction $4B_{(s)} + 3O_{2(g)} \rightarrow 2B_2O_{3(g)}$,the standard cell potential is $E^o_{cell} = 1.433 \ V$. Calculate the molar entropy $(S_m^o)$ of oxygen gas in $J/K \ mol$.
Given:
$(\Delta_fH^o)_{B_2O_3(g)} = -840 \ kJ/mol$
$(S_m^o)_{B_2O_3(g)} = 280 \ J/K \ mol$
$(S_m^o)_{B(s)} = 10 \ J/K \ mol$
Assume $\Delta_rG^o = -nFE^o_{cell}$ and $\Delta_rG^o = \Delta_rH^o - T\Delta_rS^o$ at $T = 298 \ K$.

Match List-$I$ with List-$II$.

List-$I$	List-$II$
$A$. $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$	$I$. Primary battery
$B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$	$II$. Discharging of secondary battery
$C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$	$III$. Fuel cell
$D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$	$IV$. Charging of secondary battery

Choose the correct answer from the options given below.

Match the following:

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

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

At $300 \ K$,the $E_{cell}^{\circ}$ of $A_{(s)} + B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + B_{(s)}$ is $1.0 \ V$. If $\Delta_r S^{\circ}$ of this reaction is $100 \ J \ K^{-1} \ mol^{-1}$,what is $\Delta_r H^{\circ}$ (in $kJ \ mol^{-1}$) of this reaction? $(F = 96500 \ C \ mol^{-1})$

The number of incorrect statements from the following is:
$A.$ The electrical work that a reaction can perform at constant pressure and temperature is equal to the reaction Gibbs energy.
$B.$ $E_{cell}^0$ is dependent on the pressure.
$C.$ $\frac{dE_{cell}^0}{dT} = \frac{\Delta_{r}S^0}{nF}$.
$D.$ $A$ cell is operating reversibly if the cell potential is exactly balanced by an opposing source of potential difference.