If the standard electrode potential for a cell is $2 \ V$ at $300 \ K,$ the equilibrium constant $(K)$ for the reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$ at $300 \ K$ is approximately $(R = 8 \ J \ K^{-1} \ mol^{-1}, F = 96000 \ C \ mol^{-1})$

The equilibrium constant of the reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ with $E^{\circ}_{\text{cell}} = 0.0295 \ V$ is (Given: $\frac{2.303RT}{F} = 0.059$)

$A$ hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $pH=9$ and passing hydrogen gas around the platinum wire at $1.2 \ atm$ pressure. The oxidation potential of such an electrode equals $V$.

For the electrochemical cell,$Mg_{(s)} \mid Mg^{2+}(aq, 1 \ M) \parallel Cu^{2+}(aq, 1 \ M) \mid Cu_{(s)}$,the standard emf of the cell is $2.70 \ V$ at $300 \ K$. When the concentration of $Mg^{2+}$ is changed to $x$,the cell potential changes to $2.67 \ V$ at $300 \ K$. The value of $x$ is.
(Given: $\frac{F}{R} = 11500 \ K \ V^{-1}$,where $F$ is the Faraday constant and $R$ is the gas constant; $\ln(10) = 2.30$)

In the electrochemical cell $:$
$Zn \,|\,ZnSO_4\,(0.01\,M)\,||\,CuSO_4\,(1.0\,M)\,|\,Cu$
the $emf$ of this Daniell cell is $E_1.$ When the concentration of $ZnSO_4$ is changed to $1.0\,M$ and that of $CuSO_4$ changed to $0.01\,M,$ the $emf$ changes to $E_2.$ From the followings,which one is the relationship between $E_1$ and $E_2$ $?$ (Given,$RT/F = 0.059$)

The equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is: