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,$RT/F = 0.059$)

Explain the equilibrium state in a Daniell cell and derive its equilibrium constant. Given: $E^o_{cell} = 1.1 \ V$.

For a cell involving one electron $E_{cell}^{\ominus} = 0.59 \; V$ at $298 \; K$,the equilibrium constant for the cell reaction is.
Given that $\frac{2.303 \; RT}{F} = 0.059 \; V$ at $T = 298 \; K$.

One half cell in a voltaic cell is constructed by dipping a silver rod in an $AgNO_3$ solution of unknown concentration,and the other half cell is a $Zn$ rod dipped in a $1 \text{ M}$ solution of $ZnSO_4$. $A$ voltage of $1.60 \text{ V}$ is measured at $298 \text{ K}$ for this cell. What is the concentration of $Ag^+$ ions in terms of $\log x$ (where $x = [Ag^+]$)? Given: $E^\ominus_{Zn^{2+}/Zn} = -0.76 \text{ V}$,$E^\ominus_{Ag^+/Ag} = +0.80 \text{ V}$,and $\frac{2.303RT}{F} = 0.059 \text{ V}$.

The cell potential for the following reaction is $0.03305 \ V$ at $298 \ K$. Find the value of $x$ for the reaction: $Zn | Zn^{2+} (0.1 \ M) || Cd^{2+} (x \ M) | Cd$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$,$E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $M$)

Calculate the e.m.f. of the half-cell given below: $Fe | FeSO_4$ $(a = 0.1 \ M)$ where $E^o_{OP} = 0.44 \ V$. (in $V$)