Which is the proper value of $x$ for the following to increase the cell potential of $Zn_{(s)} | Zn_{(x \ M)}^{2+} || Cu_{(0.02 \ M)}^{2+} | Cu_{(s)}$?

Which of the following relations represents the correct relation between standard electrode potential and equilibrium constant?
$I$. $\log K = \frac{nF E^o}{2.303 RT}$
$II$. $K = e^{\frac{nF E^o}{RT}}$
$III$. $\log K = -\frac{nF E^o}{2.303 RT}$
$IV$. $\log K = 0.4342 \frac{nF E^o}{RT}$
Choose the correct statement$(s)$.

The $emf$ (in $V$) of a $Daniell$ cell containing $0.1 \ M \ ZnSO_4$ and $0.01 \ M \ CuSO_4$ solutions at their respective electrodes is $\left(E_{Cu^{2+} / Cu}^{\circ}=+0.34 \ V ; E_{Zn^{2+} / Zn}^{\circ}=-0.76 \ V\right)$

The Gibbs energy change of the reaction (in $kJ \ mol^{-1}$) corresponding to the following cell $Cr | Cr^{3+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe$ is: (Given: $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$)

For the cell,$Zn_{(s)} | Zn^{2+} (1 \ M) || Ag^{+} (1 \ M) | Ag_{(s)}$. If the concentration of $Zn^{2+}$ decreases to $0.1 \ M$ at $298 \ K$,then the $EMF$ of the cell:

What will be the $pH$ of a solution which is in contact with a hydrogen electrode having an oxidation potential of $0.177 \ V$?