$E^0 = \frac{RT}{nF} \ln K_{eq}$. This is called

Consider the following $4$ electrodes:

$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$	$B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$	$D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

$E^{\circ}_{Ag^{+} / Ag} = +0.80 \ V$
Arrange the reduction potential of these electrodes in decreasing order.

Calculate $pH$ of $HCl$ solution at $298\,K$ temperature for the following cell: $Pt_{(s)} \mid H_2 \,(1\,bar) \mid HCl\,(xM) \parallel Cu^{2+}\,(0.02\,M) \mid Cu_{(s)}$. Given that the standard cell potential $E^{\circ}_{cell} = 0.34\,V$ and the measured cell potential $E_{cell} = 0.45\,V$.

The standard cell potential for $Zn \mid Zn^{2+} \parallel Cu^{2+} \mid Cu$ is $1.10 \ V$. When the cell is completely discharged,$\log ([Zn^{2+}] / [Cu^{2+}])$ is closest to $.....$

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 $\left(E_{Cu^{2+} / Cu}^{\circ}=+0.34 \ V ; E_{Zn^{2+} / Zn}^{\circ}=-0.76 \ V\right)$

At $298 \ K$,a $1 \ L$ solution containing $10 \ mmol$ of $Cr_2O_7^{2-}$ and $100 \ mmol$ of $Cr^{3+}$ shows a $pH$ of $3.0$. Given: $Cr_2O_7^{2-} \rightarrow Cr^{3+}; E^0 = 1.330 \ V$ and $\frac{2.303 RT}{F} = 0.059 \ V$. The potential for the half-cell reaction is $x \times 10^{-3} \ V$. The value of $x$ is $........$