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

$A$ hydrogen gas electrode is made by dipping platinum wire in a solution of $HCl$ of $pH = 10$ and by passing hydrogen gas around the platinum wire at $1 \ atm$ pressure. The oxidation potential of the electrode would be $.......... \ V$.

Calculate the equilibrium constant for the cell reaction at $298 \ K$: $Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$,given $E^{0}_{cell} = 0.46 \ V$.

What is the potential of the cell containing two hydrogen electrodes as represented below $V$:
$Pt | \frac{1}{2} H_{2(g)} | H^{+} (10^{-8} M) || H^{+} (10^{-3} M) | \frac{1}{2} H_{2(g)} | Pt$

For $Cr_2O_7^{2-} + 14H^{+} + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$,$E^0 = 1.33 \ V$. Given $[Cr_2O_7^{2-}] = 4.5 \ mmol$,$[Cr^{3+}] = 1.5 \ mmol$ and $E = 1.067 \ V$,calculate the $pH$ of the solution.

$Pt_{(s)} | H_{2(g)} (1 \ atm) | H^{+} (pH = 2) || H^{+} (pH = 3) | H_{2(g)} (1 \ atm) | Pt_{(s)}$ cell reaction will be