What is the potential of a hydrogen electrode in contact with a solution whose $pH$ is $1$ (in $V$)?

At $298 \ K$,some standard electrode potentials are given below:

$Pb^{2+} / Pb$	$-0.13 \ V$
$Ni^{2+} / Ni$	$-0.24 \ V$
$Cd^{2+} / Cd$	$-0.40 \ V$
$Fe^{2+} / Fe$	$-0.44 \ V$

Metal rods $X$ and $Y$ are inserted into a solution containing $0.001 \ M$ $X^{2+}$ and $0.1 \ M$ $Y^{2+}$ at $298 \ K$ and connected by a conducting wire. This results in the dissolution of $X$. The correct combination$(s)$ of $X$ and $Y$ are,respectively:
(Given: Gas constant,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,Faraday constant,$F = 96500 \ C \ mol^{-1}$)
$(A) \ Cd$ and $Ni \ \ (B) \ Cd$ and $Fe \ \ (C) \ Ni$ and $Pb \ \ (D) \ Ni$ and $Fe$

For the cell $Zn | Zn^{2+}(0.01 \, M) || Fe^{2+}(0.001 \, M) | Fe$ at $25^o C$,the $E_{cell} = 0.2905 \, V$. The equilibrium constant $K_c$ is:

For the reaction $A_{(s)} + 2B^{+}_{(aq)} \rightarrow A^{2+}_{(aq)} + 2B_{(s)}$,the value of $K_c$ is $10^{12}$. What is the value of $E^o_{cell}$ (in $, V$)?

For a cell involving a one-electron change at $25^o C$,$E^{o}_{cell} = 0.591 \ V$. The equilibrium constant for the reaction is .....

Give the equation to calculate the equilibrium constant $K_C$ of a Daniell cell.