The potential of a hydrogen electrode with $pH = 10$ with respect to a standard hydrogen electrode is:

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 $Fe$

The reduction potential of a hydrogen electrode at $25^{\circ}C$ is ............... $V$ $(P_{H_2} = 1 \ atm; [H^+] = 0.1 \ M)$.

The standard electrode potential for $Cu^{+2}/Cu$ is $0.34 \ V$. Calculate the reduction potential at $pH = 14$ for the above couple $V$ $[K_{sp}[Cu(OH)_2] = 1 \times 10^{-19}]$

For the redox reaction $Zn(s) + Cu^{2+}(0.1 \ M) \rightarrow Zn^{2+}(1 \ M) + Cu(s)$,given $E^o_{cell} = 1.10 \ V$,calculate the $E_{cell}$ value in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$) (in $V$)

The standard $emf$ of a galvanic cell involving $3$ moles of electrons in its redox reaction is $0.59 \ V$. The equilibrium constant for the reaction of the cell is: