What is the reduction potential of a hydrogen gas electrode when pure hydrogen gas is at $1 \ atm$ pressure and the platinum electrode is in contact with an $HCl$ solution of $pH$ $1$ at $298 \ K$ (in $V$)?

Consider the cell at $25^{\circ} C$:
$Zn | Zn^{2+}_{(aq)} (1 \ M) || Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} | Pt_{(s)}$
The fraction of total iron present as $Fe^{3+}$ ion at the cell potential of $1.500 \ V$ is $X \times 10^{-2}$. The value of $X$ is $.....$ (Nearest integer).
(Given $E^{0}_{Fe^{3+} / Fe^{2+}} = 0.77 \ V, E^{0}_{Zn^{2+} / Zn} = -0.76 \ V$)

Calculate the $E.M.F.$ of the following cell at $298 \ K$: $Zn_{(s)} | ZnSO_4(0.01 \ M) || CuSO_4(1.0 \ M) | Cu_{(s)}$ if $E^o_{cell} = 2.0 \ V$. (in $V$)

The standard electrode potential for the following reaction is $+1.33 \ V$. What is the potential at $pH = 2.0$ for the reaction: $Cr_2O_7^{2-} (aq, 1 \ M) + 14H^{+} (aq) + 6e^{-} \rightarrow 2Cr^{3+} (aq, 1 \ M) + 7H_2O (l)$?

The equilibrium constant for the following general reaction is $10^{30}$. Calculate $E^o$ for the cell at $298 \ K$.
$2 X_2(s)+3 Y^{2+}(a q) \rightarrow 2 X_2^{3+}(a q)+3 Y(s)$

For a cell having a standard $emf$ of $0.295 \ V$ at $25^o \ C$ involving a two-electron change,the equilibrium constant for the reaction is: