$A$ hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $pH=9$ and passing hydrogen gas around the platinum wire at $1.2 \ atm$ pressure. The oxidation potential of such an electrode equals $V$.

$A$ solution of $Fe^{2+}$ is titrated potentiometrically using $Ce^{4+}$ solution. When $80 \%$ of $Fe^{2+}$ is titrated,the $EMF$ of the system in $V$ is (Given,$E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77 \ V$ and $Fe^{2+} + Ce^{4+} \longrightarrow Fe^{3+} + Ce^{3+}$)
$(\log 2 = 0.3, \log 3 = 0.5, \log 4 = 0.6)$

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

Calculate the cell potential for $Ni_{(s)} | Ni^{2+}(0.036 \ M) || Co^{2+}(0.018 \ M) | Co_{(s)}$ at $298 \ K$.

What pressure $(bar)$ of $H_2$ would be required to make the $emf$ of a hydrogen electrode zero in pure water at $25^{\circ} C$?

Calculate $\Delta G$ and $E_{cell}$ for the following cell at $298 \ K$ temperature.
$Al_{(s)} | Al^{3+} (0.01 \ M) || Fe^{2+} (0.02 \ M) | Fe_{(s)}$ $\left[ E^o_{Al^{3+}|Al} = -1.66 \ V \right.$ and $\left. E^o_{Fe^{2+}|Fe} = -0.44 \ V \right]$