At $298 \ K$,find out the $emf$ for the cell:
$Al_{(s)} | Al^{+3} (0.1 \ M) || Fe^{+2} (0.001 \ M) | Fe_{(s)}$
Given: $E^o_{Al^{+3}/Al} = -1.66 \ V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.

In an electrochemical reaction of lead,at standard temperature,if $E^0_{(Pb^{2+}/Pb)} = m \ V$ and $E^0_{(Pb^{4+}/Pb)} = n \ V$,then the value of $E^0_{(Pb^{2+}/Pb^{4+})}$ is given by $m - x n$. The value of $x$ is $...........$. (Nearest integer)

What will be the oxidation potential for the following hydrogen half cell at $1 \ bar$ pressure and $25^{\circ} C$ temperature (in $V$)?
$Pt \mid H_{2(g)} (1 \ bar) \mid HCl_{(aq)} \ pH = 3$

The standard $e.m.f.$ of a cell,involving one electron change is found to be $0.591 \ V$ at $25^{\circ} C$. The equilibrium constant of the reaction is :
$(F=96500 \ C \ mol^{-1} ; R=8.314 \ JK^{-1} \ mol^{-1})$

$E_{cell}^{0}$ of the reaction $Mg_{(s)} + 2 Ag_{(0.0001 \ M)}^{+} \rightleftharpoons Mg_{(0.01 \ M)}^{2+} + 2 Ag_{(s)}$ is $3.17 \ V$. The $E_{cell}$ of the reaction and its cell notation respectively are :

$A$ 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 $V$.