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)

For the cell reaction,$Cu | Cu^{2+}(0.1 \ M) || Cu^{2+}(1.0 \ M) | Cu$,the emf of the cell at $25^{\circ}C$ is given that $E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$. (in $V$)

Calculate the equilibrium constant of the reaction,$Cu_{(s)} + 2 Ag^{+}_{(aq)} \longrightarrow Cu^{2+}_{(aq)} + 2 Ag_{(s)}$,given that for the reaction $E^{\circ}_{cell} = 0.46 \ V$.

For the cell, $Pt | Cl_{2(g)} (0.4 \ bar) | Cl^{-} (aq.) (0.1 \ M) || Cl^{-} (aq.) (0.01 \ M) | Cl_{2(g)} (0.2 \ bar) | Pt$, the measured potential at $298 \ K$ is .............. $V$.

The logarithm of the equilibrium constant for the reaction $Pd^{2+}{(aq)} + 4Cl^{-}{(aq)} \rightleftharpoons PdCl_4^{2-}{(aq)}$ is (Nearest integer).
Given: $\frac{2.303 RT}{F} = 0.06 \ V$
$Pd^{2+}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} \quad E^{\circ} = 0.83 \ V$
$PdCl_4^{2-}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} + 4Cl^{-}{(aq)} \quad E^{\circ} = 0.65 \ V$

The e.m.f. of the cell $Ag | Ag^{+}(0.1 \ M) || Ag^{+}(1 \ M) | Ag$ at $298 \ K$ is ........... $V$.