If a hydrogen electrode is dipped in two solutions of $pH = 3$ and $pH = 6$ and a salt bridge is connected,the e.m.f. of the resulting cell is ............ $V$.

Calculate the $E_{cell}$ for $Zn_{(s)} | Zn^{2+}_{(0.1 \ M)} || Cr^{3+}_{(0.1 \ M)} | Cr_{(s)}$ at $25^{\circ} C$ if $E^{\circ}_{cell}$ is $0.02 \ V$. (in $V$)

Calculate the cell potential for the reaction $Mg_{(s)} \mid Mg^{2+}(0.18 \ M) \parallel Ag^{+}(0.01 \ M) \mid Ag_{(s)}$. Given standard electrode potentials are $E^{\circ}_{Mg^{2+}/Mg} = -2.37 \ V$ and $E^{\circ}_{Ag^{+}/Ag} = 0.80 \ V$. (in $V$)

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

Calculate the $emf$ of the cell: $Cr | Cr^{3+}(0.1 \ M) || Fe^{2+}(0.01 \ M) | Fe$. Given: $E^0_{Cr^{3+}/Cr} = -0.75 \ V$; $E^0_{Fe^{2+}/Fe} = -0.45 \ V$. Cell reaction: $2 \ Cr_{(s)} + 3 \ Fe^{2+}_{(aq)} \rightarrow 2 \ Cr^{3+}_{(aq)} + 3 \ Fe_{(s)}$.

For a cell involving a one-electron change at $25^o C$,$E^{o}_{cell} = 0.591 \ V$. The equilibrium constant for the reaction is .....