The standard electrode potential of a $Cu^{2+} | Cu$ electrode is $0.34 \, V$ (reduction potential). What will be the electrode potential of a $0.001 \, M \, Cu^{2+}$ solution in $V$?

The $emf$ of a $Daniel$ cell at $298 \ K$ is ${E_1}$ for the cell reaction $Zn|ZnSO_4(0.01 \ M)||CuSO_4(1.0 \ M)|Cu$. When the concentration of $ZnSO_4$ is $1.0 \ M$ and that of $CuSO_4$ is $0.01 \ M$,the $emf$ changes to ${E_2}$. What is the relationship between ${E_1}$ and ${E_2}$?

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 equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is:

Consider the cell
$Pt_{(s)} \mid H_2(g, 1\,atm) \mid H^{+}(aq, 1\,M) \parallel Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} \mid Pt_{(s)}$
When the potential of the cell is $0.712\,V$ at $298\,K$,the ratio $[Fe^{2+}] / [Fe^{3+}]$ is $.......$ (Nearest integer).
Given: $Fe^{3+} + e^- \longrightarrow Fe^{2+}$,$E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.771\,V$
$\frac{2.303 RT}{F} = 0.06\,V$

In which of the following Galvanic cells is the $emf$ maximum? (Given: $E_{Mg^{2+} \mid Mg}^0 = -2.36 \ V$ and $E_{Cl_2 \mid 2 Cl^{-}}^0 = +1.36 \ V$)