What will be the $EMF$ of the cell formed by the following half-cells in $V$?
$Mg^{2+} + 2e^- \to Mg_{(s)}; E = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E = +0.33 \ V$

Consider the systems having liquid-solid interface,$(A)$ copper wire in silver nitrate solution and $(B)$ silver wire in copper sulphate solution. Predict which interface will show spontaneous reaction,if $E_{Cu^{2+}/Cu}^{\circ} = 0.34 \ V$ and $E_{Ag^{+}/Ag}^{\circ} = 0.80 \ V$?

If the $E^{0}$ values for $Mg^{+2} | Mg$,$Zn^{+2} | Zn$,and $Fe^{+2} | Fe$ are $-2.37 \ V$,$-0.76 \ V$,and $-0.44 \ V$ respectively,which statement is correct?

Given the standard electrode potentials: $E^o_{Cr^{3+}/Cr} = -0.74 \ V$,$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$,$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$,and $E^o_{Cl_2/Cl^-} = 1.36 \ V$. Based on this information,which is the strongest oxidizing agent?

Iron displaces copper from its salt solution,because

If the electrode potential of $Zn^{2+} | Zn$ is $-0.76 \, V$ and the potential of $Cu^{2+} | Cu$ is $0.34 \, V$,what will be the $EMF$ of the cell formed by both electrodes in $V$?