For $Mg^{2+} + 2e^{-} \rightarrow Mg_{(s)}$,$E^{0} = -2.37 \, V$ and for $Cu^{2+} + 2e^{-} \rightarrow Cu_{(s)}$,$E^{0} = +0.33 \, V$. The $emf$ of the cell $(E^{0}_{Cell})$ is .......... $V$.

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 $V$?

Consider the reaction $M_{(aq)}^{n+} + n e^{-} \to M_{(s)}$. The standard reduction potential values of the elements $M_1$,$M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively. The order of their reducing power will be

In a cell,the following reactions take place:
$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$ $\quad$ $E^{\circ}_{Fe^{3+} / Fe^{2+}} = 0.77 \, V$
$2I^{-} \rightarrow I_{2} + 2e^{-}$ $\quad$ $E^{\circ}_{I_{2} / I^{-}} = 0.54 \, V$
The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} \, V$ at $298 \, K$. The value of $x$ is .... (Nearest Integer)

$E^o$ values for the half-cell reactions are given below:
$Cu^{2+} + e^- \to Cu^{+} \quad E^o = 0.15 \ V$
$Cu^{2+} + 2e^- \to Cu \quad E^o = 0.34 \ V$
What will be the $E^o$ for the half-cell reaction:
$Cu^{+} + e^- \to Cu$ ? $\dots \ V$

Calculate $E_{\text{cell}}^{\circ}$ for the reaction: $Mg_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Mg_{(aq)}^{2+} + 2 Ag_{(s)}$,given that $E_{Ag^{+}/Ag}^{\circ} = 0.8 \ V$ and $E_{Mg^{2+}/Mg}^{\circ} = -2.37 \ V$. (in $V$)