If $Cu^{+} + e^- \to Cu$ ; $E^o = X_1$ and $Cu^{2+} + 2e^- \to Cu$ ; $E^o = X_2$,then the value of $E^o$ for $Cu^{2+} + e^- \to Cu^{+}$ will be

The standard electrode potentials of $Ag^{+}/Ag$,$Hg_2^{2+}/2Hg$,$Cu^{2+}/Cu$,and $Mg^{2+}/Mg$ are $0.80 \ V$,$0.79 \ V$,$0.34 \ V$,and $-2.37 \ V$,respectively. An aqueous solution containing $1 \ M$ concentration of each of these metal salts is electrolyzed. With increasing voltage,what is the correct sequence of deposition of the metals at the cathode?

The $E^{\circ}_{cell}$ of $Cu_{(s)} | Cu^{2+}_{(1M)} || Ag^{+}_{(1M)} | Ag_{(s)}$ is $0.647 \ V$. Calculate the $E^{\circ}_{Ag}$ if $E^{\circ}_{Cu}$ is $0.153 \ V$. (in $V$)

Given below are half-cell reactions:
$MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$,
$E^{o}_{MnO_{4}^{-} / Mn^{2+}} = +1.510 \, V$
$\frac{1}{2} O_{2} + 2H^{+} + 2e^{-} \rightarrow H_{2}O$,
$E^{o}_{O_{2} / H_{2}O} = +1.223 \, V$
Will the permanganate ion,$MnO_{4}^{-}$,liberate $O_{2}$ from water in the presence of an acid?

When oxidation and reduction take place in a cell,its electromotive force $(EMF)$ will be

Using the standard reduction potentials of the electrodes $Li$,$Zn$,$Mg$,and $Ni$ as $-3.05 \ V$,$-0.76 \ V$,$-2.36 \ V$,and $-0.25 \ V$ respectively,identify the correct statement.