The standard electrode potential of $M^{+} / M$ in aqueous solution does not depend on

Based on the given data,which is the strongest oxidizing agent? $E_{Li^+|Li}^o = -3.03 \ V$,$E_{Ba^{2+}|Ba}^o = -2.73 \ V$,$E_{Na^+|Na}^o = -2.71 \ V$,and $E_{Mg^{2+}|Mg}^o = -2.37 \ V$.

The standard reduction potentials of $4$ elements are given below. Which of the following will be the most suitable reducing agent?
$I = -3.04 \ V$
$II = -1.90 \ V$
$III = 0 \ V$
$IV = 1.90 \ V$

If $Cu^{2+} + 2e^{-} \rightarrow Cu, E^{0} = 0.337 \ V$ and $Cu^{2+} + e^{-} \rightarrow Cu^{+}, E^{0} = 0.153 \ V$,then for the reaction $Cu^{+} + e^{-} \rightarrow Cu$,$E^{0}_{cell} =$ .............. $V$.

Consider the cell given below:
$Ag_{(s)} | Ag^{\oplus} || Cu^{2+} | Cu_{(s)}$
Given:
$Ag^{\oplus} + e^{-} \to Ag; E^{o} = x$
$Cu^{2+} + 2e^{-} \to Cu; E^{o} = y$
The value of $E^{o}_{cell}$ is:

The reduction potential ($E^{\circ}$,in $V$) of $MnO_4^-{_{\text{(aq)}}} / Mn_{\text{(s)}}$ is. . . . . $[Given : E^{\circ}_{(MnO_4^{-(aq)} / MnO_{2(s)})} = 1.68 \ V ; E^{\circ}_{(MnO_{2(s)} / Mn^{2+}_{(aq)})} = 1.21 \ V ; E^{\circ}_{(Mn^{2+(aq)} / Mn_{(s)})} = -1.03 \ V]$