Arrange the following metals in the order in which they displace each other from the solution of their salts: $Al, Cu, Fe, Mg$ and $Zn$.

If $E^{0}_{Ag^{+} | Ag} = 0.80\, V$ and $E^{0}_{Cu^{2+} | Cu} = 0.34\, V$,then the standard cell potential of the cell formed by these electrodes is ........... $V$.

The electrode potential for $M^{2+}_{(aq)} + e^- \rightarrow M^{+}_{(aq)}$ and $M^{+}_{(aq)} + e^- \rightarrow M_{(s)}$ are $+0.15 \text{ V}$ and $+0.50 \text{ V}$ respectively. The value of $E^\circ_{M^{2+}/M}$ will be: (in $\text{ V}$)

Given the standard electrode potentials:
$K^{+} / K = -2.93 \, V$,$Ag^{+} / Ag = 0.80 \, V$
$Hg^{2+} / Hg = 0.79 \, V$
$Mg^{2+} / Mg = -2.37 \, V$,$Cr^{3+} / Cr = -0.74 \, V$
Arrange these metals in their increasing order of reducing power.

Identify the reaction from the following that occupies the top position in the $EMF$ series (standard reduction potential) at $298 \; K$.

The two half-cell reactions of an electrochemical cell are given as: $Ag^{+} + e^{-} \rightarrow Ag$; $E^{\circ}_{Ag^{+}/Ag} = 0.7995 \ V$ and $Fe^{2+} \rightarrow Fe^{3+} + e^{-}$; $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.7710 \ V$. The value of cell $EMF$ will be: (in $V$)