Bubbles of gas $z$ are evolved from a solution containing $x$ and $y$. If the order of reduction potential is $x > y > z$,then:

$2H^{+} \, (aq) + 2e^- \to H_2 \, (g)$. The standard electrode potential for the above reaction is (in volts):

The standard reduction potentials of $Zn^{2+}|Zn$,$Cu^{2+}|Cu$ and $Ag^{+}|Ag$ are respectively $-0.76 \ V$,$0.34 \ V$ and $0.80 \ V$. The following cells were constructed:
$(1)$ $Zn|Zn^{2+}||Cu^{2+}|Cu$
$(2)$ $Zn|Zn^{2+}||Ag^{+}|Ag$
$(3)$ $Cu|Cu^{2+}||Ag^{+}|Ag$
What is the correct order of $E_{\text{cell}}^{\circ}$ of these cells?

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

The $EMF$ of a cell whose half-cell reactions are given below is .......... $V$.
$Mg^{2+} + 2e^- \to Mg_{(s)}; E^o = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E^o = +0.34 \ V$ (in $V$)

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 $Cu^{+} + e^{-} \rightarrow Cu$,$E^{0}_{cell} =$ .............. $V$.