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 $V$)

Fluorine is the best oxidising agent because it has

The standard oxidation potentials for the half-reactions are given as $Zn \to Zn^{2+} + 2e^{-}; E^o = +0.76 \ V$ and $Fe \to Fe^{2+} + 2e^{-}; E^o = +0.41 \ V$. The $EMF$ for the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.

For the given reactions:
$Sn^{2+} + 2e^{-} \rightarrow Sn$
$Sn^{4+} + 4e^{-} \rightarrow Sn$
The electrode potentials are $E^{\circ}_{Sn^{2+}/Sn} = -0.140 \ V$ and $E^{\circ}_{Sn^{4+}/Sn} = 0.010 \ V$. The magnitude of standard electrode potential for $Sn^{4+}/Sn^{2+}$,i.e.,$E^{\circ}_{Sn^{4+}/Sn^{2+}}$,is $..... \times 10^{-2} \ V$. (Nearest integer)

The standard reduction potential at $298 \ K$ for the following half cells are given below :-
$NO_3^{-} + 4H^{+} + 3e^{-} \rightarrow NO_{(g)} + 2H_2O \quad E^0 = 0.97 \ V$

$V^{2+}_{(aq)} + 2e^{-} \rightarrow V_{(s)}$	$E^0 = -1.19 \ V$
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$	$E^0 = -0.04 \ V$
$Ag^{+}_{(aq)} + e^{-} \rightarrow Ag_{(s)}$	$E^0 = 0.80 \ V$
$Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au_{(s)}$	$E^0 = 1.40 \ V$

The number of metal$(s)$ which will be oxidized by $NO_3^{-}$ in aqueous solution is $....$.

Based on the data given below: $E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$. The strongest reducing agent is: