The standard electrode potential for the Daniell cell is $1.1 \ V$. What will be the value of standard Gibbs energy for the reaction?
$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
$(1 \ F = 96487 \ C \ mol^{-1})$

Given are $E^{\circ}$ values for some half reactions:
$I_2 + 2e^{-} \to 2I^{-}$ ; $E^{\circ} = 0.54 \, V$
$MnO_4^{-} + 8H^{+} + 5e^{-} \to Mn^{2+} + 4H_2O$ ; $E^{\circ} = 1.52 \, V$
$Fe^{3+} + e^{-} \to Fe^{2+}$ ; $E^{\circ} = 0.77 \, V$
$Sn^{4+} + 2e^{-} \to Sn^{2+}$ ; $E^{\circ} = 0.1 \, V$
The strongest reducant and oxidant respectively are:

For the following cell reaction,$Ag | Ag^{+} | AgCl | Cl^{-} | Cl_2, Pt$
$\Delta G_f^{\circ}(AgCl) = -109 \ kJ/mol$
$\Delta G_f^{\circ}(Cl^{-}) = -129 \ kJ/mol$
$\Delta G_f^{\circ}(Ag^{+}) = 78 \ kJ/mol$
$E^{\circ}$ of the cell is

On the basis of the following electrode potentials,which one is the strongest reducing agent?
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Br_2/Br^{-}} = 1.09 \text{ V}$,$E^0_{Zn^{2+}/Zn} = -0.76 \text{ V}$

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

Reduction potential of ions are given below:

$ClO_4^{-}$	$E^{\circ} = 1.19 \ V$
$IO_4^{-}$	$E^{\circ} = 1.65 \ V$
$BrO_4^{-}$	$E^{\circ} = 1.74 \ V$

The correct order of their oxidising power is: