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

Given:
$E^o_{Cr^{3+}/Cr} = -0.74 \ V$,$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$
$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$,$E^o_{Cl_2/Cl^{-}} = 1.36 \ V$
Based on the data given above,the strongest oxidising agent will be:

Standard electrode potential for the cell with cell reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ is $1.1 \ V$. Calculate the standard Gibbs energy change for the cell reaction. (Given $F = 96487 \ C \ mol^{-1}$)

The $E^{\ominus}$ for $Cl_{2}/Cl^{-}$ is $+1.36 \ V,$ for $I_{2}/I^{-}$ is $+0.53 \ V,$ for $Ag^{+}/Ag$ is $+0.79 \ V,$ $Na^{+}/Na$ is $-2.71 \ V$ and for $Li^{+}/Li$ is $-3.04 \ V.$ Arrange the following ionic species in decreasing order of reducing strength: $I^{-}, Ag, Cl^{-}, Li, Na$

$K,$ $Ca,$ and $Li$ metals may be arranged in the increasing order of their standard electrode potentials as

The standard electrode potential of the cell $H_2 | H^{+} || Ag^{+} | Ag$ is .......... $V$.