If $E^0_{Fe^{3+}/Fe} = x \ V$ and $E^0_{Fe^{2+}/Fe} = y \ V$,then what will be the value of $E^0_{Fe^{3+}/Fe^{2+}}$?

Given that $E_{O_2/H_2O}^o = +1.23 \ V$; $E_{S_2O_8^{2-}/SO_4^{2-}}^o = 2.05 \ V$; $E_{Br_2/Br^-}^o = +1.09 \ V$; $E_{Au^{3+}/Au}^o = 1.4 \ V$. The strongest oxidizing agent 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:

If $E_{Au^{+}/Au}^o$ is $1.69 \text{ V}$ and $E_{Au^{3+}/Au}^o$ is $1.40 \text{ V}$,then $E_{Au^{+}/Au^{3+}}^o$ will be ... $\text{V}$.

The standard electrode potential (in $V$) values for $Al^{3+}/Al$ and $Tl^{3+}/Tl$ are respectively:

The standard reduction potentials at $298 \ K$ for the following half reactions are given against each:
$Zn^{2+}(aq.) + 2e^- \rightleftharpoons Zn_{(s)}$; $E^\circ = -0.762 \ V$
$Cr^{3+}(aq.) + 3e^- \rightleftharpoons Cr_{(s)}$; $E^\circ = -0.740 \ V$
$2H^{+}(aq.) + 2e^- \rightleftharpoons H_{2(g)}$; $E^\circ = 0.00 \ V$
$Fe^{3+}(aq.) + e^- \rightleftharpoons Fe^{2+}(aq.)$; $E^\circ = 0.770 \ V$
Which is the strongest reducing agent?