What happens to the Gibbs free energy in a spontaneous electrochemical process?

For the reaction $F_2 + 2e^{-} \to 2F^{-}$,$E^{\circ} = 2.8 \, V$. What is the $E^{\circ}$ for the reaction $\frac{1}{2} F_2 + e^{-} \to F^{-}$?

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

Give the symbolic representation of the following half-cells (electrodes):
$(i)$ $2H^{+}_{(aq)} + 2e^- \to H_{2_{(g)}}$
$(ii)$ $Br_{2_{(aq)}} + 2e^- \to 2Br^{-}_{(aq)}$
$(iii)$ $2Br^{-}_{(aq)} \to Br_{2_{(aq)}} + 2e^-$

$I_2 + 2e^{-} \to 2I^{-}$; $E^{o} = 0.54 \ V$
$Cl_2 + 2e^{-} \to 2Cl^{-}$; $E^{o} = 1.36 \ V$
$Mn^{3+} + e^{-} \to Mn^{2+}$; $E^{o} = 1.50 \ V$
$Fe^{3+} + e^{-} \to Fe^{2+}$; $E^{o} = 0.77 \ V$
Which of the following is a correct statement?

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