Standard reduction potentials of the half reactions are given below $\begin{array}{ll} \mathrm{F}_2(g)+2…
Standard reduction potentials of the half reactions are given below
$\begin{array}{ll}
\mathrm{F}_2(g)+2 e^{-} \longrightarrow 2 \mathrm{~F}^{-}(a q) ; & E^{\circ}=+2.85 \mathrm{~V} \\
\mathrm{Cl}_2(g)+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-}(a q) ; & E^{\circ}=+1.36 \mathrm{~V} \\
\mathrm{Br}_2(l)+2 e^{-} \longrightarrow 2 \mathrm{Br}^{-}(a q) ; & E^{\circ}=+1.06 \mathrm{~V} \\
\mathrm{I}_2(s)+2 e^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) ; & E^{\circ}=+0.53 \mathrm{~V}
\end{array}$
The strongest oxidising and reducing agents respectively are
$\mathrm{F}_2$ and $\mathrm{I}^{-}$
$\mathrm{Br}_2$ and $\mathrm{Cl}^{-}$
$\mathrm{Cl}_2$ and $\mathrm{Br}^{-}$
$\mathrm{Cl}_2$ and $\mathrm{I}_2$
Solution
Higher the value of standard reduction potential, stronger will be the oxidising agent. Therefore, $\mathrm{F}_2$ will act as strongest oxidising agent.
Similarly lower the value of standard reduction potential, stronger will be the reducing agent. Therefore, $\mathrm{I}^{-}$will act as strongest reducing agent.