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Standard reduction potentials of the half reactions are given below:
$F_2(g) + 2e^- \rightarrow 2F^-(aq)$; $E^0 = +2.85$ V
$Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq)$; $E^0 = +1.36$ V
$Br_2(l) + 2e^- \rightarrow 2Br^-(aq)$; $E^0 = +1.06$ V
$I_2(s) + 2e^- \rightarrow 2I^-(aq)$; $E^0 = +0.53$ V
The strongest oxidising and reducing agents respectively are:
$F_2(g) + 2e^- \rightarrow 2F^-(aq)$; $E^0 = +2.85$ V
$Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq)$; $E^0 = +1.36$ V
$Br_2(l) + 2e^- \rightarrow 2Br^-(aq)$; $E^0 = +1.06$ V
$I_2(s) + 2e^- \rightarrow 2I^-(aq)$; $E^0 = +0.53$ V
The strongest oxidising and reducing agents respectively are:
A
$Cl_2$ and $I_2$
B
$F_2$ and $I^-$
C
$Br_2$ and $Cl^-$
D
$Cl_2$ and $Br^-$
Detailed Solution
A higher reduction potential means a greater tendency to be reduced, i.e. a stronger oxidising agent.
$F_2$ has the highest $E^0$ (+2.85 V), so it is the strongest oxidising agent.
The $I_2/I^-$ couple has the lowest $E^0$ (+0.53 V), so $I^-$ is most easily oxidised and is the strongest reducing agent.
$F_2$ has the highest $E^0$ (+2.85 V), so it is the strongest oxidising agent.
The $I_2/I^-$ couple has the lowest $E^0$ (+0.53 V), so $I^-$ is most easily oxidised and is the strongest reducing agent.