Standard reduction potentails of the half reactions are given below:
`F_(2)(g)+2e^(-) rarr 2F^(-)(aq.),, E^(ɵ)= +2.87`
`Cl_(2)(g)+2e^(-) rarr 2Cl^(-)(aq.),, E^(ɵ)= +1.36 V`
`Br_(2)(g)+2e^(-) rarr 2Br^(-)(aq.),, E^(ɵ)= +1.09 V`
`I_(2)(s)+2e^(-) rarr 2l^(-)(aq.),, E^(ɵ)= +0.54 V`
The strongest oxidizing and reducing agents respectively are:

Options

A. `Cl_(2)` and `I_(2)`
B. `F_(2)` and `I^(-)`
C. `Br_(2)` and `Cl^(-)`
D. `Cl_(2)` and `Br^(-)`

Answer

B

Explanation

Oxidizing agent oxidizes the other substance and gets oxidized. Therefore, strongest reducing agent has the lowest reduction potenital but highest oxidation potential:
`F_(2)(g)+2e^(-)rarr2F^(-)(aq.),E^(Theta) = +2.87V`
Reducting agent reduces the other substance and gets oxidized. Therefore, strongest reducing agent has the lowest reduction potential but highest oxidatiuon potential :
`2I^(-)(aq.)rarr 2e^(-)+I_(2)(s),E^(Theta) = -0.54 V`