Standard reduction potentials of the half reactions are given below
`{:(F_(2(g))+2e^(-) rarr 2F^(-)""_((aq)),,E^(@)=+2.85V), (Cl_(2(g))+2e^(-) rarr 2Cl^(-)""_((aq)),, E^(@)=+1.36V), (Br_(2(l))+2e^(-) rarr 2Br^(-)""_((aq)),,E^(@)=+1.06V), (l_(2(s))+2e^(-) rarr 2l^(-)""_((aq)),, E^(@)=+0.53V):}`
The strongest oxidising and reducing agents respectively are

Options

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

Answer

A

Explanation

Since `F_(2)` has the highest electrode potential, therefore, it is the strongest oxidising agent and `F^(-)` is the weakest reducing agent.
Further, since `I_(2)` has the lowest electrode potential, therefore, it is the weakest oxidising agent and conversely `I^(-)` is the strongest reducing agent. Thus, option (a) is correct.