Standrd reduction potentials of the half reactions are given below :
`F_2 (g) +re^- rarr 2 F^(-) (aq) E^@ = + 2.85 V`
` Cl_2 (g) +2e^- rarr 2 Cl^-(aq) , E^2 = + 1.36 V`
`Br _2 (i) + 2 e^- rarr 2Br (aq) , E^2 = + 1. 06 V`
`I_2 (s) + 2 e^- rarr 2I^(-) (aq) , E^2 = + . 53 V`.
The strongest oxidizing and reducing agents respectively

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 lowest electrode potential. it is the weakest oxidising agent and `I^-` is the strongest reducing agent.