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

Standard reduction potentials of the half reactions are given below :

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