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

Answer

C

Explanation

Higher the reduction potential, more easily it is reduced and hence stronger is the oxidizing agent. `F_(2)` has highest reduction potential. Hence, `F_(2)` is strongest oxidizing agent. Considering the reverse reaction (oxidation reaction), higher the oxidation potential, more easily it is oxidized and hence stronger is the reducing agent. oxidation potential of `I^(-)` ion `(-0.53V)` is highest out of -2.85,-1.36,,-1.06 and -0.53. hence, `I^(-)` is strongest reducing agent.