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Step-by-Step Explanation
Step 1: Identify the Given Reduction Potentials
The question provides the standard reduction potentials for conversions of halogens (F2, Cl2, Br2, and I2) to their respective halide ions:
$\displaystyle F_{2}(g) + 2e^- \to 2F^{-}(aq) \quad E^\circ = +2.85\text{ V}$
$\displaystyle Cl_{2}(g) + 2e^- \to 2Cl^{-}(aq) \quad E^\circ = +1.36\text{ V}$
$\displaystyle Br_{2}(l) + 2e^- \to 2Br^{-}(aq) \quad E^\circ = +1.06\text{ V}$
$\displaystyle I_{2}(s) + 2e^- \to 2I^{-}(aq) \quad E^\circ = +0.53\text{ V}$
Step 2: Determine the Strongest Oxidizing Agent
The higher the reduction potential, the greater the tendency of the species to gain electrons (i.e., the stronger it acts as an oxidizing agent). From the values given:
$E^\circ(F_{2}/F^{-}) = +2.85\text{ V}$
$E^\circ(Cl_{2}/Cl^{-}) = +1.36\text{ V}$
$E^\circ(Br_{2}/Br^{-}) = +1.06\text{ V}$
$E^\circ(I_{2}/I^{-}) = +0.53\text{ V}$
The highest reduction potential is for fluorine ($F_{2}$). Hence, $F_{2}$ is the strongest oxidizing agent among the given species.
Step 3: Determine the Strongest Reducing Agent
Conversely, the lower the reduction potential of the halide pair, the stronger its conjugate base (halide ion) is as a reducing agent (i.e., it more readily loses electrons when converted back to its element). From the data, iodide ($I^{-}$) has the lowest potential in its corresponding half-reaction:
$E^\circ(I_{2}/I^{-}) = +0.53\text{ V}$
Hence, $I^{-}$ is the strongest reducing agent among the given halides.
Step 4: Combine the Results
Based on the standard reduction potentials, the strongest oxidizing agent is $F_{2}$ and the strongest reducing agent is $I^{-}$.
Final Answer
The strongest oxidizing and reducing agents, respectively, are $F_{2}$ and $I^{-}$.
