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 "l_(2)`
B. `Br_(2)" and "Cl^(-)`
C. `Cl_(2)" and "Br^(-)`
D. `Cl_(2)" and "l_(2)`

Answer

To determine the strongest oxidizing and reducing agents from the given standard reduction potentials, we can follow these steps:

### Step 1: Understand the Concept of Oxidizing and Reducing Agents
- An **oxidizing agent** is a substance that gains electrons (is reduced) and causes another substance to be oxidized.
- A **reducing agent** is a substance that loses electrons (is oxidized) and causes another substance to be reduced.

### Step 2: Analyze the Given Standard Reduction Potentials
The standard reduction potentials (E°) for the half-reactions are:
1. \( F_2(g) + 2e^- \rightarrow 2F^-(aq) \), \( E° = +2.85 \, V \)
2. \( Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq) \), \( E° = +1.36 \, V \)
3. \( Br_2(l) + 2e^- \rightarrow 2Br^-(aq) \), \( E° = +1.06 \, V \)
4. \( I_2(s) + 2e^- \rightarrow 2I^-(aq) \), \( E° = +0.53 \, V \)

### Step 3: Identify the Strongest Oxidizing Agent
- The strongest oxidizing agent corresponds to the half-reaction with the highest standard reduction potential.
- From the values provided, \( F_2 \) has the highest reduction potential of \( +2.85 \, V \).
- Therefore, **Fluorine (F2)** is the strongest oxidizing agent.

### Step 4: Identify the Strongest Reducing Agent
- The strongest reducing agent corresponds to the half-reaction with the lowest standard reduction potential.
- From the values provided, \( I_2 \) has the lowest reduction potential of \( +0.53 \, V \).
- Therefore, **Iodine (I2)** is the strongest reducing agent.

### Step 5: Conclusion
- The strongest oxidizing agent is **Fluorine (F2)**.
- The strongest reducing agent is **Iodine (I2)**.

Thus, the answer to the question is:
**Fluorine and Iodine**.