Given that `E_(O_(2))^(ɵ)//H_(2)O=+1.23V`,
`E_(S_(2)O_(8)^(2-)//SO_(4)^(-2))=2.05V`
`E_(Br_(2)^(ɵ)//Br-=+1.09V`,
`E_(Au^(ɵ)^(3+)//Au=+1.4V`
The strongest oxidizing agent is :

Options

A. `Au^(3+)`
B. `O_(2)`
C. `S_(2)O_(8)^(2-)`
D. `Br_(2)`

Answer

To determine the strongest oxidizing agent from the given standard reduction potentials, we will follow these steps:

### Step 1: Understand the concept of oxidizing agents
An oxidizing agent is a substance that gains electrons in a chemical reaction and is reduced. The strength of an oxidizing agent is indicated by its standard reduction potential (E°). The higher the E° value, the stronger the oxidizing agent.

### Step 2: List the given standard reduction potentials
We have the following standard reduction potentials:
1. \( E^{\circ}_{O_2/H_2O} = +1.23 \, V \)
2. \( E^{\circ}_{S_2O_8^{2-}/SO_4^{2-}} = +2.05 \, V \)
3. \( E^{\circ}_{Br_2/Br^-} = +1.09 \, V \)
4. \( E^{\circ}_{Au^{3+}/Au} = +1.40 \, V \)

### Step 3: Compare the standard reduction potentials
Now we will compare the values of the standard reduction potentials:
- \( E^{\circ}_{O_2/H_2O} = +1.23 \, V \)
- \( E^{\circ}_{S_2O_8^{2-}/SO_4^{2-}} = +2.05 \, V \)
- \( E^{\circ}_{Br_2/Br^-} = +1.09 \, V \)
- \( E^{\circ}_{Au^{3+}/Au} = +1.40 \, V \)

### Step 4: Identify the highest value
From the comparison:
- The highest standard reduction potential is \( +2.05 \, V \) for \( S_2O_8^{2-}/SO_4^{2-} \).

### Step 5: Conclusion
Since \( S_2O_8^{2-} \) has the highest standard reduction potential, it is the strongest oxidizing agent among the given options.

### Final Answer
The strongest oxidizing agent is \( S_2O_8^{2-} \).

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