Given
`E_(cr^(3+)//Cr)^(0)=-0.74 V, E_(MnO_(4)//Mn^(2+))^(0)=1.51 cm`
`E_(Cr_(2)O_(7)^(2-)//Cr^(3+))^(0)=1.33 V, E_(Cl//Cl^(-))^(0)=1.36 V`
Based on the data given above, strongest oxidising agent will be:

Options

A. `Cr^(3+)`
B. `Mn^(2+)`
C. `MnO_(4)^(-)`
D. `CI^(-)`

Answer

To determine the strongest oxidizing agent from the given standard reduction potentials, we can 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. \( E^{\circ}_{Cr^{3+}/Cr} = -0.74 \, V \)
2. \( E^{\circ}_{MnO_{4}^{-}/Mn^{2+}} = +1.51 \, V \)
3. \( E^{\circ}_{Cr_{2}O_{7}^{2-}/Cr^{3+}} = +1.33 \, V \)
4. \( E^{\circ}_{Cl/Cl^{-}} = +1.36 \, V \)

### Step 3: Compare the Standard Reduction Potentials
To find the strongest oxidizing agent, we need to identify which of these potentials is the highest:
- \( -0.74 \, V \) (for Cr)
- \( +1.51 \, V \) (for MnO4)
- \( +1.33 \, V \) (for Cr2O7)
- \( +1.36 \, V \) (for Cl)

### Step 4: Identify the Highest Value
From the values listed:
- The highest standard reduction potential is \( +1.51 \, V \) for \( MnO_{4}^{-}/Mn^{2+} \).

### Step 5: Conclusion
Since \( MnO_{4}^{-} \) has the highest standard reduction potential, it is the strongest oxidizing agent among the given options.

### Final Answer
The strongest oxidizing agent is \( MnO_{4}^{-} \).

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