Using the data given below:
`E_(Cr_(2)O_(7)^(2-)|Cr^(3+))^(@)=1.33V E_(Cl_(2)|Cl^(-))^(@)=1.36V`
`E_(MnO_(4)^(-)|Mn^(2+))^(@)=1.51V E_(Cr^(3+)|Cr)=-0.74V`
Find the most stable oxidised species.

Options

A. `Cr^(3+)`
B. `MnO_4^-`
C. `Cr_2O_7^(2-)`
D. `Mn^(2+)`

Explanation

To determine the most stable oxidized species from the given standard reduction potentials, we will follow these steps:

### Step 1: Understand the concept of oxidation and reduction potentials
- Oxidation potential is the opposite of reduction potential. If we have a reduction potential \( E^\circ \), the corresponding oxidation potential can be calculated as:
\[ E^\circ_{\text{oxidation}} = -E^\circ_{\text{reduction}} \]
- The more positive the oxidation potential, the more stable the oxidized species.

### Step 2: List the given standard reduction potentials
- \( E^\circ (Cr_2O_7^{2-} | Cr^{3+}) = 1.33 \, V \)
- \( E^\circ (Cl_2 | Cl^-) = 1.36 \, V \)
- \( E^\circ (MnO_4^- | Mn^{2+}) = 1.51 \, V \)
- \( E^\circ (Cr^{3+} | Cr) = -0.74 \, V \)

### Step 3: Calculate the oxidation potentials
1. For \( Cr_2O_7^{2-} \) to \( Cr^{3+} \):
\[ E^\circ (Cr^{3+} | Cr_2O_7^{2-}) = -1.33 \, V \]

2. For \( Cl_2 \) to \( Cl^- \):
\[ E^\circ (Cl^- | Cl_2) = -1.36 \, V \]

3. For \( MnO_4^- \) to \( Mn^{2+} \):
\[ E^\circ (Mn^{2+} | MnO_4^-) = -1.51 \, V \]

4. For \( Cr \) to \( Cr^{3+} \):
\[ E^\circ (Cr^{3+} | Cr) = +0.74 \, V \]

### Step 4: Compare the oxidation potentials
- The calculated oxidation potentials are:
- \( Cr^{3+} | Cr \): \( +0.74 \, V \)
- \( Cr_2O_7^{2-} | Cr^{3+} \): \( -1.33 \, V \)
- \( Cl^- | Cl_2 \): \( -1.36 \, V \)
- \( Mn^{2+} | MnO_4^- \): \( -1.51 \, V \)

### Step 5: Identify the most stable oxidized species
- The highest oxidation potential is \( +0.74 \, V \) for the pair \( Cr^{3+} | Cr \).
- Therefore, the most stable oxidized species is \( Cr^{3+} \).

### Conclusion
The most stable oxidized species from the given data is \( Cr^{3+} \).

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