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 ion in its reduced forms

Options

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

Explanation

To determine the most stable ion in its reduced form using the provided standard reduction potentials, we will follow these steps:

### Step 1: List the given standard reduction potentials
We have the following standard reduction potentials:
1. \( E^\circ (Cr_2O_7^{2-} | Cr^{3+}) = 1.33 \, V \)
2. \( E^\circ (Cl_2 | Cl^-) = 1.36 \, V \)
3. \( E^\circ (MnO_4^- | Mn^{2+}) = 1.51 \, V \)
4. \( E^\circ (Cr^{3+} | Cr) = -0.74 \, V \)

### Step 2: Identify the highest reduction potential
The stability of an ion in its reduced form is indicated by the highest standard reduction potential. We will compare the given values:

- \( 1.33 \, V \) (for \( Cr_2O_7^{2-} \) to \( Cr^{3+} \))
- \( 1.36 \, V \) (for \( Cl_2 \) to \( Cl^- \))
- \( 1.51 \, V \) (for \( MnO_4^- \) to \( Mn^{2+} \))
- \( -0.74 \, V \) (for \( Cr^{3+} \) to \( Cr \))

### Step 3: Compare the values
Among these values, \( 1.51 \, V \) (from \( MnO_4^- \) to \( Mn^{2+} \)) is the highest.

### Step 4: Conclusion
Since \( MnO_4^- \) has the highest standard reduction potential, the most stable ion in its reduced form is \( Mn^{2+} \).

### Final Answer
The most stable ion in its reduced form is \( Mn^{2+} \).

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