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`
In which option the order of reducing power is correct?

Options

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

Explanation

To determine the order of reducing power based on the given standard reduction potentials, we can follow these steps:

### Step 1: Identify the Standard Reduction Potentials
We have the following standard reduction potentials (E° values):
- \( E_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \, V \)
- \( E_{Cl_2/Cl^-} = 1.36 \, V \)
- \( E_{MnO_4^-/Mn^{2+}} = 1.51 \, V \)
- \( E_{Cr^{3+}/Cr} = -0.74 \, V \)

### Step 2: Determine the Oxidation Potentials
The reducing power of a species is inversely related to its reduction potential. Therefore, we need to convert the reduction potentials to oxidation potentials by changing the sign:
- \( E_{Cr^{3+}/Cr} = -0.74 \, V \) (oxidation potential = +0.74 V)
- \( E_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \, V \) (oxidation potential = -1.33 V)
- \( E_{Cl_2/Cl^-} = 1.36 \, V \) (oxidation potential = -1.36 V)
- \( E_{MnO_4^-/Mn^{2+}} = 1.51 \, V \) (oxidation potential = -1.51 V)

### Step 3: Rank the Reducing Agents
The reducing power increases as the oxidation potential increases. Thus, we can rank the species based on their oxidation potentials:
1. \( Cr^{3+}/Cr \) (0.74 V) - Strongest reducing agent
2. \( Cr_2O_7^{2-}/Cr^{3+} \) (-1.33 V)
3. \( Cl_2/Cl^- \) (-1.36 V)
4. \( MnO_4^-/Mn^{2+} \) (-1.51 V) - Weakest reducing agent

### Step 4: Write the Order of Reducing Power
Based on the above analysis, the order of reducing power from strongest to weakest is:
1. \( Cr^{3+}/Cr \)
2. \( Cr_2O_7^{2-}/Cr^{3+} \)
3. \( Cl_2/Cl^- \)
4. \( MnO_4^-/Mn^{2+} \)

### Conclusion
The correct order of reducing power is:
- \( Cr^{3+} > Cr_2O_7^{2-} > Cl^- > Mn^{2+} \)