Find out 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_(2)O_(7)^(2-) lt MnO_(4)^(-)`
D. `Mn^(2+) lt Cr^(3+) lt Cl^(-) lt Cr`

Explanation

To determine the correct order of reducing power among the given species, we need to analyze their oxidation potentials. The reducing power of a species is inversely related to its oxidation potential; that is, the lower the oxidation potential, the stronger the reducing agent.

### Step-by-Step Solution:

1. **Understand Reducing Power**:
- Reducing power is the ability of a species to donate electrons. A stronger reducing agent has a lower (more negative) oxidation potential.

2. **List the Oxidation Potentials**:
- We need to look at the standard oxidation potentials (E°) for the following reactions:
- \( \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \) (E° = 0.74 V)
- \( \text{Cr}_2\text{O}_7^{2-} + 14H^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7H_2O \) (E° = 1.33 V)
- \( \text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \) (E° = -1.36 V)
- \( \text{Mn}^{2+} + 4H_2O \rightarrow \text{MnO}_4^{-} + 8H^+ + 5e^- \) (E° = -1.51 V)

3. **Identify the Oxidation Potentials**:
- From the reactions, we can summarize the oxidation potentials:
- \( \text{Cr}^{3+} \) to \( \text{Cr} \): +0.74 V
- \( \text{Cr}_2\text{O}_7^{2-} \) to \( \text{Cr}^{3+} \): +1.33 V
- \( \text{Cl}^- \) to \( \text{Cl}_2 \): -1.36 V
- \( \text{Mn}^{2+} \) to \( \text{MnO}_4^{-} \): -1.51 V

4. **Rank the Species by Oxidation Potential**:
- The order of oxidation potentials from highest to lowest (most positive to most negative) is:
1. \( \text{Cr}_2\text{O}_7^{2-} \) (1.33 V)
2. \( \text{Cr}^{3+} \) (0.74 V)
3. \( \text{Cl}^- \) (-1.36 V)
4. \( \text{Mn}^{2+} \) (-1.51 V)

5. **Determine the Order of Reducing Power**:
- Since reducing power is inversely related to oxidation potential, the order of reducing power (from strongest to weakest) will be the reverse of the oxidation potentials:
1. \( \text{Mn}^{2+} \) (strongest reducing agent)
2. \( \text{Cl}^- \)
3. \( \text{Cr}^{3+} \)
4. \( \text{Cr}_2\text{O}_7^{2-} \) (weakest reducing agent)

6. **Select the Correct Option**:
- Based on the analysis, the correct order of reducing power is:
- \( \text{Mn}^{2+} > \text{Cl}^- > \text{Cr}^{3+} > \text{Cr}_2\text{O}_7^{2-} \)
- This corresponds to option number 2.