Given `E_(Cl_(2)//Cl^(-))^(@)=1.36V,E_(Cr^(3+)//Cr)^(@)=-0.74V`
`E_("Cr_(2)O_(7)^(2-)//Cr^(3+))^(@)=1.33V,E_(MnO_(4)^(-)//Mn^(2+))^(@)=1.51V`
Among the following, the strongest reducing agent is

Options

A. `Cl^(1-)`
B. `Cr`
C. `Cr^(3+)`
D. `Mn^(2+)`

Answer

To determine the strongest reducing agent among the given options based on standard reduction potentials, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Reducing Agents**:
- A reducing agent is a substance that donates electrons in a chemical reaction and gets oxidized in the process. The strength of a reducing agent is inversely related to its standard reduction potential (E°). The lower the E° value, the stronger the reducing agent.

2. **Identify the Standard Reduction Potentials**:
- The given standard reduction potentials are:
- \( E_{Cl_2/Cl^-}^\circ = 1.36 \, V \)
- \( E_{Cr^{3+}/Cr}^\circ = -0.74 \, V \)
- \( E_{Cr_2O_7^{2-}/Cr^{3+}}^\circ = 1.33 \, V \)
- \( E_{MnO_4^-/Mn^{2+}}^\circ = 1.51 \, V \)

3. **Compare the Standard Reduction Potentials**:
- The key point is to look for the lowest value among the standard reduction potentials, as this indicates a strong tendency to lose electrons (i.e., be oxidized).
- From the values:
- \( 1.36 \, V \) (for \( Cl_2 \)) is high
- \( -0.74 \, V \) (for \( Cr^{3+}/Cr \)) is low
- \( 1.33 \, V \) (for \( Cr_2O_7^{2-}/Cr^{3+} \)) is high
- \( 1.51 \, V \) (for \( MnO_4^-/Mn^{2+} \)) is high

4. **Identify the Strongest Reducing Agent**:
- The lowest standard reduction potential is \( -0.74 \, V \) for the \( Cr^{3+}/Cr \) couple. This indicates that chromium (Cr) is the strongest reducing agent among the options provided.

5. **Conclusion**:
- Therefore, the strongest reducing agent among the given options is \( Cr^{3+} \) to \( Cr \).

### Final Answer:
The strongest reducing agent is **Chromium (Cr)**.