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

Answer

To determine the strongest reducing agent among the given half-cell reactions, we need to analyze the standard reduction potentials provided. The strongest reducing agent will be the species with the lowest standard reduction potential because it has a greater tendency to lose electrons (be oxidized).

### Step-by-Step Solution:

1. **List the Standard Reduction Potentials**:
- \( E^\circ_{Cl_2/Cl^-} = 1.36 \, V \)
- \( E^\circ_{Cr^{3+}/Cr} = -0.74 \, V \)
- \( E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \, V \)
- \( E^\circ_{MnO_4^-/Mn^{2+}} = 1.51 \, V \)

2. **Identify the Reducing Agents**:
- The reducing agent is the species that gets oxidized. Therefore, we look for the half-reaction that has the lowest standard reduction potential, as this indicates a stronger tendency to lose electrons.

3. **Compare the Standard Reduction Potentials**:
- Among the values:
- \( 1.36 \, V \) (for \( Cl_2/Cl^- \))
- \( -0.74 \, V \) (for \( Cr^{3+}/Cr \))
- \( 1.33 \, V \) (for \( Cr_2O_7^{2-}/Cr^{3+} \))
- \( 1.51 \, V \) (for \( MnO_4^-/Mn^{2+} \))
- The lowest value is \( -0.74 \, V \) for the half-reaction \( Cr^{3+}/Cr \).

4. **Conclusion**:
- Since \( Cr^{3+} \) has the lowest standard reduction potential, it is the strongest reducing agent among the given options.

### Final Answer:
The strongest reducing agent is \( Cr^{3+} \).