The standard readuction potential at 298 k for the followng half cell reaction are
`Zn^(2)+2e^(-)rarrZn(s),E^(@)=-0.7623 V`
`Cr^(3+)(aq)+3e^(-)rarrCr(s),E^(@)=-0.70 v`
`2H^(+)(aq)+2e^(-)rarrH_(2)(g),E^(@)=0.0V`
`F_(2)+2e^(-)rarr 2f^(-)(aq)=2.87V`
which of the following is the strongest reducing agent

Options

A. Cr (s)
B. Zn(g)
C. `H_(2)(g)`
D. `F_(2)`(g)

Explanation

To determine which of the given half-cell reactions represents the strongest reducing agent, we need to analyze the standard reduction potentials provided. The reducing agent is the species that gets oxidized, meaning it donates electrons to another species.

### Step-by-Step Solution:

1. **Identify the Half-Cell Reactions and Their Standard Reduction Potentials:**
- \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}(s), \quad E^\circ = -0.7623 \, \text{V} \)
- \( \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}(s), \quad E^\circ = -0.70 \, \text{V} \)
- \( 2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g), \quad E^\circ = 0.0 \, \text{V} \)
- \( \text{F}_2 + 2e^- \rightarrow 2\text{F}^-(aq), \quad E^\circ = 2.87 \, \text{V} \)

2. **Convert Reduction Potentials to Oxidation Potentials:**
- The oxidation potential is the negative of the reduction potential.
- For Zinc: \( E^\circ_{\text{oxidation}} = +0.7623 \, \text{V} \)
- For Chromium: \( E^\circ_{\text{oxidation}} = +0.70 \, \text{V} \)
- For Hydrogen: \( E^\circ_{\text{oxidation}} = 0.0 \, \text{V} \)
- For Fluorine: \( E^\circ_{\text{oxidation}} = -2.87 \, \text{V} \)

3. **Compare the Oxidation Potentials:**
- Zinc: \( +0.7623 \, \text{V} \)
- Chromium: \( +0.70 \, \text{V} \)
- Hydrogen: \( 0.0 \, \text{V} \)
- Fluorine: \( -2.87 \, \text{V} \)

4. **Determine the Strongest Reducing Agent:**
- The strongest reducing agent is the one with the highest oxidation potential.
- Comparing the values, Zinc has the highest oxidation potential of \( +0.7623 \, \text{V} \).

5. **Conclusion:**
- Therefore, the strongest reducing agent among the given options is **Zinc (Zn)**.