Which one among the followig is the strongest reducing agent
`Fe^(2+)+2e^(-)toFe(-0.44V)`
`Ni^(2+)+2e^(-)toNi(-0.25V)`
`Sn^(2+)+2e^(-)toSn(-0.14V)`
`Fe^(3+)+e^(-)toFe^(2+)(-0.77V)`.

Options

A. Fe
B. `Fe^(2+)`
C. `Ni`
D. `Sn`

Explanation

To determine which species is the strongest reducing agent among the given options, we need to analyze the standard reduction potentials provided for each half-reaction. The stronger the reducing agent, the more negative its standard reduction potential will be.

### Step-by-Step Solution:

1. **List the Given Half-Reactions and Their Potentials**:
- \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad E^\circ = -0.44 \, \text{V} \)
- \( \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \quad E^\circ = -0.25 \, \text{V} \)
- \( \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn} \quad E^\circ = -0.14 \, \text{V} \)
- \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad E^\circ = -0.77 \, \text{V} \)

2. **Identify the Reducing Agents**:
- The reducing agent is the species that gets oxidized (loses electrons). Therefore, we can consider the reverse of the given half-reactions to identify the reducing agents:
- \( \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \quad E^\circ = +0.44 \, \text{V} \)
- \( \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^- \quad E^\circ = +0.25 \, \text{V} \)
- \( \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \quad E^\circ = +0.14 \, \text{V} \)
- \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \quad E^\circ = +0.77 \, \text{V} \)

3. **Compare the Reduction Potentials**:
- The more negative the standard reduction potential, the stronger the reducing agent.
- The potentials for the reducing reactions are:
- For Fe: \( +0.44 \, \text{V} \)
- For Ni: \( +0.25 \, \text{V} \)
- For Sn: \( +0.14 \, \text{V} \)
- For Fe²⁺: \( +0.77 \, \text{V} \)

4. **Determine the Strongest Reducing Agent**:
- Among the values, \( +0.44 \, \text{V} \) (for Fe) is the highest, indicating that Fe is the strongest reducing agent among the choices given.

### Conclusion:
The strongest reducing agent among the given options is **Fe**.