`SnCl_(2)` acts as a .............................. agent.

Answer

To determine the role of SnCl₂ (tin(II) chloride) in chemical reactions, we need to analyze its oxidation states and how it interacts with other substances.

### Step-by-Step Solution:

1. **Identify the Oxidation State of Sn in SnCl₂:**
- Tin (Sn) is in group 14 of the periodic table. In SnCl₂, tin has an oxidation state of +2. This is because each chlorine (Cl) atom has an oxidation state of -1, and there are two chlorine atoms.
- Therefore, the equation for the oxidation states can be set up as:
\[ x + 2(-1) = 0 \quad \text{(where } x \text{ is the oxidation state of Sn)} \]
Solving for \( x \):
\[ x - 2 = 0 \implies x = +2 \]

2. **Consider Possible Oxidation States of Sn:**
- Tin can exhibit multiple oxidation states, primarily +2 and +4. The +4 state is achieved when tin loses both its 5s and 5p electrons.
- In the case of SnCl₂, it is in the +2 oxidation state.

3. **Determine the Behavior of SnCl₂:**
- When SnCl₂ is involved in a reaction, it can be oxidized to SnCl₄ (tin(IV) chloride) by losing additional electrons, which means it is undergoing oxidation.
- Since SnCl₂ is losing electrons (being oxidized), it can act as a reducing agent in reactions with other substances.

4. **Conclusion:**
- Therefore, SnCl₂ acts as a reducing agent because it can donate electrons to other species, facilitating their reduction while it itself is oxidized.

### Final Answer:
SnCl₂ acts as a **reducing agent**.