If the half-cell reaction `A+e^(-)toA^(-)` has a large negative reduction potentials, it follows that:

Options

A. A is readly reduced
B. A is readily oxidised
C. `A^(-)` is readily reduced
D. `A^(-)` is readily oxidised

Answer

To solve the question regarding the half-cell reaction \( A + e^- \rightarrow A^- \) having a large negative reduction potential, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Reduction Potential**:
The reduction potential indicates the tendency of a species to gain electrons. A large negative reduction potential means that the species does not favor gaining electrons and is more likely to lose them instead.

**Hint**: Recall that a more negative reduction potential indicates a stronger tendency to act as a reducing agent.

2. **Identify the Nature of the Species**:
In the given reaction \( A + e^- \rightarrow A^- \), since the reduction potential is large and negative, it suggests that \( A^- \) is not stable and prefers to revert back to \( A \) by losing an electron.

**Hint**: Think about the stability of the species involved in the half-reaction.

3. **Determine Reducing and Oxidizing Agents**:
A species with a large negative reduction potential is a good reducing agent. This means that it can readily donate electrons to other species, thereby reducing them.

**Hint**: Remember that reducing agents are oxidized in the process of reducing other species.

4. **Conclusion**:
Since \( A^- \) can easily lose an electron and revert to \( A \), it follows that \( A^- \) is readily oxidized. Therefore, the correct conclusion is that \( A^- \) acts as a reducing agent.

**Hint**: Consider the definitions of reducing and oxidizing agents in the context of electron transfer.

### Final Answer:
The correct option is that \( A^- \) is readily oxidized.