Based on the cell notation for a spontaneous reaction, at the anode:
`Ag(s)"|"AgCl(s)"|"Cl^(-)(aq)"||"Br^(-)(aq)"|"Br_2(l)"|"C(s)`

Options

A. AgCl gets reduced
B. Ag gets oxidized
C. `Br^(-)` gets oxidized
D. `Br_2` gets reduced

Answer

To solve the problem based on the given cell notation, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Cell Notation**: The cell notation provided is:
\[ \text{Ag(s)} | \text{AgCl(s)} | \text{Cl}^-(\text{aq}) || \text{Br}^-(\text{aq}) | \text{Br}_2(\text{l}) | \text{C(s)} \]
This notation indicates the components involved in the electrochemical cell.

2. **Understand the Components**:
- The left side of the notation represents the anode half-cell, where oxidation occurs.
- The right side represents the cathode half-cell, where reduction occurs.

3. **Determine the Anodic Reaction**:
- At the anode, oxidation occurs. In this case, we need to look at the species on the left side of the notation.
- The relevant species are Ag(s) and Cl^-(aq).

4. **Identify Oxidation**:
- The oxidation half-reaction can be written as:
\[ \text{Ag}(s) \rightarrow \text{Ag}^+(aq) + e^- \]
- Here, silver (Ag) is being oxidized from an oxidation state of 0 to +1.

5. **Conclusion**:
- Since oxidation occurs at the anode, we conclude that Ag is being oxidized. Therefore, the correct answer is that at the anode, Ag gets oxidized.

### Final Answer:
At the anode, Ag is being oxidized.

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