During electrolysis, `O_(2)(g)` is evolved at anode in

Options

A. Dilute `H_(2)SO_(4)` with `Pt` electrode
B. Aqueous `AgNO_(3)` with `Pt` electrode
C. Dilute `H_(2)SO_(4)` with `Cu` electrode
D. Fused `NaOH` with an `Fe` cathode and `Ni` anode

Explanation

To determine in which case oxygen gas (O₂) is evolved at the anode during electrolysis, we will analyze each option provided in the question.

### Step-by-Step Solution:

1. **Understanding Electrolysis and Anode Reactions**:
- During electrolysis, oxidation occurs at the anode. This means that substances lose electrons, and in some cases, water can be oxidized to produce oxygen gas.

2. **Analyzing Option A: H₂SO₄ with Platinum Electrode**:
- At the anode, water (from the sulfuric acid solution) can be oxidized to produce oxygen gas.
- The reaction can be represented as:
\[ 2H₂O \rightarrow O₂ + 4H^+ + 4e^- \]
- Thus, O₂ is produced in this case.

3. **Analyzing Option B: AgNO₃ with Platinum Electrode**:
- In this case, the oxidation of water also occurs at the anode:
\[ 2H₂O \rightarrow O₂ + 4H^+ + 4e^- \]
- Therefore, O₂ is also produced in this case.

4. **Analyzing Option C: H₂SO₄ with Copper Electrode**:
- Here, copper is oxidized instead of water:
\[ Cu \rightarrow Cu^{2+} + 2e^- \]
- In this case, no oxygen gas is produced at the anode.

5. **Analyzing Option D: NaOH with Iron Cathode and Nickel Anode**:
- At the anode, hydroxide ions (OH⁻) can be oxidized to produce oxygen gas:
\[ 4OH^- \rightarrow O₂ + 2H₂O + 4e^- \]
- Thus, O₂ is produced in this case as well.

6. **Conclusion**:
- From the analysis, we find that oxygen gas is evolved at the anode in options A, B, and D. Therefore, the correct answer is A, B, and D.

### Final Answer:
O₂ is evolved at the anode in options A, B, and D.