During the electrolysis of aqueous solution of `HCOOK,` the number of gases obtained at cathode, anode, and total number of gases are
`1,2,3`
`1,2,2`
`2,1,3`
`2,1,2`

Options

A. `1,2,3`
B. `1,2,2`
C. `2,1,3`
D. `2,1,2`

Explanation

To solve the question regarding the electrolysis of an aqueous solution of HCOOK (potassium formate), we need to analyze the reactions occurring at both the cathode and anode during the process.

### Step-by-Step Solution:

1. **Identify the Electrolysis Process**:
- During electrolysis, the solution will undergo reduction at the cathode and oxidation at the anode.

2. **Reaction at the Cathode**:
- At the cathode, reduction occurs. Water (H2O) is reduced to form hydrogen gas (H2) and hydroxide ions (OH⁻).
- The half-reaction can be represented as:
\[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]
- From this reaction, we see that **1 mole of hydrogen gas** is produced at the cathode.

3. **Reaction at the Anode**:
- At the anode, oxidation occurs. The formate ion (HCOO⁻) is oxidized to produce carbon dioxide (CO2) and hydrogen gas (H2).
- The half-reaction can be represented as:
\[ HCOO^- \rightarrow CO_2 + H_2 + 2e^- \]
- From this reaction, we see that **2 moles of gases** (1 mole of CO2 and 1 mole of H2) are produced at the anode.

4. **Total Number of Gases Produced**:
- At the cathode, we have 1 mole of hydrogen gas (H2).
- At the anode, we have 1 mole of carbon dioxide (CO2) and 1 mole of hydrogen gas (H2).
- Therefore, the total number of gases produced is:
- At cathode: 1 (H2)
- At anode: 2 (1 CO2 + 1 H2)
- Total: 1 + 2 = 3 gases.

5. **Conclusion**:
- The number of gases obtained at the cathode is 1, at the anode is 2, and the total number of gases is 3.
- Therefore, the correct answer is **1, 2, 3**.