Assertion: Terminal voltage of a cell is greater than emf of cell during charging of the cell.
Reason: The emf of a cell is always greater than its terminal voltage.

Explanation

To solve the problem, we need to analyze the assertion and the reason provided in the question step by step.

### Step 1: Understanding the Assertion
The assertion states that the terminal voltage of a cell is greater than the EMF of the cell during charging.

- **Explanation**: When a cell is being charged, current flows into the cell. The terminal voltage (V) can be expressed using the formula:
\[ V = E + I \cdot r \]
where \(E\) is the EMF of the cell, \(I\) is the current flowing into the cell, and \(r\) is the internal resistance of the cell. Since \(I\) is positive during charging, the terminal voltage \(V\) becomes greater than the EMF \(E\).

### Step 2: Understanding the Reason
The reason states that the EMF of a cell is always greater than its terminal voltage.

- **Explanation**: This statement is generally true when the cell is discharging. In this case, the terminal voltage can be expressed as:
\[ V = E - I \cdot r \]
Here, the terminal voltage \(V\) is less than the EMF \(E\) because the current \(I\) is flowing out of the cell, causing a drop due to the internal resistance.

### Step 3: Conclusion
Based on the analysis:
- The assertion is **correct**: During charging, the terminal voltage is indeed greater than the EMF of the cell.
- The reason is **incorrect**: The EMF of the cell is not always greater than its terminal voltage; it depends on whether the cell is charging or discharging.

### Final Answer
- **Assertion**: Correct
- **Reason**: Incorrect