The times taken by the galvanic cell which operates almost idealy under reversible conditions at a current of `10^(-16)A` to deliver 1 mole of electron is

Options

A. `19.30 xx 10^(20)s`
B. `4.825 xx 10^(20)s`
C. `9.65 xx 10^(20)s`
D. `3.4 xx 10^(11)s`

Answer

To solve the problem of determining the time taken by a galvanic cell to deliver 1 mole of electrons at a current of \(10^{-16}\) A, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Current (\(I\)) = \(10^{-16}\) A
- Charge of 1 mole of electrons (\(Q\)) = 96500 C (Coulombs)

2. **Use the Formula for Charge**:
The relationship between charge, current, and time is given by the formula:
\[ Q = I \times t \]
where \(Q\) is the total charge, \(I\) is the current, and \(t\) is the time.

3. **Rearranging the Formula to Solve for Time**:
To find the time (\(t\)), we can rearrange the formula:
\[ t = \frac{Q}{I} \]

4. **Substituting the Known Values**:
Now, substitute the values of \(Q\) and \(I\) into the equation:
\[ t = \frac{96500 \, \text{C}}{10^{-16} \, \text{A}} \]

5. **Calculating the Time**:
Performing the calculation:
\[ t = 96500 \times 10^{16} \]
\[ t = 9.65 \times 10^{20} \, \text{seconds} \]

6. **Final Answer**:
Therefore, the time taken by the galvanic cell to deliver 1 mole of electrons is:
\[ t = 9.65 \times 10^{20} \, \text{seconds} \]

### Conclusion:
The correct answer is \(9.65 \times 10^{20}\) seconds.