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 1: Understand the relationship between charge, current, and time
The relationship between charge (Q), current (I), and time (t) is given by the formula:
\[ Q = I \cdot t \]
Where:
- \(Q\) is the charge in coulombs,
- \(I\) is the current in amperes,
- \(t\) is the time in seconds.

### Step 2: Calculate the charge for 1 mole of electrons
1 mole of electrons corresponds to a charge of 1 Faraday, which is approximately \(96500\) coulombs. Therefore:
\[ Q = 96500 \, \text{C} \]

### Step 3: Rearrange the formula to solve for time
We need to find the time \(t\), so we rearrange the formula:
\[ t = \frac{Q}{I} \]

### Step 4: Substitute the values into the formula
Substituting the values of \(Q\) and \(I\):
\[ t = \frac{96500 \, \text{C}}{10^{-16} \, \text{A}} \]

### Step 5: Calculate the time
Now we perform the division:
\[ t = 96500 \times 10^{16} \]
This simplifies to:
\[ t = 9.65 \times 10^{20} \, \text{seconds} \]

### Final Answer
The time taken by the galvanic cell to deliver 1 mole of electrons at a current of \(10^{-16} A\) is:
\[ t = 9.65 \times 10^{20} \, \text{seconds} \]

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