What is the time (in sec) required to deposit all the silver present in 125 ml of 1 M `AgNO_(3)` solution by passing a current of `241.25` A?

Options

A. 10 sec
B. 50 sec
C. 1000 sec
D. 100 sec

Answer

To find the time required to deposit all the silver present in 125 ml of 1 M AgNO₃ solution by passing a current of 241.25 A, we can follow these steps:

### Step 1: Calculate the number of moles of AgNO₃
The number of moles (n) can be calculated using the formula:
\[ n = \text{Molarity} \times \text{Volume (in liters)} \]

Given:
- Molarity = 1 M
- Volume = 125 ml = 0.125 L (since 1 L = 1000 ml)

So,
\[ n = 1 \, \text{mol/L} \times 0.125 \, \text{L} = 0.125 \, \text{mol} \]

**Hint:** Remember to convert milliliters to liters by dividing by 1000.

### Step 2: Determine the moles of electrons required
Since the reduction of silver ions (Ag⁺) to solid silver (Ag) involves the transfer of one electron per silver ion, the number of moles of electrons required will be equal to the number of moles of silver ions.

Thus, for 0.125 moles of Ag⁺, we need:
\[ \text{Moles of electrons} = 0.125 \, \text{mol} \]

**Hint:** Each mole of Ag⁺ requires one mole of electrons for reduction.

### Step 3: Calculate the total charge required
Using Faraday's constant (F = 96500 C/mol), we can calculate the total charge (Q) required to deposit the silver:
\[ Q = \text{Moles of electrons} \times F \]
\[ Q = 0.125 \, \text{mol} \times 96500 \, \text{C/mol} \]
\[ Q = 12062.5 \, \text{C} \]

**Hint:** Faraday's constant tells you how much charge is needed to deposit one mole of a substance.

### Step 4: Relate charge to current and time
Using the formula:
\[ Q = I \times t \]
Where:
- Q = total charge (C)
- I = current (A)
- t = time (s)

We can rearrange this to solve for time (t):
\[ t = \frac{Q}{I} \]

Substituting the values we have:
\[ t = \frac{12062.5 \, \text{C}}{241.25 \, \text{A}} \]

**Hint:** Make sure the units are consistent when substituting values.

### Step 5: Calculate the time
Now, performing the calculation:
\[ t = \frac{12062.5}{241.25} \approx 50 \, \text{s} \]

Thus, the time required to deposit all the silver is approximately 50 seconds.

**Final Answer:** The time required to deposit all the silver present in the solution is **50 seconds**.