Silver is removed electrolytically from `200mL` of a `0.1N` solution of `AgNO_(3)` by a current of `0.1A`. How long will it take to remove half of the silver from the solution ?

Options

A. `0.1s`
B. `100s`
C. `965s`
D. `9650s`

Explanation

To solve the problem of how long it will take to remove half of the silver from a 200 mL solution of 0.1N AgNO₃ using a current of 0.1A, we can follow these steps:

### Step 1: Calculate the number of equivalents of AgNO₃ in the solution.
The number of equivalents can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume in liters} \]
Given:
- Normality (N) = 0.1N
- Volume (V) = 200 mL = 0.2 L

Calculating:
\[ \text{Number of equivalents} = 0.1 \, \text{N} \times 0.2 \, \text{L} = 0.02 \, \text{equivalents} \]

### Step 2: Determine the number of moles of Ag⁺ ions.
Since AgNO₃ dissociates into Ag⁺ and NO₃⁻, the number of moles of Ag⁺ will be equal to the number of equivalents of AgNO₃:
\[ \text{Moles of Ag} = \text{Number of equivalents} = 0.02 \, \text{moles} \]

### Step 3: Calculate the total mass of silver (Ag) in the solution.
The molar mass of silver (Ag) is approximately 108 g/mol. Therefore, the total mass of silver in the solution is:
\[ \text{Mass of Ag} = \text{Moles of Ag} \times \text{Molar mass of Ag} = 0.02 \, \text{moles} \times 108 \, \text{g/mol} = 2.16 \, \text{g} \]

### Step 4: Calculate the mass of silver to be removed (half).
To find half of the mass of silver:
\[ \text{Mass of Ag to be removed} = \frac{2.16 \, \text{g}}{2} = 1.08 \, \text{g} \]

### Step 5: Calculate the number of equivalents of silver to be removed.
Using the equivalent weight of silver:
\[ \text{Equivalent weight of Ag} = \frac{\text{Molar mass of Ag}}{\text{n-factor}} = \frac{108 \, \text{g/mol}}{1} = 108 \, \text{g/equiv} \]
Now, calculate the equivalents of silver to be removed:
\[ \text{Equivalents of Ag to be removed} = \frac{1.08 \, \text{g}}{108 \, \text{g/equiv}} = 0.01 \, \text{equivalents} \]

### Step 6: Use Faraday's laws of electrolysis to find the time required.
According to Faraday's first law:
\[ \text{Mass} = \text{Equivalent weight} \times \text{Number of equivalents} \]
We also know:
\[ Q = I \times T \]
Where:
- \( Q \) = total charge (in coulombs)
- \( I \) = current (in amperes)
- \( T \) = time (in seconds)

Using Faraday's constant \( F = 96500 \, \text{C/equiv} \):
\[ Q = \text{Equivalents} \times F = 0.01 \, \text{equivalents} \times 96500 \, \text{C/equiv} = 965 \, \text{C} \]

### Step 7: Calculate the time using the current.
Substituting into the equation \( Q = I \times T \):
\[ 965 \, \text{C} = 0.1 \, \text{A} \times T \]
Solving for \( T \):
\[ T = \frac{965 \, \text{C}}{0.1 \, \text{A}} = 9650 \, \text{s} \]

### Final Answer:
It will take **9650 seconds** to remove half of the silver from the solution.

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