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 0.1 N AgNO₃ using a current of 0.1 A, we can follow these steps:

### Step 1: Calculate the total milliequivalents of AgNO₃ in the solution.
The formula to calculate milliequivalents (mEq) is:
\[ \text{mEq} = \text{Normality} \times \text{Volume (in mL)} \]
Given:
- Normality (N) = 0.1 N
- Volume (V) = 200 mL

Calculating:
\[ \text{mEq} = 0.1 \, \text{N} \times 200 \, \text{mL} = 20 \, \text{mEq} \]

### Step 2: Determine the milliequivalents of silver to be removed.
Since we need to remove half of the silver:
\[ \text{mEq of silver to be removed} = \frac{20 \, \text{mEq}}{2} = 10 \, \text{mEq} \]

### Step 3: Use Faraday's first law of electrolysis to find time.
Faraday's first law states:
\[ \text{Equivalent mass deposited} = \frac{\text{Current} \times \text{Time}}{\text{Faraday constant}} \]
Where:
- Current (I) = 0.1 A
- Faraday constant (F) = 9600 C
- Equivalent mass in milli-equivalents = 10 mEq = \(10 \times 10^{-3}\) Eq

Rearranging the formula to find time (t):
\[ t = \frac{\text{Equivalent mass} \times F}{I} \]

Substituting the values:
\[ t = \frac{10 \times 10^{-3} \, \text{Eq} \times 9600 \, \text{C}}{0.1 \, \text{A}} \]

### Step 4: Calculate the time.
\[ t = \frac{10 \times 9600}{0.1} = 96000 \, \text{seconds} \]

### Step 5: Convert seconds to a more understandable unit if necessary.
Since the question asks for the time in seconds, we can leave it as is.

### Final Answer:
It will take **96000 seconds** to remove half of the silver from the solution.

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