How long a current of 2 A has to be passed through a solution of `AgNO_(3)` to coat a metal surface of `80cm^(2)` with `5mum` thick layer? Density of water `=10.8g//cm^(3)`

Explanation

To solve the problem of how long a current of 2 A has to be passed through a solution of AgNO₃ to coat a metal surface of 80 cm² with a 5 µm thick layer, we can follow these steps:

### Step 1: Calculate the volume of silver to be deposited
To find the volume of silver that needs to be deposited, we use the formula:
\[ \text{Volume} = \text{Area} \times \text{Thickness} \]
Given:
- Area = 80 cm²
- Thickness = 5 µm = \(5 \times 10^{-4}\) cm

Calculating the volume:
\[ \text{Volume} = 80 \, \text{cm}^2 \times 5 \times 10^{-4} \, \text{cm} = 0.04 \, \text{cm}^3 \]

### Step 2: Calculate the weight of silver to be deposited
Using the density of silver (which is given as the density of water, 10.8 g/cm³, for the sake of this problem):
\[ \text{Weight} = \text{Density} \times \text{Volume} \]
Calculating the weight:
\[ \text{Weight} = 10.8 \, \text{g/cm}^3 \times 0.04 \, \text{cm}^3 = 0.432 \, \text{g} \]

### Step 3: Calculate the equivalent weight of silver
The equivalent weight of silver (Ag) is calculated as:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]
Where:
- Molar mass of Ag = 108 g/mol
- n (valency of Ag) = 1

Thus, the equivalent weight of silver is:
\[ \text{Equivalent weight} = \frac{108 \, \text{g/mol}}{1} = 108 \, \text{g/equiv} \]

### Step 4: Calculate the number of equivalents of silver
Using the weight of silver and its equivalent weight:
\[ \text{Number of equivalents} = \frac{\text{Weight}}{\text{Equivalent weight}} = \frac{0.432 \, \text{g}}{108 \, \text{g/equiv}} = 0.004 \, \text{equiv} \]

### Step 5: Use Faraday's laws to find time
According to Faraday's first law of electrolysis:
\[ \text{Weight} = \frac{I \cdot t}{96500} \cdot \text{Equivalent weight} \]
Rearranging for time (t):
\[ t = \frac{\text{Weight} \cdot 96500}{I \cdot \text{Equivalent weight}} \]
Substituting the values:
- Weight = 0.432 g
- I = 2 A
- Equivalent weight = 108 g/equiv

Calculating time:
\[ t = \frac{0.432 \, \text{g} \cdot 96500}{2 \, \text{A} \cdot 108 \, \text{g/equiv}} = \frac{41664}{216} \approx 193 \, \text{seconds} \]

### Final Answer
The time required to pass a current of 2 A through the solution of AgNO₃ to coat the surface is approximately **193 seconds**.

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