Silver is electro-deposited on a metallic vessel of surface area 800 `cm^(2)` by passing a current of 0.2 ampere for 3 hours. Calculate the thickness of silver deposited given that its density is 10.47 g `cm^(-3)`. (At mass of Ag =107.92).

Explanation

To find the thickness of silver deposited on a metallic vessel, we will follow these steps:

### Step 1: Calculate the total charge (Q) passed through the circuit.
We can use the formula:
\[ Q = I \times t \]
Where:
- \( I \) = current in amperes (0.2 A)
- \( t \) = time in seconds

Convert time from hours to seconds:
\[ t = 3 \, \text{hours} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} = 10800 \, \text{seconds} \]

Now, calculate the charge:
\[ Q = 0.2 \, \text{A} \times 10800 \, \text{s} = 2160 \, \text{C} \]

### Step 2: Calculate the mass of silver deposited (m).
Using Faraday's laws of electrolysis, we know that:
\[ \text{mass} = \frac{Q \times M}{n \times F} \]
Where:
- \( M \) = molar mass of silver (107.92 g/mol)
- \( n \) = number of electrons transferred per ion of silver (for Ag, n = 1)
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))

Now substituting the values:
\[ m = \frac{2160 \, \text{C} \times 107.92 \, \text{g/mol}}{1 \times 96500 \, \text{C/mol}} \]

Calculating the mass:
\[ m = \frac{2160 \times 107.92}{96500} \approx 2.4156 \, \text{g} \]

### Step 3: Calculate the volume of silver deposited (V).
Using the formula:
\[ V = \frac{m}{\text{density}} \]
Where:
- Density of silver = 10.47 g/cm³

Now substituting the values:
\[ V = \frac{2.4156 \, \text{g}}{10.47 \, \text{g/cm}^3} \approx 0.2307 \, \text{cm}^3 \]

### Step 4: Calculate the thickness of silver deposited (t).
Using the formula:
\[ t = \frac{V}{A} \]
Where:
- \( A \) = surface area (800 cm²)

Now substituting the values:
\[ t = \frac{0.2307 \, \text{cm}^3}{800 \, \text{cm}^2} \approx 0.000288375 \, \text{cm} \]

To express this in a more convenient unit:
\[ t \approx 2.88 \times 10^{-4} \, \text{cm} \]

### Final Answer:
The thickness of silver deposited is approximately \( 2.88 \times 10^{-4} \, \text{cm} \).

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