A metal plate with a total surface are of `300cm^(2)` is to be benickel plate. If a current of 1.5A is used for 3 hours, find the thickness of nickel deposited. (Density of nickel `=8800 kgm^(-3), "E.C.E. of a nickel" 30.4xx10^(_8) kg C^(-1)`.

Answer

To find the thickness of nickel deposited on a metal plate, we can follow these steps:

### Step 1: Calculate the total time in seconds
The current is applied for 3 hours. We need to convert this time into seconds.
\[ T = 3 \text{ hours} = 3 \times 3600 \text{ seconds} = 10800 \text{ seconds} \]

### Step 2: Calculate the mass of nickel deposited using Faraday's law
According to Faraday's law, the mass of the substance deposited (W) can be calculated using the formula:
\[ W = Z \times I \times T \]
Where:
- \( Z = 30.4 \times 10^{-8} \text{ kg/C} \) (Electrochemical equivalent of nickel)
- \( I = 1.5 \text{ A} \) (Current)
- \( T = 10800 \text{ seconds} \) (Time)

Substituting the values:
\[ W = 30.4 \times 10^{-8} \text{ kg/C} \times 1.5 \text{ A} \times 10800 \text{ s} \]
Calculating \( W \):
\[ W = 30.4 \times 10^{-8} \times 1.5 \times 10800 = 0.000492 \text{ kg} = 0.492 \text{ g} \]

### Step 3: Convert the surface area from cm² to m²
The surface area of the plate is given as \( 300 \text{ cm}^2 \). We need to convert this to square meters:
\[ \text{Area} = 300 \text{ cm}^2 = 300 \times 10^{-4} \text{ m}^2 = 0.03 \text{ m}^2 \]

### Step 4: Calculate the volume of nickel deposited
Using the density of nickel, we can calculate the volume (V) of the nickel deposited:
\[ \text{Density} = 8800 \text{ kg/m}^3 \]
Using the formula:
\[ V = \frac{W}{\text{Density}} = \frac{0.000492 \text{ kg}}{8800 \text{ kg/m}^3} \]
Calculating \( V \):
\[ V = 5.59 \times 10^{-8} \text{ m}^3 \]

### Step 5: Calculate the thickness of nickel deposited
The volume of the deposited nickel can also be expressed in terms of the thickness (t) and the area (A):
\[ V = A \times t \]
Thus, we can rearrange this to find the thickness:
\[ t = \frac{V}{A} = \frac{5.59 \times 10^{-8} \text{ m}^3}{0.03 \text{ m}^2} \]
Calculating \( t \):
\[ t = 1.863 \times 10^{-6} \text{ m} = 1.863 \text{ micrometers} \]

### Final Answer
The thickness of nickel deposited is approximately \( 1.863 \text{ micrometers} \).

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