A very thin copper plate is electro`-` plated with gold using gold chloride in `HCl`. The current was passed for `20 mi n`. And the increase in the weight of the plate was found to be `2g.[Au=197]`. The current passed was `-`

Options

A. `0.816 amp`
B. `1.632 amp`
C. `2.448 amp`
D. `3.264 amp`

Explanation

To solve the problem of determining the current passed during the electroplating of gold onto a copper plate, we can follow these steps:

### Step 1: Identify the given data
- Increase in weight of gold deposited (W) = 2 g
- Atomic weight of gold (Au) = 197 g/mol
- Time (t) = 20 minutes = 20 × 60 seconds = 1200 seconds

### Step 2: Determine the equivalent weight of gold
The equivalent weight of gold can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Atomic weight}}{\text{Valency}} \]
For gold (Au), the valency in this electroplating process is 3 (since Au³⁺ is reduced to Au), thus:
\[ \text{Equivalent weight of Au} = \frac{197}{3} \approx 65.67 \text{ g/equiv} \]

### Step 3: Apply Faraday's law of electrolysis
According to Faraday's law, the weight of the substance deposited (W) is given by:
\[ W = Z \cdot I \cdot t \]
Where:
- \( Z \) = electrochemical equivalent = \(\frac{\text{Equivalent weight}}{96500}\)
- \( I \) = current (in amperes)
- \( t \) = time (in seconds)

### Step 4: Calculate the electrochemical equivalent (Z)
Using the equivalent weight calculated:
\[ Z = \frac{65.67}{96500} \approx 6.8 \times 10^{-4} \text{ g/C} \]

### Step 5: Substitute values into Faraday's law
Now substituting the values into the equation:
\[ 2 = (6.8 \times 10^{-4}) \cdot I \cdot 1200 \]

### Step 6: Solve for current (I)
Rearranging the equation to solve for \( I \):
\[ I = \frac{2}{(6.8 \times 10^{-4}) \cdot 1200} \]
\[ I = \frac{2}{0.816} \approx 2.448 \text{ A} \]

### Conclusion
The current passed during the electroplating process is approximately **2.448 A**.

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