A silver cup is plated with silver by passing 965 C of electricity. The amount of Ag deposited is

Options

A. 107 . 89 g
B. 9.89 g
C. 1.0002 g
D. 1.08 g

Answer

To find the amount of silver (Ag) deposited when 965 C of electricity is passed through a silver plating process, we can use Faraday's laws of electrolysis. Here’s the step-by-step solution:

### Step 1: Understand the relationship
The weight of the substance deposited during electrolysis can be calculated using the formula:
\[ \text{Weight of Ag} = \frac{\text{Equivalent weight of Ag} \times Q}{96500} \]
where:
- \( Q \) is the charge in coulombs (C),
- 96500 is the Faraday constant (the charge required to deposit 1 mole of a monovalent ion).

### Step 2: Find the equivalent weight of silver (Ag)
The equivalent weight of a substance can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Gram atomic weight}}{\text{Valency}} \]
For silver:
- The gram atomic weight of Ag is approximately 108 g/mol.
- The valency of silver (Ag) is 1.

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

### Step 3: Substitute the values into the formula
Now, substituting the values into the weight formula:
\[ \text{Weight of Ag} = \frac{108 \times 965}{96500} \]

### Step 4: Calculate the weight of Ag
Calculating the above expression:
\[ \text{Weight of Ag} = \frac{104340}{96500} \approx 1.08 \text{ grams} \]

### Final Answer
The amount of silver (Ag) deposited is approximately **1.08 grams**.