When 0.5 ampere of electricity is passed in aqueous solution of `AgNO_(3)` for 200 seconds, the amount of silver deposited on
cathode is `(Z=0.00118g per C` for Ag)

Options

A. 0.1118g
B. 0.118g
C. 0.9560g
D. 0.00956g

Explanation

To solve the problem, we will use Faraday's first law of electrolysis, which states that the mass of a substance deposited at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Current (I) = 0.5 A (amperes)
- Time (t) = 200 seconds
- Electrochemical equivalent (Z) for Ag = 0.00118 g/C (grams per coulomb)

2. **Calculate the Quantity of Electricity (Q):**
The quantity of electricity (Q) can be calculated using the formula:
\[ Q = I \times t \]
Substituting the values:
\[ Q = 0.5 \, \text{A} \times 200 \, \text{s} = 100 \, \text{C} \]

3. **Calculate the Mass of Silver Deposited (W):**
According to Faraday's first law:
\[ W = Z \times Q \]
Substituting the values:
\[ W = 0.00118 \, \text{g/C} \times 100 \, \text{C} = 0.118 \, \text{g} \]

4. **Conclusion:**
The amount of silver deposited on the cathode is **0.118 grams**.

### Final Answer:
The amount of silver deposited on the cathode is **0.118 grams**.

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