A current of 3 amperes was passed through silver nitrate solution for 125 seconds. The amount of silver deposited at cathode was 0.42 g. Calculate the equivalent mass of silver.

Answer

To calculate the equivalent mass of silver deposited at the cathode, we can use Faraday's first law of electrolysis. The formula is:

\[ \frac{\text{Weight of substance deposited}}{\text{Equivalent mass}} = \frac{I \times t}{96500} \]

Where:
- Weight of substance deposited (W) = 0.42 g
- Current (I) = 3 A
- Time (t) = 125 s
- 96500 is the Faraday constant (in coulombs per equivalent).

### Step-by-Step Solution:

1. **Identify the given values**:
- Weight of silver deposited (W) = 0.42 g
- Current (I) = 3 A
- Time (t) = 125 s

2. **Calculate the total charge (Q)**:
\[ Q = I \times t \]
\[ Q = 3 \, \text{A} \times 125 \, \text{s} = 375 \, \text{C} \]

3. **Use Faraday's law to find the equivalent mass**:
Rearranging the formula gives:
\[ \text{Equivalent mass} = \frac{W \times 96500}{Q} \]

4. **Substituting the values**:
\[ \text{Equivalent mass} = \frac{0.42 \, \text{g} \times 96500}{375 \, \text{C}} \]

5. **Calculating the equivalent mass**:
\[ \text{Equivalent mass} = \frac{40530}{375} \approx 108.08 \, \text{g} \]

6. **Final result**:
The equivalent mass of silver is approximately **108.08 g**.