A current of 2.6 ampere is passed through `CuSO_4` solution for 6 minutes 20 seconds. The amount of Cu deposited is (At. wt. of Cu=63.5, Faraday = 96500 C)-

Options

A. 6.35g
B. 0.635g
C. 0.325g
D. 3.175g

Answer

To solve the problem of how much copper is deposited when a current of 2.6 amperes is passed through a CuSO₄ solution for 6 minutes and 20 seconds, we can follow these steps:

### Step 1: Convert Time to Seconds
First, we need to convert the time from minutes and seconds to just seconds.

- Time in minutes: 6 minutes
- Time in seconds: 20 seconds
- Total time in seconds = (6 × 60) + 20 = 360 + 20 = 380 seconds

### Step 2: Calculate Total Charge (Q)
Using the formula for electric charge, we can calculate the total charge passed through the solution.

- Formula: \( Q = I \times t \)
- Where:
- \( I \) = current in amperes (2.6 A)
- \( t \) = time in seconds (380 s)

Calculating the charge:
\[ Q = 2.6 \, \text{A} \times 380 \, \text{s} = 988 \, \text{C} \]

### Step 3: Determine Equivalent Weight (E)
Next, we need to find the equivalent weight of copper (Cu).

- Atomic weight of Cu = 63.5 g/mol
- The n-factor for copper (Cu²⁺) = 2 (since it gains 2 electrons)

Using the formula for equivalent weight:
\[ E = \frac{\text{Atomic weight}}{\text{n-factor}} = \frac{63.5}{2} = 31.75 \, \text{g/equiv} \]

### Step 4: Calculate the Amount of Copper Deposited (w)
Now we can use Faraday's law of electrolysis to find the weight of copper deposited.

- Formula: \( w = \frac{E \times Q}{F} \)
- Where:
- \( E \) = equivalent weight (31.75 g/equiv)
- \( Q \) = total charge (988 C)
- \( F \) = Faraday's constant (96500 C/equiv)

Calculating the weight deposited:
\[ w = \frac{31.75 \, \text{g/equiv} \times 988 \, \text{C}}{96500 \, \text{C/equiv}} \]

Calculating the above expression:
\[ w = \frac{31353.5}{96500} \approx 0.325 \, \text{g} \]

### Conclusion
The amount of copper deposited is approximately **0.325 grams**.