Question

A 5 A current is passed through a solution of zinc sulphate for 40 min. The amount of zinc deposited at the cathode is

• A. 40.65 g
• B. 0.4065 g
• C. 4.065 g
• D. 65.04 g

Answer 1

The Correct Option is C

To solve this problem, we need to determine the amount of zinc deposited at the cathode when a 5 A current is passed through a zinc sulfate solution for 40 minutes. This can be calculated using Faraday's laws of electrolysis, specifically the first law, which states:

"The mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (or charge) passed through the electrolyte."

Step-by-step Calculation

1. First, calculate the total charge (Q) passed through the solution using the formula: Q = I \times t where I is the current in amperes and t is the time in seconds.
2. Convert the given time from minutes to seconds: t = 40 \text{ minutes} = 40 \times 60 = 2400 \text{ seconds}
3. Substitute the values into the equation: Q = 5 \, \text{A} \times 2400 \, \text{s} = 12000 \, \text{Coulombs}
4. Use the equivalent weight of zinc, which is the atomic weight divided by its valency. Zinc has an atomic weight of approximately 65.38 and a valency of 2. \text{Equivalent weight of Zn} = \frac{65.38}{2} = 32.69
5. The amount of zinc deposited can be found using the relation: m = \frac{Q \times \text{equivalent weight}}{96500} where 96500 C is the Faraday constant.
6. Substitute the known values to find the mass deposited: m = \frac{12000 \, \text{C} \times 32.69}{96500} \approx 4.065 \, \text{g}

Thus, the amount of zinc deposited at the cathode is 4.065 g.

The correct answer is therefore: 4.065 g.