Electrolysis of CuSO4 Solution

To determine the mass of Cu deposited at the cathode during electrolysis, we need to consider the following factors:

1. Faraday's Law of Electrolysis: This law states that the amount of substance deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.

2. Equation for Electrolysis of CuSO4: During electrolysis of CuSO4 solution, copper ions (Cu2+) are reduced at the cathode, while sulfate ions (SO4 2-) are oxidized at the anode. The overall equation can be represented as:

2CuSO4 + 2H2O → 2Cu + O2 + 4H+ + 4SO4 2-

3. Calculating the Amount of Substance Deposited: The amount of substance deposited can be determined using the formula:

Mass (m) = (I * t * M) / (n * F)

Where:
- I is the current (in amperes)
- t is the time (in seconds)
- M is the molar mass of the substance (in grams per mole)
- n is the number of electrons involved in the reaction
- F is the Faraday constant (96,485 C/mol)

Calculations:
Given:
- Current (I) = 1.5 A
- Time (t) = 10 minutes = 600 seconds
- Molar mass of Cu (M) = 63.55 g/mol
- Number of electrons involved in the reaction (n) = 2

Using the formula mentioned above, we can calculate the mass of Cu deposited:

Mass (m) = (1.5 A * 600 s * 63.55 g/mol) / (2 * 96,485 C/mol)

Simplifying the equation:
Mass (m) = (900 g⋅s) / (192,970 C)

Converting grams⋅seconds (g⋅s) to grams (g):
Mass (m) = 0.004653 g

Therefore, the mass of Cu deposited at the cathode is approximately 0.004653 grams, which is equivalent to 0.246 grams (rounded to three decimal places).

Answer: The mass of Cu deposited at the cathode during the electrolysis of CuSO4 solution for 10 minutes with a current of 1.5 A is approximately 0.246 grams.