A current of 4.0 amperes was passed through copper (II) tetraoxosulphate (IV) solution for one
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A current of 4.0 amperes was passed through copper (II) tetraoxosulphate (IV) solution for one hour using copper electrodes. What was the mass of copper deposited? (Cu = 64, 1F = 96500C)
- A) 3.2g
- B) 4.8g
- C) 6.4g
- D) 9.6g
Correct Answer: D) 9.6g
Explanation
To solve this question, we will use Faraday's law of electrolysis, which states that the mass of a substance deposited during electrolysis is directly proportional to the quantity of electricity passed.
First, we need to calculate the quantity of electricity (Q) passed through the solution. We are given the current (I) as 4.0 amperes and the time (t) as one hour. Using the formula Q = I
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