Question:

A current of 1.6 A is passed through a solution of CuSO$_4$. How many \( \text{Cu}^{++} \) ions are liberated in one minute? (Electronic charge \( e = 1.6 \times 10^{-19} \, \text{C} \))

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The number of ions liberated during electrolysis is related to the total charge passed and the charge of the ion.
Updated On: Apr 22, 2026
  • \( 3 \times 10^{20} \)
  • \( 3 \times 10^{19} \)
  • \( 6 \times 10^{20} \)
  • \( 6 \times 10^{19} \)
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The Correct Option is A

Solution and Explanation

Step 1: Use the formula for the number of ions.
The number of ions liberated can be calculated using the relation: \[ n = \frac{Q}{e} \] where: - \( Q \) is the total charge passed through the solution, - \( e \) is the charge of one \( \text{Cu}^{++} \) ion.

Step 2: Calculate the total charge.

The current \( I = 1.6 \, \text{A} \) is passed for \( t = 60 \, \text{seconds} \). The total charge is: \[ Q = I \times t = 1.6 \times 60 = 96 \, \text{C} \]

Step 3: Calculate the number of ions.

Now, using the formula for the number of ions: \[ n = \frac{Q}{e} = \frac{96}{1.6 \times 10^{-19}} = 6 \times 10^{20} \]

Step 4: Conclusion.

The number of \( \text{Cu}^{++} \) ions liberated in one minute is \( 6 \times 10^{20} \), which is option (1).
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