A current of 0.50 ampere is passing through a `CuSO_(4)` solution . How many `Cu^(++)` ions will be deposited on cathode in 10 seconds ?

Explanation

To find out how many \( \text{Cu}^{2+} \) ions will be deposited on the cathode in 10 seconds when a current of 0.50 A is passing through a \( \text{CuSO}_4 \) solution, we can follow these steps:

### Step 1: Calculate the total charge (Q) passed through the solution.
The total charge can be calculated using the formula:
\[ Q = I \times t \]
where:
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