How many copper will be deposited at cathode of an electrolytic cell containing `Cu^(2+)` ions by passing 2 ampere of current for 60 minutes.

Explanation

To find out how much copper will be deposited at the cathode of an electrolytic cell containing \( \text{Cu}^{2+} \) ions by passing 2 amperes of current for 60 minutes, we can follow these steps:

### Step 1: Calculate the total charge (Q) passed through the cell.
The formula to calculate charge is:
\[ Q = I \times T \]
Where:
- \( I \) is the current in amperes (A)
- \( T \) is the time in seconds (s)

Given:
- \( I = 2 \, \text{A} \)
- \( T = 60 \, \text{minutes} = 60 \times 60 \, \text{seconds} = 3600 \, \text{s} \)

Now, substituting the values:
\[ Q = 2 \, \text{A} \times 3600 \, \text{s} = 7200 \, \text{C} \]

### Step 2: Determine the number of moles of electrons transferred.
Using Faraday's law of electrolysis, we know that 1 mole of electrons corresponds to 96500 coulombs (Faraday's constant). The number of moles of electrons (\( n \)) can be calculated as:
\[ n = \frac{Q}{F} \]
Where:
- \( F = 96500 \, \text{C/mol} \)

Substituting the values:
\[ n = \frac{7200 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.0745 \, \text{mol} \]

### Step 3: Relate moles of electrons to moles of copper deposited.
The reaction at the cathode is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
From the reaction, we see that 2 moles of electrons are required to deposit 1 mole of copper. Therefore, the moles of copper deposited (\( n_{\text{Cu}} \)) can be calculated as:
\[ n_{\text{Cu}} = \frac{n}{2} = \frac{0.0745 \, \text{mol}}{2} \approx 0.03725 \, \text{mol} \]

### Step 4: Calculate the mass of copper deposited.
To find the mass of copper deposited, we use the molar mass of copper, which is approximately 63.5 g/mol:
\[ \text{mass} = n_{\text{Cu}} \times \text{molar mass} \]
Substituting the values:
\[ \text{mass} = 0.03725 \, \text{mol} \times 63.5 \, \text{g/mol} \approx 2.37 \, \text{g} \]

### Final Answer:
Thus, the mass of copper deposited at the cathode is approximately **2.37 grams**.

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