` 96500` culombs of electric current liberates from `CuSO_4` solution.

Options

A. ` 63.5 f Cu`
B. ` 96500 g Cu`
C. ` 31. 75 g Cu`
D. ` 100 g Cu`

Explanation

To solve the problem of how much copper (Cu) is liberated when 96500 coulombs of electric current is passed through a CuSO₄ solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electrochemical Reaction**:
When CuSO₄ dissociates in solution, it produces Cu²⁺ ions and SO₄²⁻ ions. The relevant half-reactions are:
- At the cathode (reduction): Cu²⁺ + 2e⁻ → Cu
- At the anode (oxidation): 2H₂O → O₂ + 4H⁺ + 4e⁻ (for example, if water is oxidized)

2. **Determine the Number of Electrons Required**:
From the cathode reaction, we see that to deposit 1 mole of Cu, 2 moles of electrons (2 Faraday) are required.

3. **Calculate the Total Charge for 1 Mole of Cu**:
The charge required to deposit 1 mole of Cu is given by:
\[ Q = n \times F \]
where \( n \) is the number of moles of electrons (2 for Cu) and \( F \) is Faraday's constant (approximately 96500 coulombs).
\[ Q = 2 \times 96500 = 193000 \text{ coulombs} \]

4. **Relate the Given Charge to Moles of Cu**:
We have 96500 coulombs available. To find out how many moles of Cu can be deposited, we set up the proportion:
\[ \frac{96500 \text{ coulombs}}{193000 \text{ coulombs/mole}} = \frac{x \text{ moles}}{1 \text{ mole}} \]
Solving for \( x \):
\[ x = \frac{96500}{193000} = 0.5 \text{ moles of Cu} \]

5. **Convert Moles of Cu to Mass**:
The molar mass of copper (Cu) is 63.5 grams/mole. Therefore, the mass of copper deposited is:
\[ \text{Mass of Cu} = 0.5 \text{ moles} \times 63.5 \text{ grams/mole} = 31.75 \text{ grams} \]

### Final Answer:
When 96500 coulombs of electric current is passed through a CuSO₄ solution, **31.75 grams of copper (Cu) is liberated**.

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