96500C of electricity liberates from `CuSO_(4)` solution.

Options

A. 63.5g of Cu
B. 31.75g of Cu
C. 96500g of Cu
D. 100g of Cu

Explanation

To solve the problem of how much copper will be liberated from a `CuSO4` solution when 96,500 C of electricity is passed through it, we can follow these steps:

### Step 1: Understand the Reaction
In the electrolysis of copper sulfate (`CuSO4`), copper ions (`Cu^2+`) gain electrons to form solid copper (`Cu`). The half-reaction can be represented as:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

This indicates that 2 moles of electrons are required to deposit 1 mole of copper.

### Step 2: Calculate the Charge Required for 1 Mole of Copper
From Faraday's laws of electrolysis, we know that:
- 1 mole of electrons carries a charge of approximately 96500 C (Faraday's constant).
- Therefore, to deposit 1 mole of copper, which requires 2 moles of electrons, the total charge required is:
\[ \text{Charge for 1 mole of Cu} = 2 \times 96500 \, \text{C} = 193000 \, \text{C} \]

### Step 3: Determine the Amount of Copper Liberated
Now, we have 96,500 C of electricity. We can find out how many moles of copper can be deposited using the charge provided:
\[ \text{Moles of Cu deposited} = \frac{\text{Charge}}{\text{Charge required for 1 mole of Cu}} \]
\[ \text{Moles of Cu deposited} = \frac{96500 \, \text{C}}{193000 \, \text{C}} = \frac{1}{2} \, \text{mole} \]

### Step 4: Convert Moles of Copper to Grams
The molar mass of copper (Cu) is approximately 63.5 g/mol. Therefore, the mass of copper deposited can be calculated as:
\[ \text{Mass of Cu} = \text{Moles of Cu} \times \text{Molar mass of Cu} \]
\[ \text{Mass of Cu} = \frac{1}{2} \, \text{mole} \times 63.5 \, \text{g/mol} = 31.75 \, \text{g} \]

### Conclusion
Thus, the amount of copper liberated from the `CuSO4` solution when 96,500 C of electricity is passed through it is **31.75 grams**.

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