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

Options

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

Explanation

To determine how many grams of copper will be liberated from a copper sulfate solution when 96,500 coulombs of electricity passes through it, we can follow these steps:

### Step 1: Understand the Electrolysis Reaction
In the electrolysis of copper sulfate (CuSO₄), copper ions (Cu²⁺) are reduced at the cathode, and sulfate ions (SO₄²⁻) are oxidized at the anode. The relevant half-reactions are:
- At the cathode: Cu²⁺ + 2e⁻ → Cu (s)
- At the anode: SO₄²⁻ → SO₂ + O₂ + 2e⁻

### Step 2: Determine the Charge Required for Copper Deposition
From the cathode reaction, we see that it takes 2 moles of electrons (2 Faradays) to deposit 1 mole of copper.

### Step 3: Use Faraday's Law
Faraday's law states that the amount of substance deposited during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. The relationship can be expressed as:
\[ \text{Mass} = \frac{Q}{nF} \times M \]
Where:
- \( Q \) = total charge (in coulombs)
- \( n \) = number of moles of electrons required to deposit 1 mole of copper (which is 2 for copper)
- \( F \) = Faraday's constant (approximately 96,500 C/mol)
- \( M \) = molar mass of copper (63.5 g/mol)

### Step 4: Calculate the Moles of Copper
Given that 96,500 C of electricity is passed through the solution:
- \( n = 2 \) (because it takes 2 moles of electrons to deposit 1 mole of copper)
- \( Q = 96,500 \, C \)

Using the formula:
\[ \text{Moles of Cu} = \frac{Q}{nF} = \frac{96,500}{2 \times 96,500} = 0.5 \, \text{moles of Cu} \]

### Step 5: Calculate the Mass of Copper
Now, we can find the mass of copper deposited:
\[ \text{Mass of Cu} = \text{Moles of Cu} \times M = 0.5 \, \text{moles} \times 63.5 \, \text{g/mol} = 31.75 \, \text{g} \]

### Final Answer
Therefore, when 96,500 coulombs of electricity passes through a copper sulfate solution, 31.75 grams of copper will be liberated.

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