How many faradays of eletricity is required to deposit 2 mole copper from `CuSO_(4)` solution

Answer

To determine how many Faradays of electricity are required to deposit 2 moles of copper from a copper sulfate (CuSO₄) solution, we can follow these steps:

### Step 1: Understand the Dissociation of Copper Sulfate
Copper sulfate dissociates in solution to form copper ions (Cu²⁺) and sulfate ions (SO₄²⁻). The relevant reaction can be written as:
\[ \text{CuSO}_4 \rightarrow \text{Cu}^{2+} + \text{SO}_4^{2-} \]

### Step 2: Determine the Reduction Reaction
At the cathode, the copper ions (Cu²⁺) are reduced to solid copper (Cu). The reduction half-reaction is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \]

### Step 3: Calculate the Number of Electrons Required
For every mole of Cu²⁺, 2 moles of electrons are required to deposit 1 mole of copper. Therefore, for 2 moles of copper:
\[ 2 \text{ moles of Cu} \times 2 \text{ moles of electrons/mole of Cu} = 4 \text{ moles of electrons} \]

### Step 4: Relate Moles of Electrons to Faradays
According to Faraday's law, 1 Faraday is the charge required to transfer 1 mole of electrons, which is approximately 96,500 coulombs. Therefore, the number of Faradays required for 4 moles of electrons is:
\[ \text{Faradays} = \text{moles of electrons} = 4 \text{ Faraday} \]

### Conclusion
Thus, to deposit 2 moles of copper from a copper sulfate solution, 4 Faradays of electricity are required.

### Summary of Steps:
1. **Dissociation of CuSO₄**: Understand the ions produced.
2. **Reduction Reaction**: Write the half-reaction for copper ion reduction.
3. **Calculate Electrons**: Determine how many electrons are needed for the desired amount of copper.
4. **Convert to Faradays**: Relate the number of electrons to Faradays.