The amount of electricity required to produce one mole of copper from copper sulphate solution will be:

Options

A. 1 faraday
B. 2.33 faraday
C. 2 faraday
D. 1.33 faraday

Explanation

To determine the amount of electricity required to produce one mole of copper from copper sulfate (CuSO₄), we can follow these steps:

### Step 1: Identify the Reaction
The electrolysis of copper sulfate solution involves the reduction of copper ions (Cu²⁺) to form copper metal (Cu). The half-reaction can be represented as:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

### Step 2: Determine the Valency Factor
The valency factor (n) indicates the number of moles of electrons required to reduce one mole of copper ions. In this case, since Cu²⁺ requires 2 electrons (2e⁻) to be reduced to Cu, the valency factor is 2.

### Step 3: Use Faraday's Law
According to Faraday's law of electrolysis, the amount of electricity (Q) required to produce a certain amount of substance is given by:
\[ Q = n \times F \]
where:
- \( n \) is the number of moles of electrons (valency factor),
- \( F \) is Faraday's constant, approximately 96500 C/mol (the charge of one mole of electrons).

### Step 4: Calculate the Electricity Required
Since we need to produce one mole of copper, and we have determined that the valency factor is 2, we can substitute the values into the equation:
\[ Q = 2 \times 96500 \, \text{C} \]
\[ Q = 193000 \, \text{C} \]

### Step 5: Convert to Faradays
Since 1 Faraday is equivalent to the charge of one mole of electrons (approximately 96500 C), we can express the amount of electricity in terms of Faradays:
\[ Q = 2 \, \text{Faradays} \]

### Conclusion
Thus, the amount of electricity required to produce one mole of copper from copper sulfate solution is **2 Faradays**.

### Final Answer
The answer is **2 Faradays** (Option C).

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