` 2.5 F` of electricity is passed through a `CuSO_4` solution. The number of gm equivalent of `Cu deposited on anode is .

Options

A. ` Zero`
B. ` 2.5`
C. ` 1.25`
D. `5.0`

Explanation

To find the number of gram equivalents of copper (Cu) deposited when 2.5 Faraday of electricity is passed through a CuSO₄ solution, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: In a CuSO₄ solution, copper ions (Cu²⁺) are reduced to solid copper (Cu) at the cathode. The half-reaction can be written as:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
This indicates that 2 moles of electrons are required to deposit 1 mole of copper.

2. **Determine the Faraday's Law of Electrolysis**: According to Faraday's laws, the amount of substance deposited is directly proportional to the quantity of electricity passed through the electrolyte. The relationship can be expressed as:
\[ \text{Gram equivalent} = \frac{\text{Electricity (in Faraday)}}{n} \]
where \( n \) is the number of moles of electrons required to deposit one mole of the substance.

3. **Identify the Value of n**: For copper, \( n = 2 \) because it takes 2 moles of electrons to deposit 1 mole of copper.

4. **Calculate the Gram Equivalents of Copper**:
- Given that 2 Faraday of electricity deposits 1 gram equivalent of copper, we can set up a proportion:
\[ \text{If 2 Faraday} \rightarrow 1 \text{ gram equivalent of Cu} \]
\[ \text{Then 1 Faraday} \rightarrow \frac{1}{2} \text{ gram equivalent of Cu} \]
- Now, for 2.5 Faraday:
\[ \text{Gram equivalent of Cu} = 2.5 \times \frac{1}{2} = 1.25 \text{ gram equivalent of Cu} \]

5. **Conclusion**: Therefore, the number of gram equivalents of copper deposited on the cathode when 2.5 Faraday of electricity is passed through the CuSO₄ solution is **1.25 gram equivalent**.