2.5 faradays of electricity is passed through solution of `CuSO_(4)`. The number of gram equivalents of copper depsoited on the cathode would be

Options

A. 1
B. 2
C. 2.5
D. 1.25

Answer

To solve the problem of how many gram equivalents of copper are deposited when 2.5 Faradays of electricity are passed through a solution of CuSO₄, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
The electrolysis of copper sulfate (CuSO₄) involves the reduction of copper ions (Cu²⁺) at the cathode:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \]
This equation shows that 1 mole of Cu²⁺ requires 2 moles of electrons (2 Faradays) to deposit 1 mole of copper.

2. **Determine the Relationship Between Faradays and Gram Equivalents**:
According to Faraday's laws of electrolysis, 1 Faraday of charge deposits 1 gram equivalent of substance. For copper, since it has a valency of 2 (Cu²⁺), 1 Faraday will deposit 0.5 gram equivalents of copper. Therefore:
\[ 1 \text{ Faraday} \rightarrow 0.5 \text{ gram equivalents of Cu} \]

3. **Calculate the Total Gram Equivalents for 2.5 Faradays**:
To find out how many gram equivalents of copper are deposited with 2.5 Faradays, we can use the relationship established in the previous step:
\[ \text{Gram equivalents of Cu} = 2.5 \text{ Faradays} \times 0.5 \text{ gram equivalents/Faraday} \]
\[ \text{Gram equivalents of Cu} = 2.5 \times 0.5 = 1.25 \text{ gram equivalents} \]

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

### Final Answer:
1.25 gram equivalents of copper are deposited on the cathode.

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