One Faraday of electricity when passed through a solution of copper sulphate deposits .

Options

A. 1 mloe of Cu
B. `1` g ewuivalent of Cu
C. `1` g ewiovalent of Cu
D. `1` molecule of Cu

Explanation

To solve the problem of how much copper is deposited when 1 Faraday of electricity is passed through a solution of copper sulfate (CuSO4), we can follow these steps:

### Step-by-Step Solution:

1. **Understand Faraday's Law of Electrolysis**:
Faraday's law states that the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the solution. The relationship can be expressed as:
\[ W = Z \cdot Q \]
where:
- \( W \) = mass of the substance deposited (in grams)
- \( Z \) = electrochemical equivalent of the substance (in grams per Faraday)
- \( Q \) = total electric charge passed (in Faradays)

2. **Determine the Electrochemical Equivalent (Z) for Copper**:
For copper (Cu), the valency is +2 (as it forms Cu²⁺ ions in solution). The equivalent weight (E) of copper can be calculated using the formula:
\[ E = \frac{M}{n} \]
where:
- \( M \) = molar mass of copper (approximately 63.5 g/mol)
- \( n \) = number of electrons transferred per ion (which is 2 for Cu²⁺)

Thus, the equivalent weight of copper is:
\[ E = \frac{63.5 \, \text{g/mol}}{2} = 31.75 \, \text{g/equiv} \]

3. **Calculate the Amount of Copper Deposited**:
When 1 Faraday of charge is passed through the solution, we can substitute \( Q = 1 \) Faraday into the equation:
\[ W = Z \cdot Q \]
Since \( Z \) for copper is equal to its equivalent weight (31.75 g/equiv), we have:
\[ W = 31.75 \, \text{g/equiv} \cdot 1 \, \text{Faraday} = 31.75 \, \text{g} \]

4. **Conclusion**:
Therefore, when 1 Faraday of electricity is passed through a solution of copper sulfate, approximately 31.75 grams of copper will be deposited.

### Final Answer:
**31.75 grams of copper will be deposited.**

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