Copper can be deposited from acidified copper sulphate and alkaline cuprous cyanide. If the same current is passed for a definite time `:`

Options

A. The amount of copper deposited from acidic copper sulphate will be higher.
B. The amount of copper deposited from alkaline cuprous cyanide will be higher
C. The same amount of copper will be deposited.
D. None of these

Explanation

To solve the problem of determining the amount of copper deposited from acidified copper sulfate and alkaline cuprous cyanide when the same current is passed for a definite time, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electrolytes:**
- The two electrolytes in question are:
- Acidified Copper Sulfate (CuSO4)
- Alkaline Cuprous Cyanide (CuCN)

2. **Determine the n-factor for Copper:**
- In the case of acidified copper sulfate (CuSO4), copper is deposited from Cu²⁺ ions. The n-factor for copper in this case is 2 because it goes from Cu²⁺ to Cu (0).
- In the case of alkaline cuprous cyanide (CuCN), copper is deposited from Cu⁺ ions. The n-factor for copper here is 1 because it goes from Cu⁺ to Cu (0).

3. **Apply Faraday's First Law of Electrolysis:**
- Faraday's First Law states that the amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity (Q) passed through the electrolyte.
- The formula can be expressed as:
\[ W = Z \cdot Q \]
- Where \( W \) is the weight of the substance deposited, \( Z \) is the electrochemical equivalent, and \( Q \) is the total charge passed.

4. **Calculate the Charge (Q):**
- The charge \( Q \) can be calculated using:
\[ Q = I \cdot t \]
- Where \( I \) is the current and \( t \) is the time.

5. **Calculate the Electrochemical Equivalent (Z):**
- The electrochemical equivalent \( Z \) can be calculated using:
\[ Z = \frac{E}{96500} \]
- Where \( E \) is the equivalent weight of the substance.

6. **Calculate the Equivalent Weight for Each Case:**
- For Acidified Copper Sulfate:
\[ W_1 = \frac{M}{n_1} \cdot \frac{I \cdot t}{96500} \]
Where \( M \) is the molar mass of copper (approximately 63.5 g/mol) and \( n_1 = 2 \).
- For Alkaline Cuprous Cyanide:
\[ W_2 = \frac{M}{n_2} \cdot \frac{I \cdot t}{96500} \]
Where \( n_2 = 1 \).

7. **Compare the Amounts of Copper Deposited:**
- From the equations:
- \( W_1 = \frac{63.5}{2} \cdot \frac{I \cdot t}{96500} \)
- \( W_2 = 63.5 \cdot \frac{I \cdot t}{96500} \)
- Since \( W_2 \) has a larger n-factor in the denominator (1 vs. 2), it will yield a greater weight of copper deposited.

8. **Conclusion:**
- Therefore, the amount of copper deposited from alkaline cuprous cyanide will be higher than that from acidified copper sulfate.

### Final Answer:
The amount of copper deposited from alkaline cuprous cyanide will be higher.

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