What weight of copper (Atomic Mass 63.5u) deposits when 1 Faraday of electricity is passed through cupric salt solution ?

Question

Accepted Answer

When 1 Faraday of electricity is passed through a cupric salt solution, it results in the deposition of copper. To understand why Option 2 (31.75 g) is the correct answer, we need to delve into the principles of electrochemistry and Faraday's laws of electrolysis.

1. **Explanation for Option 2 (31.75 g):**

- **Faraday's First Law of Electrolysis** states that the mass of a substance deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. Mathematically, this is expressed as:
     $$
     m = \frac{Q \times M}{n \times F}
     $$
     where:
     - $ m $ is the mass of the substance deposited,
     - $ Q $ is the total electric charge passed through the substance,
     - $ M $ is the molar mass of the substance,
     - $ n $ is the number of electrons involved in the reaction,
     - $ F $ is Faraday's constant (approximately 96500 C/mol).

- For copper (Cu), which has an atomic mass of 63.5 u, the reaction at the cathode in a cupric salt solution is:
     $$
     \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}
     $$
     Here, $ n = 2 $ because two electrons are required to reduce one copper ion to copper metal.

- When 1 Faraday (96500 C) of electricity is passed, the mass of copper deposited can be calculated as:
     $$
     m = \frac{96500 \times 63.5}{2 \times 96500} = \frac{63.5}{2} = 31.75 \text{ g}
     $$
     Thus, 31.75 g of copper is deposited, confirming that Option 2 is correct.

2. **Explanation for why other options are incorrect:**

- **Option 1 (63.5 g):** This option assumes that 1 Faraday deposits the entire molar mass of copper (63.5 g). However, since copper ions require two electrons for reduction, only half of the molar mass is deposited per Faraday, making this option incorrect.

- **Option 3 (127 g):** This option suggests that twice the molar mass of copper is deposited, which is incorrect because it does not align with the stoichiometry of the reaction. Only 31.75 g is deposited per Faraday, not 127 g.

- **Option 4 (1 g):** This option significantly underestimates the mass of copper deposited. Given the stoichiometry and Faraday's laws, 1 g is far too low for the deposition resulting from 1 Faraday of electricity.

In conclusion, based on Faraday's laws and the stoichiometry of the copper deposition reaction, Option 2 (31.75 g) is the correct answer.