Two faraday of electricity is passed through a solution of `CuSO_(4)`. The mass of copper deposited at the cathode is: (at mass of Cu = 63.5 amu)

Options

A. 63.5 g
B. 2 g
C. 127 g
D. 0 g

Answer

To find the mass of copper deposited at the cathode when two Faraday of electricity is passed through a solution of CuSO₄, we can follow these steps:

### Step 1: Understand the electrochemical reaction
When copper sulfate (CuSO₄) dissociates in solution, it produces copper ions (Cu²⁺) and sulfate ions (SO₄²⁻). The copper ions are reduced at the cathode, gaining electrons to form solid copper.

### Step 2: Determine the relationship between charge and moles of copper
The reduction reaction for copper can be represented as:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \]
This indicates that 1 mole of Cu²⁺ requires 2 moles of electrons (2 Faraday) to deposit 1 mole of copper.

### Step 3: Calculate the moles of electrons passed
Given that 2 Faraday of electricity is passed, we can relate this to the number of moles of electrons:
- 1 Faraday corresponds to 1 mole of electrons.
- Therefore, 2 Faraday corresponds to 2 moles of electrons.

### Step 4: Calculate the moles of copper deposited
From the stoichiometry of the reaction, we see that 2 moles of electrons will deposit 1 mole of copper:
- Since we passed 2 moles of electrons, we will deposit:
\[ \text{Moles of Cu deposited} = \frac{2 \text{ moles of electrons}}{2} = 1 \text{ mole of Cu} \]

### Step 5: Calculate the mass of copper deposited
The atomic mass of copper (Cu) is given as 63.5 g/mol. Therefore, the mass of copper deposited can be calculated as:
\[ \text{Mass of Cu} = \text{Moles of Cu} \times \text{Molar mass of Cu} \]
\[ \text{Mass of Cu} = 1 \text{ mole} \times 63.5 \text{ g/mol} = 63.5 \text{ g} \]

### Final Answer
The mass of copper deposited at the cathode is **63.5 grams**.

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