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

Explanation

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
In the electrolysis of copper(II) sulfate (CuSO₄), copper ions (Cu²⁺) are reduced at the cathode to form solid copper (Cu). The half-reaction can be represented as:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

This equation shows that 2 moles of electrons (2e⁻) are required to deposit 1 mole of copper.

### Step 2: Relate Faraday to moles of electrons
One Faraday (F) is equivalent to the charge of one mole of electrons, which is approximately 96500 coulombs. Therefore, two Faraday corresponds to:
\[ 2 \text{ F} = 2 \times 96500 \text{ C} \]

### Step 3: Calculate moles of electrons
Since 1 mole of electrons corresponds to 1 Faraday, 2 Faraday corresponds to 2 moles of electrons.

### Step 4: Determine moles of copper deposited
From the half-reaction, we know that 2 moles of electrons will deposit 1 mole of copper. Therefore, if we have 2 moles of electrons, we can deposit:
\[ \text{Moles of Cu} = \frac{2 \text{ moles of e}^-}{2} = 1 \text{ mole of Cu} \]

### Step 5: Calculate the mass of copper deposited
The mass of copper deposited can be calculated using the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar mass} \]
Given that the molar mass of copper (Cu) is 63.5 g/mol:
\[ \text{Mass of Cu} = 1 \text{ mole} \times 63.5 \text{ g/mol} = 63.5 \text{ g} \]

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

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