When 0.04 F of electricity is passed through a solution of `CaSO_(4)`, then the weight of `Ca^(2+)` metal deposited at the cathode is

Options

A. 0.2 g
B. 0.4 g
C. 0.6 g
D. 0.8 g

Answer

To solve the problem of determining the weight of calcium (Ca) deposited at the cathode when 0.04 Farad of electricity is passed through a solution of \( CaSO_4 \), we can follow these steps:

### Step 1: Understand the electrochemical reaction
When \( CaSO_4 \) is electrolyzed, calcium ions \( Ca^{2+} \) are reduced at the cathode. The half-reaction for the reduction of calcium ions is:
\[ Ca^{2+} + 2e^- \rightarrow Ca \]
This indicates that 1 mole of \( Ca^{2+} \) requires 2 moles of electrons (2 Faraday) to deposit 1 mole of calcium.

### Step 2: Calculate the moles of electrons passed
We know that 1 Faraday corresponds to the charge of 1 mole of electrons. Therefore, if 0.04 Farad of electricity is passed, the number of moles of electrons (n) can be calculated using the formula:
\[ n = \frac{\text{Charge (in Farads)}}{F} \]
Where \( F \) (Faraday's constant) is approximately 96500 C/mol. However, since we are dealing with Farads directly, we can use:
\[ \text{Moles of electrons} = 0.04 \text{ Farad} \]

### Step 3: Relate moles of electrons to moles of \( Ca^{2+} \)
From the half-reaction, we know that 2 moles of electrons are required to deposit 1 mole of calcium. Therefore, the moles of \( Ca^{2+} \) deposited can be calculated as follows:
\[ \text{Moles of } Ca^{2+} = \frac{\text{Moles of electrons}}{2} = \frac{0.04}{2} = 0.02 \text{ moles} \]

### Step 4: Calculate the mass of calcium deposited
To find the mass of calcium deposited, we use the molar mass of calcium, which is approximately 40 g/mol. The mass (m) can be calculated using the formula:
\[ m = \text{Moles} \times \text{Molar mass} \]
Substituting the values:
\[ m = 0.02 \text{ moles} \times 40 \text{ g/mol} = 0.8 \text{ g} \]

### Conclusion
The weight of calcium deposited at the cathode is **0.8 grams**.

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