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 find the weight of \( \text{Ca}^{2+} \) metal deposited at the cathode when 0.04 F of electricity is passed through a solution of \( \text{CaSO}_4 \), we can follow these steps:

### Step 1: Understand the Electrochemical Reaction
In the electrolysis of \( \text{CaSO}_4 \), the calcium ions \( \text{Ca}^{2+} \) will be reduced at the cathode. The half-reaction for the reduction of calcium ions can be written as:
\[ \text{Ca}^{2+} + 2e^- \rightarrow \text{Ca} \]
This indicates that 1 mole of \( \text{Ca}^{2+} \) requires 2 moles of electrons (or 2 Faradays) to deposit 1 mole of calcium.

### Step 2: Calculate the Moles of Calcium Deposited
Since 2 Faradays are required to deposit 1 mole of calcium, we can determine how many moles of calcium will be deposited with 0.04 Faraday of electricity:
\[ \text{Moles of } \text{Ca} = \frac{\text{Faradays used}}{2} = \frac{0.04 \, \text{F}}{2} = 0.02 \, \text{moles} \]

### Step 3: Calculate the Mass of Calcium Deposited
The molar mass of calcium (Ca) is approximately 40 g/mol. To find the mass of calcium deposited, we use the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
Substituting the values we have:
\[ \text{Mass} = 0.02 \, \text{moles} \times 40 \, \text{g/mol} = 0.8 \, \text{g} \]

### Conclusion
The weight of \( \text{Ca}^{2+} \) metal deposited at the cathode is \( 0.8 \, \text{g} \).

### Final Answer
The correct answer is \( 0.8 \, \text{g} \).

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