How much electricity in terms of Faradays is required to produce 20 g of calcium from molten CaCl₂?

1F

The correct answer is option 2. 1F.

To calculate the amount of electricity required to produce 20 g of calcium from molten \(CaCl_2\), we first need to determine the number of moles of calcium involved in the reaction.

The reaction of interest is:

\( \text{Ca}^{2+} + 2e^- \rightarrow \text{Ca} \)

From this reaction, we can see that for each calcium ion (\( \text{Ca}^{2+} \)), 2 moles of electrons (\( 2e^- \)) are required to produce 1 mole of calcium (\( \text{Ca} \)).

Now, let's find the number of moles of calcium in 20 g:

\(\text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \)

The molar mass of calcium (\( \text{Ca} \)) is approximately 40 g/mol.

\(\text{Number of moles of Ca} = \frac{20 \, \text{g}}{40 \, \text{g/mol}} = 0.5 \, \text{mol} \)

Since 1 mole of calcium requires 2 moles of electrons, the total number of moles of electrons required is:

\(\text{Number of moles of electrons} = 0.5 \, \text{mol} \times 2 = 1 \, \text{mol} \)

Now, to convert moles of electrons to Faradays, we use Faraday's constant, which is approximately \( 96485 \, \text{C/mol} \).

\(\text{Electricity (in terms of Faradays)} = \text{Number of moles of electrons} \times \text{Faraday's constant} \)

\(\text{Electricity (in terms of Faradays)} = 1 \, \text{mol} \times 96485 \, \text{C/mol} \)

\( \text{Electricity (in terms of Faradays)} = 96485 \, \text{C} \)

Therefore, the amount of electricity required to produce 20 g of calcium from molten \(CaCl_2\) is approximately 96485 C or 1 Faraday.

So, the correct option is 1F.