During electrolysis of molten `CaCl_(2)` 0.005 A current is passed through the cell for 200 s. The mass of product formed at cathode (molar mass of Ca =40 `g mol^(-1)`) will be__________.

Options

A. 0.0002073 g of Ca
B. 0.0004145 g of Ca
C. 0.0003678 g of `Cl_(2)`
D. 0.0007357g of `Cl_(2)`

Answer

To find the mass of calcium deposited at the cathode during the electrolysis of molten \( \text{CaCl}_2 \), we can use Faraday's laws of electrolysis. Here’s the step-by-step solution:

### Step 1: Identify the reaction at the cathode
During the electrolysis of molten \( \text{CaCl}_2 \), calcium ions (\( \text{Ca}^{2+} \)) are reduced at the cathode to form solid calcium (\( \text{Ca} \)):
\[ \text{Ca}^{2+} + 2e^- \rightarrow \text{Ca} \]

### Step 2: Determine the number of moles of electrons transferred
Using the formula:
\[ Q = I \times T \]
where:
- \( Q \) is the total charge in coulombs,
- \( I \) is the current in amperes (0.005 A),
- \( T \) is the time in seconds (200 s).

Calculate \( Q \):
\[ Q = 0.005 \, \text{A} \times 200 \, \text{s} = 1 \, \text{C} \]

### Step 3: Calculate the number of moles of electrons
Using Faraday's constant (\( F \approx 96500 \, \text{C/mol} \)):
\[ \text{Number of moles of electrons} = \frac{Q}{F} = \frac{1 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.00001036 \, \text{mol} \]

### Step 4: Determine the number of moles of calcium produced
From the reaction, 2 moles of electrons produce 1 mole of calcium. Therefore, the number of moles of calcium produced is:
\[ \text{Moles of Ca} = \frac{\text{Number of moles of electrons}}{2} = \frac{0.00001036}{2} \approx 0.00000518 \, \text{mol} \]

### Step 5: Calculate the mass of calcium deposited
Using the molar mass of calcium (40 g/mol):
\[ \text{Mass of Ca} = \text{Moles of Ca} \times \text{Molar mass of Ca} = 0.00000518 \, \text{mol} \times 40 \, \text{g/mol} \approx 0.0002072 \, \text{g} \]

### Final Answer
The mass of calcium deposited at the cathode is approximately:
\[ \text{Mass} \approx 0.0002072 \, \text{g} \]

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