Na-Amalgam is prepared by electrolysis of aq. `NaCI` solution using 10gm Hg cathode. How many Faraday's of electricity is required to prepare `18.7%` Na-Amalgam (current Eff `= 50%)`

Options

A. `0.081`
B. `0.16F`s
C. `0.2F`s
D. `0.1F`

Explanation

To solve the problem of how many Faraday's of electricity is required to prepare an 18.7% Na-Amalgam from the electrolysis of an aqueous NaCl solution using a 10g Hg cathode, we can follow these steps:

### Step 1: Understand the Composition of Na-Amalgam
Na-Amalgam is formed when sodium ions (Na⁺) react with mercury (Hg) to form sodium amalgam (Na-Hg). The percentage composition of sodium in the amalgam is given as 18.7%.

### Step 2: Calculate the Mass of Sodium in the Amalgam
To find the mass of sodium in the amalgam, we can use the formula for percentage composition:
\[ \text{Mass of Na} = \text{Percentage of Na} \times \text{Total mass of amalgam} \]
Let the total mass of the amalgam be \( m \). The mass of sodium can be expressed as:
\[ \text{Mass of Na} = 0.187 \times m \]

### Step 3: Determine the Total Mass of the Amalgam
The total mass of the amalgam consists of the mass of mercury and the mass of sodium:
\[ m = \text{Mass of Hg} + \text{Mass of Na} \]
Given that the mass of mercury (Hg) is 10g, we can substitute:
\[ m = 10g + 0.187m \]
Rearranging gives:
\[ m - 0.187m = 10g \implies 0.813m = 10g \implies m = \frac{10g}{0.813} \approx 12.3g \]

### Step 4: Calculate the Mass of Sodium
Now we can calculate the mass of sodium:
\[ \text{Mass of Na} = 0.187 \times 12.3g \approx 2.3g \]

### Step 5: Convert Mass of Sodium to Moles
Using the molar mass of sodium (Na = 23g/mol):
\[ \text{Moles of Na} = \frac{2.3g}{23g/mol} \approx 0.1 \text{ moles} \]

### Step 6: Calculate the Charge Required
According to Faraday's first law of electrolysis, the charge (Q) required to deposit 1 mole of sodium is 1 Faraday (F). Therefore, for 0.1 moles of sodium:
\[ \text{Charge (Q)} = 0.1 \text{ F} \]

### Step 7: Adjust for Current Efficiency
Since the current efficiency is given as 50%, we need to adjust the charge required:
\[ \text{Effective charge required} = \frac{Q}{\text{Efficiency}} = \frac{0.1 \text{ F}}{0.5} = 0.2 \text{ F} \]

### Final Answer
Thus, the total charge required to prepare the 18.7% Na-Amalgam is **0.2 Faraday**.

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