Cadmium amalgam is prepared by electrolysis of a sodium of `CdCl_2` using a mercury cathode. How long should a current of 4 A be passed in order to prepare 10% by mass Cd in Cd-Hg amalgam on cathode of 4.5 g Hg? (atomic mass of Cd=112)

Options

A. 400sec
B. 215.40 sec
C. 861.6 sec
D. 4308.8 sec

Answer

To solve the problem of how long a current of 4 A should be passed to prepare a 10% by mass of cadmium (Cd) in a cadmium-mercury amalgam on a cathode of 4.5 g of mercury (Hg), we can follow these steps:

### Step 1: Calculate the mass of cadmium required
Given that the amalgam should contain 10% by mass of cadmium and the total mass of mercury is 4.5 g, we can calculate the mass of cadmium (W) needed.

\[ \text{Mass of Cd} = \frac{10}{100} \times (\text{Mass of Hg} + \text{Mass of Cd}) \]

Let \( x \) be the mass of Cd. Then:

\[ x = 0.1 \times (4.5 + x) \]

Solving for \( x \):

\[ x = 0.45 + 0.1x \]
\[ 0.9x = 0.45 \]
\[ x = \frac{0.45}{0.9} = 0.5 \text{ g} \]

### Step 2: Calculate the equivalent weight of cadmium
The equivalent weight (E) of cadmium can be calculated using its atomic mass and the number of electrons involved in the reduction reaction. The atomic mass of cadmium is 112 g/mol, and since cadmium is reduced from \( \text{Cd}^{2+} \) to \( \text{Cd} \), it involves 2 electrons.

\[ E = \frac{\text{Molecular weight}}{\text{Valency}} = \frac{112}{2} = 56 \text{ g/equiv} \]

### Step 3: Use Faraday's first law of electrolysis
According to Faraday's first law, the mass of the substance deposited is proportional to the charge passed through the electrolyte.

\[ W = Z \cdot Q \]

Where:
- \( W \) = mass of substance deposited (in grams)
- \( Z \) = electrochemical equivalent (in g/C)
- \( Q \) = total charge (in coulombs)

The electrochemical equivalent \( Z \) can be calculated as:

\[ Z = \frac{E}{F} \]

Where \( F \) (Faraday's constant) is approximately \( 96500 \, \text{C/mol} \).

\[ Z = \frac{56 \, \text{g/equiv}}{96500 \, \text{C/mol}} = 5.8 \times 10^{-4} \, \text{g/C} \]

### Step 4: Calculate the total charge required
Using the mass of cadmium required:

\[ Q = \frac{W}{Z} = \frac{0.5 \, \text{g}}{5.8 \times 10^{-4} \, \text{g/C}} \approx 862.07 \, \text{C} \]

### Step 5: Calculate the time required using the formula \( Q = I \cdot t \)
Rearranging the formula gives:

\[ t = \frac{Q}{I} \]

Substituting the values:

\[ t = \frac{862.07 \, \text{C}}{4 \, \text{A}} \approx 215.52 \, \text{s} \]

### Final Calculation
The final answer for the time required to deposit 0.5 g of cadmium is approximately:

\[ t \approx 215.52 \, \text{s} \]

### Summary
To prepare a 10% by mass of cadmium in cadmium-mercury amalgam on a cathode of 4.5 g of mercury, a current of 4 A should be passed for approximately **215.52 seconds**.

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