To solve the problem of how long a current of 4 A should be passed to prepare a 10% by mass cadmium (Cd) in a cadmium-mercury amalgam on a cathode of 4.5 g of mercury (Hg), we can follow these steps:
### Step 1: Determine the mass of cadmium required
Given that the amalgam should be 10% by mass cadmium, we first need to find the total mass of the amalgam. The mass of mercury is given as 4.5 g.
Let the mass of cadmium be \( m_{Cd} \) and the mass of mercury be \( m_{Hg} \).
The total mass of the amalgam is:
\[
m_{total} = m_{Cd} + m_{Hg}
\]
Since cadmium constitutes 10% of the total mass:
\[
m_{Cd} = 0.10 \times m_{total}
\]
We can express the total mass in terms of the mass of mercury:
\[
m_{total} = m_{Hg} + m_{Cd} = 4.5 \, \text{g} + m_{Cd}
\]
Substituting \( m_{Cd} \):
\[
m_{total} = 4.5 \, \text{g} + 0.10 \times m_{total}
\]
Rearranging gives:
\[
m_{total} - 0.10 \times m_{total} = 4.5 \, \text{g}
\]
\[
0.90 \times m_{total} = 4.5 \, \text{g}
\]
\[
m_{total} = \frac{4.5 \, \text{g}}{0.90} = 5 \, \text{g}
\]
Now, we can find the mass of cadmium:
\[
m_{Cd} = 0.10 \times 5 \, \text{g} = 0.5 \, \text{g}
\]
### Step 2: Calculate the equivalent weight of cadmium
The atomic mass of cadmium is given as 112 g/mol. The equivalent weight (Z) of cadmium can be calculated using the formula:
\[
Z = \frac{\text{Molar mass}}{n}
\]
where \( n \) is the number of electrons transferred per atom of cadmium during the reduction process. For cadmium, \( n = 2 \) (as it goes from Cd²⁺ to Cd).
Thus, the equivalent weight of cadmium is:
\[
Z = \frac{112 \, \text{g/mol}}{2} = 56 \, \text{g/equiv}
\]
### Step 3: Calculate the total charge required to deposit cadmium
Using the formula:
\[
\text{Weight deposited} = \frac{Z \times I \times T}{F}
\]
where:
- \( I \) = current (4 A)
- \( T \) = time in seconds
- \( F \) = Faraday's constant (96500 C/equiv)
We need to find the charge required to deposit 0.5 g of cadmium. Rearranging the formula gives:
\[
T = \frac{\text{Weight deposited} \times F}{Z \times I}
\]
Substituting the values:
\[
T = \frac{0.5 \, \text{g} \times 96500 \, \text{C/equiv}}{56 \, \text{g/equiv} \times 4 \, \text{A}}
\]
Calculating:
\[
T = \frac{0.5 \times 96500}{56 \times 4}
\]
\[
T = \frac{48250}{224}
\]
\[
T \approx 215.5 \, \text{s}
\]
### Final Answer
The time required to prepare the cadmium amalgam is approximately **215.5 seconds**.
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