To solve the problem of how much time will be required to precipitate 2 g of copper from a solution of CuSO₄ using a current of 0.5 amperes, we can follow these steps:
### Step 1: Determine the molar mass of copper (Cu)
The molar mass of copper (Cu) is approximately 63.5 g/mol.
### Step 2: Calculate the number of moles of copper to be precipitated
To find the number of moles of copper in 2 g, we use the formula:
\[
\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \, \text{g}}{63.5 \, \text{g/mol}} \approx 0.0315 \, \text{mol}
\]
### Step 3: Determine the total charge required to deposit the copper
The reaction for the deposition of copper from Cu²⁺ ions is:
\[
\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)}
\]
This indicates that 2 moles of electrons are required to deposit 1 mole of copper. Therefore, for 0.0315 moles of copper, the number of moles of electrons required is:
\[
\text{Moles of electrons} = 0.0315 \, \text{mol} \times 2 = 0.0630 \, \text{mol}
\]
### Step 4: Calculate the total charge (Q) required
Using Faraday's constant (approximately 96500 C/mol), the total charge required can be calculated as:
\[
Q = \text{moles of electrons} \times \text{Faraday's constant} = 0.0630 \, \text{mol} \times 96500 \, \text{C/mol} \approx 6080 \, \text{C}
\]
### Step 5: Use the formula to find time (t)
The relationship between charge (Q), current (I), and time (t) is given by:
\[
Q = I \times t
\]
Rearranging this gives:
\[
t = \frac{Q}{I}
\]
Substituting the values we have:
\[
t = \frac{6080 \, \text{C}}{0.5 \, \text{A}} = 12160 \, \text{s}
\]
### Step 6: Convert seconds to a more convenient unit (if necessary)
12160 seconds can be converted to hours or minutes if needed, but in this case, we can leave it in seconds.
### Final Answer
The time required to precipitate 2 g of copper using a current of 0.5 amperes is approximately **12160 seconds**.
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