A current of 5.99 A is passed through a Cu(NO3)2 solution. How long, in hours, would this current have to be applied to plate out 7.20 g of copper?

Steps to Solve

1. Calculate moles of copper deposited

To find the moles of copper deposited, we use the molar mass of copper, which is approximately $63.55 , \text{g/mol}$.

The formula for moles is:

$$ \text{moles} = \frac{\text{mass}}{\text{molar mass}} $$

Substituting the values:

$$ \text{moles} = \frac{7.20 , \text{g}}{63.55 , \text{g/mol}} \approx 0.113 , \text{mol} $$

2. Use Faraday's first law of electrolysis

According to Faraday's laws, the amount of substance deposited is directly proportional to the charge passed through the solution, expressed as:

$$ Q = n \cdot F $$

Where:

• $Q$ is the charge (in Coulombs)
• $n$ is the number of moles
• $F$ is Faraday's constant, approximately $96485 , \text{C/mol}$

Calculating $Q$:

$$ Q = 0.113 , \text{mol} \times 96485 , \text{C/mol} \approx 10979.55 , \text{C} $$

3. Calculate the time using the relationship between charge and current

We can find the time $t$ in seconds by rearranging the formula:

$$ t = \frac{Q}{I} $$

Where:

• $I$ is the current in Amperes (5.99 A in this case)

Plugging in the numbers:

$$ t = \frac{10979.55 , \text{C}}{5.99 , \text{A}} \approx 1831.21 , \text{s} $$

4. Convert time from seconds to hours

To convert seconds into hours, use the conversion factor:

$$ \text{hours} = \frac{\text{seconds}}{3600} $$

Calculating the time in hours:

$$ \text{hours} = \frac{1831.21 , \text{s}}{3600} \approx 0.51 , \text{hours} $$