To solve the problem of finding the charge needed to plate copper on an area of \(10 \times 10 \, \text{cm}^2\) to a thickness of \(10^{-2} \, \text{cm}\) using a \( \text{CuSO}_4 \) solution, we will follow these steps:
### Step 1: Calculate the Volume of Copper Deposited
The volume \(V\) of copper deposited can be calculated using the formula:
\[
V = \text{Area} \times \text{Thickness}
\]
Given:
- Area = \(10 \, \text{cm} \times 10 \, \text{cm} = 100 \, \text{cm}^2\)
- Thickness = \(10^{-2} \, \text{cm}\)
Calculating the volume:
\[
V = 100 \, \text{cm}^2 \times 10^{-2} \, \text{cm} = 1 \, \text{cm}^3
\]
### Step 2: Calculate the Mass of Copper Deposited
Using the density of copper, we can find the mass \(m\) of copper deposited:
\[
m = \text{Density} \times V
\]
Given:
- Density of copper = \(8.94 \, \text{g/mL} = 8.94 \, \text{g/cm}^3\)
Calculating the mass:
\[
m = 8.94 \, \text{g/cm}^3 \times 1 \, \text{cm}^3 = 8.94 \, \text{g}
\]
### Step 3: Calculate the Number of Moles of Copper
To find the number of moles \(n\) of copper deposited, we use the formula:
\[
n = \frac{m}{M}
\]
where \(M\) is the molar mass of copper (63.6 g/mol).
Calculating the number of moles:
\[
n = \frac{8.94 \, \text{g}}{63.6 \, \text{g/mol}} \approx 0.140 \, \text{mol}
\]
### Step 4: Calculate the Charge Required
From Faraday's laws of electrolysis, the charge \(Q\) required to deposit \(n\) moles of copper can be calculated using the formula:
\[
Q = n \times F \times z
\]
where:
- \(F\) (Faraday's constant) = \(96500 \, \text{C/mol}\)
- \(z\) (valency of copper) = 2 (since copper ions \(Cu^{2+}\) are reduced to copper metal)
Calculating the charge:
\[
Q = n \times F \times z = 0.140 \, \text{mol} \times 96500 \, \text{C/mol} \times 2
\]
\[
Q = 0.140 \times 96500 \times 2 \approx 27071 \, \text{C}
\]
### Final Answer
The charge needed to plate the copper is approximately:
\[
Q \approx 27172 \, \text{C} \quad \text{(or } 2.7 \times 10^4 \text{ C)}
\]
