A layer of chromium metal 0.25 mm thick is to be plated on an auto bumper with a total area of 032`m^2` from a solution cantaining `CrO_4^(2-)`? What current flow is required for this electroplating if the bumper is to be plated in 60s ? The density of chromium metal is 7.20g/`cm^3`
a. `4.9xx10^3`A
b. `1.78xx10^3`A
c. `5.3xx10^4`A
d. `10.69xx10^4`A

Options

A. `4.9xx10^3`A
B. `1.78xx10^3`A
C. `5.3xx10^4`A
D. `10.69xx10^4`A

Answer

To solve the problem of determining the current required for electroplating a layer of chromium metal on an auto bumper, we can follow these steps:

### Step 1: Calculate the Volume of Chromium to be Plated
The thickness of the chromium layer is given as 0.25 mm, which we need to convert to meters:
\[ 0.25 \text{ mm} = 0.25 \times 10^{-3} \text{ m} = 0.00025 \text{ m} \]
The area of the bumper is given as \(0.32 \text{ m}^2\). The volume \(V\) of chromium can be calculated using the formula:
\[ V = \text{Area} \times \text{Thickness} = 0.32 \text{ m}^2 \times 0.00025 \text{ m} = 0.08 \times 10^{-3} \text{ m}^3 \]

### Step 2: Calculate the Mass of Chromium
We know the density of chromium is \(7.20 \text{ g/cm}^3\). To use consistent units, we convert this to kg/m³:
\[ 7.20 \text{ g/cm}^3 = 7.20 \times 10^{3} \text{ kg/m}^3 \]
Now, we can calculate the mass \(m\) of chromium using the formula:
\[ m = \text{Density} \times \text{Volume} = 7.20 \times 10^{3} \text{ kg/m}^3 \times 0.08 \times 10^{-3} \text{ m}^3 = 0.576 \text{ kg} = 576 \text{ g} \]

### Step 3: Use Faraday's Law to Find the Current
According to Faraday's law, the mass of the metal deposited is given by:
\[ m = \frac{M \times I \times t}{n \times F} \]
Where:
- \(m\) = mass of chromium (576 g)
- \(M\) = molar mass of chromium (52 g/mol)
- \(I\) = current (A)
- \(t\) = time (60 s)
- \(n\) = number of electrons transferred (6 for Cr)
- \(F\) = Faraday's constant (96500 C/mol)

Rearranging the formula to solve for current \(I\):
\[ I = \frac{m \times n \times F}{M \times t} \]
Substituting the values:
\[ I = \frac{576 \text{ g} \times 6 \times 96500 \text{ C/mol}}{52 \text{ g/mol} \times 60 \text{ s}} \]

### Step 4: Calculate the Current
Calculating the numerator:
\[ 576 \times 6 \times 96500 = 3.34 \times 10^{7} \text{ g C/mol} \]
Calculating the denominator:
\[ 52 \times 60 = 3120 \text{ g s} \]
Now substituting back into the equation:
\[ I = \frac{3.34 \times 10^{7}}{3120} \approx 10700 \text{ A} = 10.69 \times 10^{4} \text{ A} \]

### Conclusion
The required current for electroplating is approximately \(10.69 \times 10^{4} \text{ A}\). Thus, the correct answer is option **d**.

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