Chromium plating is applied by electrolysis to objects suspended in a dichlromate solution , according to following `(` unbalanced `)` hald reaction `:`
`Cr_(2)O_(7)^(2-)(aq) +e^(-) +H^(o+)(aq) rarr Cr(s)+H_(2)O(l)`
How many hours would it take to apply a chromium plating of thickness `2.0xx10^(-2)mm` to a car bumper of suface area `0.25m^(2)` in an electrolysis cell carrying a current of `75.0A?`
`[` Density of chromium is `7.19g cm^(-3)]`

Options

A. `2.2h`
B. `1.5h`
C. `3.0h`
D. `0.2`

Explanation

To solve the problem of how long it will take to apply a chromium plating of thickness \(2.0 \times 10^{-2} \, \text{mm}\) to a car bumper with a surface area of \(0.25 \, \text{m}^2\) using an electrolysis cell carrying a current of \(75.0 \, \text{A}\), we will follow these steps:

### Step 1: Calculate the Volume of Chromium Plating
The volume \(V\) of the chromium plating can be calculated using the formula:
\[ V = \text{Surface Area} \times \text{Thickness} \]
Convert the thickness from mm to cm:
\[ \text{Thickness} = 2.0 \times 10^{-2} \, \text{mm} = 2.0 \times 10^{-3} \, \text{cm} \]
Convert the surface area from \(m^2\) to \(cm^2\):
\[ \text{Surface Area} = 0.25 \, \text{m}^2 = 0.25 \times 10^4 \, \text{cm}^2 = 2500 \, \text{cm}^2 \]
Now, calculate the volume:
\[ V = 2500 \, \text{cm}^2 \times 2.0 \times 10^{-3} \, \text{cm} = 5.0 \, \text{cm}^3 \]

### Step 2: Calculate the Mass of Chromium
Using the density of chromium (\(7.19 \, \text{g/cm}^3\)), we can find the mass \(m\):
\[ m = \text{Density} \times \text{Volume} = 7.19 \, \text{g/cm}^3 \times 5.0 \, \text{cm}^3 = 35.95 \, \text{g} \]

### Step 3: Calculate the Moles of Chromium
The molar mass of chromium is approximately \(52 \, \text{g/mol}\). Thus, the number of moles \(n\) of chromium is:
\[ n = \frac{m}{\text{Molar Mass}} = \frac{35.95 \, \text{g}}{52 \, \text{g/mol}} \approx 0.69 \, \text{mol} \]

### Step 4: Balance the Half-Reaction
The balanced half-reaction for chromium plating is:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 12 e^- \rightarrow 2 \text{Cr} + 7 \text{H}_2\text{O} \]
From this, we see that 2 moles of chromium require 12 moles of electrons (or Faraday).

### Step 5: Calculate the Total Charge Required
For \(0.69 \, \text{mol}\) of chromium, the number of moles of electrons required is:
\[ \text{Electrons required} = \frac{12}{2} \times 0.69 = 4.14 \, \text{mol} \]
The total charge \(Q\) in coulombs can be calculated using Faraday's constant (\(96500 \, \text{C/mol}\)):
\[ Q = \text{Moles of electrons} \times 96500 \, \text{C/mol} = 4.14 \, \text{mol} \times 96500 \, \text{C/mol} \approx 3998100 \, \text{C} \]

### Step 6: Calculate the Time Required
Using the formula \(Q = I \times t\), where \(I\) is the current:
\[ t = \frac{Q}{I} = \frac{3998100 \, \text{C}}{75.0 \, \text{A}} \approx 53241.33 \, \text{s} \]
Convert seconds to hours:
\[ t \approx \frac{53241.33 \, \text{s}}{3600 \, \text{s/hour}} \approx 14.77 \, \text{hours} \]

### Final Answer
It would take approximately **14.77 hours** to apply the chromium plating.

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