Breathalyzer is used to detect the alcohol content in the suspected drunk drivers. The ethanol in the exhaled breath is oxidized to ethanoic acid with an acidic solution of `K_(2)Cr_(2)O_(7)` as follows `:`
`underset(Ehanol)(3CH_(3)CH_(2)OH(aq))+2Cr_(2)O_(7)^(2-)(aq) +16H^(o+)(aq) rarr underset(Ethanoic ac i d)(3CH_(3)COH(aq))+4Cr^(3+)(aq)+11H_(2)O(l)`
The breathalyzer measures the colour change and produces a metre reading calibrated in the terms of blood alcohol content.
What is the ethanol ethanoic acid ratio if the breathalyzer records `1.33V` and other species are at `1M ?`

Options

A. `10^(4)`
B. `10^(-4)`
C. `10^(12)`
D. `10^(-12)`

Explanation

To solve the problem of finding the ethanol to ethanoic acid ratio based on the given conditions, we can follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ 3 \text{CH}_3\text{CH}_2\text{OH} + 2 \text{Cr}_2\text{O}_7^{2-} + 16 \text{H}^+ \rightarrow 3 \text{CH}_3\text{COOH} + 4 \text{Cr}^{3+} + 11 \text{H}_2\text{O} \]
This indicates that 3 moles of ethanol (C2H5OH) are oxidized to produce 3 moles of ethanoic acid (CH3COOH).

### Step 2: Identify the Standard Reduction Potential
The standard reduction potential (E°) for the half-reaction involving Cr2O7^2- is typically around 1.27 V.

### Step 3: Use the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \]
where:
- \( E \) is the cell potential (1.33 V in this case),
- \( E^\circ \) is the standard potential (1.27 V),
- \( n \) is the number of electrons transferred (12 for this reaction),
- \([\text{Products}]\) and \([\text{Reactants}]\) are the concentrations of the products and reactants.

### Step 4: Set Up the Equation
Substituting the values into the Nernst equation gives:
\[ 1.33 = 1.27 - \frac{0.059}{12} \log \frac{[\text{CH}_3\text{COOH}]^3 [\text{Cr}^{3+}]^4 [\text{H}_2\text{O}]^{11}}{[\text{CH}_3\text{CH}_2\text{OH}]^3 [\text{Cr}_2\text{O}_7^{2-}]^2 [\text{H}^+]^{16}} \]

### Step 5: Simplify the Concentrations
Given that all species except ethanol and ethanoic acid are at 1 M concentration, we can simplify the equation:
\[ 1.33 = 1.27 - \frac{0.059}{12} \log \frac{[\text{CH}_3\text{COOH}]^3}{[\text{CH}_3\text{CH}_2\text{OH}]^3} \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ 1.33 - 1.27 = -\frac{0.059}{12} \log \frac{[\text{CH}_3\text{COOH}]^3}{[\text{CH}_3\text{CH}_2\text{OH}]^3} \]
\[ 0.06 = -\frac{0.059}{12} \log \frac{[\text{CH}_3\text{COOH}]^3}{[\text{CH}_3\text{CH}_2\text{OH}]^3} \]

### Step 7: Solve for the Logarithm
Multiply both sides by -12/0.059:
\[ \log \frac{[\text{CH}_3\text{COOH}]^3}{[\text{CH}_3\text{CH}_2\text{OH}]^3} = -\frac{0.06 \times 12}{0.059} \]
Calculating the right side gives approximately:
\[ \log \frac{[\text{CH}_3\text{COOH}]^3}{[\text{CH}_3\text{CH}_2\text{OH}]^3} \approx -12.24 \]

### Step 8: Exponentiate to Find the Ratio
Taking the antilog:
\[ \frac{[\text{CH}_3\text{COOH}]}{[\text{CH}_3\text{CH}_2\text{OH}]} = 10^{-12.24} \]
This means:
\[ \frac{[\text{CH}_3\text{CH}_2\text{OH}]}{[\text{CH}_3\text{COOH}]} = 10^{12.24} \]

### Step 9: Final Ratio
Thus, the ratio of ethanol to ethanoic acid is approximately:
\[ \text{Ethanol:Ethanoic Acid} \approx 10^{12.24} : 1 \]