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.
The `EMF` of the reaction when the concentration of all the species are `1.0M` and `pH` is `4.0` is

Options

A. `1.64`
B. `0.31`
C. `-1.01V`
D. `0.95V`

Explanation

To find the EMF of the reaction involving the oxidation of ethanol to ethanoic acid using potassium dichromate, we will follow these steps:

### Step 1: Write the balanced redox reaction
The balanced reaction is given as:
\[ 3 \text{C}_2\text{H}_5\text{OH} + 2 \text{Cr}_2\text{O}_7^{2-} + 16 \text{H}^+ \rightarrow 3 \text{C}_2\text{H}_5\text{COOH} + 4 \text{Cr}^{3+} + 11 \text{H}_2\text{O} \]

### Step 2: Identify the number of electrons transferred
In this reaction, each chromium in \(\text{Cr}_2\text{O}_7^{2-}\) is reduced from +6 to +3, which involves the transfer of 6 electrons per chromium atom. Since there are 2 chromium atoms, the total number of electrons transferred is:
\[ n = 2 \times 6 = 12 \]

### Step 3: Use the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ} - \frac{0.0591}{n} \log Q \]
where \(E^{\circ}\) is the standard electrode potential, \(n\) is the number of electrons transferred, and \(Q\) is the reaction quotient.

### Step 4: Determine the standard electrode potential \(E^{\circ}\)
The standard electrode potentials for the half-reactions are:
- For the reduction of \(\text{Cr}_2\text{O}_7^{2-}\) to \(\text{Cr}^{3+}\): \(E^{\circ} = +1.33 \, \text{V}\)
- For the oxidation of ethanol to ethanoic acid: \(E^{\circ} = -0.07 \, \text{V}\)

Thus,
\[ E^{\circ} = 1.33 - 0.07 = 1.26 \, \text{V} \]

### Step 5: Calculate the reaction quotient \(Q\)
Given that the concentrations of all species are \(1.0 \, \text{M}\) and the pH is \(4.0\), we can find the concentration of \(H^+\):
\[ [H^+] = 10^{-pH} = 10^{-4} \, \text{M} \]

Now, substituting into the expression for \(Q\):
\[ Q = \frac{[\text{Cr}^{3+}]^4 \cdot [H_2O]^{11}}{[\text{Cr}_2\text{O}_7^{2-}]^2 \cdot [H^+]^{16}} \]
Since the concentration of water is constant and can be omitted in the expression, we have:
\[ Q = \frac{(1)^4}{(1)^2 \cdot (10^{-4})^{16}} = \frac{1}{(10^{-64})} = 10^{64} \]

### Step 6: Substitute values into the Nernst equation
Now we substitute \(E^{\circ}\), \(n\), and \(Q\) into the Nernst equation:
\[ E_{\text{cell}} = 1.26 - \frac{0.0591}{12} \log(10^{64}) \]
Calculating the logarithm:
\[ \log(10^{64}) = 64 \]
Thus,
\[ E_{\text{cell}} = 1.26 - \frac{0.0591}{12} \times 64 \]
Calculating the second term:
\[ \frac{0.0591 \times 64}{12} = 0.3168 \]
So,
\[ E_{\text{cell}} = 1.26 - 0.3168 = 0.9432 \, \text{V} \]

### Step 7: Final answer
Rounding to two decimal places, we find:
\[ E_{\text{cell}} \approx 0.95 \, \text{V} \]