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.
If `E^(c-)._(CH_(3)COOH|C_(2)H_(5)OH)=0.06 V` and `E^(c-)._(Cr_(2)O_(7)^(2-)|Cr^(3+))=1.33V`,
then `E^(c-)._(cell)` of the reaction taking place in alcohol metre is
a 1.39 V
b1.27 V −
c1.39 V −
d1.51 V

Options

A. `1.39V`
B. `1.27V`
C. `-1.39 V`
D. `-1.51V`

Explanation

To solve the problem, we need to calculate the standard cell potential \( E^\circ_{\text{cell}} \) for the reaction taking place in the breathalyzer. The reaction involves the oxidation of ethanol to ethanoic acid and the reduction of dichromate ions to chromium ions.

### Step-by-step Solution:

1. **Identify the half-reactions:**
- Oxidation half-reaction (anode):
\[ \text{C}_2\text{H}_5\text{OH} \rightarrow \text{CH}_3\text{COOH} + 2\text{H}^+ + 2e^- \]
- Reduction half-reaction (cathode):
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

2. **Standard reduction potentials:**
- For the reduction of dichromate:
\[ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V} \]
- For the oxidation of ethanol:
\[ E^\circ_{\text{CH}_3\text{COOH}/\text{C}_2\text{H}_5\text{OH}} = 0.06 \, \text{V} \]

3. **Determine the standard cell potential:**
The standard cell potential \( E^\circ_{\text{cell}} \) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, the cathode is the reduction half-reaction (dichromate) and the anode is the oxidation half-reaction (ethanol).
\[ E^\circ_{\text{cell}} = 1.33 \, \text{V} - 0.06 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 1.27 \, \text{V} \]

4. **Final Answer:**
The standard cell potential \( E^\circ_{\text{cell}} \) for the reaction taking place in the alcohol meter is:
\[ \boxed{1.27 \, \text{V}} \]