is this the right answer for this question? Blood alcohol content (BAC) is often

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is this the right answer for this question?

Blood alcohol content (BAC) is often reported in weight-volume percent (w/v%). For example, a BAC of 0.08% corresponds to 0.08 g CH3CH2OH per 100 mL of blood. In Ontario, and the rest of Canada, the maximum legal BAC for fully licensed drivers is 0.08%. Estimates of BAC can be obtained from breath samples using a number of commercially available instruments, including The Breathalyzer. The original Breathalyzer instrument was based on the following oxidation-reduction reaction. (The equation below is

not balanced! Your first task is to balance it.)

CH3CH2OH(aq) + Cr2O72?(aq)---->CH3COOH(aq) + Cr3+(aq)
it is in acidic solution


A Breathalyzer instrument contains two ampules, each of which contains 0.75 mg K2Cr2O7 dissolved in 3 mL of 9 mol L?1 H2SO4(aq). One of the ampules is used as reference. When a person exhales into the tube of the Breathalyzer, the breath is directed into one of the ampules, and ethyl alcohol (CH3CH2OH) in the breath converts Cr2O72? into Cr3+. The instrument compares the colours of the solutions in the two ampules to estimate the blood alcohol content. The estimate of the BAC rests upon the assumption that 2100 mL of air exhaled from the lungs contains the same amount of alcohol as one millilitre of blood.
The Breathalyzer instrument typically tests 52.5 mL of exhaled breath.

(1a) If the test of such a breath sample yields a reading of 0.08%, what are the partial pressure and mole fraction of CH3CH2OH in the breath of the person who was tested? Assume the atmospheric pressure and temperature are 101 kPa and 21.0 oC, respectively. (b) A BAC of 0.40% is considered the lethal limit for humans (i.e. for a majority of adults, a BAC value of 0.40% causes death). By performing an appropriate calculation, determine whether or not the Breathalyzer instrument, as described above, can be used to measure BAC as high as 0.40%.

Explanation / Answer

For these chemical equations, I like to group the same elements to make things more aesthetic. Therefore: C2H6O + Cr2O7(-2) + H+ --> C2H4O2 + Cr(3+) + H2O You can see that there is an oxidation of 4 e- (2 e- for each carbon) from C2H6O to C2H4O2 and a reduction of 6 e- (3 e- for each chromium atom) from Cr2O7 to Cr3+. I will use the Half-Reaction Method here. Oxidation reaction: C2H6O -> C2H4O2 C2H6O + H2O --> C2H4O2 C2H6O + H2O --> C2H4O2 + 4H(+) + 4e- Reduction reaction: Cr2O7(-2) --> Cr(+3) Cr2O7(-2) --> 2Cr(+3) --> 2Cr(+3) + 7H2O Cr2O7(-2) + 14H+ + 6e(-) --> 2Cr(3+) + 7H2O So far, you have: C2H6O + H2O --> C2H4O2 + 4H(+) + 4e- --(O) Cr2O7(-2) + 14H+ + 6e(-) --> 2Cr(3+) + 7H2O --(R) Since the electrons in equations O and R are different, their number must be equalised by multiplying O by 3 and R by 2. Therefore, you will get: 3C2H6O + 3H2O + 2Cr2O7(-2) + 28H+ + 12e(-) --> 3C2H4O2 + 12H(+) + 12e- + 4Cr(3+) + 14H2OAfter elimination of the electrons (same amount on each side) and simplification of H2O: 3C2H6O + 2Cr2O7(-2) + 28H+ --> 3C2H4O2 + 12H+ + 4Cr(3+) + 11H2O Let's check the charges. For the reactants, charge = 2(-2) + 28 = 24 Products' charge = 12 + 4(3) = 24. Since charges on both sides equal, this chemical equation has been balanced. I

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