(BAC) is often reported in weight-volume percent (w/v%). For example, a BAC of 0

Question

(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

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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