Question

One type of breathalyzer employs a fuel cell to measure the quantity of alcohol in the breath. When a suspect blows into the breathalyzer, ethyl alcohol is oxidized to acetic acid at the anode: CH3CH2OH(g)+4OH−(aq)→HC2H3O2(g)+3H2O(l)+4e− At the cathode, oxygen is reduced: O2(g)+2H2O(l)+4e−→4OH−(aq) The overall reaction is the oxidation of ethyl alcohol to acetic acid and water. When a suspected drunk driver blows 185 mL of his breath through this breathalyzer, the breathalyzer produces an average of 325 mA of current for 10 s.

Assuming a pressure of 1.0 atm and a temperature of 24 ∘C, what percent (by volume) of the driver's breath is ethanol?

Answer 1

The balanced equation for the breath analyser is CH,CH2OH(g) +o2 (g) - > HC2H3O2 (g) + H2O (g) The half reactions are At anode: CH3CH20h(g) + 4 OH-(aq) > HC2H302 (g) + 3 H2O (l) + 4e- > 40H -(aq) At cathode: The number of electrons transfered = 4 electrons per 1 mole of ethyl alcohol We know that charge -current x time 325 mA x 10 s 0.325 A x 10 s 3.25 Coulombs We know that 1 F (96485 Coulombs) of charge has1 mole of electrons So 3.25 Coulombs of charge would have 1* 3.25/96485 0.00003368 mol of electrons Number moles of alcohol 0.00003368 mol/4 0.000005496 mol 8.42x 10-6 mol Given that P 1.0 atm Volume of alcohol is calculated by Ideal gas relation PV nRT T = 24 °C 25 + 273-297 K nRT 8.42 x10-6 molx0.0821 L.atm.mol K-1x297K -2.05x10 L 1.0 atm Volume of ethanol-2.05 × 10-4 L-2.05x 10ㅢ × 1 000 mL-0.205 mL Volume of breath -185 mL 0.205 mL percent by volume - 185 mL x100 = 0.11081%-10.11 %!