One type of breathalyzer employs a fuel cell to measure the quantity of alcohol

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)+4e4OH(aq)
The overall reaction is the oxidation of ethyl alcohol to acetic acid and water. When a suspected drunk driver blows 191 mL of his breath through this breathalyzer, the breathalyzer produces an average of 323 mA of current for 10 s.

Assuming a pressure of 1.0 atm and a temperature of 25 C, what percent (by volume) of the driver's breath is ethanol?

Explanation / Answer

V = 191 mL of breath, I = 323 mA, t = 10 s

Total charge used = I*t = (323*10^-3)(10) = 3.23 Couloumbs

mol of e- = 3.23 /96500 = 0.00003347 mol of e- flowed

from the reaction

CH3CH2OH(g)+4OH(aq)HC2H3O2(g)+3H2O(l)+4e

4 mol of e- = 1 mol of alcohol

0.00003347 mol of e- = 0.00003347 /4 = 0.0000083675 mol of alcohol

So

at P1 = 1 atm, T = 25°C

PV = nRT

V = nRT/P = (0.0000083675)(0.082)(25+273)/(1) = 0.000204468 Liters = 0.000204468*10^ 3 = 0.204468 mL of ethanol

% = 0.204468 / 191 * 100 = 0.107 %

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