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Ch.19 - Electrochemistry
All textbooksTro 4th EditionCh.19 - ElectrochemistryProblem 118
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Chapter 19, Problem 118

When a suspected drunk driver blows 188 mL of their breath through the fuel-cell breathalyzer described in Section 19.7, the breathalyzer produces an average of 324 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?

Verified step by step guidance
1
Convert the volume of breath from mL to L by dividing by 1000.
Use the ideal gas law, PV = nRT, to calculate the number of moles of gas in the breath sample. Use P = 1.0 atm, V = volume in L, R = 0.0821 L·atm/mol·K, and T = 298 K (25 °C).
Calculate the total charge (Q) passed through the breathalyzer using the formula Q = I × t, where I is the current in amperes (324 mA converted to A) and t is the time in seconds (10 s).
Determine the number of moles of electrons transferred using the relationship between charge and moles of electrons: moles of electrons = Q / F, where F is Faraday's constant (96485 C/mol).
Assuming complete oxidation of ethanol (C2H5OH) to acetic acid (CH3COOH), use the stoichiometry of the reaction to find the moles of ethanol. Then, calculate the volume percent of ethanol in the breath by dividing the moles of ethanol by the total moles of gas and multiplying by 100.
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