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 186 mL of his breath through this breathalyzer, the breathalyzer produces an average of 320 mA of current for 10 s.

Required:
Assuming a pressure of 1.0 atm and a temperature of 26C, what percent (by volume) of the driver's breath is ethanol?

Answer 1

Answer:

The percent of ethanol is 0.1093%

Explanation:

Given:

t = time = 10 s

I = current = 320 mA

F = Faraday's constant = 96485.3365 C mol⁻¹

n = number of electrons = 4

Molecular weight of ethanol = 46 g/mol

Question: What percent (by volume) of the driver's breath is ethanol, %E = ?

First, it is necessary to calculate the mass of ethanol:

$W=\frac{\frac{46}{4} *0.32*10}{96485.3365} =3.814x10^{-4} g$

The moles of ethanol:

$n_{ethanol} =3.814x10^{-4} g*\frac{1mol}{46} =8.291x10^{-6} moles$

Applying the equation of ideal gas:

$V=\frac{nRT}{P}$

Here:

T = 26°C = 299 K

P = 1 atm

Substituting values:

$V=\frac{8.291x10^{-6}*0.082*299 }{1} =2.033x10^{-4} L=0.2033mL$

The percent of ethanol:

$E=\frac{0.2033}{186} *100=0.1093$%