Ethanol is dissolved in blood and distributed to organs in the body. As a volatile compound, ethanol — Physical Chemistry — Kinetics Chemistry Question
Breath analysis
Ethanol is dissolved in blood and distributed to organs in the body. As a volatile compound, ethanol can be vaporized quite easily. In lungs, ethanol can change its phase from liquid to gaseous and, hence, it can be exhaled with air. Since the concentration of alcohol vapor in lungs is directly related to its concentration in blood, blood alcohol concentration can be measured using a device called a breathalyzer. In one of the older versions of breathalyzer, a suspect breathes into the device and exhaled air is allowed to pass through a solution of potassium dichromate which oxidizes ethanol to acetic acid. This oxidation is accompanied by a color change from orange to green and a detector records the change in intensity, hence, the change in color, which is used to calculate the percentage of alcohol in breath. When the oxidation of alcohol by potassium dichromate is carried out in an electrochemical cell, either the electrical current generated by this reaction or the change in the electromotive force can be measured and used for the estimation of alcohol content of blood.
Write a balanced equation for the oxidation of ethanol by the dichromate ion in acidic solution.
Model Answer
3 CH3CH2OH(g) + 2 Cr2O72-(aq) + 16 H+(aq) → 3 CH3COOH(l) + 4 Cr3+(aq) + 11 H2O(l)
If the standard potential for the reduction of Cr2O72– to Cr3+ is 1.330 V and reduction of acetic acid to ethanol is 0.058 V, calculate the standard electromotive force E° for the overall reaction and show that overall reaction is spontaneous at 25 °C and 1.0 bar.
Model Answer
E° = 1.330 – 0.058 = 1.272 V
Since E° > 0, the cell reaction is spontaneous under standard conditions.
In a breathalyzer which uses oxidation of ethanol, the volume of solution is 10.0 cm3. When a suspect breathes into the device, 0.10 A of current is recorded for 60 s. Calculate the mass of alcohol per volume of exhaled breath.
Model Answer
Q = I . t = 0.1 A × 60 s = 6.0 A s = 6.0 C
According to the balanced reaction equation, 6 mol e- produces 2 mol Cr3+.
Therefore: n(Cr3+) = (6.0 C / (6 × 96485 C)) × 2 mol = 2.07 ⋅ 10-5 mol Cr3+
4 mol Cr3+ ⇒ 3 mol CH3CH2OH
n(alcohol per volume of exhaled breath) = 1.55 ⋅ 10–5 mol
m(alcohol per volume of exhaled breath) = 1.55 ⋅ 10–5 × 46.0 g mol–1 = 7.15 ⋅ 10−4 g
In calculating the alcohol content in blood from the amount of alcohol in a breath, the “2100:1 partition ratio” needs to be considered. The ratio states that 2100 cm3 of expired air (breath) contains the same amount of ethanol as 1 cm3 of blood. Alternatively, each milliliter of blood has 2100 times the amount of ethanol as each milliliter of expired air. If the volume of expired air described in part 12.3 is 60.0 cm3, calculate the amount of alcohol per cm3 of blood.
Model Answer
7.15 ⋅ 10−4 g of alcohol in 60 cm3 expired air
1.19 ⋅ 10–5 g/ cm3 expired air
1.19 ⋅ 10–5 g/ cm3 expired air × 2100 = 0.025 g alcohol in 1 cm3 blood.
Cr3+ precipitates in basic solution as Cr(OH)3. The solubility product of chromium(III) hydroxide is 6.3 ⋅ 10-31 at 25 °C. Calculate the standard potential for the reduction of Cr(OH)3 to Cr. Standard potential for the reduction of Cr3+ to Cr is –0.74 V.
Model Answer
Cell reaction should be: Cr(OH)3(s) → Cr3+(aq) + OH–(aq)
Suitable cell for this reaction: Cr(s)Cr3+(aq) OH–(aq) Cr(OH)3(s) Cr(s)
∆G° = -3 × 96485 × E°cell = -8.314 × 298 ln Ks
E°cell = -(8.314 × 298) / (3 × 96485) ln(6.31 × 10-31) = 0.595 V
0.595 = E°cathode - (-0.74) = E°cathode + 0.74
E°cathode = -1.34 V