[VISUAL] A student uses a galvanic cell to determine the concentration of ethanol, C2H5OH, in an aqu — Electrochemistry Chemistry Question
Problem Context
[VISUAL]
A student uses a galvanic cell to determine the concentration of ethanol, C2H5OH, in an aqueous solution. The cell is based on the half-cell reactions represented in the table above.
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An alternative approach to determine the concentration of C2H5OH(aq) in a solution is based on the reaction represented below.
3 C2H5OH(aq) + 2 Cr2O7 2−(aq) + 16 H+(aq) → 4 Cr3+(aq) + 3 CH3COOH(aq) + 11 H2O(l)
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A solution has an initial Cr2O7 2−(aq) concentration of 1.0 × 10−3 M and an initial C2H5OH(aq) concentration of 0.500 M. The solution contains enough strong acid to keep the pH essentially constant throughout the reaction. The student places a sample of the solution in a cuvette that has a path length of 0.50 cm and places it in a spectrophotometer set to measure absorbance at 440 nm. (Cr2O7 2−(aq) is the only species in the reaction mixture that absorbs light at this wavelength.) The absorbance of Cr2O7 2−(aq) in the solution is monitored as the reaction proceeds; the table below shows the absorbance as a function of time for the first trial.
[VISUAL]
Model Answer
C2H5OH(aq) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
1 point is earned for the correct reactants and products.
1 point is earned for balancing the equation.
Model Answer
E° = −(−0.085 V) + 1.229 V = +1.314 V
1 point is earned for a correct answer that is consistent with the equation in part (a).
Model Answer
I = q / t
q = I * t = 0.10 amp * 20. s = 2.0 C
1 point is earned for the correct charge. || 2.0 C * (1 mol e- / 96,485 C) * (1 mol C2H5OH / 12 mol e-) = 1.7 * 10^-6 mol C2H5OH
1.7 * 10^-6 mol C2H5OH / (10.0 mL * (1.0 L / 1000 mL)) = 1.7 * 10^-4 M
1 point is earned for the number of moles of C2H5OH.
1 point is earned for the initial molarity of C2H5OH.
Model Answer
Absorbance is proportional to [Cr2O7 2-].
(1.0 * 10^-3 M / 0.782) = (x / 0.553)
x = 7.1 * 10^-4 M
OR
Initial condition: A = abc
0.782 = a * (0.50 cm) * (1.0 * 10^-3 M)
a = 1564 cm^-1 M^-1
At 1.50 min:
0.553 = (1564 cm^-1 M^-1) * (0.50 cm) * c
c = 7.1 * 10^-4 M
1 point is earned for the correct concentration.
Model Answer
A = abc
If path length (b) is doubled, and molar absorptivity (a) is constant, the initial concentration of Cr2O7 2- (c) must be halved to keep the initial absorbance (A) constant.
1 point is earned for reference to a factor that affects absorbance.
1 point is earned for the correct adjustment.
Model Answer
Absorbance is proportional to concentration of Cr2O7 2-. Absorbance is halved after 3.00 min and again after another 3.00 min. Thus the half-life of the reaction is constant, so the reaction must be first order with respect to Cr2O7 2-.
OR
Demonstration that the rate of change in ln(A) over time is constant.
1 point is earned for a correct explanation that uses the data.