Exam 4B, Spring 2000
Exam 4B, Spring 20001. Complete and balance the following equations. Give the oxidation numbers for all of the atoms in both products and reactants except for hydrogen and oxygen. Predict the sign of ΔS° for each reaction.
a. H3PO4(aq) + N2H4(aq) → PH3(g) + N2(g)
b. Mn(s) + MnO4–(aq) → MnO2(s) + Mn(OH)2(s)
a. H3PO4(aq) + N2H4(aq) → PH3(g) + N2(g)
Oxidation: N2H4(aq) → N2(g) + 4 H+(aq) + 4 e–
Reduction: H3PO4(aq) + 8 H+(aq) + 8 e– → PH3(g) + 4 H2O(l)
Net: H3PO4(aq) + 2 N2H4(aq) → PH3(g) + 2 N2(g) + 4 H2O(l)
Oxidation Numbers: H3PO4, P is +5; PH3, P is –3; N2H4, N is –2; N2, N is 0.
ΔS° is > 0 because there are more moles of gas phase products.
b. Mn(s) + MnO4–(aq) → MnO2(s) + Mn(OH)2(s)
Mn(OH)2 indicates that the reaction is run in base.
Oxidation: Mn(s) + 2 H2O(l) → MnO2(s) + 4 H+(aq) + 4 e–
Reduction: MnO4–(aq) + 6 H+(aq) + 5 e– → Mn(OH)2(s) + 2 H2O(l)
Net in acid: 5 Mn(s) + 4 MnO4–(aq) + 2 H2O(l) + 4 H+(aq) → 5 MnO2(s) + 4 Mn(OH)2(s)
Neutralization: 4 H2O(l) → 4 H+(aq) + 4 OH–(aq)
Net in base: 5 Mn(s) + 4 MnO4–(aq) + 6 H2O(l) → 5 MnO2(s) + 4 Mn(OH)2(s) + 4 OH–(aq)
OR
Oxidation: Mn(s) + 2 H2O(l) → Mn(OH)2(s) + 2 H+(aq) + 2 e–
Reduction: MnO4–(aq) + 4 H+(aq) + 3 e– → MnO2(s) + 2 H2O(l)
Net in acid: 3 Mn(s) + 2 MnO4–(aq) + 2 H2O(l) + 2 H+(aq) → 3 MnO2(s) + 2 Mn(OH)2(s)
Neutralization: 2 H2O(l) → 2 H+(aq) + 2 OH–(aq)
Net in base: 3 Mn(s) + 2 MnO4–(aq) + 4 H2O(l) → 3 MnO2(s) + 2 Mn(OH)2(s) + 2 OH–(aq)
Either way, the oxidation numbers are: Mn, Mn is 0; MnO4–, Mn is +7; MnO2, Mn is +4; Mn(OH)2, Mn is +2.
Either way, ΔS° < 0 because the Mn products are solids.
2. Consider the reaction between solid magnesium chloride and aqueous carbonic acid.
a. Write the balanced reaction.
b. Find ΔH° for the reaction in units of kJ.
c. Find ΔS° for the reaction in units of J/K.
d. Find ΔG° for the reaction at 25 °C in units of kJ.
e. Find ΔG° for the reaction at 75 °C in units of kJ.
a. MgCl2(s) + H2CO3(aq) + 2 H2O(l) → MgCO3(s) + 2 H3O+(aq) + 2 Cl–(aq)
b. ΔH° = [(–1096) + 2(–285.8) + 2(–167.2)] – [(–641.3) + (–698.7) + 2(–285.8)] = –90. kJ/mol
c. ΔS° = [(65.7) + 2(69.91) + 2(56.5)] – [(89.62) + (191) + 2(69.91)] = –102. J/mol·K
d. 25 °C = 298 K, so ΔG° = –90. – (298)(–0.102) = –60. kJ/mol.
e. 75 °C = 348 K, so ΔG° = –90. – (348)(–0.102) = –55 kJ/mol.
3. Estimate Ka1 for carbonic acid using thermochemical data.
H2CO3(aq) + H2O(l) → ← H3O+(aq) + HCO3–(aq)
ΔH° = [(–285.8) + (–691.11)] – [(–698.7) + (–285.8)] = 7.6 kJ/mol
ΔS° = [(69.91) + (95.0)] – [(191) + (69.91)] = –96 J/mol·K
ΔG° (at 25 °C) = 7.6 – (298)(–0.096) = 36.2 kJ/mol = 36200 J/mol
ΔG° = –RTln Ka1, so Ka1 = e(–ΔG°/RT) = e–36200/(8.314×298) = 4.5×10–7
This compares to Ka1 = 4.4×10–7 from the Table of Acid Ionization Constants
4. Estimate the cell potential at standard condition and 25 °C for the cell
Pt(s) | H3AsO3(aq) | H2AsO4–(aq) | H+(aq) || H3AsO4(aq) | H3AsO3(aq) | H+(aq) | Pt(s)
Oxidation: H3AsO3(aq) + H2O(l) → H2AsO4–(aq) + 3 H+(aq) + 2 e–
Reduction: H3AsO4(aq) + 2 H+(aq) + 2 e– → H3AsO3(aq) + H2O(l)
Net: H3AsO4(aq) → H+(aq) + H2AsO4–(aq)
This is the first acid ionization of H3AsO4, for which Ka1 = 6.0×10–3
Thus, E° = (RT/nF)ln Ka1 = (8.314×298/2×96485)ln(6.0×10–3) = –0.066 V
5. Consider the cell: Au(s) | Au3+(aq) || Fe2+(aq) | Fe(s)
a. What is the balanced oxidation half-reaction?
b. What is the balanced reduction half-reaction?
c. What is the balanced net reaction?
d. What is the standard potential for the cell?
e. Is the reaction spontaneous or nonspontaneous? Why?
a. Au(s) → Au3+(aq) + 3 e–
b. Fe2+(aq) + 2 e– → Fe(s)
c. 2 Au(s) + 3 Fe2+(aq) → 2 Au3+(aq) + 3 Fe(s)
d. E° = E°ox + E°red = –1.52 + –0.440 = –1.96 V
e. Nonspontaneous because E° < 0.