Exam 4B, Spring 1999
Exam 4B, Spring 19991. Predict the sign of ΔS° for the following reactions.
a. K(g) → K(s)
b. Cl2(g) + 2 H2O(l) → Cl–(aq) + HClO(aq) + H3O+(aq)
c. CuCO3(s) + 2 HNO3(aq) → CO2(g) + H2O(l) + Cu2+(aq) + 2 NO3–(aq)
a. K(g) → K(s)
ΔS° < 0 : gas phase reactant
b. Cl2(g) + 2 H2O(l) → Cl–(aq) + HClO(aq) + H3O+(aq)
ΔS° <. 0 : gas phase reactant with no gas phase products
c. CuCO3(s) + 2 HNO3(aq) → CO2(g) + H2O(l) + Cu2+(aq) + 2 NO3–(aq)
ΔS° > 0 : gas phase product
2. Complete and balance the following reactions:
a. HPO42–(aq) + NaOH(aq)
b. CaCl2(aq) + Na2SO4(aq)
c. CrO42–(aq) + Fe(s) → Cr3+(aq) + Fe2+(aq) pH = 1
d. MnO4–(aq) + NO2–(aq) → MnO2(s) + NO3–(aq) pH = 14
a. HPO42–(aq) + NaOH(aq)
HPO42–(aq) + NaOH(aq) → H2O(l) + Na+(aq) + PO43–(aq)
b. CaCl2(aq) + Na2SO4(aq)
CaCl2(aq) + Na2SO4(aq) → CaSO4(s) + 2 Na+(aq) + 2 Cl–(aq)
c. CrO42–(aq) + Fe(s) → Cr3+(aq) + Fe2+(aq) pH = 1
Oxidation: Fe(s) → Fe2+(aq) + 2 e–
Reduction: CrO42–(aq) + 8 H+(aq) + 3 e– → Cr3+(aq) + 4 H2O(l)
Net: 2 CrO42–(aq) + 3 Fe(s) + 16 H+(aq) → 2 Cr3+(aq) + 3 Fe2+(aq) + 8 H2O(l)
d. MnO4–(aq) + NO2–(aq) → MnO2(s) + NO3–(aq) pH = 14
Oxidation: NO2–(aq) + H2O(l) → NO3–(aq) + 2 H+(aq) + 2 e–
Reduction: MnO4–(aq) + 4 H+(aq) + 3 e– → MnO2(s) + 2 H2O(l)
Net in acid: 2 MnO4–(aq) + 3 NO2–(aq) + 2 H+(aq) → 2 MnO2(s) + 3 NO3–(aq) + H2O(l)
Neutralization: 2 H2O(l) → 2 H+(aq) + 2 OH–(aq)
Net in base: 2 MnO4–(aq) + 3 NO2–(aq) + H2O(l) → 2 MnO2(s) + 3 NO3–(aq) + 2 OH–(aq)
3. Consider the formation of the complex ion Cd(NH3)42+(aq).
a. Find ΔG° for the reaction at 25 °C.
b. The following cell is constructed: Cd(s)|Cd2+(aq)|| Cd(NH3)42+(aq)|NH3(aq)|Cd(s); what is E° for the cell?
a. Cd2+(aq) + 4 NH3(aq) → ← [Cd(NH3)4]2+(aq) Kf = 1.3×107
ΔG° = –RTln Kf = –(8.314)(298)ln(1.3×107) = –(8.314)(298)(16.38)= –40600 J/mol = –40.6 kJ/mol
b. Oxidation: Cd(s) → Cd2+(aq) + 2 e–
Reduction: Cd(NH3)42+(aq) + 2 e– → 4 NH3(aq) + Cd(s)
Net: [Cd(NH3)4]2+(aq) → ← Cd2+(aq) + 4 NH3(aq)
This is the reverse of the complex ion formation reaction, so ΔG° = +40600 J/mol
E° = –ΔG°/nF = –(40600)/(2)(96485) = –0.210 V
4. Consider the reaction: CaO(s) + HCl(aq) → H2O(l) + Ca2+(aq) + Cl–(aq).
a. Find ΔH° for the reaction.
b. Find ΔS° for the reaction.
c. Find ΔG° for the reaction at 25 °C.
d. Is the reaction spontaneous?
Balance the reaction: CaO(s) + 2 HCl(aq) → H2O(l) + Ca2+(aq) + 2 Cl–(aq)
a. ΔH° = [(–285.8) + (–542.96) + 2(–167.2)] – [(–635.1) + 2(–167.2)] = –193.7 kJ/mol
b. ΔS° = [(69.91) + (–55.2) + 2(56.5)] – [(39.75) + 2(56.48)] = –25.0 J/mol·K
c. ΔG° = –193.7 –(298)(–0.0250) = –186.3 kJ/mol
d. Is the reaction spontaneous? Yes, ΔG° < 0
5. The following cell was constructed: Zn(s) | Zn2+(aq) || I2(s) | I–(aq) | Pt.
a. Write the balanced reaction for the cell.
b. Find E° for the cell.
c. Find the thermodynamic equilibrium constant for the cell at 25 °C.
d. The concentration of ions in the anode half-cell are 0.10 M and the concentration of ions in the cathode half-cell are 0.050 M; find the cell potential at 25 °C.
a. Write the balanced reaction for the cell.
Oxidation: Zn(s) → Zn2+(aq) + 2 e– E°ox = +0.763 V
Reduction: I2(s) + 2 e– → 2 I–(aq) E°red = +0.535 V
Net: Zn(s) + I2(s) → Zn2+(aq) + 2 I–(aq)
b. E° = +0.763 + 0.535 = +1.298 V
c. E° = (RT)/(nF)ln Kc, so ln Kc = nFE°/RT = (2)(96485)(1.298)/(8.314)(298) = 101.
Kc = e101. = 7×1043
d. The concentration of ions in the anode half-cell are 0.10 M and the concentration of ions in the cathode half-cell are 0.050 M; find the cell potential at 25 °C.
E = E° – (RT/nF)ln Q, where Q = [Zn2+][I–]2
Q = [0.10][0.050]2 = 2.5×10–4
E = 1.298 – (8.314·298/2·96485)ln(2.5×10–4) = 1.298 – (0.0128)(–8.29) = 1.404 V