Exam 4A, Spring 1999

1. Predict the sign of ΔS° for the following reactions.

a. Na(s) → Na(g)

b. Cl₂(g) + 2 H₂O(l) → Cl^–(aq) + HClO(aq) + H₃O⁺(aq)

c. CO₂(g) + 3 H₂O(l) + Ca(NO₃)₂(aq) → CaCO₃(s) + 2 H₃O⁺(aq) + 2 NO₃^–(aq)

Answer

a. Na(s) → Na(g)

ΔS° > 0 : gas phase product

b. Cl₂(g) + 2 H₂O(l) → Cl^–(aq) + HClO(aq) + H₃O⁺(aq)

ΔS° <. 0 : gas phase reactant with no gas phase products

c. CO₂(g) + 3 H₂O(l) + Ca(NO₃)₂(aq) → CaCO₃(s) + 2 H₃O⁺(aq) + 2 NO₃^–(aq)

ΔS° < 0 : gas phase reactant with no gas phase products

2. Complete and balance the following reactions:

a. HCO₃^–(aq) + KOH(aq)

b. AgNO₃(aq) + NaCl(aq)

c. CrO₄^2–(aq) + Cu(s) → Cr³⁺(aq) + Cu²⁺(aq) pH = 1

d. Mn²⁺(aq) + H₂O₂(aq) → MnO₂(s) + H₂O(l) pH = 14

Answer

a. HCO₃^–(aq) + KOH(aq)

HCO₃^–(aq) + KOH(aq) → H₂O(l) + K⁺(aq) + CO₃^2–(aq)

b. AgNO₃(aq) + NaCl(aq)

AgNO₃(aq) + NaCl(aq) → AgCl(s) + Na⁺(aq) + NO₃^–(aq)

c. CrO₄^2–(aq) + Cu(s) → Cr³⁺(aq) + Cu²⁺(aq) pH = 1

Oxidation: Cu(s) → Cu²⁺(aq) + 2 e^–

Reduction: CrO₄^2–(aq) + 8 H⁺(aq) + 3 e^– → Cr³⁺(aq) + 4 H₂O(l)

Net: 2 CrO₄^2–(aq) + 3 Cu(s) + 16 H⁺(aq) → 2 Cr³⁺(aq) + 3 Cu²⁺(aq) + 8 H₂O(l)

d. Mn²⁺(aq) + H₂O₂(aq) → MnO₂(s) + H₂O(l) pH = 14

Oxidation: Mn²⁺(aq) + 2 H₂O(l) → MnO₂(s) + 4 H⁺(aq) + 2 e^–

Reduction: H₂O₂(aq) + 2 H⁺(aq) + 2 e^– → 2 H₂O(l)

Net in acid: Mn²⁺(aq) + H₂O₂(aq) → MnO₂(s) + 2 H⁺(aq)

Neutralization: 2 H⁺(aq) + 2 OH^–(aq) → 2 H₂O(l)

Net in base: Mn²⁺(aq) + H₂O₂(aq) + 2 OH^–(aq) → MnO₂(s) + 2 H₂O(l)

3. Consider the formation of the complex ion Ag(NH₃)₂⁺(aq).

a. Find ΔG° for the reaction at 25 °C.

b. The following cell is constructed: Ag(s)|Ag⁺(aq)|| Ag(NH₃)₂⁺(aq)|NH₃(aq)|Ag(s); what is E° for the cell?

Answer

a. Ag⁺(aq) + 2 NH₃(aq) → ← [Ag(NH₃)₂]⁺(aq) K_f = 1.6×10⁷

ΔG° = –RTln K_f = –(8.314)(298)ln(1.6×10⁷) = –(8.314)(298)(16.59)= –41100 J/mol = –41.1 kJ/mol

b. Oxidation: Ag(s) → Ag⁺(aq) + e^–

Reduction: Ag(NH₃)₂⁺(aq) + e^– → 2 NH₃(aq) + Ag(s)

Net: [Ag(NH₃)₂]⁺(aq) → ← Ag⁺(aq) + 2 NH₃(aq)

This is the reverse of the complex ion formation reaction, so ΔG° = +41100 J/mol

E° = –ΔG°/nF = –(41100)/(1)(96485) = –0.426 V

4. Consider the reaction: MgO(s) + HCl(aq) → H₂O(l) + Mg²⁺(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?

Answer

Balance the reaction: MgO(s) + 2 HCl(aq) → H₂O(l) + Mg²⁺(aq) + 2 Cl^–(aq)

a. ΔH° = [(–285.8) + (–466.9) + 2(–167.2)] – [(–601.7) + 2(–167.2)] = –151.0 kJ/mol

b. ΔS° = [(69.91) + (–138.1) + 2(56.5)] – [(26.94) + 2(56.48)] = –95.1 J/mol·K

c. ΔG° = –151.0 –(298)(–0.0951) = –122.7 kJ/mol

d. Is the reaction spontaneous? Yes, ΔG° < 0

5. The following cell was constructed: Sn(s)|Sn²⁺(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.

Answer

a. Write the balanced reaction for the cell.

Oxidation: Sn(s) → Sn²⁺(aq) + 2 e^– E°_ox = +0.137 V

Reduction: I₂(s) + 2 e^– → 2 I^–(aq) E°_red = +0.535 V

Net: Sn(s) + I₂(s) → Sn²⁺(aq) + 2 I^–(aq)

b. E° = +0.137 + 0.535 = +0.672 V

c. E° = (RT)/(nF)ln K_c, so ln K_c = nFE°/RT = (2)(96485)(0.672)/(8.314)(298) = 52.3

K_c = e^52.3 = 5×10²²

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 = [Sn²⁺][I^–]²

Q = [0.10][0.050]² = 2.5×10^–4

E = 0.672 – (8.314·298/2·96485)ln(2.5×10^–4) = 0.672 – (0.0128)(–8.29) = 0.778 V