Exam 4A, Spring 2004

1. Complete and balance the following reactions:

a. Ca(OH)2(s) + C6H5COOH(aq)

b. Fe(NO3)3(aq) + Na2CO3(aq)

c. Cd2+(aq) + NH3(aq)

d. MnO4–(aq) + HClO3(aq) → Mn2+(aq) + HClO4(aq)

e. MnO4–(aq) + Mg(s) → MnO2(s) + Mg(OH)2(s)

Answer

a. Ca(OH)2(s) + 2 C6H5COOH(aq) → Ca(C6H5COO)2(s) + 2 H2O(l)

b. 2 Fe(NO3)3(aq) + 3 Na2CO3(aq) → Fe2(CO3)3(s) + 6 Na+(aq) + 6 NO3–(aq)

c. Cd2+(aq) + 4 NH3(aq) → ← [Cd(NH3)4]2+(aq)

d. MnO4–(aq) + HClO3(aq) → Mn2+(aq) + HClO4(aq)

Oxidation: HClO3(aq) + H2O(l) → HClO4(aq) + 2 H+(aq) + 2 e–

Reduction: MnO4–(aq) + 8 H+(aq) + 5 e– → Mn2+(aq) + 4 H2O(l)

Net: 2 MnO4–(aq) + 5 HClO3(aq) + 6 H+(aq) → 2 Mn2+(aq) + 5 HClO4(aq) + 3 H2O(l)

e. MnO4–(aq) + Mg(s) → MnO2(s) + Mg(OH)2(s)

Oxidation: Mg(s) + 2 H2O(l) → Mg(OH)2(s) + 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 Mg(s) + 2 H2O(l) + 2 H+(aq) → 2 MnO2(s) + 3 Mg(OH)2(s)

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

Net in base: 2 MnO4–(aq) + 3 Mg(s) + 4 H2O(l) → 2 MnO2(s) + 3 Mg(OH)2(s) + 2 OH–(aq)

2. Identify the oxidation number for:

a. As in H3AsO4

b. Fe in Fe2O3

Answer

a. As is +5 in H3AsO4

b. Fe is +3 in Fe2O3

3. Co(OH)2 is a sparingly soluble salt. Determine if the solubility will increase, decrease, or stay the same in the following solutions. Briefly (15 words or less) explain your answer.

a. 1 M HCl solution.

b. 1 M Co(NO3)2 solution.

c. 1 M NH3 solution.

Answer

a. The solubility of Co(OH)2 will increase in 1 M HCl solution because of the acid/base reaction between the strong acid and the hydroxide ion.

b. The solubility of Co(OH)2 will decrease in 1 M Co(NO3)2 solution because of the common Co2+ ion.

c. The solubility of Co(OH)2 will increase in 1 M NH3 solution because of the complex ion formation.

4. Consider the reaction of solid barium chloride with aqueous sodium hydroxide solution.

a. Write the reaction.

b. Find ΔH° at 25 °C.

c. Find ΔS° at 25 °C.

d. Find ΔG° at 25 °C.

Answer

a. BaCl2(s) + 2 NaOH(aq) → Ba(OH)2(s) + 2 Na+(aq) + 2 Cl–(aq)

b. ΔH° = [(–946.0) + 2(–240.1) + 2(–167.2)] – [(–858.1) + 2(–469.2)] = 35.9 kJ/mol

c. ΔS° = [(107) + 2(59.0) + 2(56.5)] – [(123.7) + 2(48.1)] = 118 J/mol·K

d. ΔG° = 35.9 – (298)(0.118) = 0.7 kJ/mol

5. Consider the cell:

Pt(s) | Ni(OH)3(s) | Ni(OH)2(s) | pH = 14 || ClO3–(aq) | Cl –(aq) | pH = 14 | Pt(s)

a. Write the oxidation half–reaction.

b. Write the reduction half–reaction.

c. Write the net reaction.

d. Find the cell potential under standard conditions.

Answer

a. Ni(OH)2(s) + OH–(aq) → Ni(OH)3(s) + e–

b. ClO3–(aq) + 3 H2O(l) + 6 e– → Cl –(aq) + 6 OH–(aq)

c. 6 Ni(OH)2(s) + ClO3–(aq) + 3 H2O(l) → 6 Ni(OH)3(s) + Cl –(aq)

d. E° = –0.48 + 0.622 = +0.14 V