Exam 4B, Spring 2018

1. Complete and balance the following in aqueous solution.

a. Ni(NO3)2(aq) + Cs2S(aq)

b. HNO3(aq) + Cu(OH)2(s)

c. ClO2(aq) → ClO(aq) + ClO4(aq)

d. BrO3(aq) + [Cr(OH)4](aq) → CrO42–(aq) + Br(aq)

Answer

a. Ni(NO3)2(aq) + Cs2S(aq) → NiS(s) + 2 Cs+(aq) + 2 NO3(aq)

b. 2 HNO3(aq) + Cu(OH)2(s) → 2 H2O(l) + Cu2+(aq) + 2 NO3(aq)

c. ClO2(aq) → ClO(aq) + ClO4(aq)

Oxidation: ClO2(aq) + 2 H2O(l) → ClO4(aq) + 4 H+(aq) + 4 e

Reduction: ClO2(aq) + 2 H+(aq) + 2 e → ClO(aq) + H2O(l)

Net: 3 ClO2(aq) → 2 ClO(aq) + ClO4(aq)

d. BrO3(aq) + [Cr(OH)4](aq) → CrO42–(aq) + Br(aq)

The hydroxide species indicate that the reaction is run in base.

Oxidation: [Cr(OH)4](aq) → CrO42–(aq) + 4 H+(aq) + 3 e

Reduction: BrO3(aq) + 6 H+(aq) + 6 e → Br(aq) + 3 H2O(l)

Net in acid: BrO3(aq) + 2 [Cr(OH)4](aq) → 2 CrO42–(aq) + Br(aq) + 3 H2O(l) + 2 H+(aq)

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

Net in base: BrO3(aq) + 2 [Cr(OH)4](aq) + 2 OH(aq) → 2 CrO42–(aq) + Br(aq) + 5 H2O(l)

2. Consider the reaction between solid calcium hydroxide and aqueous H3PO4.

a. Write the balanced reaction.

b. Find ΔH° for the reaction at 25 °C.

c. Find ΔS° for the reaction at 25 °C.

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

e. Find ΔG° for the reaction at 50 °C.

f. Does the spontaneity for the reaction increase or decrease as the temperature is raised?

Answer

a. 3 Ca(OH)2(s) + 2 H3PO4(aq) → Ca3(PO4)2(s) + 6 H2O(l)

b. ΔH° = [(–4121) + 6(–285.8)] – [3(–986.1) + 2(–1288.3)] = –301 kJ/mol

c. ΔS° = [(236) + 6(69.91)] – [3(83.39) + 2(158)] = 89 J/mol·K

d. ΔG° = –301 – (298)(0.089) = –328 kJ/mol

e. ΔG° = –301 – (323)(0.089) = –330. kJ/mol

f. Spontaneity increases as the temperature increases because ΔG° becomes more negative.

3. Give the oxidation numbers for each atom in (NH4)2[PbI4].

Answer

N: –3

H: +1

Pb: +2

I: –1

4. Consider the cell: Ag(s) | Ag+(aq) || O2(g), pH = 0 | Pt(s)

a. Write the oxidation half-reaction.

b. Write the reduction half-reaction.

c. Write the net reaction.

d. Find the cell potential at standard conditions and 25 °C.

e. Is this a voltaic cell or an electrolytic cell? Why?

f. Will the cell potential increase, decrease, or stay the same as the pH is increased? Why?

Answer

a. Ag(s) → Ag+(aq) + e

b. O2(g) + 4 H+(aq) + 4 e → 2 H2O(l)

c. 4 Ag(s) + O2(g) + 4 H+(aq) → 4 Ag+(aq) + 2 H2O(l)

d. E° = 1.229 + (–0.800) = +0.429 V

e. Voltaic, because the reaction is spontaneous, E° > 0.

f. H+ is a reactant and increasing the pH reduces the concentration of H+, so the reaction will shift towards reactants, which is the nonspontaneous direction, so the potential becomes more negative.