Exam 4A, Spring 2006
Exam 4A, Spring 20061. Complete and balance the following reactions:
a. aqueous perchloric acid plus solid magnesium hydroxide
b. SrCl2(aq) + CO32–(aq)
c. HClO2(aq) → HClO(aq) + HClO4(aq)
d. Cu(s) + BrO3–(aq) → Cu(OH)2(s) + Br–(aq)
a. 2 HClO4(aq) + Mg(OH)2(s) → Mg2+(aq) + 2 ClO4–(aq) + 2 H2O(l)
b. SrCl2(aq) + 2 CO32–(aq) → SrCO3(s) + 2 Cl–(aq)
c. HClO2(aq) → HClO(aq) + HClO4(aq)
Oxidation: HClO2(aq) + 2 H2O(l) → HClO4(aq) + 4 H+(aq) + 4 e–
Reduction: HClO2(aq) + 2 H+(aq) + 2 e– → HClO(aq) + H2O(l)
Net: 3 HClO2(aq) → 2 HClO(aq) + HClO4(aq)
d. Cu(s) + BrO3–(aq) → Cu(OH)2(s) + Br–(aq)
Oxidation: Cu(s) + 2 H2O(l) → Cu(OH)2(s) + 2 H+(aq + 2 e–
Reduction: BrO3–(aq) + 6 H+(aq) + 6 e– → Br–(aq) + 3 H2O(l)
Net in acid: 3 Cu(s) + BrO3–(aq) + 3 H2O(l) → 3 Cu(OH)2(s) + Br–(aq)
Neutralization: None
Net in base: 3 Cu(s) + BrO3–(aq) + 3 H2O(l) → 3 Cu(OH)2(s) + Br–(aq)
2. Metal oxides react with acids similarly to metal hydroxides, i.e. the oxide ion acts as a hydrogen ion acceptor (twice) to form water as a product. Consider the following at 25° C.
a. Complete and balance: BaO(s) + HNO3(aq)
b. Calculate ΔH° for the reaction in part a.
c. Calculate ΔS° for the reaction in part a.
d. Calculate ΔG° for the reaction in part a.
e. Predict the spontaneity of the reaction in part a. Briefly explain your reasoning.
f. Assuming ΔH° and ΔS° were constant at all temperatures (a poor assumption, of course), would you expect the spontaneity to be the same for the reaction in part a at all temperatures? Briefly explain your reasoning.
a. BaO(s) + 2 HNO3(aq) → Ba2+(aq) + 2 NO3–(aq) + H2O(l)
b. ΔH° = [(–537.6) + 2(–205.0) + (–285.8)] – [(–548.1) + 2(–207.4)] = –270.5 kJ/mol
c. ΔS° = [(9.6) + 2(146.4) + (69.91)] – [(72.09) + 2(146.4)] = 7.4 J/mol·K
d. ΔG° = (–270.5) – (298)(0.0074) = –272.7 kJ/mol
e. Spontaneous because ΔG° < 0.
f. Spontaneous at all temperatures because ΔG° will be negative at all temperatures..
3. Consider the following electrochemical cell at standard conditions and 25° C.
Pt(s) | O3(g) | O2(g) | H+(aq) || SO2(g) | SO42–(aq) | H+(aq) | Pt(s)
a. Write the reduction half–reaction and give the standard reduction potential.
b. Write the oxidation half–reaction and give the standard oxidation potential.
c. Write the net reaction and give the standard potential.
d. Is the reaction spontaneous or nonspontaneous? Explain (very briefly).
e. Consider the cell if it were used under atmospheric conditions rather than standard conditions so that the partial pressure of oxygen gas were 0.2 atm rather than 1.0 atm. Assuming all other components remained at standard conditions, would the cell potential, increase, decrease, or remain the same? Explain (very briefly).
f. Ozone (O3) is a form of the element oxygen but not in its standard state. What is the oxidation number of O in ozone?
a. SO42–(aq) + 4 H+(aq) + 2 e– → SO2(g) + 2 H2O(l) E°red = +0.17 V
b. O2(g) + H2O(l) → O3(g) + 2 H+(aq) + 2 e– E°ox = –2.075 V
c. SO42–(aq) + O2(g) + 2 H+(aq) → SO2(g) + O3(g) + H2O(l) E° = 0.17 – 2.075 = –1.91 V
d. Nonspontaneous, E° < 0.
e. Reducing the partial pressure of oxygen removes reactants, so the reaction would shift towards reactants, which is the spontaneous direction. Thus, the cell potential would become more positive.
f. The total charge on ozone is zero, which is equally distributed over all three atoms. Hence, the oxidation number for each O atom is 0.