Exam 4A, Spring 2018
Exam 4A, Spring 20181. Complete and balance the following in aqueous solution.
a. Fe(NO3)2(aq) + Rb2S(aq)
b. HNO3(aq) + Zn(OH)2(s)
c. ClO2–(aq) → ClO–(aq) + ClO4–(aq)
d. BrO3–(aq) + [Cr(OH)4]–(aq) → CrO42–(aq) + Br–(aq)
a. Fe(NO3)2(aq) + Rb2S(aq) → FeS(s) + 2 Rb+(aq) + 2 NO3–(aq)
b. 2 HNO3(aq) + Zn(OH)2(s) → 2 H2O(l) + Zn2+(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?
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[HgCl4].
N: –3
H: +1
Hg: +2
Cl: –1
4. Consider the cell: Pt(s) | Ce4+(aq) | Ce3+(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?
a. Ce3+(aq) → Ce4+(aq) + e–
b. O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
c. 4 Ce3+(aq) + O2(g) + 4 H+(aq) → 4 Ce4+(aq) + 2 H2O(l)
d. E° = 1.229 + (–1.44) = –0.21 V
e. Electrolytic, because the reaction is nonspontaneous, 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.