Exam 4A, Spring 2005
Exam 4A, Spring 20051. Complete and balance the following reactions:
a. Mn(OH)2(s) + H3PO4(aq)
b. Na2CO3(aq) + Fe(NO3)3(aq)
c. As2S3(s) + H2O2(aq) → H3AsO4(aq) + SO42–(aq)
d. XeF6(s) + OH–(aq) → O2(g) + XeO64–(aq) + F–(aq) + Xe(g)
e. The disproportionation of chlorate ion into chlorite ion and perchlorate ion in basic solution.
a. 3 Mn(OH)2(s) + 2 H3PO4(aq) → Mn3(PO4)2(s) + 6 H2O(l)
b. 3 Na2CO3(aq) + 2 Fe(NO3)3(aq) → Fe2(CO3)3(s) + 6 Na+(aq) + 6 NO3–(aq)
c. As2S3(s) + H2O2(aq) → H3AsO4(aq) + SO42–(aq)
Oxidation: As2S3(s) + 20 H2O(l) → 2 H3AsO4(aq) + 3 SO42–(aq) + 34 H+(aq) + 28 e–
Reduction: H2O2(aq) + 2 H+(aq) + 2 e– → 2 H2O(l)
Net: As2S3(s) + 14 H2O2(aq) → 2 H3AsO4(aq) + 3 SO42–(aq) + 8 H2O(l) + 6 H+(aq)
d. XeF6(s) + OH–(aq) → O2(g) + XeO64–(aq) + F–(aq) + Xe(g)
Oxidation: OH–(aq) + H2O(l) → O2(g) + 3 H+(aq) + 4 e–
Reduction: 2 XeF6(s) + 6 H2O(l) + 4 e– → XeO64–(aq) + 12 F–(aq) + Xe(g) + 12 H+(aq)
Net in acid: 2 XeF6(s) + OH–(aq) + 7 H2O(l) → O2(g) + XeO64–(aq) + 12 F–(aq) + Xe(g) + 15 H+(aq)
Neutralization: 15 H+(aq) + 15 OH–(aq) → 15 H2O(l)
Net in base: 2 XeF6(s) + 16 OH–(aq) → O2(g) + XeO64–(aq) + 12 F–(aq) + Xe(g) + 8 H2O(l)
e. The disproportionation of chlorate ion into chlorite ion and perchlorate ion in basic solution.
Oxidation: ClO3–(aq) + H2O(l) → ClO4–(aq) + 2 H+(aq) + 2 e–
Reduction: ClO3–(aq) + 2 H+(aq) + 2 e– → ClO2–(aq) + H2O(l)
Net in acid: 2 ClO3–(aq) → ClO4–(aq) + ClO2–(aq)
Neutralization: None
Net in base: 2 ClO3–(aq) → ClO4–(aq) + ClO2–(aq)
2. Identify the oxidation number of:
a. V in (VO)(SO4)
b. Fe in Fe3O4
a. V in (VO)(SO4) has an oxidation number of +4
b. Fe in Fe3O4 has an oxidation number of +8/3 (two Fe are +3 and one Fe is +2)
3. Can oxygen gas be used to oxidize bromide ion in acidic solution? Show chemical equations to justify your answer.
Proposed oxidation: 2 Br–(aq) → Br2(l) + 2 e– E° = –1.065 V
Proposed reduction: O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l) E° = +1.229 V
Net proposed reaction: O2(g) + 4 Br–(aq) + 4 H+(aq) → 2 Br2(l) + 2 H2O(l)
E° = +1.229 – 1.065 = +0.164 V > 0, so the reaction is spontaneous.
4. Consider the electrochemical cell shown below.
Pt | Br–(aq) | BrO3–(aq) | pH = 14 || O2(g) | pH = 14 | Pt
a. Write the oxidation half-reaction.
b. Write the reduction half-reaction.
c. Write the net reaction.
d. Calculate the cell potential at standard conditions.
e. Is the reaction spontaneous, nonspontaneous, or at equilibrium? Why?
f. If you were to calculate the thermodynamic equilibrium constant for the reaction, what value would you use for n?
g. Write the expression for the reaction quotient for the cell.
h. If the pH were to be changed to 10 in each half-cell, would the standard potential increase, decrease, or stay the same? Why?
a. Br–(aq) + 6 OH–(aq) → BrO3–(aq) + 3 H2O(l) + 6 e–
b. O2(g) + 2 H2O(l) + 4 e– → 4 OH–(aq)
c. 2 Br–(aq) + 3 O2(g) → 2 BrO3–(aq)
d. E° = +0.401 – 0.584 = –0.183 V
e. Nonspontaneous, E° < 0.
f. n = 12
g. Q = [BrO3–]2 /( PO23[Br–]2 )
h. Neither hydroxide ion nor hydronium ion is present in the mass action expression so changing the pH would have no effect on the cell potential.