CHM112 Home Page Reference Materials Homework Assignments Lectures Exams & Quizzes Grades Study Aids URI Home Page


J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 18.31

Write balanced equations for (a) the reaction of oxalic acid (HOOCCOOH) and permanganate ion in acid solution to produce manganese(II) ion and carbon dioxide gas, (b) the reaction of Cr2O72– and UO2+ to produce UO22+ and Cr3+ in an acidic aqueous environment, and (c) the reaction in basic solution of nitrate ion and zinc to produce zinc(II) ion and gaseous ammonia.




Answer:

(a) The reaction described is: MnO4(aq) + H2C2O4(aq) Mn2+(aq) + CO2(g) in acid.

Oxidation

H2C2O4(aq) 2 CO2(g)

H2C2O4(aq) 2 CO2(g) + 2 H+(aq)

H2C2O4(aq) 2 CO2(g) + 2 H+(aq) + 2 e

Reduction

MnO4(aq) Mn2+(aq) + 4 H2O(l)

MnO4(aq) + 8 H+(aq) Mn2+(aq) + 4 H2O(l)

MnO4(aq) + 8 H+(aq) + 5 eMn2+(aq) + 4 H2O(l)

Equalizing the number of electrons

[H2C2O4(aq) 2 CO2(g) + 2 H+(aq) + 2 e] × 5

[MnO4(aq) + 8 H+(aq) + 5 e Mn2+(aq) + 4 H2O(l)] × 2

Combining and eliminating species common to both sides

2 MnO4(aq) + 5 H2C2O4(aq) + 6 H+(aq) 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l)



(b) The reaction described is: Cr2O72–(aq) + UO2+(aq) Cr3+(aq) + UO22+(aq) in acid.

Oxidation

UO2+(aq) + H2O(l)UO22+(aq)

UO2+(aq) + H2O(l)UO22+(aq) + 2 H+(aq)

UO2+(aq) + H2O(l)UO22+(aq) + 2 H+(aq) + 2 e

Reduction

Cr2O72–(aq) 2 Cr3+(aq)

Cr2O72–(aq) 2 Cr3+(aq) + 7 H2O(l)

Cr2O72–(aq) + 14 H+(aq) 2 Cr3+(aq) + 7 H2O(l)

Cr2O72–(aq) + 14 H+(aq) + 6 e2 Cr3+(aq) + 7 H2O(l)

Equalizing the number of electrons

[UO2+(aq) + H2O(l) UO22+(aq) + 2 H+(aq) + 2 e] × 3

Cr2O72–(aq) + 14 H+(aq) + 6 e 2 Cr3+(aq) + 7 H2O(l)

Combining and eliminating species common to both sides

Cr2O72–(aq) + 3 UO2+(aq) + 8 H+(aq) 2 Cr3+(aq) + 3 UO22+(aq) + 4 H2O(l)



(c) The reaction described is: NO3(aq) + Zn(s) Zn2+(aq) + NH3(g) in base.

The reaction cannot occur as described, however. (We'll fix it at the end.)

Oxidation

Zn(s) Zn2+(aq) + 2 e

Reduction

NO3(aq) NH3(g) + 3 H2O(l)

NO3(aq) + 9 H+(aq) NH3(g) + 3 H2O(l)

NO3(aq) + 9 H+(aq) + 8 eNH3(g) + 3 H2O(l)

Equalizing the number of electrons

[Zn(s) Zn2+(aq) + 2 e] × 4

NO3(aq) + 9 H+(aq) + 8 e NH3(g) + 3 H2O(l)

Combining and eliminating species common to both sides, giving the net reaction in acid

NO3(aq) + 4 Zn(s) + 9 H+(aq) 4 Zn2+(aq) + NH3(g) + 3 H2O(l)

Use an acid/base reaction to eliminate hydrogen ions

9 H2O(l) 9 H+(aq) + 9 OH(aq)

Adding the last two reactions and eliminating species common to both sides

NO3(aq) + 4 Zn(s) + 6 H2O(l) 4 Zn2+(aq) + NH3(g) + 9 OH(aq)

However, Zn2+ is not stable in this environment. Depending upon the concentration of hydroxide present, one of two further reactions must occur.

precipitation: Zn2+(aq) + 2 OH(aq) Zn(OH)2(s)

complex ion formation: Zn2+(aq) + 4 OH(aq) Zn(OH)42–(aq)

With the given stoichiometry, there is not enough hydroxide present to form the complex ion, so the most likely reaction is precipitation:

4 Zn2+(aq) + 8 OH(aq) 4 Zn(OH)2(s)

Combining:

NO3(aq) + 4 Zn(s) + 6 H2O(l) 4 Zn(OH)2(s) + NH3(g) + OH(aq)