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

Problem 18.29

Use the half–reaction method to balance the following equations in basic solution.

(a) Fe(OH)2(s) + O2(g)Fe(OH)3(s)

(b) S8(s)S2O32–(aq) + S2–(aq)

(c) CrI3(s) + H2O2(aq)CrO42–(aq) + IO4(aq)






Answer:

(a) Fe(OH)2(s) + O2(g)Fe(OH)3(s)

Oxidation

Fe(OH)2(s) + H2O(l)Fe(OH)3(s)

Fe(OH)2(s) + H2O(l) Fe(OH)3(s) + H+(aq)

Fe(OH)2(s) + H2O(l) Fe(OH)3(s) + H+(aq) + e

Reduction

Since this reaction is run in basic solution, the reasonable reduction product from oxygen gas is hydroxide:

O2(g)OH(aq)

O2(g)OH(aq) + H2O(l)

O2(g) + 3 H+(aq)OH(aq) + H2O(l)

O2(g) + 3 H+(aq)+ 4 eOH(aq) + H2O(l)

Combine the half-reactions using an equal number of electrons

[Fe(OH)2(s) + H2O(l)Fe(OH)3(s) + H+(aq) + e] × 4

O2(g) + 3 H+(aq)+ 4 eOH(aq) + H2O(l)

Net reaction in acid

4 Fe(OH)2(s) + O2(g) + 3 H2O(l)4 Fe(OH)3(s) + OH(aq) + H+(aq)

Eliminate the hydrogen ion with an acid/base reaction

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

Combine the last two reactions and eliminate common species

4 Fe(OH)2(s) + O2(g) + 2 H2O(l)4 Fe(OH)3(s)



(b) S8(s)S2O32–(aq) + S2–(aq)

Oxidation

S8(s)4 S2O32–(aq)

S8(s) + 12 H2O(l)4 S2O32–(aq)

S8(s) + 12 H2O(l)4 S2O32–(aq) + 24 H+(aq)

S8(s) + 12 H2O(l)4 S2O32–(aq) + 24 H+(aq) + 16 e

Reduction

S8(s)8 S2–(aq)

S8(s) + 16 e8 S2–(aq)

Both half–reactions have the same number of electrons, so combining gives the net reaction in acid

2 S8(s) + 12 H2O(l)4 S2O32–(aq) + 8 S2–(aq) + 24 H+(aq)

Eliminate the hydrogen ion with an acid/base reaction

24 OH(aq) + 24 H+(aq)24 H2O(l)

Combine the last two reactions and eliminate common species

2 S8(s) + 24 OH(aq)4 S2O32–(aq) + 8 S2–(aq) + 12 H2O(l)


This is an example of a disproportionation reaction, where a single species (in this case, S8) is both oxidized and reduced.



(c) CrI3(s) + H2O2(aq)CrO42–(aq) + IO4(aq)

Oxidation

CrI3(s)CrO42–(aq) + 3 IO4(aq)

CrI3(s) + 16 H2O(l)CrO42–(aq) + 3 IO4(aq)

CrI3(s) + 16 H2O(l)CrO42–(aq) + 3 IO4(aq) + 32 H+(aq)

CrI3(s) + 16 H2O(l)CrO42–(aq) + 3 IO4(aq) + 32 H+(aq) + 27 e

Reduction

H2O2(aq)2 H2O(l)

H2O2(aq) + 2 H+(aq)2 H2O(l)

H2O2(aq) + 2 H+(aq) + 2 e2 H2O(l)

Combine the half–reactions using an equal number of electrons

[CrI3(s) + 16 H2O(l)CrO42–(aq) + 3 IO4(aq) + 32 H+(aq) + 27 e] × 2

[H2O2(aq) + 2 H+(aq) + 2 e 2 H2O(l)] × 27

Net reaction in acid, after eliminating species common to both sides

2 CrI3(s) + 27 H2O2(aq)2 CrO42–(aq) + 6 IO4(aq) + 10 H+(aq) + 22 H2O(l)

Eliminate the hydrogen ion with an acid/base reaction

10 OH(aq) + 10 H+(aq)10 H2O(l)

Combine the last two reactions and eliminate common species

2 CrI3(s) + 27 H2O2(aq) + 10 OH(aq)2 CrO42–(aq) + 6 IO4(aq) + 32 H2O(l)