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

Problem 18.30

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

(a) CrO42–(aq) + AsH3(g) Cr(OH)3(s) + As(s)

(b) CH3OH(aq) + MnO4(aq) HCOO(aq) + MnO2(s)

(c) [Fe(CN)6]3–(aq) + N2H4(aq) [Fe(CN)6]4–(aq) + N2(g)






Answer:


(a) CrO42–(aq) + AsH3(g)Cr(OH)3(s) + As(s)

Oxidation:

AsH3(g)As(s) + 3 H+(aq) + 3 e

Reduction:

CrO42–(aq) + 5 H+(aq) + 3 eCr(OH)3(s) + H2O(l)

Net (in acid):

CrO42–(aq) + AsH3(g) + 2 H+(aq)Cr(OH)3(s) + As(s) + H2O(l)

Neutralization:

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

Net (in base):

CrO42–(aq) + AsH3(g) + H2O(l)Cr(OH)3(s) + As(s) + 2 OH(aq)



(b) CH3OH(aq) + MnO4(aq)HCOO(aq) + MnO2(s)

Oxidation:

[ CH3OH(aq) + H2O(l)HCOO(aq) + 5 H+(aq) + 4 e ] × 3

Reduction:

[ MnO4(aq) + 4 H+(aq) + 3 eMnO2(s) + 2 H2O(l) ] × 4

Net (in acid):

3 CH3OH(aq) + 4 MnO4(aq) + H+(aq)3 HCOO(aq) + 4 MnO2(s) + 5 H2O(l)

Neutralization:

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

Net (in base):

3 CH3OH(aq) + 4 MnO4(aq)3 HCOO(aq) + 4 MnO2(s) + 4 H2O(l) + OH(aq)



(c) [Fe(CN)6]3–(aq) + N2H4(aq)[Fe(CN)6]4–(aq) + N2(g)

Oxidation:

N2H4(aq)N2(g) + 4 H+(aq) + 4 e

Reduction:

[ [Fe(CN)6]3–(aq) + e[Fe(CN)6]4–(aq) ] × 4

Net (in acid):

4 [Fe(CN)6]3–(aq) + N2H4(aq)4 [Fe(CN)6]4–(aq) + N2(g) + 4 H+(aq)

Neutralization:

4 H+(aq) + 4 OH(aq)4 H2O(l)

Net (in base):

4 [Fe(CN)6]3–(aq) + N2H4(aq) + 4 OH(aq)4 [Fe(CN)6]4–(aq) + N2(g) + 4 H2O(l)