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

Problem 18.87

Balance the following redox equations by the half–reaction method.

(a) B2Cl4(aq) + OH(aq) BO2(aq) + Cl(aq) + H2O(l) + H2(g)

(b) CH3CH2ONO2(aq) + Sn(s) + H+(aq)CH3CH2OH(aq) + NH2OH(aq) + Sn2+(aq) + H2O(l)

(c) F5SeOF(aq) + OH(aq)SeO42–(aq) + F(aq) + H2O(l) + O2(g)

(d) As2S3(s) + OH(aq) + H2O2(aq)AsO43–(aq) + SO42–(aq) + H2O(l)

(e) XeF6(s) + OH(aq)XeO64–(aq) + F(aq) + H2O(l) + Xe(g) + O2(g)




Answer:

(a) B2Cl4(aq) + OH(aq) BO2(aq) + Cl(aq) + H2O(l) + H2(g)

The final reaction is done in base but first balance in acid:

B2Cl4(aq) + 4 H2O(l)2 BO2(aq) + 4 Cl(aq) + 8 H+(aq) + 2 e

2 H+(aq) + 2 eH2(g)

net: B2Cl4(aq) + 4 H2O(l)2 BO2(aq) + 4 Cl(aq) + 6 H+(aq) + H2(g)

Remove H+(aq) with: 6 H+(aq) + 6 OH(aq)6 H2O(l)

Net: B2Cl4(aq) + 6 OH(aq)2 BO2(aq) + 4 Cl(aq) + 2 H2O(l) + H2(g)



(b) CH3CH2ONO2(aq) + Sn(s) + H+(aq)CH3CH2OH(aq) + NH2OH(aq) + Sn2+(aq) + H2O(l)

CH3CH2ONO2(aq) + 6 H+(aq) + 6 e CH3CH2OH(aq) + NH2OH(aq) + H2O(l)

[Sn(s)Sn2+(aq) + 2 e] × 3

Net:

CH3CH2ONO2(aq) + 3 Sn(s) + 6 H+(aq)CH3CH2OH(aq) + NH2OH(aq) + 3 Sn2+(aq) + H2O(l)



(c) F5SeOF(aq) + OH(aq)SeO42–(aq) + F(aq) + H2O(l) + O2(g)

The final reaction is done in base but first balance in acid:

2 H 2O(l)O2(g) + 4 H +(aq) + 4 e

[F5SeOF(aq) + 3 H2O(l) + 2 eSeO42–(aq) + 6 F(aq) + 6 H+(aq)] × 2

net: 2 F5SeOF(aq) + 8 H2O(l)2 SeO42–(aq) + 12 F(aq) + 16 H+(aq) + O2(g)

Remove H+(aq) with: 16 H+(aq) + 16 OH(aq)16 H2O(l)

Net: 2 F5SeOF(aq) + 16 OH(aq)2 SeO42–(aq) + 12 F(aq) + 8 H2O(l) + O2(g)



(d) As2S3(s) + OH(aq) + H2O2(aq)AsO43–(aq) + SO42–(aq) + H2O(l)

The final reaction is done in base but first balance in acid:

[H2O2(l) + 2 H+(aq) + 2 e2 H2O(l)] × 14

As2S3(s) + 20 H2O(l)2 AsO43–(aq) + 3 SO42–(aq) + 40 H+(aq) + 28 e

net: As2S3(s) + 14 H 2O2(l) 2 AsO43–(aq) + 3 SO42–(aq) + 12 H+(aq) + 8 H2O(l)

Remove H+(aq) with: 12 H+(aq) + 12 OH(aq)12 H2O(l)

Net: As2S3(s) + 12 OH(aq) + 14 H2O2(aq)2 AsO43–(aq) + 3 SO42–(aq) + 20 H2O(l)



(e) XeF6(s) + OH(aq) XeO64–(aq) + F(aq) + H2O(l) + Xe(g) + O2(g)

The final reaction is done in base but first balance in acid:

2 H2O(l) O2(g) + 4 H+(aq) + 4 e

2 XeF6(s) + 6 H2O(l) + 4 eXeO64–(aq) + 12 F(aq) + Xe(g) + 12 H+(aq)

net: 2 XeF6(s) + 8 H2O(l)XeO64–(aq) + 12 F(aq) + Xe(g) + 16 H+(aq) + O2(g)

Remove H+(aq) with: 16 H+(aq) + 16 OH(aq)16 H2O(l)

Net: 2 XeF6(s) + 16 OH(aq)XeO64–(aq) + 12 F(aq) + 8 H2O(l) + Xe(g) + O2(g)