Problem 16.2 Balance the following overall reactions: A. CH,OH + 02-co, + H20 (a

Question

Problem 16.2 Balance the following overall reactions: A. CH,OH + 02-co, + H20 (at high pH) B. HS +MnO S(s) +MnO,(s) (at high pH) C. 10, (iodate)+ I (iodide) I, iodine) (at low phH) Purpose of problem Balancing overall reactions Relevant section(s) of text Section 16.3.2 Solution For each half reaction, balance all but H and O, balance O with HO, balance H with H', balance charge with e, and add H,O-H + OH to eliminate H. Then add half reactions to cancel e: A: CH,OH +O, CO+ H,O Oxidation half reaction: CH,OH +60H CO, +5H,O +6e Reduction half reaction: O, +2H,O +4e 40H Overall: 2CH,OH+30, 2C0, +4H, B: HS +MnO,S(s) + MnO (s) Oxidation half reaction: HS. + OH' .. S(s) + H2O + 2e Reduction half reaction: MnO, +2H,O +3e MnO(s) +40H Overall: 3HS +2MnO, +H,O 3S(s) +2MnO,(s)+ 50H C: 10, +I 1, Oxidation half reaction: 21 1,+ 2e Reduction half reaction: 210," +12H. +10e ·12 + 6H,0 Overall: 10, +51 +6H 31,+3H,O (a comproportionation reaction) A Problem-Solving Approach to Aquatic Chemistry J.N. Jensen 6/2002

Explanation / Answer

1)

CH3OH + O2--------------- CO2   + H2O   at high PH menas basic medium

-2             0                   +4            -2

oxidation state of an atom is increases is called as oxidation

oxidation state of an atom is decreases is called as reduction

oxidation half reaction

CH3OH----------------- CO2

CH3OH + H2O ---------------- CO2

CH3OH + H2O + 6 OH- ----------------- CO2 + 6H2O

CH3OH + H2O +6OH------------------- CO2 + 6H2O + 6e-

Reduction half reaction

O2------------ H2O

O2 --------- H2O +H2O

O2 + 4 H2O--------------- 2H2O + 4 OH-

O2 + 4H2O + 4e- ---------------- 2H2O + 4 OH-

[CH3OH + H2O +6OH------------------- CO2 + 6H2O + 6e-]x4

[O2 + 4H2O + 4e- ---------------- 2H2O + 4 OH-]x6

4CH3OH +4 H2O +24OH-------------------4 CO2 + 24H2O + 24e-

6O2 + 24H2O + 24e- ---------------- 12H2O + 24OH-

-------------------------------------------------------------------------------------------------

4CH3OH + 6O2 ------------------------ 4CO2 + 8H2O

---------------------------------------------------------------------------------------

The net eqaution is 2CH3OH + 3O2 ------------------------ 2CO2 + 4H2O

b)

HS- + MnO4- -------------- S + MnO2

-2      +7                     0       +4

oxidation half reaction

HS- --------------- S

HS-    + OH- --------------- S+ H2O

HS- + OH- --------------- S+ H2O + 2e-

reduction half reaction

MnO4- ------------------------ MnO2

MnO4- --------------- MnO2 + 2 H2O

MnO4- + 4H2O ---------- MnO2 + 2H2O + 4 OH-

MnO4- + 4 H2O + 3e- ------------- MnO2 + 2H2O + 4 OH-

[HS- + OH- --------------- S+ H2O + 2e-]x3

[MnO4- + 4 H2O + 3e- ------------- MnO2 + 2H2O + 4 OH-]x2

3HS- +3 OH- --------------- 3S+3 H2O + 6e-

2MnO4- + 8H2O + 6e- -------------2 MnO2 + 4H2O + 8 OH-

--------------------------------------------------------------------------------------------------

3HS- + 2 MnO4- +H2O ------------- 3S + 2 MnO2 + 5 OH-

------------------------------------------------------------------------------------------

c) this is acidic medium because low PH

IO3- + I- ----------------- I2

+5       -1                    0

oxidation half reaction

I- -------------- I2

2I- ------------- I2

2 I- ---------- I2 + 2e-

reduction half reaction

IO3- ------------- I2

2IO3- ------------ I2

2 IO3- ------------- I2 + 6 H2O

2 IO3- + 12H+ --------------- I2 + 6 H2O

2IO3- + 12 H+ + 10e- ----------- I2 + 6 H2O

[2 I- ---------- I2 + 2e-]x10

[2IO3- + 12 H+ + 10e- ----------- I2 + 6 H2O]x2

10 I- --------------- 5 I2 + 10e-

2IO3- + 12 H+ + 10e- ----------- I2 + 6 H2O

----------------------------------------------------------------

10 I- + 2 IO3- + 12H+ ---------- 6 I2 + 6 H2O

---------------------------------------------------------------------

The net eqauiton is

IO3- + 5 I- + 6 H+ ----------------- 3 I2 + 3 H2O.

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