Balancing redox reactions in neutral solution

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What you do is balance it in acidic solution, since that's easier than basic solution. All hydrogen ions will cancel out at the end.

Example #1: NH3 + NO2 ---> N2O + H2O

Solution:

1) Split into half-reactions:

NH3 ---> N2O
NO2 ---> N2O

2) Balance in acidic solution:

H2O + 2NH3 ---> N2O + 8H+ + 8e¯
6e¯ + 6H+ + 2NO2 ---> N2O + 3H2O

3) Equalize electrons:

3H2O + 6NH3 ---> 3N2O + 24H+ + 24e¯ <--- multiplied by a factor of 3
24e¯ + 24H+ + 8NO2 ---> 4N2O + 12H2O <--- multiplied by a factor of 4

4) Add and eliminate like items:

6NH3 + 8NO2 ---> 7N2O + 9H2O

Example #2: Fe + V2O3 ---> Fe2O3 + VO

Solution:

1) Half-reactions:

Fe ---> Fe2O3
V2O3 ---> VO

2) Balance in acidic:

2Fe + 3H2O ---> Fe2O3 + 6H+ + 6e¯
2e¯ + 2H+ + V2O3 ---> 2VO + H2O

3) Equalize electrons:

2Fe + 3H2O ---> Fe2O3 + 6H+ + 6e¯
6e¯ + 6H+ + 3V2O3 ---> 6VO + 3H2O

4) Add and eliminate like items:

2Fe + 3V2O3 ---> Fe2O3 + 6VO

6e¯, 3H2O, and 6H+ were eliminated


Example #3: CO2 + NH2OH ---> CO + N2 + H2O

Solution:

1) Half-reactions

CO2 ---> CO
NH2OH ---> N2

2) Balance in acid:

2e¯ + 2H+ + CO2 ---> CO + H2O
2NH2OH ---> N2 + 2H2O + 2H+ + 2e¯

3) Add and eliminate like items:

CO2 + 2NH2OH ---> CO + N2 + 3H2O

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