Balancing redox reactions in neutral solution

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Sometimes, an acid or basic solution can be inferred from context. However, there are times when you cannot determine if the reaction takes place in acidic or basic solution. What you do then is balance the reaction in acidic solution, since that's easier than basic solution. All hydrogen ions will cancel out at the end.

By the way, you can see I have only a few examples. The far, far greater situation is that the reaction occurring in acidic or basic solution will be given to you in the problem.


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

Example #4: SO32¯(aq) + N2(g) ---> SO42¯(aq) + NH3(g)

1) Write half-reactions:

SO32¯(aq) ---> SO42¯(aq)
N2(g) ---> NH3(g)

2) Balance half-reactions. Neither acid or basic was specified. Pick acidic, because it's easier:

H2O + SO32¯(aq) ---> SO42¯(aq) + 2H+ + 2e¯
6e¯ + 6H+ + N2(g) ---> 2NH3(g)

3) Equalize electrons:

3H2O + 3SO32¯(aq) ---> 3SO42¯(aq) + 6H+ + 6e¯
6e¯ + 6H+ + N2(g) ---> 2NH3(g)

4) Add and cancel:

3H2O + 3SO32¯(aq) + N2(g ---> 3SO42¯(aq) + 2NH3(g)

Notice how all the hydrogen ion canceled out.


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