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 you will be told in the problem that the reaction is occurring in acidic or basic solution.

August 2018 - I added an example (it's #5) where you cannot assume either acidic or basic and then everything cancels out at the end. Be sure to take a look at it, it's a nice 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.


Example #5: Na2CO3 + C + Sb2S3 ---> Sb + Na2S + CO2

Solution:

1) We need to determine the half-reactions.

(a) the reduction half-reaction (going from +3 to zero):
Sb3+ ---> Sb

(b) the oxidation half-reaction (going from zero to +4):

C ---> CO2

2) However, we have a bit of a problem. There is no indication of this reaction being in acidic or basic solution. If we choose either one, we won't be able to cancel out any hydrogen ions, hydroxide ions or water. So, how are we going to balance the oxygens in the CO2? Answer: we will use the carbonate!

3) Balance the half-reactions:

3e¯ + Sb3+ ---> Sb
2CO32¯ + C ---> 3CO2 + 2e¯

4) Equalize electrons:

6e¯ + 2Sb3+ ---> 2Sb
6CO32¯ + 3C ---> 9CO2 + 6e¯

5) Add:

6CO32¯ + 3C + 2Sb3+ ---> 2Sb + 9CO2

6) Add 12 sodium ions:

6Na2CO3 + 3C + 2Sb3+ ---> 2Sb + 9CO2 + 12Na+

7) Add 6 sulfide ions:

6Na2CO3 + 3C + 2Sb3+ + 6S2¯ ---> 2Sb + 9CO2 + 6Na2S

8) Make antimony(III) sulfide (there will be a problem):

6Na2CO3 + 3C + Sb2S3 + 3S2¯ ---> 2Sb + 9CO2 + 6Na2S

9) To solve the lack of sufficient antimony, I propose to add two more antimony atoms to each side:

6Na2CO3 + 3C + 2Sb2S3 ---> 4Sb + 9CO2 + 6Na2S

Done!

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