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 assuming acidic or basic with the first set of half-reactions means hydrogen ion/hydroxide ion/water does not cancel out at the end. This is a problem because none of those species are in the given equation. You must modify one of the half-reactions to make it work. 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 include carbonate in the half-reaction!

3) Balance the half-reactions:

3e¯ + Sb³⁺ ---> Sb
2CO₃²¯ + C ---> 3CO₂ + 4e¯

4) Equalize electrons:

12e¯ + 4Sb³⁺ ---> 4Sb
6CO₃²¯ + 3C ---> 9CO₂ + 12e¯

5) Add:

6CO₃²¯ + 3C + 4Sb³⁺ ---> 4Sb + 9CO₂

6) Add 12 sodium ions:

6Na₂CO₃ + 3C + 4Sb³⁺ ---> 4Sb + 9CO₂ + 12Na⁺

7) Add 6 sulfide ions:

6Na₂CO₃ + 3C + 4Sb³⁺ + 6S²¯ ---> 4Sb + 9CO₂ + 6Na₂S

8) Make antimony(III) sulfide:

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

Example #6: S₂O₃²¯ + I₃¯ ---> I¯ + S₄O₆²¯

Solution:

1) Half-reactions:

S₂O₃²¯ ---> S₄O₆²¯
I₃¯ ---> I¯

2) Balance:

2S₂O₃²¯ ---> S₄O₆²¯ + 2e¯
2e¯ + I₃¯ ---> 3I¯

3) Add (since electrons are already equalized):

2S₂O₃²¯ + I₃¯ ---> S₄O₆²¯ + 3I¯

Example #7: Solid zinc reacts with solid manganese(IV) oxide producing solid zinc oxide and solid manganese(III) oxide.

Discussion:

Solid phase reactions are not really a big thing in chemistry. So, from that perspective, this reaction is totally fake and does not happen in the real world.

That does not stop us from balancing it.

Solution:

1) Write the whole reaction:

Zn + MnO₂ ---> ZnO + Mn₂O₃

2) Separate into half-reactions:

Zn ---> ZnO
MnO₂ ---> Mn₂O₃

3) Acidic/basic solution not indicated? Use acidic:

H₂O + Zn ---> ZnO + 2H⁺ + 2e¯
2e¯ + 2H⁺ + 2MnO₂ ---> Mn₂O₃ + H₂O

4) Electrons are equalized. Add the half-reactions:

Zn + 2MnO₂ ---> ZnO + Mn₂O₃

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