Balancing half-reactions in basic solution
Fifteen Examples

Reminder: a redox half-reaction MUST be balanced both for atoms and charge in order to be correct. It is VERY easy to balance for atoms only, forgetting to check the charge.

DON'T FORGET TO CHECK THE CHARGE.

Example #1: Here is the half-reaction to be considered:

PbO₂ ---> PbO [basic soln]

Example #2: Here is a second half-reaction:

MnO₄¯ ---> MnO₂ [basic soln]

As I go through the steps below using Example #1, try and balance Example #2 as you go from step to step. The answer will appear at the end of the file. Before looking at the balancing technique, the fact that it is in basic solution can be signaled to you in several different ways:

1) It is explicitly said in the problem.
2) A base (usually a strong base) is included as one of the reactants.
3) An OH¯ is written just above the reaction arrow.
4) A term like "base" or "basic soln" is written after the half-reaction. Usually in parentheses or square brackets.

Here's the last point before going over the solving technique. There are three other chemical species available in a basic solution besides the ones shown above. They are:

H₂O
OH¯
e¯

The water is present because the reaction is taking place in solution, the hydroxide ion is available because it is in basic solution and electrons are available because that's what is transfered in redox reactions.

Remember, these three are always available, even if not shown in the unbalanced half-reaction presented to you in the problem.

Step One to Four: Balance the half-reaction AS IF it were in acid solution. I hope you got that. The half-reaction is actually in basic solution, but we are going to start out as if it were in acid solution. Here are the 4 acid steps:

1) Balance the atom being reduced/oxidized.
2) Balance the oxygens (using H₂O).
3) Balance the hydrogens (using H⁺).
4) Balance the charge.

When you do that to the above half-reaction, you get this sequence:

PbO₂ ---> PbO
PbO₂ ---> PbO + H₂O
2H⁺ + PbO₂ ---> PbO + H₂O
2e¯ + 2H⁺ + PbO₂ ---> PbO + H₂O

Step Five: Convert all H⁺ to H₂O. Do this by adding OH¯ ions to both sides. The side with the H⁺ will determine how many hydroxide to add. In our case, the left side has 2 hydrogen ions, while the right side has none, so:

2e¯ + 2H₂O + PbO₂ ---> PbO + H₂O + 2OH¯

Notice that, when the two hydroxide ions on the left were added, they immediately reacted with the hydrogen ion present. The reaction is:

H⁺ + OH¯ ---> H₂O

Step Six: Remove any duplicate molecules or ions. In our example, there are two water molecules on the left and one on the right. This means one water molecule may be removed from each side, giving:

2e¯ + H₂O + PbO₂ ---> PbO + 2OH¯

The half-reaction is now correctly balanced.

By the way, notice the 2OH¯. Be careful to read that as two hydroxide ions (2 OH¯) and NOT twenty hydride ions (2O H¯). People have been known to do that.

How did you do with the other one? If you didn't do it, go back and try it, then click for the answer.

Example #3: Or you could examine another example (in basic solution), then click for the permanganate answer.

NH₃ ---> N₂H₄

Solution:

1) Balanced as if in acid solution; there were no oxygens to balance.

2NH₃ ---> N₂H₄ + 2H⁺ + 2e¯

Note that the nitrogen also was balanced.

2) Add two hydroxides to each side; this is the final answer, there are no duplicates to strike out.

2OH¯ + 2NH₃ ---> N₂H₄ + 2H₂O + 2e¯

Example #4: Sometimes, the "fake acid" method can be skipped. For example, this half-reaction:

Fe ---> Fe(OH)₃

might show up. Balancing it directly in basic seems fairly easy:

Fe + 3OH¯ ---> Fe(OH)₃ + 3e¯

And yet another comment: there is an old-school method of balancing in basic solution, one that the ChemTeam learned in high school, lo these many years ago. Even though the "fake acid" method is much easier, there are still teachers that insist on the old-school method (which uses hydroxide directly and does not rely on hydrogen ion), but they are increasingly rare.

Example #5: NiO₂ ---> Ni(OH)₂

Solution:

1) Balance the half-reaction AS IF it were in acid solution:

2e¯ + 2H⁺ + NiO₂ ---> Ni(OH)₂

The Ni and O were already balanced. Only H⁺ and e¯ were needed.

2) Convert all H⁺ to H₂O:

2e¯ + 2H₂O + NiO₂ ---> Ni(OH)₂ + 2OH¯

3) Remove any duplicate molecules or ions:

There are none to be removed in this example. Other examples below will have duplicates.

Example #6: BrO₄¯ ---> Br¯

Solution:

1) Balance the half-reaction AS IF it were in acid solution:

8e¯ + 8H⁺ + BrO₄¯ ---> Br¯ + 4H₂O

2) Convert all H⁺ to H₂O:

8e¯ + 8H₂O + BrO₄¯ ---> Br¯ + 4H₂O + 8OH¯

3) Remove any duplicate molecules or ions:

8e¯ + 4H₂O + BrO₄¯ ---> Br¯ + 8OH¯

Example #7: SbO₃¯ ---> SbO₂¯

Solution:

2e¯ + H₂O + SbO₃¯ ---> SbO₂¯ + 2OH¯

Example #8: Cu₂O ---> Cu

Solution:

2e¯ + H₂O + Cu₂O ---> 2Cu + 2OH¯

Example #9: S₂O₃²¯ ---> SO₃²¯

Solution:

1) Balance in acid:

3H₂O + S₂O₃²¯ ---> 2SO₃²¯ + 6H⁺ + 4e¯

Note the 2 in front of the SO₃²¯

2) Add six hydroxide to each side:

6OH¯ + 3H₂O + S₂O₃²¯ ---> 2SO₃²¯ + 6H₂O + 4e¯

3) Eliminate duplicates:

6OH¯ + S₂O₃²¯ ---> 2SO₃²¯ + 3H₂O + 4e¯

Example #10: Tl⁺ ---> Tl₂O₃

Solution:

6OH¯ + 2Tl⁺ ---> Tl₂O₃ + 3H₂O + 4e¯

Example #11: Al ---> AlO₂¯

Solution:

4OH¯ + Al ---> AlO₂¯ + 2H₂O + 3e¯

Example #12: Sn ---> HSnO₂¯

Solution:

3OH¯ + Sn ---> HSnO₂¯ + H₂O + 2e¯

Example #13: CrO₄²¯ ---> Cr(OH)₃

3e¯ + 4H₂O + CrO₄²¯ ---> Cr(OH)₃ + 5OH¯

Example #14: HfO(OH)₂ ---> Hf

Solution:

1) The results of the "fake acid" method are:

4e¯ + 4H⁺ + HfO(OH)₂ ---> Hf + 3H₂O

2) Convert to base with 4 hydroxides on each side; eliminate three water molecules for the final answer:

4e¯ + H₂O + HfO(OH)₂ ---> Hf + 4OH¯

Example #15: H₂AiO₃¯ ---> Al

However, balance by adding hydroxide (not water) to balance the oxygen, then adding hydrogen ion to balance the H.

Solution:

1) Add three hydroxide:

H₂AlO₃¯ ---> Al + 3OH¯

2) Add hydrogen ion:

H⁺ + H₂AlO₃¯ ---> Al + 3OH¯

3) Convert to basic by adding one more hydroxide:

H₂O + H₂AlO₃¯ ---> Al + 4OH¯

4) Balance the charge:

3e¯ + H₂O + H₂AlO₃¯ ---> Al + 4OH¯