Balancing redox half-reactions in acidic solution
Fifteen Examples

Balancing redox half-reactions in basic solution     Balancing redox equations in acidic solution
Return to redox menu     Balancing redox equations in basic solution

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:

MnO4¯ ---> Mn2+

It is to be balanced in acidic solution.

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

Cr2O72¯ ---> Cr3+ [acidic soln]

As I go through the steps below using the first half-reaction, try and balance the second half-reaction 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 acid solution can be signaled to you in several different ways:

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

Someday, as you learn more about redox, you will be able to tell just by knowing the characteristic products of various reactants. For example, in the above reaction, permanganate ion is reduced to Mn2+. This is characteristic behavior for permanganate in acid solution. In basic solution (discussed in another tutorial), permanganate forms a different product.

Here's the last point before teaching the technique. There are three other chemical species available in an acidic solution besides the ones shown above. They are:

H2O
H+

The water is present because the reaction is taking place in solution, the hydrogen ion is available because it is in acid 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: Balance the atom being reduced/oxidized. In our example, there is already one Mn on each side of the arrow, so this step is already done. (Hint: not so for the dichromate example you are working in parallel.)

MnO4¯ ---> Mn2+

Step Two: Balance the oxygens. Do this by adding water molecules (as many as are needed) to the side needing oxygen. In our case, the left side has 4 oxygens, while the right side has none, so:

MnO4¯ ---> Mn2+ + 4H2O

Notice that, when the water is added, hydrogens also come along. There is nothing that can be done about this; we'll take care of it in the next step. A common question is: "Why can't I just add 4 oxygen atoms to the right side?" Quick answer: you can only add things that actually exist in solution. Free oxygen atoms (O) do not exist in solution. Free oxide ions (O2¯) do not exist in solution.

Step Three: Balance the hydrogens. Do this by adding hydrogen ions (as many as are needed) to the side needing hydrogen. In our example, we need 8 (notice the water molecule's formula, then consider 4 x 2 = 8).

8H+ + MnO4¯ ---> Mn2+ + 4H2O

Step Four: Balance the total charge. This will be done using electrons. It is ALWAYS the last step.

First, a comment. You do not need to look at the oxidation number for each atom. You only need to look at the charge on the ion or molecule, then sum those up.

Left side of the reaction, total charge is +7. There are 8 H+, giving 8 x +1 = +8 and a minus one from the permanganate. (A very typical wrong answer for the left side is zero. The person sees only the +1 and the -1, they forget the 8. When you do this step in the parallel example, don't forget to multiply 2 times 3. I'll leave you to figure out where in the problem that is.)

Right side of the reaction, total charge is +2. The water molecule is neutral (zero charge) and the single Mn is +2.

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

Five electrons reduces the +7 to a +2 and the two sides are EQUAL in total charge. The half-reaction is now correctly balanced.

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 acid solution), then click for the dichromate answer.

SO2 ---> SO42¯

By the way, a tip off that this is acid solution is the SO2. Oxides of nonmetals make acidic solutions (and oxides of metals make basic solutions).

Solution:

SO2 ---> SO42¯ (the sulfur is already balanced)

2H2O + SO2 ---> SO42¯ (now there are 4 oxygens on each side)

2H2O + SO2 ---> SO42¯ + 4H+ (2 x 2 from the water makes 4 hydrogens)

2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯ (zero charge on the left; +4 from the hydrogens and −2 from the sulfate, so 2 electrons gives the −2 charge required to make zero on the right)


Example #4: Re ---> ReO2

Solution:

Step One: Balance the atom being reduced/oxidized: Re ---> ReO2
Step Two: Balance the oxygens: 2H2O + Re ---> ReO2
Step Three: Balance the hydrogens: 2H2O + Re ---> ReO2 + 4H+
Step Four: Balance the total charge: 2H2O + Re ---> ReO2 + 4H+ + 4e¯


Example #5: Cl2 ---> HClO

Solution:

1) Cl2 ---> 2HClO (chlorine is balanced)
2) 2H2O + Cl2 ---> 2HClO (now there are 2 oxygens on each side)
3) 2H2O + Cl2 ---> 2HClO + 2H+ (2H in HClO plus 2H+ makes 4 hydrogens)
4) 2H2O + Cl2 ---> 2HClO + 2H+ + 2e¯ (zero charge on the left; +2 from the hydrogen ions so 2 electrons gives the −2 charge required to make zero on the right)

Example #6: NO3¯ ---> HNO2

Solution:

balance oxygen ---> NO3¯ ---> HNO2 + H2O

balance hydrogen ---> 3H+ + NO3¯ ---> HNO2 + H2O

balance charge ---> 2e¯ + 3H+ + NO3¯ ---> HNO2 + H2O

Note that the nitrogen was already balanced.


Example #7: H2GeO3 ---> Ge

Answer:

4e¯ + 4H+ + H2GeO3 ---> Ge + 3H2O

Example #8: H2SeO3 ---> SeO42¯

Answer:

H2O + H2SeO3 ---> SeO42¯ + 4H+ + 2e¯

Example #9: Au ---> Au(OH)3 (this one is a bit odd!)

Answer:

3H2O + Au ---> Au(OH)3 + 3H+ + 3e¯

Note that Au(OH)3 is actually auric acid, H3AuO3. Another one commonly written this way is B(OH)3 for boric acid, H3BO3.


Example #10: H3AsO4 ---> AsH3

Answer:

8e¯ + 8H+ + H3AsO4 ---> AsH3 + 4H2O

Example #11: H2MoO4 ---> Mo

Answer:

6e¯ + 6H+ + H2MoO4 ---> Mo + 4H2O

Example #12: NO ---> NO3¯

Answer:

2H2O + NO ---> NO3¯ + 4H+ + 3e¯

Example #13: H2O2 ---> H2O

Answer:

2e¯ + 2H+ + H2O2 ---> 2H2O

Example #14: VO2+(aq) ---> V3+(aq)

Solution:

Balance oxygen: VO2+ ---> V3+ + H2O

Balance hydrogen: 2H+ + VO2+ ---> V3+ + H2O

Balance charge: e¯ + 2H+ + VO2+ ---> V3+ + H2O


Example #15: N2O5 ---> NH4+

Solution:

N2O5 ---> 2NH4+

N2O5 ---> 2NH4+ + 5H2O

18H+ + N2O5 ---> 2NH4+ + 5H2O

16e¯ + 18H+ + N2O5 ---> 2NH4+ + 5H2O


Balancing redox half-reactions in basic solution     Balancing redox equations in acidic solution
Return to redox menu     Balancing redox equations in basic solution