Balancing half-reactions in acidic solution

Balancing half-reactions in basic solution

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Reminder: a 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.


Here is the half-reaction to be considered:

MnO4¯ ---> Mn2+

It is to be balanced in acid solution.

Here is a second half-reaction:

Cr2O72¯ ---> Cr3+ [acid 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, the permanganate ion is reduced to the Mn2+. This is characteristic behavior in acid solution. In basic solution (discussed in another tutorial), a different product is formed from permanganate.

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:


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. All three will be used in getting the final answer.

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: don't do it, it's wrong. The "why" will be left to another day.

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.

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). Here are the steps:

1) SO2 ---> SO42¯ (the sulfur is already balanced)
2) 2H2O + SO2 ---> SO42¯ (now there are 4 oxygens on each side)
3) 2H2O + SO2 ---> SO42¯ + 4H+ (2 x 2 from the water makes 4 hydrogens)
4) 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)

10 More Examples: Balance each half-reaction, the reaction being in acidic solution.

1) Re ---> ReO2

2) Cl2 ---> HClO

3) NO3¯ ---> HNO2

4) H2GeO3 ---> Ge

5) H2SeO3 ---> SeO42¯

6) Au ---> Au(OH)3 (this one is a bit odd!)

7) H3AsO4 ---> AsH3

8) H2MoO4 ---> Mo

9) NO ---> NO3¯

10) H2O2 ---> H2O

Answers to Ten More Examples

Balancing half-reactions in basic solution

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