Balancing redox reactions in basic solutions

By Tutor GuyNo Comments


When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how to balance a redox reaction in a basic solution.

Example : Iodide ions are oxidized in a basic solution by permanganate ions to produce iodine and Mn2+ ions. Show the balanced reaction.

Solution: I’ve made this example a little harder than in my other redox reaction posts because we first need to figure out from the example what the half reactions look like. This is not as complicated as it sounds. Just follow the English and turn it into chemical formulas. Our problem should therefore look like this:

I + MnO4 → I2 + Mn2+

Now comes the tricky part. It turns out it isn’t very easy to balance a redox reaction in basic solutions by adding OH ions, because when we balance the hydrogen atoms, we throw off the balance of the oxygen atoms. So we employ a clever technique. We first balance the reaction as though it were an acidic solution, then we change the solution from acidic to basic. This is a five-step process, and you’ll notice the first four steps are identical to the process for balancing a redox reaction in an acidic solution:

  1. Determine the number of electrons transferred in each half reaction, and balance the electrons.
  2. Balance the oxygen atoms by adding water molecules.
  3. Balance the hydrogen atoms by adding H+ ions.
  4. Check you answer by verifying that the charges are balanced.
  5. Cancel out the H+ ions by adding an equal number of OH ions to both sides, which changes the H+ ions into water.

Let’s see how this works on our example. The iodine atom goes from an oxidation number of 1- on the left to 0 on the right, so each iodine atom loses one electron. Since there are two iodine atoms in I2, we think of this half-reaction as losing two electrons. The manganese goes from an oxidation number of 7+ in the permanganate ion to an oxidation number of 2+. It has gained five electrons. To balance the electrons, we need a total transfer of ten electrons. Multiply the iodine compounds by five and the manganese compounds by two. (Note that the iodide ion was multiplied by 10 rather than 5 to balance the number of iodine atoms):

10 I + 2 MnO4 → 5 I2 + 2 Mn2+

Now we balance the oxygen atoms. There are 8 oxygen atoms on the left and none on the right, so we add 8 water molecules to the right side:

10 I + 2 MnO4 → 5 I2 + 2 Mn2+ + 8 H2O

Next we balance the hydrogen atoms by adding 16 H+ ions to the left (we pretend we have an acidic solution):

10 I + 2 MnO4 + 16 H+ → 5 I2 + 2 Mn2+ + 8 H2O

The equation should be fully balanced, and we check our work by verifying the charges balance. 10- (the iodide ions), 2- (the permanganate ions) and 16+ (the hydrogen ions) gives a net charge on the left of 4+. On the right side, we also have a total charge of 4+.

Finally, we are ready to change the reaction from acidic to basic. We cancel out the 16 H+ ions by adding 16 OH ions to both sides:

10 I + 2 MnO4 + 16 H+ + 16 OH → 5 I2 + 2 Mn2+ + 8 H2O + 16 OH

Because H+ and OH ions react to form water molecules, we simplify the left side:

10 I + 2 MnO4 + 16 H2→ 5 I2 + 2 Mn2+ + 8 H2O + 16 OH

Our last step is to simplify the equation by subtracting the 8 extra water molecules from the right side. We have to subtract the same number from the left side to keep the reaction balanced:

10 I + 2 MnO4 + 8 H2O → 5 I2 + 2 Mn2+ + 16 OH

(Look for other tips on this website if you need help with balancing simple redox reactions or redox reactions in acidic solutions.)

Blue Taste Theme created by Jabox