Balancing Redox Reactions in Neutral, Acidic, and Basic Solutions
Balancing redox reactions, as stated in the conclusion of the previous page, is much different than traditional stoichiometry; counting atoms is still involved, but the main focus is on electron transfer and atomic charge. This is where oxidation states come into play: since an oxidation state effectively tells us the charge of an individual atom, comparing the reactant and product oxidation states of an atom can tell whether it has gained or lost electrons, and how many were involved. That allows us to balance the number of electrons gained by a particular atom to the amount lost by another.
The simplest (and, unfortunately, slowest) way to balance a redox reaction is by half-reaction. For example, let's try to balance the reaction below:
Ag+(aq) + AsH3(g) + OH-(aq) Ag(s) +
H3AsO3(aq) + H2O(l)
First, let's split it up into half-reactions:
AsH3(g) + OH-(aq) H3AsO3(aq) + H2O(l)
Now, it's obvious that the first half-reaction can be balanced by adding an electron to the left side. Don't go through the work of checking oxidation states if the answer is already apparent. Let's focus on the oxidation states of the second half-reaction:
AsH3(g) + OH-(aq) H3AsO3(aq) + H2O(l)
Since hydrogen has a higher electronegativity than arsenic, they will "steal" all of its electrons, giving it an oxidation number of +3; each hydrogen has -1. The O in the hydroxide has -2, while the H has +1. Next, the O's in the H3AsO3 molecule have the highest electronegativities, giving them each a -2 oxidation state (for a total of -6). To offset this, each H has a +1 state, while the poor As still has a +3 state. Finally, the O in water has a +2 charge, while the H's have +1 each.
We can see by comparing oxidation states that three hydrogen atoms transition from a -1 to a +1 in the reaction; the O's, As, and other H do not change. Therefore, this reaction must give off 6 electrons. Notice that there are 3 oxygen atoms on the right, while there is just one on the left. The total charge is not balanced, either. Let's add 6 hydroxides to the left to make charge balance; the H3AsO3 will take three oxygen atoms, while the rest of the oxygen and hydrogen will form water. Since the oxidation states of the hydrogen and oxygen in hydroxide and water do not change, we can add as many as we like without interfering with the electron transfer.
The reduction of silver requires 1 electron, so it must happen 6 times to use up the total electron supply. Our two balanced half-reactions are shown below; adding them up gives the final reaction:
6 Ag+(aq) + 6 e- 6 Ag(s)
AsH3(g) + 6 OH-(aq) H3AsO3(aq) + 3 H2O(l) + 6 e-
6 Ag+(aq) + AsH3(g) + 6 OH-(aq) 6 Ag(s) + H3AsO3(aq) + 3 H2O(l)
Let's check charges: zero on both sides; counting atoms gives 6 silver, one arsenic, 9 hydrogen, and 6 oxygen. We're done! As you can see, there's a lot more to redox reactions than just counting atoms; you have to count electrons, too!
The example we just did was also an example of balancing a redox reaction in a basic solution, because we could add OH- molecules to the reaction. If the problem notes that the reaction occurs in a basic solution, you may add hydroxide if additional hydrogen, oxygen, or negative charge is needed. In addition, if you get H+ as a product in a basic solution, you must add enough hydroxide to both sides to convert the H+ into water.
You will also encounter problems asking you to balance redox reactions in acidic solutions. In this case, add H+ ions as needed to balance hydrogen or charge. Likewise, if OH- appears as a product in these reactions, you must convert it into water by adding H+ (or hydronium, H3O+, if you want to write it out).
In addition, if the problem does not specify acidic or basic, either hydroxide or hydronium may appear as products; you cannot add one to the reactants unless it already appears.
There is a faster (and sometimes trickier) way of balancing redox reactions, introduced to the authors by Mr. Larry Gulberg. Instead of bothering to separate a reaction into half-reactions, just consider the oxidation states immediately. We typically write the oxidation state of each atom above it, for easy comparing. When an oxidation state changes, draw a line from the reactant side to the product side and note how many electrons are consumed or produced. Then, change the coefficients as needed to make the electrons balance. Finally, balance atoms stoichiometrically, and add ions if necessary to make charge or atoms balance. It's a good idea to develop your own notation for this method, in whatever format you're comfortable with. This method is quicker than the half-reaction scheme, since less writing is involved, and with practice you'll find it just as simple.
You can balance the sample problems below using whatever method and notation you prefer; however, be sure to note the acidity or alkalinity of the solution. If you're not sure about which atom in a molecule has the highest electronegativity, check in our "Reference" section. In fact, it's a good idea to keep a browser window open or a printout handy while balancing, just to be sure.
Believe it or not, this is the trickiest part of electrochemistry! Now, let's go on to a much simpler subject, reduction potentials, which discusses just how much atoms want those electrons they keep stealing.