A redox reaction is a combination of a reduction reaction and an oxidation reaction. In a reduction reaction, a substance gains electrons. In an oxidation reaction, a substance loses electrons. The two reactions together result in a complete transfer of some number of electrons and a transformation of two chemicals. In this article we will demonstrate how to balance the chemical equations of redox reactions.
This article focuses only on balancing the equations for redox reactions occurring in neutral solution. This means that there are no hydrogen or hydroxide ions influencing the charges of the chemicals directly participating in the oxidation or reduction. Solutions that do have these ions are either acidic or basic and will not be covered here. However, many galvanic cells contain water as a vessel of transfer for the free electrons.
Many half-reactions are already balanced on their own as is. If the coefficient of the main chemical on both sides of the equation and the net charge of both sides of the equation are the same, then the half-reaction is balanced. Otherwise certain steps are needed to balance it.
Example 1: The oxidation half-reaction
$$Cr \rightarrow Cr^{3+} + 3e^{-}$$
is balanced, because both the chemical coefficients and the charge are balanced.
Here are some examples where the half-reaction is not balanced. They illustrate why they are not balanced and how to balance it from there.
Example 2: The reduction half-reaction
$$Si + e^{-} \rightarrow Si^{4-}$$
is not balanced. This is because the charge of the two sides differ. When the charge differs, add electrons to one side of the equation to balance it. Adding three more electrons balances the charges while each side of the equation still only has one silicon:
$$Si + 4e^{-} \rightarrow Si^{4-}$$
Example 3: Consider this reaction that occurs in a laboratory setting:
$$Fe_2O_3 + KMnO_4 \rightarrow K_2O + Fe(MnO_4)_2$$
Given that the permanganate ion, \(MnO_4^{-}\), has a \(-1\) charge, we can see that the charge of the iron cation on the reactant side is \(+3\), but on the product side it is \(+2\). This means that the iron cation was reduced, represented by the following reaction:
$$Fe^{3+} + e^{-} \rightarrow Fe^{2+}$$
The atoms are balanced, and so are the charges, so this reduction reaction equation is balanced.
It is incorrect to say
$$2Fe^{3+} + e^{-} \rightarrow Fe^{2+}$$
This mistake may be made if you just copy the coefficients of iron from the original chemical equation, but in fact here both the atoms and the charge are unbalanced. Changing the equation back to
$$Fe^{3+} + e^{-} \rightarrow Fe^{2+}$$
solves both problems.
If we have a balanced reduction reaction and a balanced oxidation reaction, there are special rules for combining the two equations to get a final balanced equation. In general, both half-reaction equations need to be multiplied by a coefficient such that when the equations are combined, the free electrons disappear from the reaction. It may remind you of solving linear systems of equations with the elimination technique, and that is perfectly fine, because the concept is the same.
Example 4: A galvanic cell is comprised of the following two half-reactions:
$$Cr^{3+} + 3e^{-} \rightarrow Cr$$ $$Cl^{-} \rightarrow Cl_2 + e^{-}$$
We can find an overall equation for the reaction. First, however, realize that the oxidation half-reaction equation is not balanced. For starters, the amount of chlorine is imbalanced. Fix that with a coefficient of "2" in front of the chloride ion:
$$2Cl^{-}\rightarrow Cl_2 + e^{-}$$
Now the charge is still not balanced. Since the right side's charge is 1 greater than the left side, add one more electron to the right side:
$$2Cl^{-} \rightarrow Cl_2 + 2e^{-}$$
Remember to check the other half-reaction; we see that it is balanced.
Now we need to multiply each half-reaction by a coefficient such that the electrons will be eliminated. Therefore we want the electron coefficient in the two equations to be the same. We use the least common multiple of \(2\) and \(3\), which is \(6\). Multiply the oxidation half-reaction equation by \(3\) and the reduction half-reaction equation by \(2\):
$$6Cl^{-} \rightarrow 3Cl_2 + 6e^{-}$$ $$2Cr^{3+} + 6e^{-} \rightarrow 2Cr$$
Add the two equations together:
$$6Cl^{-} + 2Cr^{3+} + 6e^{-} \rightarrow 3Cl_2 + 2Cr + 6e^{-}$$
Subtract the electrons away from the two sides of the equation:
$$6Cl^{-} + 2Cr^{3+} \rightarrow 3Cl_2 + 2Cr$$
Example 5: Find the overall redox reaction for a galvanic cell where the two half-reactions are
$$Li \rightarrow Li^{+} + e^{-}$$ $$Re^{4+} + e^{-} \rightarrow Re$$
Solution: The oxidation half-reaction is balanced as is. However, the reduction half-reaction has an imbalance in charge. The left side has a net charge of \(+3\), and the right side has a net charge of \(0\), so add 3 electrons to the left side:
$$Re^{4+} + 4e^{-} \rightarrow Re$$
Now we must match the coefficients of the electrons. Multiplying both sides of the oxidation half-reaction equation by \(4\) does just that.
$$4Li \rightarrow 4Li^{+} + 4e^{-}$$ $$Re^{4+} + 4e^{-} \rightarrow Re$$
Now combine the two equations:
$$4Li + Re^{4+} + 4e^{-} \rightarrow 4Li^{+} + Re + 4e^{-}$$
Finally subtract away the electrons:
$$4Li + Re^{4+} \rightarrow 4Li^{+} + Re$$
If you have freestanding electrons on either side of a full redox reaction chemical equation, then you have made a mistake. They should all cancel each other out.
The concept of balancing redox reactions is crucial for electrochemistry, where we analyze the relationship between half-reactions further. Also it is important to remember the basics from balancing other chemical equations because the same concept of the Law of Conservation of Matter still applies here: we are unable to create or remove new matter from a system. That goes for electrons, chemicals, and protons alike.