Oxidation-reduction (redox) reactions are another important type of reaction that you will see questions about on the SAT II Chemistry test. The test writers will expect you to be able to identify elements that are oxidized and reduced, know their oxidation numbers, identify half-cells, and balance redox reactions. The following is a brief overview of the basics.
Oxidation-Reduction
Oxidation-reduction reactions involve the transfer of electrons between substances. They take place simultaneously, which makes sense because if one substance loses electrons, another must gain them. Many of the reactions we’ve encountered thus far fall into this category. For example, all single-replacement reactions are redox reactions. Before we go on, let’s review some important terms you’ll need to be familiar with.
Electrochemistry: The study of the interchange of chemical and electrical energy.
Oxidation: The loss of electrons. Since electrons are negative, this will appear as an increase in the charge (e.g., Zn loses two electrons; its charge goes from 0 to +2). Metals are oxidized.
Oxidizing agent (OA): The species that is reduced and thus causes oxidation.
Reduction: The gain of electrons. When an element gains electrons, the charge on the element appears to decrease, so we say it has a reduction of charge (e.g., Cl gains one electron and goes from an oxidation number of 0 to -1). Nonmetals are reduced.
Reducing agent (RA): The species that is oxidized and thus causes reduction.
Oxidation number: The assigned charge on an atom. You’ve been using these numbers to balance formulas.
Half-reaction: An equation that shows either oxidation or reduction alone.
Rules for Assigning Oxidation States
A reaction is considered a redox reaction if the oxidation numbers of the elements in the reaction change in the course of the reaction. We can determine which elements undergo a change in oxidation state by keeping track of the oxidation numbers as the reaction progresses. You can use the following rules to assign oxidation states to the components of oxidation-reduction reactions:
- The oxidation state of an element is zero, including all elemental forms of the elements (e.g., N2, P4, S8, O3).
- The oxidation state of a monatomic ion is the same as its charge.
- In compounds, fluorine is always assigned an oxidation state of -1.
- Oxygen is usually assigned an oxidation state of -2 in its covalent compounds. Exceptions to this rule include peroxides (compounds containing the
group), where each oxygen is assigned an oxidation state of -1, as in hydrogen peroxide (H2O2).
- Hydrogen is assigned an oxidation state of +1. Metal hydrides are an exception: in metal hydrides, H has an oxidation state of -1.
- The sum of the oxidation states must be zero for an electrically neutral compound.
- For a polyatomic ion, the sum of the oxidation states must equal the charge of the ion.
Now try applying these rules to a problem.
Example
Assign oxidation numbers to each element in the following:
- H2S
- MgF2
Explanation
- The sum of the oxidation numbers in this compound must be zero since the compound has no net charge. H has an oxidation state of +1, and since there are two H atoms, +1 times 2 atoms = +2 total charge on H. The sulfur S must have a charge of -2 since there is only one atom of sulfur, and +2 - 2 = 0, which equals no charge.
- F is assigned an oxidation state of -1 (according to rule 3), and there are two atoms of F, so this gives F a total charge of -2. Mg must have a +2 oxidation state since +2 - 2 = 0 and the compound is electrically neutral.
- This time the net charge is equal to -3 (the charge of the polyatomic ion—according to rule 7). Oxygen is assigned a -2 oxidation state (rule 4). Multiply the oxidation number by its subscript: -2
4 = -8. Since there is only 1 phosphorus, just use those algebra skills: P + -8 = -3. Phosphorus must have a +5 charge.
