D40.2 Redox Reactions and Oxidation Number
Some redox reactions involve transfer of electrons between reactant atoms to yield ionic products. For example, formation of a crystal lattice of sodium chloride (a lattice of Na+ ions and Cl− ions) requires transfer of electrons from sodium atoms to chlorine atoms:
It is useful to consider the electron transfer as two concurrent processes. The processes are called half-reactions, one in which electrons are lost and one in which electrons are gained. The half-reactions for sodium reacting with chlorine are:
| 2Na(s) | ⟶ | 2Na+(s) + 2e− |
| Cl2(g) + 2e− | ⟶ | 2Cl−(s) |
The half-reactions show that the number of electrons lost equals the number of electrons gained—we cannot create or destroy electrons in a chemical reaction. The half-reactions also show which species gains electrons (Cl atoms in Cl2) and which species loses electrons (Na atoms). The species that loses electrons is said to be oxidized and the loss of electrons is called oxidation; the species that gains electrons is said to be reduced and the gain of electrons is called reduction.
The species that causes reduction to occur is called the reducing agent (or reductant). In this reaction, sodium is the reducing agent because it causes Cl2 to gain electrons. The species that causes oxidation to occur is called the oxidizing agent (or oxidant). In this reaction, chlorine functions as an oxidant because it causes sodium to lose electrons.
Some redox processes do not involve obvious transfer of electrons because no ions are involved in the reaction. For example:
To systematically classify redox reactions of all types, we define oxidation number (or oxidation state) of an element in a compound as the charge its atoms would possess if the compound was ionic, that is, if all electrons in each polar covalent bond were assigned to the more electronegative atom.
The steps below can be used to assign oxidation numbers to each element in a compound.
- The oxidation number of an atom in an elemental substance is zero.
- The oxidation number of a monoatomic ion is equal to the ion’s charge.
- The sum of oxidation numbers over all atoms in a neutral compound is zero.
- The sum of the oxidation numbers over all atoms in a polyatomic ion equals the ion’s charge. If a compound includes more than one polyatomic ion, the oxidation number of an atom in one ion can differ from its oxidation number in the other polyatomic ion.
- Atoms of some elements have the same oxidation number in almost all compounds.
- Fluorine always has oxidation number −1 when present in a compound
- Atoms of alkali metals (Li, Na, K, Rb, Cs) have oxidation number +1 in nearly all compounds
- Atoms of alkaline earth metals (Be, Mg, Ca, Sr, Ba) have oxidation number +2 in nearly all compounds
- Hydrogen has oxidation number +1 when combined with nonmetals, −1 when combined with metals
- Apply these two rules only if rules 1-5 have not determined all oxidation numbers.
- Oxygen has oxidation number −2 unless rules 1-5 have already given O a different oxidation number
- In binary compounds of nonmetals, the more electronegative element is given a negative oxidation number equal to the charge on its monoatomic ion; for example in PCl3, the more electronegative Cl is assigned oxidation number −1 and P is assigned +3 (by rule 3).
Activity: Oxidation Numbers
Use the guidelines above to assign an oxidation number to each element in each formula:
- KNO3
- AlH3
- NH4NO3
- H2PO4−
Write in your notebook, then left-click here for an explanation.
- Rule 5b: oxidation number of K is +1. Neither O or N has yet been defined, so we can apply rule 6a: oxidation number of O is -2. Then Rule 3: oxidation numbers must sum to zero, and we have (+1) + (x) + 3(−2) = 0, so x = +5 and oxidation number of N is +5.
- Rule 5d: oxidation number of H is −1 because Al is a metal. Then Rule 3: oxidation numbers must sum to zero, and we have (x) + 3(−1) = 0 and oxidation number of Al is +3.
- This is an ionic compound containing NH4+ and NO3− ions. Rule 4: Assign oxidation numbers to atoms in each ion separately and the oxidation numbers sum to the ionic charge. For NH4+, Rule 5d gives H oxidation number +1; then (x) + 4(+1) = +1, x = −3 and oxidation number of N is −3. For NO3−, Rule 6a says O is −2; then (x) + 3(−2) = −1 and N is +5. (Note that the oxidation number of N is different in the two polyatomic ions.)
- Rule 4: oxidation numbers sum to −1. Rule 5d: H is +1. Rule 6a: O is −2. Thus, 2(+1) + (x) + 4(−2) = −1 and P is +5.
Exercise: Assigning Oxidation Numbers
Using oxidation numbers, we can identify redox reactions by looking for one or more elements whose oxidation numbers have changed during the course of the reaction. When its oxidation number increases, an element has been oxidized; when its oxidation number decreases, the element has been reduced.
In the reaction:
sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl2 to −1 in NaCl).
In the reaction:
hydrogen is oxidized (its oxidation number increases from 0 in H2 to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl2 to −1 in HCl).
There are several subclasses of redox reactions. One example is combustion reactions in which the reductant (fuel) and oxidant (often, O2) react vigorously and produce significant temperature increase, often in the form of a flame. Another class of redox reaction is a rocket propellant reaction such as when solid aluminum is oxidized by ammonium perchlorate:
Exercise: Recognizing Oxidation-Reduction Reactions
Exercise: Recognizing Oxidizing and Reducing Agents
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