M18Q1: Redox Reactions; Oxidation Numbers and Definition (Review)

Learning Objectives

  • Determine the oxidation numbers of elements in compounds and ions.
  • Identify the oxidizing agent and the reducing agent in an oxidation-reduction reaction.

| Key Concepts and Summary | Glossary | End of Section Exercises |

A photograph is shown of a parked car plugged into a charging station in a paved parking area. The parking area is situated in a wooded area. People are walking in the background in the park-like atmosphere.
Figure 1: Electric vehicles use electric motors for propulsion, powered by batteries that are recharged and used over and over again. (credit: modification of work by Robert Couse-Baker)

Electrochemistry is the study of chemical reactions that produce electricity and the changes associated with the passage of electrical current through matter. The reactions involve electron transfer and are classified as oxidation-reduction (or redox) reactions.

Oxidation-Reduction Reactions

In oxidation-reduction reactions, electrons are transferred from one reactant species to another.

  • Oxidation is the process where electrons are lost by a reactant. After the electrons are lost, the reactant is considered oxidized.
  • Reduction is the process where electrons are gained by a reactant. After the electrons are gained, the reactant is considered reduced.
  • The reactant being oxidized is providing electrons for the reduction reaction and is called the reducing agent.
  • The reactant being reduced is accepting electrons from the oxidation reaction and is called the oxidizing agent.

Let’s consider the reaction between sodium and chlorine to yield sodium chloride to illustrate this concept and how terms are applied:

2 Na(s) + Cl2(g)  →  2 NaCl(s)

It can be helpful to view the process with regard to each individual reactant in the form of an equation called a half-reaction, where the oxidation and reduction portions of the reaction are separated and electrons exchanged are included:

Half-reaction What is Happening Type of Reaction Terms
2 Na  →  2 Na+ + 2 e– Electrons are lost (electrons are on the product side of the equation) Oxidation Na is being oxidized. Na is the reducing agent
Cl2 + 2 e–  →  2 Cl– Electrons are gained (electrons are on the reactant side of the equation) Reduction Cl is being reduced. Cl2 is the oxidizing agent.

These equations show that Na atoms lose electrons while Cl atoms (in the Cl2 molecule) gain electrons. The Na half-reaction has been multiplied by two so that an equal number of electron are gained and lost during the process.

Oxidation Numbers

In order to keep track of electrons being exchanged during a reaction, scientists use a bookkeeping system of oxidation numbers (or oxidation states.) The following guidelines are used to assign oxidation numbers to each element in a molecule or ion.

  1. The oxidation number of an atom in an elemental substance is zero.
  2. The oxidation number of a monatomic ion is equal to the ion’s charge.
  3. Oxidation numbers for common nonmetals are usually assigned as follows:
    • Halogens: -1 for F always, -1 for other halogens except when combined with oxygen or other more electronegative halogens.
    • Hydrogen: +1 when combined with nonmetals, -1 when combined with metals
    • Oxygen: -2 in most compounds, sometimes -1 (so-called peroxides, O22-), very rarely -½ (so-called superoxides, O2–), positive values when combined with F (values vary)
  4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.

The proper convention for reporting charge is to write the number first, followed by the sign (e.g., 2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these two related properties.

More information or practice problems about this topic can be found at: Oxidation-Reduction Reactions (M3Q5-6)

Example 1

Assigning Oxidation Numbers
Follow the guidelines in this section to assign oxidation numbers to all the elements in the following species:

  1. H2S
  2. SO32−
  3. Na2SO4

Solution

(a) According to guideline 3, the oxidation number for H is +1. Since there are two hydrogens giving a total of +2 and the oxidation numbers must sum to zero, sulfur must have an oxidation number of -2.

(b) According to guideline 3, the oxidation number for O is -2. Since there are three oxygens giving a total of -6 and the oxidation numbers must sum to -2, sulfur must have an oxidation number of +4.

(c) For ionic compounds, it’s convenient to assign oxidation numbers for the cation and anion separately. According to guideline 2, the oxidation number for sodium is +1. Assuming the usual oxidation number for oxygen (-2 per guideline 3), the four oxygens give a total of -8, and the oxidation numbers of sulfate must sum to -2, the sulfur must have an oxidation number of +6.

Check Your Learning
Assign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:

  1. KNO3
  2. AlH3
  3. NH4+
  4. H2PO4−

Answer:

(a) N, +5; (b) Al, +3; (c) N, −3; (d) P, +5

Using the oxidation number concept, another definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more element(s) undergo a change in oxidation number. Oxidation occurs when the oxidation number for an element increases from reactant to product. Reduction occurs when there is a decrease in the oxidation number from reactant to product.

Demonstration: Redox with Copper and Hydrogen

Set up. A piece of copper metal is heated to a very high temperature, until after the flame glows green and color of the copper goes from an orange-copper color to almost a tarnished black. Then, the flame is turned off and H2(g) is blown over the copper metal using a giant cone. The H2(g) is removed and air replaces it. This process is repeated several times.

Prediction. What reaction is taking place that causes the copper to go from a metallic orange to a tarnished black? What role does the H2(g) play?

Explanation: As the copper is heated, the pure copper metal reacts with the oxygen in the air to produce a copper oxide, CuO, layer. The copper goes from an oxidation number of zero in pure copper metal, to +2 in copper oxide.

2 Cu(s) + O2(g)  ⇌  2 CuO(s)

The hydrogen serves as a reducing agent, reacting with the CuO as follows:

CuO(s) + H2(g)  ⇌  H2O(ℓ) + Cu(s)

The hydrogen is oxidized, from an oxidation state of 0 to +1, while the copper is reduced, from an oxidation state of +2 to 0.

 

Watch a brief video showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at 3 seconds (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 seconds (yellow flame) use a solid mixture.

Example 2

Describing Redox Reactions
Identify which equations represent redox reactions,  For all redox reactions, name the element that is being oxidzed and the element being reduced.

  1. ZnCO3(s)  →  ZnO(s) + CO2(g)
  2. 2 Ga(ℓ) + 3 Br2(ℓ)  →  2 GaBr3(s)
  3. 2 H2O2(aq)  →  2 H2O(ℓ) + O2(g)
  4. BaCl2(aq) + K2SO4(aq)  →  BaSO4(s) + 2 KCl(aq)
  5. C2H4(g) + 3 O2(g)  →  2 CO2(g) + 2 H2O(ℓ)

Solution

(a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.

(b) This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in Ga(l) to +3 in GaBr3(s). The reducing agent is Ga(ℓ). Bromine is reduced, its oxidation number decreasing from 0 in Br2(l) to −1 in GaBr3(s). The oxidizing agent is Br2(ℓ).

(c) This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from −1 in H2O2(aq) to 0 in O2(g). Oxygen is also reduced, its oxidation number decreasing from −1 in H2O2(aq) to −2 in H2O(ℓ). For disproportionation reactions, the same substance functions as an oxidizing agent and a reducing agent.

(d) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.

(e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from −2 in C2H4(g) to +4 in CO2(g). The reducing agent (fuel) is C2H4(g). Oxygen is reduced, its oxidation number decreasing from 0 in O2(g) to −2 in H2O(ℓ). The oxidizing agent is O2(g).

Check Your Learning
This equation describes the production of tin(II) chloride:

Sn(s) + 2 HCl(g)  →  SnCl2(s) + H2(g)

Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidizing and reducing agents.

Answer:

Yes, this is a redox reaction since the oxidation number of Sn goes from 0 to +2, and H goes from +1 to 0. Sn(s) is the reducing agent, HCl(g) is the oxidizing agent.

Key Concepts and Summary

Electrochemistry is the study of reactions that produce electricity and involve the transfer of electrons between reactants, often referred to as redox reactions. The reactant that loses electrons is oxidized and referred to as the reducing agent. The reactant that gains electrons is reduced and referred to as the oxidizing agent. Redox reactions often include elements in their standard states and/or transition metals, but assigning oxidation numbers to all elements in the chemical reaction will help you to determine if the reaction is a redox reaction and which chemical species is reduced and which is oxidized.

Glossary

electrochemistry
the study of chemical reactions that produce electricity and the changes associated with the passage of electrical current through matter. The reactions involve electron transfer and are classified as oxidation-reduction reactions.
half-reaction
equation where the oxidation and reduction portions of the reaction are separated and electrons exchanged
oxidation
the process where electrons are lost by a reactant
oxidation numbers (or states)
a bookkeeping system in order to keep track of electrons being exchanged during a reaction
oxidizing agent
the reactant that is reduced and accepts electrons from the oxidation reaction
reducing agent
the reactant that is oxidized and provides electrons for the reduction reaction
reduction
process where electrons are gained by a reactant

Chemistry End of Section Exercises

  1. Consider the possible oxidation states of carbon. Complete the following table with the oxidation states for each element:
    Compound C H O
    Cgraphite – –
    C60 (buckminsterfullerene) – –
    CO2 –
    CH4 (methane)
    CH3OH (methanol)
    H2CO (methanal)
    HCOOH (methanoic acid)
    C6H12O6 (glucose)
  2. Consider the possible oxidation states of nitrogen. Complete the following table with the oxidation states for each element:
    Compound N H O
    N2
    HNO3
    NO2
    NO
    NH3
    NH4+
  3. Indicate the oxidation states for each element in the following compounds:
    1. H2SO4
    2. HClO4
    3. Fe2O3
    4. Pb3(PO4)4
    5. Cu(NO3)2
    6. H2O2
    7. NaH
    8. KO2 (a ‘superoxide’ used in rebreathers for fire fighting and mine rescue work to create O2 gas from CO2 gas)
  4. For each of the following half-reactions, determine whether an oxidation or reduction is occurring.
    1. Fe3+(aq) + 3 e–  →  Fe(s)
    2. Cr(s)  →  Cr3+(aq) + 3 e–
    3. MnO42-(aq)  →  MnO4–(aq) + e–
    4. Li+(aq) + e–  →  Li(s)
  5. For each of the following unbalanced half-reactions, determine whether an oxidation or reduction is occurring.
    1. 2 Cl–(aq)  →  Cl2(g)
    2. Mn2+(aq)  →  MnO2(s)
    3. H2(g)  →  2 H+(aq)
    4. NO3–(aq)  →  NO(g)

Answers to Chemistry End of Section Exercises

  1. Compound C H O
    Cgraphite 0 – –
    C60 (Buckminsterfullerene) 0 – –
    CO2 +4 – -2
    CH4 (methane) -4 +1 –
    CH3OH (methanol) -2 +1 -2
    HCOH (methanal) 0 +1 -2
    HCOOH (methanoic acid) +2 +1 -2
    C6H12O6 (glucose) 0 +1 -2
  2. Compound N H O
    N2 0 – –
    HNO3 +5 +1 -2
    NO2 +4 – -2
    NO +2 – -2
    NH3 -3 +1 –
    NH4+ -3 +1 –
  3. (a) H: +1; S: +6; O: -2
    (b) H: +1; Cl: +7; O: -2
    (c) Fe: +3; O: -2
    (d) Pb: +4; P: +5; O: -2 (Hint: Begin by breaking this ionic compound into cation and anion (Pb4+ and PO43-) and begin assigning oxidation numbers)
    (e) Cu: +2; N: +5; O: -2
    (f) H: +1; O: -1
    (g) Na: +1; H: -1
    (h) K: +1; O: -½
  4. (a) reduction; (b) oxidation; (c) oxidation; (d) reduction
  5. (a) oxidation; (b) oxidation; (c) oxidation; (d) reduction
Comments
Please use this form to report any inconsistencies, errors, or other things you would like to change about this page. We appreciate your comments. 🙂

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Chem 103/104 Resource Book Copyright © by Chem 103 Textbook Team and Chem 104 Textbook Team is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.