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ChemWiki: The Dynamic Chemistry E-textbook > Analytical Chemistry > Electrochemistry > Redox Chemistry > Oxidizing and Reducing Agents

Oxidizing and Reducing Agents

Oxidizing and reducing agents are key terms used in describing the reactants in redox reactions that involve transferring electrons between reactants to form products. Here, we will look at what defines an oxidizing and reducing agent, how to determine an oxidizing and reducing agent in a chemical reaction, and the importance of this concept in real world applications.

Oxidizing and Reducing Agents

An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. Also known as the electron acceptor, the oxidizing agent is normally in one of its higher possible oxidation states because it will gain electrons and be reduced. Examples of oxidizing agents include halogens, potassium nitrate, and nitric acid.

 

reducingoxidizingdiagram.png

A reducing agent, or reductant, loses electrons and is oxidized in a chemical reaction. A reducing agent typically is in one of its lower possible oxidation states and is known as the electron donor. A reducing agent is oxidized because it loses electrons in the redox reaction. Examples of reducing agents include the earth metals, formic acid, and sulfite compounds.

 

Definitions
  • A reducing agent reduces other substances and lose electrons; therefore, its oxidation state will increase.
  • An oxidizing agent oxidizes other substances and gains electrons therefore, its oxidation state will decrease.

To help eliminate confusion, here is a mnemonic device to help you remember how to determine oxidizing and reducing agents.

OIL RIG

Oxidation Is Loss and Reduction Is Gain of electrons

Example 1: Identify reducing and oxidizing agents

Identify the reducing and oxidizing agents in the balanced redox reaction:

\[ Cl_2 (aq) + 2Br^- (aq) \rightarrow 2Cl^- (aq) + Br_2 (aq)\]

Oxidation half reaction

\[2 Br^- (aq) \rightarrow Br_2 (aq)\]
Oxidation States:   -1      0

Reduction Half Reaction

\[Cl_2 (aq) \rightarrow 2 Cl^- (aq)\]
Oxidation States: 0     -1

Overview

  • \(Br^-\) loses an electron; it is being oxidized from \(Br^-\) to \(Br_2\), thus \(Br^-\) is the reducing agent.
  • \(Cl_2\) gains one electron; it is being reduced from \(Cl_2\) to 2 \(Cl^-\), thus \(Cl_2\) is the oxidizing agent.

 

Common oxidizing agents
Common reducing agents
O2
H2
O3
CO
F2
Fe
Br2
Zn
H2SO4
Li
Halogen metals (halogen metals tend to gain an electron to get to noble gas configuration)
Alkali metals (alkali metals tend to lose an electron to get to noble gas configuration)

Applications

Oxidizing and reducing agents are important in industrial applications. They are used in processes such as purifying water, bleaching fabrics and storing energy (such as in batteries and gasoline). Oxidizing and reducing agents are especially crucial in biological processes such as metabolism and photosynthesis. For example, organisms utilize electron acceptors such as NAD+ to harvest energy from redox reactions as in the hydrolysis of glucose:

\[C_6H_{12}O_6 + 2ADP + 2P + 2NAD^+ \rightarrow 2CH_3COCO_2H + 2ATP + 2NADH\]

All combustion reactions are also examples of redox reactions. A combustion reaction occurs when a substance reacts with oxygen to create heat. One example is the combustion of octane, the principle component of gasoline:

\[2 C_8H_{18} (l) + 25 O_2 (g)  \rightarrow 16 CO_2 (g) + 18 H_2O (g)\]

Combustion reactions are a major source of energy for modern industry.

Summary

  Oxidizing Agents Reducing Agents
Oxidation State Decreases Increases
# of Electrons Gained Lost
Substance is... Reduced Oxidized

By looking at each element's oxidation state on the reactant side of a chemical equation in comparison to the same element's oxidation state on the product side, one can determine if the element is being reduced or oxidized. Thus, one is able to conclude the oxidizing and reducing agents of a chemical reaction.

Problems

  1. Identify the oxidizing agent and the reducing agent in the following redox reaction \[ MnO_2(s) + 4 H^+(aq) + 2 Cl^-(aq) \rightarrow Mn^{2+} (aq) + 2 H_2O (l) + Cl_2(g)\]
  2. For the reaction, \(2 NO_2(g) + 7 H_2(g) \rightarrow 2 NH_3(g) + 4 H_2O(g)\), is hydrogen an oxidizing agent or a reducing agent?  Explain.
  3. An element that is oxidized is a(n) __________ agent and an element that is reduced is a(n) __________ agent.
  4. What is the reducing agent in the following redox reaction: \[ 5 SO_3^{2-} + 2 MnO_4^- + 6 H^+ \rightarrow 5 SO_4^{2-} + 2 Mn^{2+} + 3 H_2O\]
  5. What is the oxidizing agent in the following redox reaction: \[Zn (s) + Cu^{2+} (aq) \rightarrow Zn^{2+} (aq) + Cu(s)\]
  6. Determine the oxidizing and reducing agent of the following chemical equation for aerobic respiration: \[ C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (l)\]
  7. For a general redox reaction involving species \(A\) and \(B\) with \(A\) losing electrons and \(B\) gaining electrons: Is A the oxidizing or reducing agent? Is B the oxidizing or reducing agent? Which one is reduced and which one is oxidized?
  8. In a redox reaction, there must be
    1. an oxidizing agent and no reducing agent
    2. a reducing agent and no oxidizing agent
    3. a reducing agent and an oxidizing agent
    4. no reducing or oxidizing agent
  9. Which of the following will be a strong reducing agent, which of the following will be a strong oxidizing agent?

\(NO_3^-\), \(NO\), \(N_2H_4\), \(NH_3\)

Solutions

  1. \(Cl^-\) is the reducing agent because it is oxidized and loses one electron (starting with an oxidation state of -1 in the \(Cl^-\) ions and increasing to 0 in \(Cl_2\)). Remember that gaining electrons means it is "reduced".  \(MnO_2\) is the oxidizing agent because it reduced by gaining two electrons (starting with \(Mn\) in an oxidation state of +4 in \(MnO_2\) and decreasing to +2 in free \(Mn^{2+}\) ions). Keep in mind that losing electrons means it is "oxidized". 
  2. In this reaction, hydrogen loses one electron.  Hydrogen is oxidized, thus making it the reducing agent.    
  3. An element that is oxidized is a reducing agent because the element loses electrons and an element that is reduced is an oxidizing agent because the element gains electrons.
  4. \(SO_3^{2-}\) is the reducing agent because it lost two electrons, sulfur goes from an oxidation state of +4 in \(SO_3^{2-}\) to an oxidation state of +6 in\(SO_4^{2-}\).
  5. \(Cu^{2+} (aq)\) is the oxidizing agent because it gains two electrons, going from an oxidation state of +2 in \(Cu^{2+} (aq)\) to an oxidation state of 0 in Cu(s).
  6. The oxidizing agent is oxygen and the reducing agent is glucose. Oxygen is being reduced, so it is an oxidizing agent. The glucose is being oxidized, so it is a reducing agent.
  7. When \(A\) loses electrons, it is being oxidized, thus is a reducing agent. When \(B\) gains electron, it is being reduced, thus is an oxidizing agent. \(A\) is oxidized and \(B\) is reduced.
  8. The answer is C: In a redox reaction, there is always an oxidizing and reducing agent
  9. \(NO_3^-\) is most likely to be a strong oxidizing agent. \(NH_3\) is most likely to be a strong reducing agent. You determine the likelihood for oxidation or reduction by comparing the oxidation numbers of nitrogen. Since \(NO_3^-\) has the highest oxidation number of +5, compared to the other molecules, it will most likely be the oxidizing agent. Since nitrogen in \(NH_3\) has an oxidation state of -3, it has the lowest oxidation state and will most likely be the reducing agent.

References

  1. Gerhart, Karen. The Origins and Essentials of Life. Dubuque: Kendall/Hunt Publishing Company, 2009.
  2. Pettrucci, Ralph H. General Chemistry: Principles and Modern Applications. 9th. Upper Saddle River: Pearson Prentice Hall, 2007.
  3. Oxtoby, David W., H.P. Gillis, and Alan Campion. Principles of Modern Chemistry. 6th. Belmont: Thomson Brooks/Cole, 2008. 

Contributors

  • Diana Pearson, Connie Xu, Luvleen Brar (UCD)
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Last Modified
13:35, 5 Apr 2014

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