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# 10.1: Redox Principles and Balancing Redox Reactions

Learning Objectives

• Describe the concepts of oxidation and reduction
• Compute the oxidation states for elements in compounds
• Identify species in a reaction that are being oxidized and those that are being reduced
• Balance redox reactions

## Oxidation-Reduction Reactions

Earth’s atmosphere contains about 20% molecular oxygen, O2, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving O2, but its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification.

Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:

$\ce{2Na}(s)+\ce{Cl2}(g)\rightarrow \ce{2NaCl}(s)$

It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction:

\begin{align*} \ce{2Na}(s) &\rightarrow \ce{2Na+}(s)+\ce{2e-} \\[4pt] \ce{Cl2}(g)+\ce{2e-} &\rightarrow \ce{2Cl-}(s) \end{align*}

These equations show that Na atoms lose electrons while Cl atoms (in the Cl2 molecule) gain electrons, the “s” subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:

\begin{align} \textbf{oxidation}&=\textrm{loss of electrons}\\ \textbf{reduction}&=\textrm{gain of electrons} \end{align}

In this reaction, then, sodium is oxidized and chlorine undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium.

\begin{align} \textbf{reducing agent}&=\textrm{species that is oxidized}\\ \textbf{oxidizing agent}&=\textrm{species that is reduced} \end{align}

Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding $$\ce{NaCl}$$:

$\ce{H2}(g)+\ce{Cl2}(g)\rightarrow \ce{2HCl}(g)$

The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic.

## Assigning Oxidation Numbers (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:
• Hydrogen: +1 when combined with nonmetals, −1 when combined with metals
• Oxygen: −2 in most compounds, sometimes −1 (so-called peroxides, $$\ce{O2^2-}$$), very rarely $$-\dfrac{1}{2}$$ (so-called superoxides, $$\ce{O2-}$$), positive values when combined with F (values vary)
• Halogens: generally −1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in those cases, varying values), −1 for F always,
4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.

Example $$\PageIndex{3}$$: Assigning Oxidation Numbers

Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:

1. H2S
2. $$\ce{SO3^2-}$$
3. Na2SO4

Solution

(a) According to guideline 1, the oxidation number for H is +1.

Using this oxidation number and the compound’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:

$$\ce{charge\: on\: H2S}=0=(2\times +1)+(1\times x)$$

$$x=0-(2\times +1)=-2$$

(b) Guideline 3 suggests the oxidation number for oxygen is −2.

Using this oxidation number and the ion’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:

$$\ce{charge\: on\: SO3^2-}=-2=(3\times -2)+(1\times x)$$
$$x=-2-(3\times -2)=+4$$

(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 oxidation number for sulfur is calculated as directed by guideline 4:

$$\ce{charge\: on\: SO4^2-}=-2=(4\times -2)+(1\times x)$$

$$x=-2-(4\times -2)=+6$$

Exercise $$\PageIndex{3}$$

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

1. KNO3
2. AlH3
3. $$\mathrm{\underline{N}H_4^+}$$
4. $$\mathrm{\sideset{ }{_{\large{4}}^{-}}{H_2\underline{P}O}}$$
Answer a

N, +5

Answer b

Al, +3

Answer c

N, −3

Answer d

P, +5

Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a change in oxidation number. While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist as shown below\). Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:

\begin{align} \textbf{oxidation}&=\textrm{increase in oxidation number}\\ \textbf{reduction}&=\textrm{decrease in oxidation number} \end{align}

Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, 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 between molecular hydrogen and chlorine, 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).

Several subclasses of redox reactions are recognized, including combustion reactions in which the reductant (also called a fuel) and oxidant (often, but not necessarily, molecular oxygen) react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions such as the one depicted below are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation:

$\ce{10Al}(s)+\ce{6NH4ClO4}(s)\rightarrow \ce{4Al2O3}(s)+\ce{2AlCl3}(s)+\ce{12H2O}(g)+\ce{3N2}(g)$

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 s (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture.

Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals:

$\ce{Zn}(s)+\ce{2HCl}(aq)\rightarrow \ce{ZnCl2}(aq)+\ce{H2}(g)$

Metallic elements may also be oxidized by solutions of other metal salts; for example:

$\ce{Cu}(s)+\ce{2AgNO3}(aq)\rightarrow \ce{Cu(NO3)2}(aq)+\ce{2Ag}(s)$

This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu2+ ions dissolve in the solution to yield a characteristic blue color (Figure $$\PageIndex{4}$$). Figure $$\PageIndex{4}$$: (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Displacement of dissolved silver ions by copper ions results in (c) accumulation of gray-colored silver metal on the wire and development of a blue color in the solution, due to dissolved copper ions. (credit: modification of work by Mark Ott)

Example $$\PageIndex{4}$$: Describing Redox Reactions

Identify which equations represent redox reactions. For those reactions identified as redox, name the oxidizing agent and the reducing agent.

1. $$\ce{ZnCO3}(s)\rightarrow \ce{ZnO}(s)+\ce{CO2}(g)$$
2. $$\ce{2Ga}(l)+\ce{3Br2}(l)\rightarrow \ce{2GaBr3}(s)$$
3. $$\ce{2H2O2}(aq)\rightarrow \ce{2H2O}(l)+\ce{O2}(g)$$
4. $$\ce{BaCl2}(aq)+\ce{K2SO4}(aq)\rightarrow \ce{BaSO4}(s)+\ce{2KCl}(aq)$$
5. $$\ce{C2H4}(g)+\ce{3O2}(g)\rightarrow \ce{2CO2}(g)+\ce{2H2O}(l)$$

Solution

Redox reactions are identified per definition if one or more elements undergo a change in oxidation number.

1. This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
2. 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).
3. 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 oxidant and a reductant.
4. This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
5. 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).

Exercise $$\PageIndex{4}$$

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

$\ce{Sn}(s)+\ce{2HCl}(g)\rightarrow \ce{SnCl2}(s)+\ce{H2}(g) \nonumber$

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

Answer

Yes, a single-replacement reaction. Sn(s) is the reductant, HCl(g) is the oxidant.

## Balancing Redox Reactions via the Half-Reaction Method

Redox reactions that take place in aqueous media often involve water, hydronium ions (or protons), and hydroxide ions as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical change in other ways (e.g., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very difficult to balance by inspection, so systematic approaches have been developed to assist in the process. One very useful approach is to use the method of half-reactions, which involves the following steps:

1. Write the two half-reactions representing the redox process.
2. Balance all elements except oxygen and hydrogen.
3. Balance oxygen atoms by adding H2O molecules.
4. Balance hydrogen atoms by adding H+ ions.
5. Balance charge1 by adding electrons.
6. If necessary, multiply each half-reaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each.
7. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.
8. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps:
• Add OH ions to both sides of the equation in numbers equal to the number of H+ ions.
• On the side of the equation containing both H+ and OH ions, combine these ions to yield water molecules.
• Simplify the equation by removing any redundant water molecules.
9. Finally, check to see that both the number of atoms and the total charges2 are balanced.

Example $$\PageIndex{5}$$: Balancing Redox Reactions in Acidic Solution

Write a balanced equation for the reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution.

$\ce{Cr2O7^2- + Fe^2+ \rightarrow Cr^3+ + Fe^3+} \nonumber$

Solution

Write the two half-reactions.

Each half-reaction will contain one reactant and one product with one element in common.

$$\ce{Fe^2+ \rightarrow Fe^3+}$$
$$\ce{Cr2O7^2- \rightarrow Cr^3+}$$

Balance all elements except oxygen and hydrogen. The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms.

$$\ce{Fe^2+ \rightarrow Fe^3+}$$
$$\ce{Cr2O7^2- \rightarrow 2Cr^3+}$$

Balance oxygen atoms by adding H2O molecules. The iron half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven water molecules are added to the right side.

$$\ce{Fe^2+ \rightarrow Fe^3+}$$
$$\ce{Cr2O7^2- \rightarrow 2Cr^3+ + 7H2O}$$

Balance hydrogen atoms by adding H+ions. The iron half-reaction does not contain H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side.

$$\ce{Fe^2+ \rightarrow Fe^3+}$$
$$\ce{Cr2O7^2- + 14H+ \rightarrow 2Cr^3+ + 7H2O}$$

Balance charge by adding electrons. The iron half-reaction shows a total charge of 2+ on the left side (1 Fe2+ ion) and 3+ on the right side (1 Fe3+ ion). Adding one electron to the right side bring that side’s total charge to (3+) + (1−) = 2+, and charge balance is achieved.

The chromium half-reaction shows a total charge of (1 × 2−) + (14 × 1+) = 12+ on the left side ($$\ce{1 Cr2O7^2-}$$ ion and 14 H+ ions). The total charge on the right side is (2 × 3+) = 6 + (2 Cr3+ ions). Adding six electrons to the left side will bring that side’s total charge to (12+ + 6−) = 6+, and charge balance is achieved.

$$\ce{Fe^2+ \rightarrow Fe^3+ + e-}$$
$$\ce{Cr2O7^2- + 14H+ + 6e- \rightarrow 2Cr^3+ + 7H2O}$$

Multiply the two half-reactions so the number of electrons in one reaction equals the number of electrons in the other reaction. To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or destruction) of electrons, the iron half-reaction’s coefficient must be multiplied by 6.

$$\ce{6Fe^2+ \rightarrow 6Fe^3+ + 6e-}$$
$$\ce{Cr2O7^2- + 6e- + 14H+ \rightarrow 2Cr^3+ + 7H2O}$$

Add the balanced half-reactions and cancel species that appear on both sides of the equation.

$\ce{6Fe^2+ + Cr2O7^2- + 6e- + 14H+ \rightarrow 6Fe^3+ + 6e- + 2Cr^3+ + 7H2O}$

Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here:

$\ce{6Fe^2+ + Cr2O7^2- + 14H+ \rightarrow 6Fe^3+ + 2Cr^3+ + 7H2O}$

A final check of atom and charge balance confirms the equation is balanced.

Reactants Products
Fe 6 6
Cr 2 2
O 7 7
H 14 14
charge 24+ 24+

Exercise $$\PageIndex{5}$$

In acidic solution, hydrogen peroxide reacts with Fe2+ to produce Fe3+ and H2O. Write a balanced equation for this reaction.

Answer

$\ce{H2O2}(aq)+\ce{2H+}(aq)+\ce{2Fe^2+} \rightarrow \ce{2H2O}(l)+\ce{2Fe^3+} \nonumber$

## Summary

Chemical reactions are classified according to similar patterns of behavior. Redox reactions involve a change in oxidation number for one or more reactant elements. Writing balanced equations for some redox reactions that occur in aqueous solutions is simplified by using a systematic approach called the half-reaction method.

### Footnotes

1. 1 The requirement of “charge balance” is just a specific type of “mass balance” in which the species in question are electrons. An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must be balanced.
2. 2 The requirement of “charge balance” is just a specific type of “mass balance” in which the species in question are electrons. An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must be balanced.

### Glossary

combustion reaction
vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light
half-reaction
an equation that shows whether each reactant loses or gains electrons in a reaction.
oxidation
process in which an element’s oxidation number is increased by loss of electrons
oxidation-reduction reaction
(also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements
oxidation number
(also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic
oxidizing agent
(also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced
reduction
process in which an element’s oxidation number is decreased by gain of electrons
reducing agent
(also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized
single-displacement reaction
(also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species

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