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Chapter 20

Oxidation-Reduction

Reactions

20.1 The Meaning of Oxidation

and Reduction

20.2 Oxidation Numbers

20.3 Describing Redox

Equations

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Why does a sparkler have such a bright light?

CHEMISTRY & YOU

If you have ever seen or held a sparkler, then you know that sparklers give off very bright light. They are like handheld fireworks.

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Assigning Oxidation Numbers

What is the general rule for assigning oxidation numbers?

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Assigning Oxidation Numbers

An oxidation number is a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction.

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Assigning Oxidation Numbers

As a general rule, a bonded atom’s oxidation number is the charge that it would have if the electrons in the bond were assigned to the atom of the more electronegative element.

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Assigning Oxidation Numbers

In binary ionic compounds, the oxidation numbers of the atoms equal their ionic charges.

The compound sodium chloride is composed of sodium ions (Na¹⁺) and chloride ions (Cl^1–).
The oxidation number of sodium is +1.
That of chlorine is –1.

Notice that the sign is put before the oxidation number.

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Assigning Oxidation Numbers

Because water is a molecular compound, no ionic charges are associated with its atoms.

Oxygen is reduced in the formation of water.
Oxygen is more electronegative than hydrogen.
The two shared electrons in the H–O bond are shifted toward oxygen and away from hydrogen.

The oxidation number of oxygen is –2.
The oxidation number of each hydrogen is +1.

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Assigning Oxidation Numbers

Oxidation numbers are often written above the chemical symbols in a formula.

H₂O

+1 –2

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Assigning Oxidation Numbers

Rules for Assigning Oxidation Numbers
1.	The oxidation number of a monatomic ion is equal in magnitude and sign to its ionic charge. For example, the oxidation number of the bromide ion (Br^1–) is –1; that of the Fe³⁺ ion is +3.
2.	The oxidation number of hydrogen in a compound is +1, except in metal hydrides, such as NaH, where it is –1.
3.	The oxidation number of oxygen in a compound is –2, except in peroxides, such as H₂O₂, where it is –1, and in compounds with the more electronegative fluorine, where it is positive.
4.	The oxidation number of an atom in uncombined (elemental) form is 0. For example, the oxidation number of the potassium atoms in potassium metal (K) or of the nitrogen atoms in nitrogen gas (N₂) is 0.
5.	For any neutral compound, the sum of the oxidation numbers of the atoms in the compound must equal 0.
6.	For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion.

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Sample Problem 20.2

Assigning Oxidation Numbers to Atoms

What is the oxidation number of each kind of atom in the following ions and compounds?

a. SO₂ c. Na₂SO₄

b. CO₃^2– d. (NH₄)₂S

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Use the set of rules you just learned to assign and calculate oxidation numbers.

Analyze Identify the relevant concepts.

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Sample Problem 20.2

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Solve Apply concepts to this situation.

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Sample Problem 20.2

SO₂

+4 –2

a. There are two oxygen atoms, and the oxidation number of each oxygen is –2 (Rule 3).

The sum of the oxidation numbers for the neutral compound must be 0 (Rule 5).

Therefore, the oxidation number of sulfur is +4, because +4 + (2 × (–2)) = 0.

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Sample Problem 20.2

b. The oxidation number of each oxygen is –2 (Rule 3).

CO₃^2–

? –2

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Sample Problem 20.2

CO₃^2–

+4 –2

b. The sum of the oxidation numbers of the carbon and oxygen atoms must equal the ionic charge, –2 (Rule 6).

The oxidation number of carbon must be +4, because +4 + (3 × (–2)) = –2.

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Sample Problem 20.2

c. The oxidation number of each sodium ion, Na⁺, is the same as its ionic charge, +1 (Rule 1).

The oxidation number of oxygen is –2 (Rule 3).

Na₂SO₄

+1 ? –2

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Sample Problem 20.2

c. For the sum of the oxidation numbers in the compound to be 0 (Rule 5), the oxidation number of sulfur must be +6, because (2 × (+1)) + (+6) + (4 × (–2)) = 0.

Na₂SO₄

+1 +6 –2

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Sample Problem 20.2

d. Ammonium ions, NH₄⁺, have an ionic charge of +1, so the sum of the oxidation numbers of the atoms in the ammonium ion must be +1.

The oxidation number of hydrogen is +1 in this ion.

So, the oxidation number of nitrogen must be –3.

NH₄⁺

? +1

? + 4(+1) = +1

–3 + 4(+1) = +1

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Sample Problem 20.2

d. Two ammonium ions have a total charge of +2.

Since the compound (NH₄)₂S is neutral, sulfur must have a balancing oxidation number of –2.

(NH₄)₂S

–3 +1 –2

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The results are consistent with the rules for determining oxidation numbers.
Also, addition of the oxidation numbers correctly gives the final overall charge for the ion and the three neutral compounds.

Evaluate Do the results make sense?

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Sample Problem 20.2

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Chromium in its uncombined state is a dull silvery color. Orange potassium dichromate (K₂Cr₂O₇) and purple chromium(III) potassium sulfate (CrK(SO₄)₂·12H₂O) are both compounds of chromium. What is the oxidation number of chromium in each compound?

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Chromium in its uncombined state is a dull silvery color. Orange potassium dichromate (K₂Cr₂O₇) and purple chromium(III) potassium sulfate (CrK(SO₄)₂·12H₂O) are both compounds of chromium. What is the oxidation number of chromium in each compound?

K₂Cr₂O₇

+1 +6 –2

CrK(SO₄)₂·12H₂O

+3 +1 +6 –2

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Oxidation-Number Changes in Chemical Reactions

How are oxidation and reduction defined in terms of a change in oxidation number?

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Oxidation-Number Changes in Chemical Reactions

The figure below shows what happens when copper wire is placed in a solution of silver nitrate.

The oxidation number of silver decreases from +1 to 0 as each silver ion (Ag¹⁺) gains an electron and is reduced to silver metal (Ag⁰).

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Oxidation-Number Changes in Chemical Reactions

The figure below shows what happens when copper wire is placed in a solution of silver nitrate.

Copper’s oxidation number increases from 0 to +2 as each atom of copper metal (Cu⁰) loses two electrons and is oxidized to copper(II) ion (Cu²⁺).

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Oxidation-Number Changes in Chemical Reactions

The figure below shows what happens when copper wire is placed in a solution of silver nitrate.

2AgNO₃(aq) + Cu(s) → Cu(NO₃)₂(aq) + 2Ag(s)

+1 +5 –2 0 +2 +5 –2 0

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Oxidation-Number Changes in Chemical Reactions

This figure illustrates a redox reaction that shows what occurs when a shiny iron nail is dipped into a solution of copper(II) sulfate.

The iron reduces Cu²⁺ions in solution and is simultaneously oxidized to Fe²⁺.
The iron becomes coated with metallic copper.

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Oxidation-Number Changes in Chemical Reactions

You can define oxidation and reduction in terms of a change in oxidation number.

An increase in the oxidation number of an atom or ion indicates oxidation.

A decrease in the oxidation number of an atom or ion indicates reduction.

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What happens to the oxidation numbers of metals as they burn in a sparkler?

CHEMISTRY & YOU

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What happens to the oxidation numbers of metals as they burn in a sparkler?

CHEMISTRY & YOU

As the metals burn, they gain oxygen or undergo oxidation. A substance that undergoes oxidation has an increase in oxidation number. Therefore, as metals burn, their oxidation numbers increase.

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Sample Problem 20.3

Identifying Oxidized and Reduced Atoms

Use changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reactions. Also identify the oxidizing agent and the reducing agent.

a. Cl₂(g) + 2HBr(aq) → 2HCl(aq) + Br₂(l)

b. C(s) + O₂(g) → CO₂(g)

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An increase in oxidation number indicates oxidation.
A decrease in oxidation number indicates reduction.
The substance that is oxidized in a redox reaction is the reducing agent.
The substance that is reduced is the oxidizing agent.

Analyze Identify the relevant concepts.

1

Sample Problem 20.3

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Sample Problem 20.3

a. Use the rules to assign oxidation numbers to each atom in the equation.

0 +1 –1 +1 –1 0

Cl₂(g) + 2HBr(aq) → 2HCl(aq) + Br₂(l)

The oxidation number of each chlorine in Cl₂ is 0 because of Rule 4.

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Sample Problem 20.3

a. Then use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced.

The element chlorine is reduced because its oxidation number decreases (0 to –1).
The bromide ion from HBr(aq) is oxidized because its oxidation number increases (–1 to 0).

0 +1 –1 +1 –1 0

Cl₂(g) + 2HBr(aq) → 2HCl(aq) + Br₂(l)

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Sample Problem 20.3

a. Finally, identify the oxidizing and reducing agents.

Chlorine is reduced, so Cl₂ is the oxidizing agent.
The bromide ion from HBr(aq) is oxidized, so Br^– is the reducing agent.

0 +1 –1 +1 –1 0

Cl₂(g) + 2HBr(aq) → 2HCl(aq) + Br₂(l)

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Sample Problem 20.3

b. Use the rules to assign oxidation numbers to each atom in the equation.

0 0 +4 –2

C(s) + O₂(g) → CO₂(g)

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Sample Problem 20.3

b. Then use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced.

The element carbon is oxidized because its oxidation number increases (0 to +4).
The element oxygen is reduced because its oxidation number decreases (0 to –2).

0 0 +4 –2

C(s) + O₂(g) → CO₂(g)

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Sample Problem 20.3

b. Finally, identify the oxidizing and reducing agents.

Carbon is oxidized, so C is the reducing agent.
Oxygen is reduced, so O₂ is the oxidizing agent.

0 0 +4 –2

C(s) + O₂(g) → CO₂(g)

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It makes sense that what is oxidized in a chemical reaction is the reducing agent because it loses electrons—it becomes the agent by which the atom that is reduced gains electrons.
Conversely, it makes sense that what is reduced in a chemical reaction is the oxidizing agent because it gains electrons—it is the agent by which the atom that is oxidized loses electrons.

Evaluate Do the results make sense?

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Sample Problem 20.3

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Sample Problem 20.4

Identifying Oxidized and Reduced Atoms

Use changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reaction. Also identify the oxidizing agent and the reducing agent.

Zn(s) + 2MnO₂(s) + 2NH₄Cl(aq) → ZnCl₂(aq) + Mn₂O₃(s) + 2NH₃(g) + H₂O(l)

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An increase in oxidation number indicates oxidation.
A decrease in oxidation number indicates reduction.
The substance that is oxidized in a redox reaction is the reducing agent.
The substance that is reduced is the oxidizing agent.

Analyze Identify the relevant concepts.

1

Sample Problem 20.4

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Sample Problem 20.4

Use the rules to assign oxidation numbers to each atom in the equation.

0 +4 –2 –3 +1 –1 +2 –1 +3 –2 –3 +1 +1 –2

Zn(s) + 2MnO₂(s) + 2NH₄Cl(aq) → ZnCl₂(aq) + Mn₂O₃(s) + 2NH₃(g) + H₂O(l)

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Sample Problem 20.4

Then use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced.

The element zinc is oxidized because its oxidation number increases (0 to +2).
The manganese ion is reduced because its oxidation number decreases (+4 to +3).

0 +4 –2 –3 +1 –1 +2 –1 +3 –2 –3 +1 +1 –2

Zn(s) + 2MnO₂(s) + 2NH₄Cl(aq) → ZnCl₂(aq) + Mn₂O₃(s) + 2NH₃(g) + H₂O(l)

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Sample Problem 20.4

Finally, identify the oxidizing and reducing agents.

Zinc is oxidized, so Zn is the reducing agent.
Manganese (in MnO₂) is reduced, so Mn⁴⁺ is the oxidizing agent.

0 +4 –2 –3 +1 –1 +2 –1 +3 –2 –3 +1 +1 –2

Zn(s) + 2MnO₂(s) + 2NH₄Cl(aq) → ZnCl₂(aq) + Mn₂O₃(s) + 2NH₃(g) + H₂O(l)

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Use changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reaction.

2HNO₃(aq) + 3H₂S(g) → 2NO(g) + 4H₂O(l) + 3S(s)

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Use changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reaction.

2HNO₃(aq) + 3H₂S(g) → 2NO(g) + 4H₂O(l) + 3S(s)

+1 +5 –2 +1 –2 +2 –2 +1 –2 0

Sulfur is oxidized because its oxidation number increases (–2 to 0). Nitrogen is reduced because its oxidation number decreases (+5 to +2).

2HNO₃(aq) + 3H₂S(g) → 2NO(g) + 4H₂O(l) + 3S(s)

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Key Concepts

As a general rule, a bonded atom’s oxidation number is the charge that it would have if the electrons in the bond were assigned to the atom of the more electronegative element.

An increase in the oxidation number of an atom or ion indicates oxidation. A decrease in the oxidation number of an atom or ion indicates reduction.

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Glossary Term

oxidation number: a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction; the oxidation number of an uncombined element is zero

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Redox reactions are identified by changes in oxidation number.

BIG IDEA

Reactions

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END OF 20.2

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