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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 cut fruit turn brown?

CHEMISTRY & YOU

You have probably noticed that the flesh of an apple turns brown after you remove the skin. The apple is still safe to eat; it just doesn’t look as appetizing.

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Identifying Redox Reactions

What are the two classes of chemical reactions?

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Identifying Redox Reactions

All chemical reactions can be assigned to one of two classes.

One class of chemical reactions is oxidation-reduction (redox) reactions, in which electrons are transferred from one reacting species to another.

The other class includes all other reactions, in which no electron transfer occurs.

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Identifying Redox Reactions

Many single-replacement reactions, combination reactions, decomposition reactions, and combustion reactions are redox reactions.

Potassium metal reacts violently with water to produce hydrogen gas (which ignites) and potassium hydroxide.

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Identifying Redox Reactions

Many single-replacement reactions, combination reactions, decomposition reactions, and combustion reactions are redox reactions.

Zinc metal reacts vigorously with hydrochloric acid to produce hydrogen gas and zinc chloride.

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Identifying Redox Reactions

Examples of reactions that are not redox reactions include double-replacement reactions and acid-base reactions.

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Identifying Redox Reactions

During an electrical storm, oxygen molecules and nitrogen molecules in air react to form nitrogen monoxide.

N₂(g) + O₂(g) → 2NO(g)

How can you tell if this is a redox reaction?

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Identifying Redox Reactions

During an electrical storm, oxygen molecules and nitrogen molecules in air react to form nitrogen monoxide.

N₂(g) + O₂(g) → 2NO(g)

How can you tell if this is a redox reaction?

The oxidation number of nitrogen increases from 0 to +2.
The oxidation number of oxygen decreases from 0 to –2.
The reaction between nitrogen and oxygen to form nitrogen monoxide is a redox reaction.

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Identifying Redox Reactions

Many reactions in which color changes occur are redox reactions.

An example is shown below.

MnO₄^–(aq) + Br^–(aq) → Mn²⁺(aq) + Br₂(aq)

Permanganate ion (purple)

Bromide ion (colorless)

Manganese(III) ion (colorless)

Bromine (brown)

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Some fruits, including apples, turn brown when you cut them. What do you think is happening on the surface of the fruit that causes it to turn brown?

CHEMISTRY & YOU

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Some fruits, including apples, turn brown when you cut them. What do you think is happening on the surface of the fruit that causes it to turn brown?

CHEMISTRY & YOU

Oxygen in air reacts with chemicals on the surface of the cut fruit. The oxygen oxidizes the chemicals in the fruit, causing a redox reaction and therefore the color change.

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Sample Problem 20.5

Identifying Redox Reactions

Use the change in oxidation number to identify whether each reaction is a redox reaction or a reaction of some other type. If a reaction is a redox reaction, identify the element reduced, the element oxidized, the reducing agent, and the oxidizing agent.

a. Cl₂(g) + 2NaBr(aq) → 2NaCl(aq) + Br₂(aq)

b. 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

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If changes in oxidation number occur, the reaction is a redox reaction.
The element whose oxidation number increases is oxidized and is the reducing agent.
The element whose oxidation number decreases is reduced and is the oxidizing agent.

Analyze Identify the relevant concepts.

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Sample Problem 20.5

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Solve Apply concepts to this situation.

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Sample Problem 20.5

a. Assign oxidation numbers.

0 +1 –1 +1 –1 0

Cl₂(g) + 2NaBr(aq) → 2NaCl(aq) + Br₂(aq)

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Solve Apply concepts to this situation.

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Sample Problem 20.5

a. Interpret the change (or lack of change) in oxidation numbers to identify if the reaction is a redox reaction.

This is a redox reaction.
The chlorine is reduced.
The bromide ion is oxidized.
Chlorine is the oxidizing agent; the bromide ion is the reducing agent.

0 +1 –1 +1 –1 0

Cl₂(g) + 2NaBr(aq) → 2NaCl(aq) + Br₂(aq)

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Solve Apply concepts to this situation.

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Sample Problem 20.5

b. Assign oxidation numbers.

+1 –2 +1 +1 +6 –2 +1 +6 –1 +1 –2

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

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Solve Apply concepts to this situation.

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Sample Problem 20.5

b. Interpret the change (or lack of change) in oxidation numbers to identify if the reaction is a redox reaction.

None of the elements change in oxidation number.
This is not a redox reaction.

+1 –2 +1 +1 +6 –2 +1 +6 –1 +1 –2

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

This is an acid-base (neutralization) reaction.

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Which of the following are redox reactions?

A. NH₃ + HCl → NH₄Cl

B. SO₃ + H₂O → H₂SO₄

C. NaOH + HCl → NaCl + H₂O

D. H₂S + NHO₃ → H₂SO₄ + NO₂ + H₂O

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Which of the following are redox reactions?

A. NH₃ + HCl → NH₄Cl

B. SO₃ + H₂O → H₂SO₄

C. NaOH + HCl → NaCl + H₂O

D. H₂S + NHO₃ → H₂SO₄ + NO₂ + H₂O

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Balancing Redox Equations

What are two different methods for balancing a redox equation?

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Balancing Redox Equations

Two different methods for balancing redox equations are the oxidation-number-change method and the half-reaction method.

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Balancing Redox Equations

Two different methods for balancing redox equations are the oxidation-number-change method and the half-reaction method.

These two methods are based on the fact that the total number of electrons gained in reduction must equal the total number of electrons lost in oxidation.

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Balancing Redox Equations

In the oxidation-number-change method, you balance a redox equation by comparing the increases and decreases in oxidation numbers.

Using Oxidation-Number Changes

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Balancing Redox Equations

Using Oxidation-Number Changes

To use this method, start with the skeleton equation for the redox reaction.

Fe₂O₃(s) + CO(g) → Fe(s) + CO₂(g) (unbalanced)

In a blast furnace, air is blown through a combination of iron ore and coke. The carbon monoxide produced from the oxidation of coke reduces the Fe³⁺ ions to metallic iron.

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Balancing Redox Equations

Using Oxidation-Number Changes

Step 1: Assign oxidation numbers to all the atoms in the equation.

Write the numbers above the atoms.
The oxidation number is stated per atom.

Fe₂O₃(s) + CO(g) → Fe(s) + CO₂(g)

+3 –2 +2 –2 0 +4 –2

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Balancing Redox Equations

Using Oxidation-Number Changes

Step 2: Identify which atoms are oxidized and which are reduced.

Iron is reduced.
Carbon is oxidized.

Fe₂O₃(s) + CO(g) → Fe(s) + CO₂(g)

+3 –2 +2 –2 0 +4 –2

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Balancing Redox Equations

Using Oxidation-Number Changes

Step 3: Use one bracketing line to connect the atoms that undergo oxidation and another such line to connect those that undergo reduction.

–3 (reduction)

Fe₂O₃(s) + CO(g) → Fe(s) + CO₂(g)

+3 –2 +2 –2 0 +4 –2

+2 (oxidation)

Write the oxidation-number change at the midpoint of each line.

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Balancing Redox Equations

Using Oxidation-Number Changes

Step 4: Make the total increase in oxidation number equal to the total decrease in oxidation number by using appropriate coefficients.

2 × (–3) = –6

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

+3 –2 +2 –2 0 +4 –2

3 × (+2) = +6

The oxidation-number increase should be multiplied by 3 and the decrease by 2.

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Balancing Redox Equations

Using Oxidation-Number Changes

Step 5: Finally, make sure the equation is balanced for both atoms and charge.

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

If necessary, finish balancing the equation by inspection.

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Sample Problem 20.6

Balancing Redox Equations by Oxidation-Number Change

Balance this redox equation by using the oxidation-number-change method.

K₂Cr₂O₇(aq) + H₂O(l) + S(s) → KOH(aq) + Cr₂O₃(s) + SO₂(g)

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You can balance redox equations by determining changes in oxidation numbers and applying the five steps.

Analyze Identify the relevant concepts.

1

Sample Problem 20.6

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Solve Apply concepts to this situation.

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Step 1: Assign oxidation numbers.

Sample Problem 20.6

K₂Cr₂O₇(aq) + H₂O(l) + S(s) → KOH(aq) + Cr₂O₃(s) + SO₂(g)

+1 +6 –2 +1 –2 0 +1–2+1 +3 –2 +4 –2

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Solve Apply concepts to this situation.

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Step 2: Identify the atoms that are oxidized and reduced.

Sample Problem 20.6

Cr is reduced.
S is oxidized.

K₂Cr₂O₇(aq) + H₂O(l) + S(s) → KOH(aq) + Cr₂O₃(s) + SO₂(g)

+1 +6 –2 +1 –2 0 +1–2+1 +3 –2 +4 –2

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Step 3: Connect the atoms that change in oxidation number. Indicate the signs and magnitudes of the changes.

Sample Problem 20.6

K₂Cr₂O₇(aq) + H₂O(l) + S(s) → KOH(aq) + Cr₂O₃(s) + SO₂(g)

+6 0 +3 +4

–3

+4

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Solve Apply concepts to this situation.

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Step 4: Balance the increase and decrease in oxidation numbers.

Sample Problem 20.6

2K₂Cr₂O₇(aq) + H₂O(l) + 3S(s) → KOH(aq) + 2Cr₂O₃(s) + 3SO₂(g)

+6 0 +3 +4

(4)(–3) = –12

(3)(+4) = +12

Four chromium atoms must be reduced (4 × (–3) = –12 decrease) for every three sulfur atoms that are oxidized (3 × (+4) = +12 increase). Put the coefficient 3 in front of S and SO₂, and the coefficient 2 in front of K₂Cr₂O₇ and Cr₂O₃.

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Step 5: Check the equation and balance by inspection if necessary.

Sample Problem 20.6

2K₂Cr₂O₇(aq) + 2H₂O(l) + 3S(s) →

4KOH(aq) + 2Cr₂O₃(s) + 3SO₂(g)

The coefficient 4 in front of KOH balances potassium. The coefficient 2 in front of H₂O balances hydrogen and oxygen.

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Balancing Redox Equations

A half-reaction is an equation showing just the oxidation or just the reduction that takes place in a redox reaction.

Using Half-Reactions

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Balancing Redox Equations

In the half-reaction method, you write and balance the oxidation and reduction half-reactions separately before combining them into a balanced redox equation.

The procedure is different, but the outcome is the same as with the oxidation-number-change method.

Using Half-Reactions

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Balancing Redox Equations

Sulfur is an element that can have several different oxidation numbers.

Using Half-Reactions

The oxidation of sulfur by nitric acid in aqueous solution is one example of a redox reaction that can be balanced by following the steps of the half-reaction method.

Oxidation Numbers of Sulfur in Different Substances
Substance	Oxidation number
SO₃	+6
SO₂	+4
Na₂S₂O₃	+2
S₂Cl₂	+1
S	0
H₂S	–2

S(s) + HNO₃(aq) → SO₂(g) + NO(g) + H₂O(l)

(unbalanced)

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Balancing Redox Equations

Using Half-Reactions

S(s) + H⁺(aq) + NO₃^–(aq) → SO₂(g) + NO(g) + H₂O(l)

Step 1: Write the unbalanced equation in ionic form.

Only HNO₃ is ionized.
The products are covalent compounds.

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Balancing Redox Equations

Using Half-Reactions

Step 2: Write separate half-reactions for the oxidation and reduction processes.

Sulfur is oxidized.
Nitrogen is reduced.
H⁺ ions and H₂O are not included because they are neither oxidized nor reduced.

Oxidation half-reaction: S(s) → SO₂(g)

Reduction half-reaction: NO₃^–(aq) → NO(g)

0 +4

+5 +2

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Balancing Redox Equations

Using Half-Reactions

Step 3: Balance the atoms in the half-reactions.

The half-reaction method is very useful in balancing equations for reactions that take place in acidic or basic solutions.

In acidic solutions, H₂O and H⁺(aq) can be used to balance oxygen and hydrogen as needed.
In basic solution, H₂O and OH^– are used to balance these species.

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Balancing Redox Equations

Sulfur is already balanced, but oxygen is not.
This reaction takes place in acid solution, so H₂O and H⁺(aq) are present and can be used to balance oxygen and hydrogen as needed.

Using Half-Reactions

Step 3: Balance the atoms in the half-reactions.

a. Balance the oxidation half-reaction.

2H₂O(l) + S(s) → SO₂(g) + 4H⁺(aq)

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Balancing Redox Equations

Nitrogen is already balanced.
Add two molecules of H₂O on the right to balance the oxygen.
Four hydrogen ions must be added to the left to balance hydrogen.

Using Half-Reactions

Step 3: Balance the atoms in the half-reactions.

b. Balance the reduction half-reaction.

4H⁺(aq) + NO₃^–(aq) → NO(g) + 2H₂O(l)

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Balancing Redox Equations

Neither half-reaction is balanced for charge.
Four electrons are needed on the right side in the oxidation half-reaction.

Using Half-Reactions

Step 4: Add enough electrons to one side of each half-reaction to balance the charges.

Oxidation: 2H₂O(l) + S(s) → SO₂(g) + 4H⁺(aq) + 4e^–

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Balancing Redox Equations

Neither half-reaction is balanced for charge.
Three electrons are needed on the left side in the reduction half-reaction.

Using Half-Reactions

Step 4: Add enough electrons to one side of each half-reaction to balance the charges.

Reduction: 4H⁺(aq) + NO₃^–(aq) + 3e^– → NO(g) + 2H₂O(l)

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Balancing Redox Equations

The number of electrons lost in oxidation must equal the number of electrons gained in reduction.

Using Half-Reactions

Step 5: Multiply each half-reaction by an appropriate number to make the numbers of electrons equal in both.

Oxidation: 6H₂O(l) + 3S(s) → 3SO₂(g) + 12H⁺(aq) + 12e^–

Reduction: 16H⁺(aq) + 4NO₃^–(aq) + 12e^– →

4NO(g) + 8H₂O(l)

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Balancing Redox Equations

Using Half-Reactions

Step 6: Add the balanced half-reactions to show an overall equation.

6H₂O(l) + 3S(s) + 16H⁺(aq) + 4NO₃^–(aq) + 12e^– →

3SO₂(g) + 12H⁺(aq) + 12e^– + 4NO(g) + 8H₂O(l)

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Balancing Redox Equations

Then subtract terms that appear on both sides of the equation.

Using Half-Reactions

Step 6: Add the balanced half-reactions to show an overall equation.

3S(s) + 4H⁺(aq) + 4NO₃^–(aq) →

3SO₂(g) + 4NO(g) + 2H₂O(l)

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Balancing Redox Equations

Recall that spectator ions are present but do not participate in or change during a reaction.
There are no spectator ions in this particular example.

Using Half-Reactions

Step 7: Add the spectator ions and balance the equation.

3S(s) + 4HNO₃(aq) → 3SO₂(g) + 4NO(g) + 2H₂O(l)

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Sample Problem 20.7

Balancing Redox Equations by Half-Reactions

Balance this redox equation using the half-reaction method.

KMnO₄(aq) + HCl(l) → MnCl₂(aq) + Cl₂(g) + KCl(aq)

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You can use the seven steps of the half-reaction method.

Analyze Identify the relevant concepts.

1

Sample Problem 20.7

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Solve Apply concepts to this problem.

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Step 1: Write the equation in ionic form.

Sample Problem 20.7

K⁺(aq) + MnO₄^–(aq) + H⁺(aq) + Cl^–(aq) →

Mn²⁺(aq) + 2Cl^–(aq) + Cl₂(g) + H₂O(l) + K⁺(aq) + Cl^–(aq)

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Solve Apply concepts to this problem.

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Step 2: Write half-reactions. Determine the oxidation and reduction process.

Sample Problem 20.7

Oxidation half-reaction: Cl^– → Cl₂

Reduction half-reaction: MnO₄^– → Mn²⁺

–1 0

+7 +2

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Step 3: Balance the atoms in each half-reaction.

Sample Problem 20.7

Oxidation: 2Cl^–(aq) → Cl₂(g) (atoms balanced)

Reduction: MnO₄^–(aq) + 8H⁺(aq) →

Mn²⁺(aq) + 4H₂O(l) (atoms balanced)

The solution is acidic, so use H₂O and H⁺ to balance the oxygen and hydrogen.

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Step 4: Balance the charges by adding electrons.

Sample Problem 20.7

Oxidation: 2Cl^–(aq) → Cl₂(g) + 2e^– (charges balanced)

Reduction: MnO₄^–(aq) + 8H⁺(aq) + 5e^– →

Mn²⁺(aq) + 4H₂O(l) (charges balanced)

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Step 5: Make the numbers of electrons equal.

Sample Problem 20.7

Oxidation: 10Cl^–(aq) → 5Cl₂(g) + 10e^–

Reduction: 2MnO₄^–(aq) + 16H⁺(aq) + 10e^– →

2Mn²⁺(aq) + 8H₂O(l)

Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2.

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Step 6: Add the half-reactions. Then, subtract the terms that appear on both sides.

Sample Problem 20.7

10Cl^–(aq) + 2MnO₄^–(aq) + 16H⁺(aq) + 10e^– →

5Cl₂(g) + 10e^–+ 2Mn²⁺(aq) + 8H₂O(l)

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Step 7: Add the spectator ions, making sure the charges and atoms are balanced.

Sample Problem 20.7

10Cl^– + 2MnO₄^– + 2K⁺ + 16H⁺ + 6Cl^– →

5Cl₂ + 2Mn²⁺+ 4Cl^– + 8H₂O + 2K⁺ + 2Cl^–

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Solve Apply concepts to this problem.

2

Combine the spectator and nonspectator Cl^– on each side.

Sample Problem 20.7

16Cl^–(aq) + 2MnO₄^–(aq) + 2K⁺(aq) + 16H⁺(aq) →

5Cl₂(g) + 2Mn²⁺(aq)+ 6Cl^–(aq) + 8H₂O(l) + 2K⁺(aq)

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Show the balanced equation for the substances given in the question (rather than for ions).

Sample Problem 20.7

2KMnO₄(aq) + 16HCl(aq) →

2MnCl₂(aq) + 5Cl₂(g) + 8H₂O(l) + 2KCl(aq)

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Use the half-reaction method to balance the following redox equation.

FeCl₃ + H₂S → FeCl₂ + HCl + S

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Use the half-reaction method to balance the following redox equation.

FeCl₃ + H₂S → FeCl₂ + HCl + S

Oxidation: H₂S → 2H⁺ + S + 2e^–

Reduction: 2Fe³⁺ + 2e^– → 2Fe²⁺

2FeCl₃ + H₂S → 2FeCl₂ + 2HCl + S

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Key Concepts

One class of chemical reactions is oxidation reduction (redox) reactions, in which electrons are transferred from one reacting species to another. The other class includes all other reactions, in which no electron transfer occurs.

To balance a redox equation using the oxidation-number-change method, the total increase in oxidation number of the species oxidized must be balanced by the total decrease in the oxidation number of the species reduced.

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Key Concepts

To balance a redox reaction using half-reactions, write separate half-reactions for the oxidation and the reduction. After you balance atoms in each half-reaction, balance electrons gained in the reduction with electrons lost in the oxidation.

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Glossary Terms

oxidation-number-change method: a method of balancing a redox equation by comparing the increases and decreases in oxidation numbers
half-reaction: an equation showing either the oxidation or the reduction that takes place in a redox reaction
half-reaction method: a method of balancing a redox equation by balancing the oxidation and reduction half-reactions separately before combining them into a balanced redox equation

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Redox equations can be balanced by two methods, the oxidation-number-change method and balancing the oxidation and reduction half-reactions.

BIG IDEA

Reactions

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