Chapter 20 Oxidation-Reduction Reactions 20.3 Describing Redox

Chapter 20 Oxidation-Reduction Reactions 20.3 Describing Redox
20.1 The Meaning of Oxidation and Reduction 20.2 Oxidation Numbers 20.3 Describing Redox Equations Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Why does cut fruit turn brown?
CHEMISTRY & YOU Why does cut fruit turn brown? 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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Identifying Redox Reactions
What are the two classes of chemical reactions? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU 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? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU 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? 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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze Identify the relevant concepts.
Sample Problem 20.5 Analyze Identify the relevant concepts. 1 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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solve Apply concepts to this situation.
Sample Problem 20.5 Solve Apply concepts to this situation. 2 a. Assign oxidation numbers. – – Cl2(g) + 2NaBr(aq) → 2NaCl(aq) + Br2(aq) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.5 Solve Apply concepts to this situation. 2 a. Interpret the change (or lack of change) in oxidation numbers to identify if the reaction is a redox reaction. – – Cl2(g) + 2NaBr(aq) → 2NaCl(aq) + Br2(aq) 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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.5 Solve Apply concepts to this situation. 2 b. Assign oxidation numbers. +1 – – – –2 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.5 Solve Apply concepts to this situation. 2 b. Interpret the change (or lack of change) in oxidation numbers to identify if the reaction is a redox reaction. +1 – – – –2 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) None of the elements change in oxidation number. This is not a redox reaction. This is an acid-base (neutralization) reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Balancing Redox Equations
What are two different methods for balancing a redox equation? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Oxidation-Number Changes In the oxidation-number-change method, you balance a redox equation by comparing the increases and decreases in oxidation numbers. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Write the oxidation-number change at the midpoint of each line. –3 (reduction) Fe2O3(s) + CO(g) → Fe(s) + CO2(g) +3 – – –2 +2 (oxidation) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. The oxidation-number increase should be multiplied by 3 and the decrease by 2. 2 × (–3) = –6 Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) +3 – – –2 3 × (+2) = +6 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Oxidation-Number Changes Step 5: Finally, make sure the equation is balanced for both atoms and charge. If necessary, finish balancing the equation by inspection. Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

K2Cr2O7(aq) + H2O(l) + S(s) → KOH(aq) + Cr2O3(s) + SO2(g)
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) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.6 Analyze Identify the relevant concepts. 1 You can balance redox equations by determining changes in oxidation numbers and applying the five steps. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.6 Solve Apply concepts to this situation. 2 Step 1: Assign oxidation numbers. K2Cr2O7(aq) + H2O(l) + S(s) → KOH(aq) + Cr2O3(s) + SO2(g) – – – – –2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.6 Solve Apply concepts to this situation. 2 Step 2: Identify the atoms that are oxidized and reduced. K2Cr2O7(aq) + H2O(l) + S(s) → KOH(aq) + Cr2O3(s) + SO2(g) – – – – –2 Cr is reduced. S is oxidized. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.6 Solve Apply concepts to this situation. 2 Step 3: Connect the atoms that change in oxidation number. Indicate the signs and magnitudes of the changes. K2Cr2O7(aq) + H2O(l) + S(s) → KOH(aq) + Cr2O3(s) + SO2(g) –3 +4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

2K2Cr2O7(aq) + H2O(l) + 3S(s) → KOH(aq) + 2Cr2O3(s) + 3SO2(g)
Sample Problem 20.6 Solve Apply concepts to this situation. 2 Step 4: Balance the increase and decrease in oxidation numbers. 2K2Cr2O7(aq) + H2O(l) + 3S(s) → KOH(aq) + 2Cr2O3(s) + 3SO2(g) (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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.6 Solve Apply concepts to this situation. 2 Step 5: Check the equation and balance by inspection if necessary. 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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Half-Reactions A half-reaction is an equation showing just the oxidation or just the reduction that takes place in a redox reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Half-Reactions 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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Half-Reactions Oxidation Numbers of Sulfur in Different Substances Substance Oxidation number SO3 +6 SO2 +4 Na2S2O3 +2 S2Cl2 +1 S H2S –2 Sulfur is an element that can have several different oxidation numbers. 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. S(s) + HNO3(aq) → SO2(g) + NO(g) + H2O(l) (unbalanced) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Half-Reactions Step 1: Write the unbalanced equation in ionic form. Only HNO3 is ionized. The products are covalent compounds. S(s) + H+(aq) + NO3–(aq) → SO2(g) + NO(g) + H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Half-Reactions Step 3: Balance the atoms in the half-reactions. a. Balance the oxidation half-reaction. 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. 2H2O(l) + S(s) → SO2(g) + 4H+(aq) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Half-Reactions Step 3: Balance the atoms in the half-reactions. b. Balance the reduction half-reaction. 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. 4H+(aq) + NO3–(aq) → NO(g) + 2H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Half-Reactions Step 4: Add enough electrons to one side of each half-reaction to balance the charges. Neither half-reaction is balanced for charge. Four electrons are needed on the right side in the oxidation half-reaction. Oxidation: 2H2O(l) + S(s) → SO2(g) + 4H+(aq) + 4e– Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Half-Reactions Step 4: Add enough electrons to one side of each half-reaction to balance the charges. Neither half-reaction is balanced for charge. Three electrons are needed on the left side in the reduction half-reaction. Reduction: 4H+(aq) + NO3–(aq) + 3e– → NO(g) + 2H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Half-Reactions Step 5: Multiply each half-reaction by an appropriate number to make the numbers of electrons equal in both. The number of electrons lost in oxidation must equal the number of electrons gained in reduction. Oxidation: 6H2O(l) + 3S(s) → 3SO2(g) + 12H+(aq) + 12e– Reduction: 16H+(aq) + 4NO3–(aq) + 12e– → 4NO(g) + 8H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Half-Reactions Step 6: Add the balanced half-reactions to show an overall equation. Then subtract terms that appear on both sides of the equation. 3S(s) + 4H+(aq) + 4NO3–(aq) → 3SO2(g) + 4NO(g) + 2H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Half-Reactions Step 7: Add the spectator ions and balance the equation. 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. 3S(s) + 4HNO3(aq) → 3SO2(g) + 4NO(g) + 2H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

KMnO4(aq) + HCl(l) → MnCl2(aq) + Cl2(g) + KCl(aq)
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) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solve Apply concepts to this problem.
Sample Problem 20.7 Solve Apply concepts to this problem. 2 Step 1: Write the equation in ionic form. K+(aq) + MnO4–(aq) + H+(aq) + Cl–(aq) → Mn2+(aq) + 2Cl–(aq) + Cl2(g) + H2O(l) + K+(aq) + Cl–(aq) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.7 Solve Apply concepts to this problem. 2 Step 2: Write half-reactions. Determine the oxidation and reduction process. Oxidation half-reaction: Cl– → Cl2 Reduction half-reaction: MnO4– → Mn2+ – Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.7 Solve Apply concepts to this problem. 2 Step 3: Balance the atoms in each half-reaction. The solution is acidic, so use H2O and H+ to balance the oxygen and hydrogen. Oxidation: 2Cl–(aq) → Cl2(g) (atoms balanced) Reduction: MnO4–(aq) + 8H+(aq) → Mn2+(aq) + 4H2O(l) (atoms balanced) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.7 Solve Apply concepts to this problem. 2 Step 4: Balance the charges by adding electrons. Oxidation: 2Cl–(aq) → Cl2(g) + 2e– (charges balanced) Reduction: MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l) (charges balanced) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.7 Solve Apply concepts to this problem. 2 Step 5: Make the numbers of electrons equal. Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2. Oxidation: 10Cl–(aq) → 5Cl2(g) + 10e– Reduction: 2MnO4–(aq) + 16H+(aq) + 10e– → 2Mn2+(aq) + 8H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.7 Solve Apply concepts to this problem. 2 Step 6: Add the half-reactions. Then, subtract the terms that appear on both sides. 10Cl–(aq) + 2MnO4–(aq) + 16H+(aq) + 10e– → 5Cl2(g) + 10e– + 2Mn2+(aq) + 8H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.7 Solve Apply concepts to this problem. 2 Step 7: Add the spectator ions, making sure the charges and atoms are balanced. 10Cl– + 2MnO4– + 2K+ + 16H+ + 6Cl– → 5Cl2 + 2Mn2+ + 4Cl– + 8H2O + 2K+ + 2Cl– Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.7 Solve Apply concepts to this problem. 2 Combine the spectator and nonspectator Cl– on each side. 16Cl–(aq) + 2MnO4–(aq) + 2K+(aq) + 16H+(aq) → 5Cl2(g) + 2Mn2+(aq) + 6Cl–(aq) + 8H2O(l) + 2K+(aq) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 20.7 Solve Apply concepts to this problem. 2 Show the balanced equation for the substances given in the question (rather than for ions). 2KMnO4(aq) + 16HCl(aq) → 2MnCl2(aq) + 5Cl2(g) + 8H2O(l) + 2KCl(aq) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

BIG IDEA Reactions Redox equations can be balanced by two methods, the oxidation-number-change method and balancing the oxidation and reduction half-reactions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 20 Oxidation-Reduction Reactions 20.3 Describing Redox

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Chapter 20 Oxidation-Reduction Reactions 20.3 Describing Redox

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