Chapter 17
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Chapter 17. Oxidation and Reduction by Christopher Hamaker. Oxidation–Reduction Reactions. Oxidation–reduction reactions are reactions involving the transfer of electrons from one substance to another. We have seen several oxidation–reduction reactions so far.

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Chapter 17

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Chapter 17

Chapter 17

Oxidation and

Reduction

by Christopher Hamaker

Chapter 17


Oxidation reduction reactions

Oxidation–Reduction Reactions

  • Oxidation–reduction reactions are reactions involving the transfer of electrons from one substance to another.

  • We have seen several oxidation–reduction reactions so far.

  • Whenever a metal and a nonmetal react, electrons are transferred.

    2 Na(s) + Cl2(g) → 2 NaCl(s)

  • Combustion reactions are also examples of oxidation–reduction reactions.

Chapter 17


Example of oxidation reduction

Example of Oxidation–Reduction

The rusting of iron is also an example of an oxidation–reduction reaction.

Iron metal reacts with oxygen in air to produce the ionic compound iron(III) oxide, which is composed of Fe3+ and O2- ions.

4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)

Iron loses electrons and is oxidized.

Fe → Fe3+ + 3 e-

Oxygen gains electrons and is reduced.

O2 + 4 e-→ 2 O2-

Chapter 17


Oxidation numbers

Oxidation Numbers

  • The oxidation number describes how many electrons have been lost or gained by an atom.

  • Oxidation numbers are assigned according to seven rules:

    • A metal or a nonmetal in the free state has an oxidation number of 0.

    • A monoatomic ion has an oxidation number equal to its ionic charge.

    • A hydrogen atom is usually assigned an oxidation number of +1.

Chapter 17


Rules for oxidation numbers

Rules for Oxidation Numbers

4.An oxygen atom is usually assigned an oxidation number of –2.

5.For a molecular compound, the more electronegative element is assigned a negative oxidation number equal to its charge as an anion.

6.For an ionic compound, the sum of the oxidation numbers for each of the atoms in the compound is equal to 0.

7.For a polyatomic ion, the sum of the oxidation numbers for each of the atoms in the compound is equal to the ionic charge on the polyatomic ion.

Chapter 17


Assigning oxidation numbers

Assigning Oxidation Numbers

  • What is the oxidation number for magnesium metal, Mg?

    • Mg = 0 according to Rule #1.

  • What is the oxidation number for sulfur in the sulfide ion, S2-?

    • S = –2 (Rule #2).

  • What is the oxidation number for barium and chloride in BaCl2?

    • Ba is present as Ba2+, so Ba = +2 (Rules #2 and #6).

    • Cl is present as Cl-, so Cl = –1 (Rules #2 and #6).

Chapter 17


Oxidation numbers in compounds

Oxidation Numbers in Compounds

  • What are the oxidation numbers for each element in oxalic acid, H2C2O4?

    – H = +1 (Rule #3).

    – O = 2 (Rule #4).

  • To find the oxidation number for carbon, recall that the sum of the oxidation numbers equals zero.

    2(+1) + 2(ox no C) + 2(–2) = 0

    2 + 2(ox no C) – 8 = 0

    2(ox no C) = +6

    C = +3

Chapter 17


Oxidation numbers in compounds continued

Oxidation Numbers in Compounds, Continued

  • What are the oxidation numbers for each element in carbon tetrachloride, CCl4?

    – Cl = –1 (Rule #5).

  • To find the oxidation number for carbon, recall that the sum of the oxidation numbers equals zero.

    (ox no C) + 4(–1) = 0

    (ox no C) – 4 = 0(ox no C) = +4

    C = +4

Chapter 17


Oxidation numbers in polyatomic ions

Oxidation Numbers in Polyatomic Ions

  • What are the oxidation numbers for chlorine and oxygen in the perchlorate ion, ClO4-?

    – O = –2 (Rule #4).

  • To find the oxidation number for carbon, recall that the sum of the oxidation numbers equals the charge on the ion (Rule #7).

    (ox no Cl) + 4(–2) = –1

    (ox no Cl) – 8 = –1(ox no Cl) = +7

    Cl = +7

Chapter 17


Redox reactions

Redox Reactions

  • Recall that a chemical reaction involving the transfer of electrons is an oxidation–reduction reaction, or a redox reaction.

  • For example, iron metal is heated with sulfur to produce iron(II) sulfide.

    Fe(s) + S(s) → FeS(s)

  • The sulfur changes from 0 to –2 and the iron changes from 0 to +2.

Chapter 17


Oxidation and reduction

Oxidation and Reduction

The iron loses electrons and is oxidized.

Fe → Fe2+ + 2 e-

The sulfur gains electrons and is reduced.

S + 2 e- → S2-

Chapter 17


Oxidizing and reducing agents

Oxidizing and Reducing Agents

Oxidation is the loss of electrons, and reduction is the gain of electrons.

An oxidizing agent is a substance that causes oxidation by accepting electrons. The oxidizing agent is reduced.

  • A reducing agent is a substance that causes reduction by donating electrons. The reducing agent is oxidized.

Chapter 17


Redox reactions1

Redox Reactions

In a redox reaction, one substance must be oxidized and one substance must be reduced.

The total number of electrons lost is equal to the total electrons gained.

Chapter 17


Redox reactions continued

Redox Reactions, Continued

Identify the reducing agent, the oxidizing agent, and the oxidation and reduction in the following reaction:

CuS(s) + H2(g) → Cu(s) + H2S(g)

Cu is reduced from +2 to 0.

H is oxidized from 0 to +1.

Chapter 17


Ionic equations

Ionic Equations

  • Redox reactions in aqueous solutions are most often shown in the ionic form.

  • Ionic equations readily show us the change in oxidation number.

    5 Fe2+(aq)+MnO4-(aq)+8 H+(aq)→5 Fe3+(aq)+Mn2+(aq)+4 H2O(l)

  • We can easily tell that the oxidation number of iron changes from +2 to +3; iron is oxidized.

  • Manganese is reduced from +7 in MnO4- to +2 in Mn2+; manganese is reduced.

Chapter 17


Ionic equations continued

Ionic Equations, Continued

We can map the reaction to show the oxidation and reduction processes and to determine the oxidizing and reducing agents:

Chapter 17


Balancing redox reactions

Balancing Redox Reactions

  • When we balance redox reactions, the number of electrons lost must equal the number of electrons gained.

  • We will balance redox reactions using the oxidation number method, which has three steps:

    • Inspect the reaction and the substances undergoing a change in oxidation number.

      • Write the oxidation number above each element.

      • Diagram the number of electrons lost by the oxidized substance and the number of electrons gained by the reduced substance.

Chapter 17


Oxidation number method

Oxidation Number Method

2.Balance each element in the equation using a coefficient. Remember that the electrons lost must equal the electrons gained. If they are not the same, balance the electrons as follows:

  • In front of the oxidized substance, place a coefficient equal to the number of electrons gained by the reduced substance.

  • In front of the reduced substance, place a coefficient equal to the number of electrons lost by the oxidized substance.

Chapter 17


Oxidation number method continued

Oxidation Number Method, Continued

3.After balancing the equation, verify that the coefficients are correct.

  • Place a checkmark above the symbol for each element to verify that the number of atoms is the same on both sides.

  • For ionic equations, verify that the total charge on the left side of the equation is the same as the total charge on the right side of the equation.

Chapter 17


Balancing a redox reaction

Balancing a Redox Reaction

Balance the following redox reaction using the oxidation number method:

Fe2O3(l) + CO(g) → Fe(l) + CO2(g)

  • Since the total electrons gained and lost must be equal, we must find the lowest common multiple. For this reaction, it is 6.

Chapter 17


Balancing a redox reaction continued

Balancing a Redox Reaction, Continued

  • Each iron gains three electrons, so place a 2 in front of the Fe. There are two iron atoms in Fe2O3, so no coefficient is necessary.

  • Each carbon loses two electrons, so place a 3 in front of CO and CO2.

    Fe2O3(l) + 3 CO(g) → 2 Fe(l) + 3 CO2(g)

  • Check to see that the number of each type of atom is the same on both sides:

    • There are 2 Fe atoms, 6 O atoms, and 3 C atoms on each side.

Chapter 17


Balancing redox equations

Balancing Redox Equations

  • An alternative method for balancing redox reactions is the half-reaction method.

  • A half-reaction shows the oxidation or reduction process of a redox reaction separately.

  • The steps are as follows:

    • Write the half-reaction for both the oxidation and reduction processes.

Chapter 17


Half reaction method

Half-Reaction Method

2.Balance the atoms in each half-reaction using coefficients.

  • Balance all elements except oxygen and hydrogen.

  • Balance oxygen using H2O.

  • Balance hydrogen using H+.

  • For reactions in basic solution, add one OH- to each side for each H+ and combine H+ and OH- to H2O.

  • Balance the ionic charges using electrons.

Chapter 17


Half reaction method continued

Half-Reaction Method, Continued

3.Multiply each half-reaction by a whole number so that the total number of electrons in each is the same.

4.Add the two half-reactions together and cancel the identical species, including electrons.

5.After balancing, verify that the coefficients are correct by making sure there are the same number of each atom on each side of the reaction and that the overall charge is the same on both sides.

Chapter 17


Balancing a redox equation

Balancing a Redox Equation

  • Balance the following redox reaction using the half-reaction method:

    Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)

  • The two unbalanced half-reactions are:

    Fe2+ → Fe3+

    MnO4- → Mn2+

  • We balance the two half-reactions as follows:

    Fe2+ → Fe3+ + e-

    5 e- + 8 H+ + MnO4- → Mn2+ + 4 H2O

Chapter 17


Balancing a redox equation continued

Balancing a Redox Equation, Continued

  • Since Fe2+ loses one electron and MnO4- gains five electrons, we have to multiply the iron half-reaction by 5:

    5 Fe2+ → 5 Fe3+ + 5 e-

    5 e- + 8 H+ + MnO4- → Mn2+ + 4 H2O

  • We add the two half-reactions together and cancel out the five electrons on each side to get the following balanced equation:

    5 Fe2+ + 8 H+ + MnO4- →Fe2+ + Mn2+ + 4 H2O

Chapter 17


Spontaneous redox reactions

Spontaneous Redox Reactions

  • Chemical reactions that occur without any input of energy are spontaneous.

  • The reaction of zinc metal with aqueous copper sulfate is spontaneous:

    Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

  • Cu2+ has a greater tendency to gain electrons than Zn2+.

  • We can compare metals and arrange them in a series based on their ability to gain electrons.

Chapter 17


Reduction potentials

Reduction Potentials

The tendency for a substance to gain electrons is its reduction potential.

The strongest reducing agent is the most easily oxidized.

At right is a table of reduction potentials for several metals.

Chapter 17


Spontaneous reactions

Spontaneous Reactions

A species can be reduced by any reducing agent lower in the table.

Any metal below H2 can react with acid and be oxidized.

A species can be oxidized by any oxidizing agent above it on the table.

Chapter 17


Predicting spontaneous reactions

Predicting Spontaneous Reactions

weaker oxidizing agent

weaker reducing agent

stronger reducing agent

stronger oxidizing agent

  • The reaction is nonspontaneous as written.

  • A reaction will be spontaneous when the stronger oxidizing and reducing agents are the reactants, and the weaker oxidizing and reducing agents are the products.

  • Predict whether the following reaction will be spontaneous:

    Ni2+(aq) + Sn(s) → Ni(s) + Sn2+(aq)

Chapter 17


Voltaic cells

Voltaic Cells

  • The conversion of chemical energy to electrical energy in a redox reaction is electrochemistry.

  • If we can physically separate the oxidation and reduction half-reactions, we can use the electrons from the redox reaction to do work. This is called an electrochemical cell.

  • Let’s look at the following reaction of zinc metal with copper(II) sulfate:

    Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)

Chapter 17


Voltaic cells continued

Voltaic Cells, Continued

We place a zinc electrode in aqueous ZnSO4 and a copper electrode in aqueous CuSO4. The electrodes are connected by a wire to allow the flow of electrons.

A salt bridge is used to complete the circuit.

  • Zinc metal is oxidized and copper ions are reduced in each half cell.

Chapter 17


Voltaic cells continued1

Voltaic Cells, Continued

Oxidation occurs at the anode of an electrochemical cell.

Reduction occurs at the cathode of an electrochemical cell.

Electrons flow through the wire from the anode to the cathode in a voltaic cell.

Negatively charged ions travel through the salt bridge away from the cathode and toward the anode in a voltaic cell.

Chapter 17


Batteries

Batteries

A battery is any electrochemical cell that spontaneously produces electrical energy.

A battery has stored chemical energy that can be converted to electrical energy by a chemical reaction.

A battery is made up of one or more voltaic cells.

An automobile battery is made up of six cells.

A battery without an electrolyte solution is a dry cell.

Chapter 17


Electrolytic cells

Electrolytic Cells

Electrolytic cells are electrochemical cells that do not operate spontaneously. The process is referred to as electrolysis.

A source of electricity is required to drive an electrolytic cell.

An example of an electrolysis reaction is the recharging of the battery in a cell phone.

Chapter 17


Chemistry connection hybrid vehicles

Chemistry Connection: Hybrid Vehicles

Hybrid vehicles are designed to operate on both gasoline and lightweight voltaic cells.

Hybrids generate energy while braking or coasting, and store it in batteries.

The General Motors Volt is estimated to get better than 100 miles per gallon.

Chapter 17


Chapter summary

Chapter Summary

A redox reaction is a reaction involving the transfer of electrons from one substance to another.

The oxidation number describes how many electrons have been lost or gained by an atom.

Oxidation is the loss of electrons.

Reduction is the gain of electrons.

Chapter 17


Chapter summary continued

Chapter Summary, Continued

An oxidizing agent is a substance that causes oxidation by accepting electrons. The oxidizing agent is reduced.

A reducing agent is a substance that causes reduction by donating electrons. The reducing agent is oxidized.

In redox reactions, the number of electrons lost must equal the number of electrons gained.

Chapter 17


Chapter summary continued1

Chapter Summary, Continued

  • There are two methods to balance redox reactions:

    • The oxidation number method

    • The half-reaction method

  • The tendency for a substance to gain electrons is its reduction potential.

  • The conversion of chemical energy to electrical energy in a redox reaction is electrochemistry.

  • We can physically separate oxidation and reduction half-reactions and use the electrons from the redox reaction to do work in an electrochemical cell.

  • Chapter 17


    Chapter summary continued2

    Chapter Summary, Continued

    Chapter 17


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