Chapter 11 - Electrochemistry and Electron Transfer Reactions (2023)

Introduction

Redox or electron transfer reactions form one of the largest and most important classes of reactions in chemistry. All reactions involving molecular oxygen, such as combustion and corrosion, are electron transfer reactions. Biological processes such as respiration, photosynthesis, and the breakdown of food molecules consist of sequences of electron transfer reactions that serve to transport and use solar energy. Batteries are devices that allow us to harness free energy from electron transfer reactions.

11.1 Electron transfer or redox reactions

Introduction

We begin our study of electron transfer reactions by introducing some terms and definitions and examining the electron transfer process.

requirements

• •

  1.8 Electromagnetism and Coulomb's Law

• •

  4.2 Orbital occupations of ions Exercise(Determine the occupation of the electrons of an ion from the occupation of its atom and vice versa).

• •

  4.4 Oxidation states(Determine the oxidation states of the atoms in a compound or ion.)

• •

  9.7 Free energy(Describe the factors that determine whether a chemical process occurs spontaneously.)

• •

  2.3 Bohr model

Goals

• •

  Describe an electron transfer reaction.

• •

  Define oxidation and reduction.

• •

  Define oxidizing and reducing agents.

• •

  Indicate whether a substance can act as an oxidizing agent, a reducing agent, or both.

• •

  Identify the oxidizing and reducing agents in a redox reaction.

• •

  Determine the number of electrons transferred in a balanced chemical equation for an electron transfer reaction.

• •

  Identify the donor and acceptor orbitals in a simple redox reaction.

• •

  Explain the influence of orbital energy on electron transfer.

• •

  Identify the factor responsible for oxidation and reduced strength.

• •

  Describe a redox pair and write the abbreviation for a given pair.

11.1-1. Introduction to electron transfer video

11.1-2. electron transfer

The reaction that occurs when iron (steel wool) is added to a CuSO solution₄is shown in the following table.

		→
Steel wool is mainly composed of Fe atoms.	copper2+ions gives a CuSO4solution its blue color.	→	Steel wool is coated with metallic Cu where it has been dipped in CuSO4Solution.	The solution loses its color because the Cu2+were replaced by colorless Fe2+ions

Table 11.1: An electron transfer reaction

We can make the following observations:

• 1

  The deep blue color of CuSO₄solution, due to the presence of Cu²⁺ions, is lost.

• 2

  A brown solid is formed. Analysis shows that the solid is metallic copper.

• 3

  Steel wool decomposes when the Fe atoms disappear.

• 4

  The analysis shows that Fe²⁺Ions are generated in solution.

and draw the following conclusions:

• 1

  copper²⁺converted to Cu.

• 2

  Faith became faith²⁺.

During the reaction, the oxidation state of copper changes from +2 to Cu²⁺Ions in solution at 0 in atoms containing metallic copper. each copper²⁺The ion needs to gain two electrons to become a Cu atom. Likewise, the oxidation state of iron changes from 0 in the atoms that make up steel wool to +2 in Fe.²⁺ions in solution. Each Fe atom has to lose two electrons to become an Fe²⁺Ion. Thus, each Fe atom donates two electrons, while each Cu atom²⁺takes two electrons, i.e. two electrons are transferred from iron atoms to Cu²⁺ions in solution. This is an example of aelectron transfer reaction🇧🇷 The reaction is written as

11.1-3. oxidation and reduction

ReductionIt is an electron collector. The added electrons "lower" the oxidation state of the substance. copper²⁺The ions gain two electrons and are reduced to copper atoms. Note that the two-electron reduction reduces the oxidation state of copper from +2 in the ion to 0 in the atom.

Oxidationis a loss of electrons. The loss of negative charge causes an increase in the oxidation state of the substance. Fe atoms lose two electrons and are oxidized to Fe.²⁺ions Note that the oxidation of two electrons increases the oxidation state of iron from 0 in the atom to +2 in the ion.redox reactionsThey are those involving oxidation and reduction. Electron transfer reactions always involve oxidation and reduction, as you can't gain electrons if you don't lose any.

11.1-4. Oxidizing and reducing agents

Electron transfer results from a combination of oxidation and reduction. A species cannot be oxidized unless another species accepts electrons and is reduced. That is, oxidation causes reduction andand viceversa🇧🇷 Consequently, the species that is oxidized by the reaction is calledreducing agentoto reduce, and the species that is reduced by the reaction is calledoxidizing agentooxidizing agent.The reducing agent contains the electrons that are transferred during the reaction, so it is in its reduced form, which we call red.₁🇧🇷 Electron transfer converts it to its oxidized form, which we will call Ox₁🇧🇷 Likewise, the oxidant has empty orbitals that can accept the donated electrons, so it is in its oxidized form, Ox₂🇧🇷 Accepting the electrons converts it to its reduced red form.₂🇧🇷 A typical redox reaction can be expressed as follows.

That's why,reducing agent, red₁, can be identified as the reduced form (lower oxidation state) of species 1, while the oxidant Ox₂is the oxidized form (higher oxidation state form) of species 2.

11.1-5. Requirements for reducing and oxidizing agents

To function as a reducing agent, a substance must be able to donate electrons and reach a higher oxidation state, so reducing agents must contain atoms that can be oxidized. Also, oxidants must be able to accept electrons to reach a lower oxidation state, so oxidants must contain atoms that can be reduced. For example, the nitrogen atom in NH₃it has an oxidation state of -3, which is the lowest oxidation state that nitrogen can have. therefore NH₃it can be oxidized but not reduced, so it can be a reducing agent. The nitrogen atom in NO₃^1-It is in the +5 oxidation state, the highest it can be, so nitrate ions cannot be oxidized or act as a reducing agent, but can be reduced and act as an oxidizing agent.

11.1-6. Oxidizing and reducing agents in reactions.

To determine the oxidizing and reducing agents in a redox reaction, you must first identify the atoms whose oxidation state is changing. Changes in oxidation state are generally attributed to a single atom of a polyatomic molecule or ion involved in a redox reaction, but the molecule or ion is the oxidizing or reducing agent. The following points will help you to identify oxidizing and reducing agents:

• 1

  Any atom present asElementoxidizes or reduces.

• 2

  O and H are commonly encountered in redox reactions, but remember from Chapter 4 that the oxidation states of O and H are –2 and +1, respectively, in most of their compounds, so they rarely change. That's why,H₂OH¹⁺, OH^1-rarely participate in electron transfer🇧🇷 Exceptions occur when either is elementary (OR₂s/o h₂).

• 3

  Transition metals and polyatomic ions often participate in redox reactions.

Identify the oxidizing agent, reducing agent, and number of electrons transferred in the following redox reaction.

We identify elemental copper (Item 1) as a redox reagent. Point 2 can be used to ignore O and H. Copper oxidizes from 0 to +2, soWhat is the reducing agent.

We identified the nitrate ion (item 3) as the redox reagent. Point 2 can be used to neglect O and H. The oxidation state of nitrogen in the nitrate ion is determined using the rules developed inSection 4.4e

Likewise, the oxidation state of nitrogen in NO is +2, so nitrogen is reduced from +5 to +2. However, the oxidizing agent is the molecule or ion, not the atom, soNitrate ions are the oxidizing agent.🇧🇷 In fact, the oxidizing agent is nitric acid due to the presence of H¹⁺ions The simplest definition of an acid is a substance that produces H¹⁺ions in the water. Acids are discussed in detail inChapter 12.

11.1-7. transfer electrons

The number of electrons transferred in a reaction, i. h the number gained by the oxidant or lost by the reductant is an important feature of a balanced redox equation. Consider the following redox equation:

Each chromium atom loses three electrons when its oxidation state changes from 0 to +3. There is only one chromium atom in the equation, as well as the number of electrons lost by the chromium atoms in the above equation.

Each silver ion gains an electron when its oxidation state changes from +1 to 0. However, there are three silver ions, so the number of electrons lost is the same

The balanced equation implies a transfer of three electrons, that isnorte= 3.

11.1-8. Example of transferred electrons

Next, the method for determining the number of electrons transferred in oxidation and reduction processes is demonstrated.Chemical oxidation-reduction reaction:

Oxidation:

The oxidation state of copper changes from 0 to +2, corresponding to a loss of two electrons for each copper atom. So there are three copper atoms in the equation.

This is a six-electron oxidation.Reduction:

The nitrogen atom in the nitrate ion is in a +5 oxidation state, but the N atom in NO is +2. Thus, each nitrogen atom receives three electrons. So there are two nitrogen atoms on each side of the equation.

This is a six-electron reduction.As needed, the number of electrons gained in reduction equals the number lost in oxidation. We conclude that the reaction is a six-electron transfer reaction.

11.1-9. exercise

11.1-10. giver/acceptor

In a redox reaction, electrons are transferred from a series of orbitals to the electron donor, called El.donor orbitalsin a series of orbitals in the acceptor calledacceptor orbitals. To beredox electronsthey are in the donor orbitals, so the donor must be in a reduced form of the substance called red₁🇧🇷 The acceptor bits can accept the redox electrons, so the acceptor must be in an oxidized form called Ox₂.Electron transfer removes electrons from the reductant donor orbitals, converting them to acceptor orbitals and turning the substance into an oxidant. Likewise, the transfer places electrons in the acceptor orbitals of the oxidant, transforming them into donor orbitals and transforming the substance into a reductant.The red spontaneous process₁+ boi₂ →Boi₁+ rot₂will be considered below.reagentsReagent 1 is in its reduced form (Red₁), then it is a donor. Reagent 2 is in the oxidized form (Ox₂), that is, an acceptor. As we'll see in the next section, the fact that the reaction is spontaneous usually means that the donor orbitals have a higher energy than the acceptor orbitals.

productsThe electron transfer depletes the redox orbital of reactant 1, leaving product 1 in its oxidized form (Ox₁🇧🇷 Electron transfer fills the redox orbital of reagent 2, leaving it in its reduced form (Red₂). Boi₁it has an empty orbital so now it is an acceptor (oxidant) and red₂it has one orbital filled, so now it's a donor (reducing agent). That's why,Electron transfer converts the oxidizing agent into a reducing agent and the reducing agent into an oxidizing agent.

11.1-11. low energy

Remember ofSection 9.7that a reaction proceeds spontaneously at constant temperature and pressure to minimize its free energy. Also, a reaction is extensive if its standard free energy change is negative. Therefore, electrons are transferred spontaneously from a donor to an acceptor when the transfer reduces their free energy and are transferred extensively when the transfer reduces the standard free energy. We will soon see that there are several factors that contribute to the change in standard free energy, but the most important is usually the difference in orbital energy. That's why,The driving force behind many redox reactions arises because the acceptor orbitals are much lower in energy than the donor orbitals..In extended electron transfer, the donor orbitals on the reactants are higher in energy than the acceptor orbitals, but the acceptor orbitals on the products are higher in energy. Consequently, the oxidant produced is weaker than the oxidant that reacts (the acceptor orbital is smaller in the reactants). Furthermore, the produced reductant is weaker than the reactive reductant (the donor orbital is higher in reactants). We conclude from thisSpontaneous transfer of electrons occurs between stronger oxidizing and reducing agents to produce weaker oxidizing and reducing agents.Weak reagents do not react to create strong ones.

Electron transfer between Fe and Cu²⁺is large because occupied Fe orbitals have much higher energy than vacant Cu orbitals²⁺.Electron transfer from Cu to Fe²⁺is not extensive (Cu does not reduce Fe²⁺) because electrons in Cu have much less energy than empty orbitals in Fe²⁺🇧🇷 Electrons are not transferred extensively at higher energies.

11.1-12. Strengths of oxidizing and reducing agents.

Strong reducing agents have high energy electrons and strong oxidizing agents have low energy empty orbitals.When a strong oxidizing agent comes into contact with a strong reducing agent, high-energy electrons in the donor orbitals spontaneously flow into vacant orbitals in the acceptor with much lower energy. As we'll see shortly, the free energy released can be used to do work on batteries. When the donor orbitals are much smaller than the acceptor orbitals, the reducing and oxidizing agents are too weak and electrons are not transferred.

11.1-13. Redox-Paar

Electron transfer converts a reducing agent (red₁) into an oxidizing agent (Ox₁) and converts an oxidizing agent (Ox₂) in a reducing agent (red₂🇧🇷 The oxidizing agent and the reducing agent form a redox couple. Two redox couples are involved in redox reactions. A redox couple is written as the oxidized form, as a bar, and then as the reduced form (Ox/Red). For example, the two redox couples in the reaction of copper(II) and iron are expressed as follows:

• •

  Reduction Torque: Cu²⁺/Cu

• •

  Oxidation: Fe²⁺/Fe

11.2 The half-reaction

Introduction

Redox reactions can be divided into two half-reactions, an oxidation and a reduction, which explicitly show the loss and gain of electrons. The total reaction is the sum of the two half-reactions. Using half-reactions simplifies writing balanced redox reactions and helps us to better quantify the driving force behind a redox reaction.

Goals

• •

  Identify the half-reaction involved in a redox reaction.

• •

  Write a balanced redox reaction with a table of reduction half-reactions.

11.2-1. the half reaction

A redox reaction can be divided into two parts.the half reaction: an oxidation half-reaction and a reduction half-reaction. Gained electrons are shown as reactants in the reduction half-reaction, and lost electrons are shown as products in the oxidation half-reaction. The net redox reaction is simply the sum of the two half-reactions. The two half-reactions in the redox reaction between Fe and Cu.²⁺they are:

Note that the electrons gained in reduction equal the number lost in oxidation, so the sum of the electrons gives the net reaction. ThisThe net reaction never contains electrons because the number of electrons gained in reduction must equal the number of electrons lost in oxidation..

Exercise 11.5:

Consider the following reaction:

3 Sn + 2 Cr³⁺ →3 Sn²⁺+ 2 Cr

Represent the oxidation half-reaction aszPutrefaction→ zboi +nortemi^1-.Specify subscripts with an underscore (_) and superscripts with a carat (^). Example: NH_4^1+ for NH₄¹⁺🇧🇷 Include all coefficients other than 1. Omit all spaces.

zPutrefaction→ zboi +nortemi^1-

o_3Sn_s The reducing agent is 3Sn.

→

o_3Sn^2+_s The oxidized form is 3Sn²⁺.

+

o_6e^1-_s Thus, three Sn atoms each lose two electrons.

norte= 3 × 2 = 6

electrons

Represent the reduction half-reaction asjboi +nortemi^1- → jPutrefaction.Specify subscripts with an underscore (_) and superscripts with a carat (^). Example: NH_4^1+ for NH₄¹⁺🇧🇷 Include all coefficients other than 1. Omit all spaces.

jboi +nortemi^1- → jPutrefaction

o_2Cr^3+_s The oxidizing agent is 2Cr³⁺.

+

o_6e^1-_s of the cr³⁺The ions gain three electrons each, so

norte= 2 × 3 = 6

electrons

→

o_2Cr_s The reduced form is 2Cr.

11.2-2. Sum of reactions to the medium

Net redox reactions are usually constructed from tabulated half-reactions, which are always in the form of reductions along this course. The procedure is as follows:

• 1

  Identify oxidation and reduction half-reactions.

• 2

  Invert the tabulated reduction half-reaction corresponding to the oxidation pair.

• 3

  Find the least common multiple (LCM) of the number of electrons gained in reduction and lost in oxidation.

• 4

  Multiply each half reaction by the integer required for the electrons gained or lost to equal the LCM determined in step 3.

• 5

  Add the two half reactions to get the net redox reaction.

Use the following two reduction half-reactions to establish a chemical equation that explains what happens when metallic aluminum is converted to a strong acid (H¹⁺) for the production of H₂Gas.

reduction half-reaction

H¹⁺is a reactant and H₂is a product, soWe use Reaction 2 as is.

Protons are reduced to hydrogen gas in the reduction half-reaction. Strong acids are the source of H¹⁺Ions, so strong acids can act as oxidizing agents. Indeed,this half-reaction represents the redox reaction of most strong acids.The nitrate ion in nitric acid is a stronger oxidizing agent than H¹⁺, so this half-reaction is not used for nitric acid.

oxidationhalbreaktion

Metallic aluminum is oxidized to aluminum ions, so the reduction half-reaction is given byReaction 1 must be reversed to be an oxidation..

Aluminum is a good reducing agent and reacts violently with acid. Most metals oxidize their ions with acid and some with water.

transfer electrons

The reduction half-reaction involves a gain of two electrons, while the oxidation half-reaction involves a loss of three electrons. Thisso mcm is 6🇧🇷 The net equation is a transfer of six electrons.

electrons gained = electrons lost

The reduction half-reaction is multiplied by three to be a gain of six electrons, and the oxidation half-reaction is multiplied by two to be a loss of six electrons.

half sum reactions

The sum of the two half-reactions gives the net reaction equation. The number of electrons gained is equal to the number of electrons lost, so the total number of electrons is zero. Note that both the number of atoms and the total charge, which is +6 on each side, are balanced.

11.2-3. Practice writing redox equations

11.3 Galvanic Cells

Introduction

We have seen that electrons are transferred spontaneously from donor to acceptor when the transfer lowers their free energy. Furthermore, the released free energy can be used for work, as in a battery. In this section, we show how the relative free energies of electrons in different redox pairs are determined.

requirements

• •

  9.7 Free energy(Relate free energy change to spontaneity.)

Goals

• •

  set agalvanic celland name your components.

• •

  Explain the purpose of an electrode and differentiate between them.activeypassiveelectrodes

• •

  Explain the function of the anode and write the half-reaction of the anode for a given galvanic cell.

• •

  Describe the function of the cathode and write the cathode half-reaction of a galvanic cell.

• •

  Explain the role of crossover.

• •

  Explain what a voltmeter measures and relate the cell potential to the two half-cell potentials.

• •

  Describe the movement of ions and electrons in a galvanic cell.

11.3-1. The definition

We now combine electron transfer reactions and electrical conduction through a circuit, which is the domain ofElectrochemistry🇧🇷 We started calling backSection 9.7a -DGRAMSis the maximum amount of work (W) that can be obtained from any reaction at constant temperatureTyPAG🇧🇷 Extending this to redox reactions, we can write:

When redox reactants are in direct contact, the released free energy is lost as heat. However, free energy can be harnessed by separating the reactants into aelectrochemical celland forcing the electrons through an external circuit. Electrochemical cells in which electrons flow spontaneously.

(DGRAMS< 0)

The work done bynortemolar electrons (a charge ofnorteℱ Coulomb) transferred by an electric potential ℰ is given as

norteℱ is the amount of charge in coulombs that is transferred and ℰ is the electric potential difference through which electrons move, expressed in volts. ONEVolt(V) is one joule per coulomb (1 V = 1 J/C), so Equation 11.2 gives the work done by electrons in joules. The rearrangement shows that the electric potential is equal to the work done by the electrons divided by the number of coulombs. In other words, ℰ is the work each charged coulomb can do. Combining Equation 11.2 with Equation 11.1, we obtain the following relationship between the free energy of the redox reaction and the voltage that would be measured in the corresponding electrochemical cell:

Equation 11.3 shows that the cell potential (ℰ) becomes more positive as the change in free energy of the redox reaction becomes more negative, which means that the amount of work each electron can do increases as the cell potential becomes more positive.

11.3-2. half cells

Just as a redox reaction consists of two half-reactions, an electrochemical cell is divided into two half-cells:

• •

  anode: the half cell where oxidation takes place

• •

  cathode: the half cell where the reduction takes place

These two cells are separate but connected by an electrical circuit, allowing electrons to flow from the anode to the cathode while the two reactants remain separate.Each of the two half-cells develops an electrical potential known ashalf cell potentialand denoted by ℰ_anodeand ℰ_cathode🇧🇷 Cell potentials are defined as the final electric potential of the electron minus its initial electric potential; to mean.

In a redox process, electrons flow from the species to be oxidized at the anode to the species to be reduced at the cathode, i.e. ℰ_Initial= ℰ_anodeand ℰ_Final= ℰ_cathode, and we can write

That is, the potential of the cell is the potential difference of half a cell. If all substances in the cell are in their standard state, then all potentials are standard potentials, and the cell standard potential can be expressed as the difference between the standard potentials of half the cell, as shown in Eq. 11.4.

The extent of the reaction depends on this.DGRAMS°, which can be determined with equation 11.5.

DGRAMS° < 0 yes ℰ°_cell> 0, then a reaction is large if the half-cell potential of the standard cathode (ℰ°_cathode) is greater than the normal half-cell anode potential (ℰ°_anode🇧🇷 Therefore, we could predict the extent of a redox reaction if we knew the relative values ​​of the standard half-cell potentials of the redox pairs.

11.3-3. cell description

A typical galvanic cell consists of four components:

• •

  the anode room,

• •

  the cathode room,

• •

  a perfect connection and

• •

  A load.

Each compartment contains the reagents in solution and a piece of metal immersed in a solution. the metal is onelectrodes🇧🇷 The electrodes provide a surface for electrons to move between the circuit and the reagent.active electrodesparticipate in the reaction (atoms are reactive), whilepassive electrodesthey only provide the surface and do not participate in the reaction.

11.3-4. anode

To beanodeis the oxidation half cell. A typical anodic half-reaction has the following form:

The following points must be observed for the anode framed in red in figure 11.6.

• 1

  (Video) 14. Prof. Rudolph Marcus - Electron Transfer Theory and Its Evolution (Feb 3, 2022)
  The half-cell reaction is Fe→Faith²⁺+ 2 mi^1-.

• 2

  Electrons are lost through the iron atoms in the electrode and enter the external circuit.

• 3

  The loss of electrons from iron atoms creates ferrous ions, which go into solution and increase Fe.²⁺Concentration.

• 4

  The negative charge that is carried out of the compartment with each electron is compensated for by a flow of Cl^1-ion in the salt bridge compartment.

• 5

  The iron electrode is an active electrode that slowly dissolves as the iron atoms are converted to Fe.²⁺ions entering the solution.

11.3-5. cathode

To becathodeis the reduction half cell. The cathode half-reaction can be expressed as

The following points must be observed for the cathode marked in red in Fig. 11.7.

• 1

  The half-cell reaction is Cu²⁺+ 2 mi^1- →Cu.

• 2

  Electrons flowing into the cell from the anode are gained by Cu²⁺Ions that are in solution but on the surface of the electrode.

• 3

  The acceptance of electrons by Cu²⁺The ions create copper atoms that become part of the electrode.

• 4

  The negative charge created when each electron enters the compartment is balanced by the flow of a cation, a K¹⁺Ion here, in the salt bridge compartment.

• 5

  The copper electrode is an active electrode that slowly increases in size like Cu²⁺The ions become copper atoms.

11.3-6. ponte

As the reaction proceeds, Cu²⁺Ions are consumed when electrons enter the cathode and Fe²⁺Ions are created when electrons leave the anode. However, under these conditions, all reaction would cease, as the two chambers would become electrically charged. The liquid connection maintains electrical neutrality by allowing ions to pass between the two chambers while the oxidant and reductant are separated. ONEsalt bridgeIt is a liquid compound that contains a saturated solution of a strong electrolyte such as KCl.The flow of electrons to the cathode must be balanced by a flow of anions out of the chamber or cations into the chamber. The charge balance in the cell shown in the figure can be maintained by sulfate ions leaving the cathode and entering the salt bridge or by potassium ions entering the cathode from the bridge. Note that two K¹⁺you must enter a Cu for each²⁺that is consumedThe flow of electrons out of the anode is balanced by a flow of anions into the compartment or cations out of the compartment. Therefore, chloride ions migrate from the salt bridge towards the anode or Fe²⁺Ions migrate from the anode to the salt bridge.

11.3-7. charge

The load can be a light bulb filament, a power tool, a starter motor, a toy, or anything that requires a battery to run. In the experiment discussed here, the load is a voltmeter. A voltmeter measures the electrical potential difference between the two electrodes. The two terminals of a voltmeter are labeled as follows:

• •

  Red = Hello = +. The red terminal is assumed to be at the highest or most positive potential.

• •

  Black = Low = -. The black terminal is assumed to be at the lowest or most negative potential.

Note that a negative cell potential simply means that the connections have been reversed; that is, the red terminal is actually connected to the cell with the most negative potential. In this case, electrons flow from the Hi (+) terminal to the Lo (-) terminal.The voltage measured by the voltmeter (cell potential) is defined as follows:

Remember that electrons move from a lower to a higher potential (from a more negative to a more positive charge) and from the anode to the cathode. Consequently, the anode must be connected to the black terminal (Lo or -) and the cathode to the red terminal (Hi or +) to obtain a positive voltage. Therefore, the cell potential can also be expressed as follows.

11.3-8. example

11.4 Standard reduction potential

Introduction

The standard reduction potential of a redox pair is a measure of the electrical potential of the redox electron in that pair relative to the potentials of redox electrons in other pairs under standard conditions. Therefore, standard reduction potentials can be used to determine cell potentials and predict the spontaneity of redox processes.

Goals

• •

  Compare the movement of electrons and ions in a cell when the SHE is connected to a positive standard reduction pair with one where it is connected to a negative standard reduction potential pair.

11.4-1. Standard reduction potential video

11.4-2. reference electrodes

Although we cannot measure the potential of a half-cell, we can measure the potential difference between two half-cells. Relative half-cell potentials can be obtained by defining a half-cell as a reference against which all other half-cells can be measured. Which half-reaction we choose for the reference half-cell and what value we assign to its potential is arbitrary. The decision taken many decades ago was to use theStandard hydrogen electrode (SHE) as a reference and assign it a value of exactly 0V. The SHE half reaction is

boi +nortemi1-		Putrefaction	ℰ° (V)
2 standard1++ 2 mi1-		H2	0,00

Normally, the half cell to be measured is connected to the "Hi" connection, so it isof coursebe the cathode or the reduction half-reaction. The SHE is then connected to the "Lo" terminal, so this is considered the anode or oxidation reaction medium. If both cells are in their normal states, the resulting cell potential is obtainedStandard reduction potentialthe half cell. It is a "reduction potential" as the half cell to be measured is connected to the "Hi" terminal. Consider the setup in Figure 11.10.

• •

  Yes ℰ°_METRO> 0, then M²⁺+H₂ →M + 2H¹⁺.

• •

  Yes ℰ°_METRO< 0, so M + 2H¹⁺ →METRO²⁺+H₂.

None of the reactants or products of the SHE reaction is solid, so an inert metal is used as the electrode. In the figure, platinum is used as an electrode. Electrons move in and out of solution at the Pt surface.

• •

  ℰ°_cell= ℰ°_ola– ℰ°_low= ℰ°_Metal– ℰ°_UDS

• •

  ℰ°_UDS= 0 V, also ℰ°_cell= ℰ°_Metal, which is the standard reduction potential of M.

• •

  ℰ°_cell> 0 means that ℰ°_Metal> ℰ°_UDS🇧🇷 Since the metal has a higher potential, electrons flow there. Consequently, electrons flow from "Hi" to "Lo" in the circuit and metal ions are reduced to H₂.

• •

  ℰ°_cell< 0 means that ℰ°_Metal< ℰ°_UDS, then the metal has a lower potential, and electrons flow from there to the higher potential of the SHE. Therefore, electrons flow from "Lo" to "Hi" in the circuit when the metal is oxidized by H¹⁺.
  ℰ°_cellit is still the metal's standard reduction potential, even though the metal will oxidize because it was connected to the "Hi" terminal; that is, the cell was connected as if the metal were the cathode. The negative sign of the cell potential simply indicates that electrons are flowing through the circuit in the opposite direction of the junction.

11.4-3. Determination of standard reduction potentials

Standard Copper Reduction Potential

Determination of the standard Cu reduction potential²⁺:

• •

  The copper half cell is connected to the "Hi" or "+" terminal, so the copper half reaction is considered to be reduction.

• •

  All reactants and products are in their standard state, so the measured voltage (+0.34 V) is the standard reduction potential of Cu²⁺/ With couple.

• •

  The reduction potential of Cu²⁺/Cu torque is greater than that of H¹⁺/H₂pair, so that electrons spontaneously move from H₂zu Cu²⁺.

• •

  Electrons spontaneously move from the highest to the lowest free energy, i.e., thetransferred electrons have lower free energies in copper.

boi +nortemi1-		Putrefaction	ℰ° is V
copper2++ 2 mi1-		copper	+0,34

and he with²⁺The /Cu pair implies a half-reaction at a potential below 0.34 V, electrons move from the lower potential reagent and reduce Cu.²⁺to Cu, but when the pair engages in a half-reaction at a higher potential, the electrons leave Cu and reduce the species at a higher potential.

Standard Iron Reduction Potential

Determination of the standard Fe reduction potential²⁺:

• •

  The iron half cell is connected to the "Hi" or "+" terminal, i.e. the iron half cell reactionof coursebe the cut

• •

  All reactants and products are in their standard state, so the measured voltage (-0.44 V) is the standard reduction potential of Fe²⁺/Fe By.

• •

  Fe reduction potential²⁺The pair /Fe is smaller than that of H¹⁺/H₂coupled so that electrons spontaneously move from Fe to H¹⁺.

• •

  Electrons spontaneously move from the highest to the lowest free energy, i.e., thetransferred electrons have higher free energy in iron.

• •

  As always, the spontaneous flow of electrons is to the compartment with the highest potential, but in this case to the compartment connected to the "Lo" terminal. The apparent contradiction arises because the cell is designed to measure its potential when Fe²⁺/Fe even is the cathode, but Fe²⁺The /Fe pair is actually the anode in the spontaneous cell. It was connected that way because thereduction potentialthere faith²⁺The /Fe pair was desirable even when Fe is oxidized. It is important that the Faith²⁺The /Fe pair is at a lower potential (more negative voltage) than the H¹⁺/H₂Par.

boi +nortemi1-		Putrefaction	ℰ° is V
Faith2++ 2 mi1-		Faith	–0,44

if faith²⁺/Fe is associated with a half-reaction at a potential more negative than -0.44 V, electrons move from the lower potential reagent and reduce Fe²⁺for Fe. When the pair engages in a higher potential half reaction, the electrons leave the Fe to reduce the higher potential species. The fairy²⁺/faith by(-0.44 V) is more negative than H¹⁺/H₂torque (0 V), then the electrons are transferred to the H¹⁺/H₂match and convert H¹⁺ah₂.

11.4-4. Calculation of cell potentials

A standard cell potential is the potential difference between the cathode and anode. The relative potentials of the two half-reactions are given by their standard reduction potentials, so standard reduction potentials can be used to determine the difference. Consequently, the standard cell potential of any electrochemical cell can be expressed as

It's

• •

  ℰ°_cathode=Standard reduction potentialof the reduction torque.

• •

  ℰ°_anode=Standard reduction potentialof the oxidation pair.

Note that the standard reduction potential is a measure of the relative electrical potential of the half-cell. as a consequence,Standard reduction potentials are independent of the number of electrons transferred and the direction in which they are transferred.🇧🇷 faith dies²⁺The /Fe pair is -0.44 V, regardless of whether the iron atoms are reduced or the iron(II) ions are oxidized, ie, cathode or anode.Determination of cell potential for Fe + Cu²⁺ →Faith²⁺+ Reaction Cu, follow these steps.

11.4-5. reduction potentials

The default mitigations considered so far are listed in the table of default mitigation potentials below.

ox + are not1-		Putrefaction	ℰ° (V)
Faith2++ 2 mi1-		Faith	–0,44
2 standard1++ 2 mi1-		H2	0
copper2++ 2 mi1-		copper	+0,34

Table 11.2

Note that the half reactions are numbered so that ℰ° is more negative at the top and more positive at the bottom. This choice puts the electrons with the highest free energy at the top. Donor electrons are on reducing agents (Red), while acceptor orbitals are on oxidizing agents (Ox). Spontaneous transfer of electrons occurs from a higher free energy donor (lower or more negative potential) to a lower free energy acceptor (higher or more positive potential). Thus, a redox reaction occurs spontaneously when the reducing agent is greater than the oxidizing agent in a table of ordered standard reduction potentials.

11.5 Writing Redox Reactions

Introduction

We now use our understanding of half-reactions and standard reduction potentials to write net equations for redox reactions.

Goals

• •

  Predict the relative oxidation and reduction powers as a function of pair position in a table of standard reduction potentials.

• •

  Predict whether a redox reaction is extensive based on the position of reagent pairs in a standard table of reduction potentials.

• •

  Use a standard reduction potential table to predict whether a redox reaction will occur when two substances are mixed.

• •

  Write balanced chemical equations for redox reactions from tabulated half-reactions.

11.5-1. Video on reactivity and standard reduction potentials

11.5-2. Video on writing REDOX chemical equations

11.5-3. Standard reduction potential

Electron transfer is spontaneous when the reactants are the strongest reducing and oxidizing agents and the products are the weakest reducing and oxidizing agents. In this discussion, we will express a general redox reaction between species 1 and 2 as:

The two half-reactions have relative positions in a standard reduction potential table shown in Figure 11.13.

Boi₁it's about ox₂in the table of standard reduction potentials, then we can infer that

• •

  ℰ°₂> ℰ°₁

• •

  Redox electrons in red.₁they have more energy than those in red₂.

• •

  Putrefaction₁it is a better reducing agent than red₂.

• •

  Boi₂is a better oxidizing agent than Ox₁.

• •

  Red electron transfer₁for the ox₂it is extensive because red₁it's about ox₂, which implies that electron transfer would occur at a higher electric potential and a lower free energy.

• •

  ℰ°_cell= ℰ°_cathode– ℰ°_anode= ℰ°₂– ℰ°₁> 0. The positive value of ℰ°_cellalso indicates that electron transfer is extensive.

we closed that too

• •

  Boi₁is under ox₂in the table, i.e. ℰ°₂< ℰ°₁.

• •

  Redox electrons in red.₁have a higher electric potential and lower energy than those in red₂.

• •

  Red electron transfer₂for the ox₁NOT extended because electrons do not transfer extensively at lower potential or higher free energy (lower right to upper left in table).

• •

  ℰ°_cell= ℰ°_cathode– ℰ°_anode= ℰ°₂– ℰ°₁< 0. The negative value of ℰ°_cellalso indicates that electron transfer is NOT extensive.

11.5-4. response prediction exercise

11.5-5. liquid equations

11.6 Common Batteries

Introduction

Batteries are devices that use the free energy of spontaneous redox reactions to generate electrical energy, ie. h are electrochemical cells. Three of the most common batteries are described in this section.

Goals

• •

  Describe the common alkaline battery.

• •

  Depicts a button on a silver oxide battery.

• •

  Describe a car battery.

11.6-1. alkaline battery

Alkaline battery is commonly used in flashlights. The half-reactions are:

• •

  Anode:zinc→zinc²⁺+ 2 mi^1-

• •

  Cathode:2 MnO₂+H₂the +2 and^1- →Minnesota₂o₃+2OH^1-

ℰ_cell= 1,5 volts

The cathodic reaction shown above is simply representative of a much more complicated process involving manganese species. The zinc base is an active anode and a graphite rod serves as a passive cathode. the mnO₂is present in a wet electrolytic paste (KOH, H₂Of e KOH).

11.6-2. Silver (button) battery

The button cell is used in calculators, cameras, clocks, etc. The half-reactions are:

• •

  Anode:zinc→zinc²⁺+ 2 mi^1-

• •

  Cathode:Agriculture₂O + H₂the +2 and^1- →2Ag + 2OH^1-

Note: Hg is sometimes used instead of Ag, in this case it is the cathode reaction

HgO+H₂the +2 and^1- →Hg + 2OH^1-.

11.6-3 lead-acid battery

Almost all cars start with a lead-acid battery. A 12 V battery actually contains six identical electrochemical cells, each producing 2.0 V. The two half-reactions for each cell are

• •

  Anode:Pb + SO₄^2– →PbSO₄+ 2 mi^1-

• •

  Cathode:PbO₂+ 4 pattern¹⁺+ SO₄^2–+ 2 mi^1- →2 PbSO₄+ 2 pattern₂o

The electrodes, sponge lead anodes and PbO powder₂Cathodes are immersed in ~4.5 MH₂AS SOON AS₄.One of the main advantages of the lead-acid battery is that it can be recharged using an external energy source, reversing the half-reactions. A car's alternator, powered by the car's engine, produces electrical energy that is applied in reverse through the battery. Electrical energy is used to drive the reactions in reverse. In this way, the raw materials are regenerated for the next time the car is started. We will go into more detail about the recharging process in our consideration of electrolytic cells.

11.7 Corrosion

Introduction

Corrosion is the unwanted oxidation of a metal. Approximately 25% of new steel production is used to replace corroded steel. This very expensive process consists of a series of electrochemical reactions that take place on the iron and the water in it.

Goals

• •

  Distinguish between galvanized and passivated.

• •

  Use standard reduction potentials to select a metal that will protect a specific metal from corrosion.

11.7-1. definition and example

CorrosionIt is the undesirable oxidation of a metal. Consider the following observations about iron corrosion.

• 1

  Areas of corrosion (holes in the surface of the iron) and rust are usually separated.

• 2

  water is needed. Iron does not rust in dry climates.

• 3

  Oxygen is needed. Iron does not rust in water that does not contain O.₂.

• 4

  The acid promotes oxidation.

These observations are explained in the figure which shows iron oxidation as an electrochemical cell in which iron is an active anode and a passive cathode.Note that oxygen and other gases in the atmosphere dissolve in raindrops, and some of these gases react with water to produce H¹⁺, causing the droplets to become acidic. For example OS₃, an industrial pollutant, reacts with water to form sulfuric acid (H₂AS SOON AS₄) and company₂reacts in a similar way to form carbonic acid (H₂CO₃).

• •

  Anode (Grube): Fe→Faith²⁺+ 2 mi^1-🇧🇷 The ion dissolves in water. The loss of Fe atoms causes corrosion on the iron surface.

• •

  Cathode: 4H¹⁺+ Ö₂+ 4 mi^1- →2 standard₂O. This reaction explains the need for oxygen and acid.

• •

  ℰ°_cell= ℰ°_cathode– ℰ°_anode= 1.23 - (-0.44) = +1.67 V, a long response.

• •

  Oxygen in solution further oxidizes Fe²⁺in solution for Fe³⁺, which forms insoluble Fe₂o₃(face).

11.7-2. corrosion protection

11.8 Electrolytic cells

Introduction

We have seen that spontaneous redox reactions can be used to generate electrical energy in an electrochemical cell, and in this section we demonstrate the reversal of this process by using electrical energy to drive non-spontaneous redox reactions.

Goals

• •

  Distinguish between electrolytic and galvanic cells.

• •

  Explain why the lead-acid battery can be either a galvanic cell or an electrolytic cell.

11.8-1. electrolytic cells

So far we have dealt with galvanic cells, that is, cells in which electron transfer occurs spontaneously. However, a great advantage of electrochemistry is that we can vary the free energy of the electrons at the electrodes and force the electrons to transfer free energy upwards. The cell potential is a measure of the work we can get from a galvanic cell (ℰ_cell> 0), but when negative it indicates the amount of work an external power supply must provide for an electrolytic cell (ℰ_cell< 0) to force the answer. That is, galvanic cells convert chemical potential energy into electrical potential energy, while electrolytic cells convert electrical potential energy into chemical potential energy.This is called forcing a non-spontaneous response by applying a voltage from an external source.electrolysis🇧🇷 Electrolysis is used to extract many metals from their ores and to coat one metal (usually iron) with another in a process called plating (silver and gold jewelry and chrome or nickel-plated sink fixtures).

For example, the redox electrons of Na have a much higher potential energy than that of Cl.^1-🇧🇷 Consequently, the reaction of Na with Cl₂in an electrochemical cell, it would generate a large voltage when the chemical potential energy of electrons is converted to electrical potential energy. Furthermore¹⁺y cl^1-contact, no reaction will occur unless electrical potential energy is supplied from an energy source to drive electrons from the lower potential energy of the chloride ion to the higher potential energy of the sodium atom. The spontaneous reaction is called electrolysis. In this example, NaCl is electrolyzed into Na and Cl.₂.

11.8-2. lead acid batteries

Battery charging is also a function of electrolytic cells. The lead-acid battery is one of the most common examples. The crankshaft rating is given in amps, which is a measure of the rate at which electrons flow through the circuit, and every two moles of electrons flowing through the circuit creates two moles of PbSO₄🇧🇷 A battery with an initial power of 550 amps produces 1.7 g of PbSO₄per second. At this rate, the battery would run out of reagents after just a few starts. However, cars are equipped with generators (or alternators) that force the reversal of the spontaneous discharge reaction while the car is in motion, ie. h charge the battery. This is what lead-acid batteries look like.galvanicwhile a car starts andelectrolyticwhile it is running.

11.9 Exercises and solutions

Select the links to view end-of-chapter exercises or solutions to odd-numbered exercises.

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