《基礎(chǔ)化學(xué)》英文教學(xué)課件：chapter_8

1、Chapter 8. Redox Reaction and Electrode Potential, 8-1. Redox reaction 8-2. Primary cell and electrode potential 8-3. Electromotive force of cell and Gibbs free energy 8-4. Nernst equation of electrode potential 8-5. Determination of pH,Voltaic pile 伏打電堆,Daniells battery,we will look at the general

2、principles involved in setting up a chemical reaction as a battery. We will answer such questions as the following: What voltage can you expect from a particular battery? How can you relate the battery voltage to the equilibrium constant for the reaction?,1. Redox Reaction,1. Redox Reaction,Oxidatio

3、n number (IUPAC definition) is the apparent (approximate) charge of an atom, which is assigned by designating the bonding electrons to the atom with greater electronegativity.(氧化值: 某元素一個原子的表觀荷電數(shù)，這種荷電數(shù)是假設(shè)把每個化學(xué)鍵中的電子指定給電負性更大的原子而求得。),For example:,I. Oxidation numbers (氧化值),Na+ Cl- H :Cl CH4C(-4), H(+1)

4、CO2 C(+4), O(-2),1. Redox Reaction,II. Oxidation-number rules,1. Redox Reaction,II. Oxidation-number rules,1. Redox Reaction,Example 8-1: Use the rules in last slides to obtain the oxidation number of the chlorine atom in each of the following: (1) HClO4 (perchloric acid), (2) ClO3- (chlorate ion).,

5、Solution:,(1) For HClO4, Rule 6 gives the equation:,Using Rules 3 and 4, you obtain:,Therefore,(2) For ClO3-,compound OF2 Cr2O72- Fe3O4 CH3OH Na2S2O3 C6H6O6 C6H8O6,1. Redox Reaction,Examples of oxidation number,Element O Cr Fe C S C C,Oxidation number +2 +6 +8/3 -2 +2 +1 +4/6,III. Redox reaction (氧化

6、還原反應(yīng)),1. Redox Reaction,An oxidation-reduction reaction (or redox reaction): A reaction in which electrons are transferred between species or in which atoms change oxidation numbers.,Zn + Cu2+ Zn2+ + Cu (0) (+2) (+2) (0),A redox reaction always involves both the loss of electrons and the gain of ele

7、ctrons.,1. Redox Reaction,A half-reaction is one of two parts of a redox reaction, one part of which involves a loss of electrons (or increase of oxidation number) and the other a gain of electrons (or decrease of oxidation number).,1. Redox Reaction,Oxidation (氧化反應(yīng)) is the half-reaction in which th

8、ere is a loss of electrons by a species (or an increase of oxidation number of an atom).,Reduction (還原反應(yīng)) is the half-reaction in which there is a gain of electrons by a species (or a decrease in the oxidation number of an atom).,Zn has been oxidized.,Cu2+ has been reduced.,An oxidizing agent (Cu2+)

9、 (氧化劑) is a species that oxidizes another species; it is itself reduced; A reducing agent (Zn) (還原劑) is a species that reduces another species; it is itself oxidized.,1. Redox Reaction,Zn has been oxidized by ?; Cu2+ has been reduced by ?.,Oxidizing agents and reducing agents are referred with respe

10、ct to the reactants of a reaction.,1. Redox Reaction,IV. Redox electric couple,Ox + ne- Red,Redox electric couple: (氧化還原電對),Ox/Red,For example, Cu2+/Cu, Zn2+/Zn,1. Redox Reaction,Zn(s) Zn2+(aq) + 2e-,Cu2+(aq) + 2e- Cu(s),Ox: Oxidized form (higher ON) in a half-reaction; Red: Reduced form (lower ON)

11、in a half-reaction.,V. Some common redox reactions,1. Combination reactions,1. Redox Reaction,A combination reaction (化合反應(yīng)) is a reaction in which two substances combine to form a third substance.,Two elements react to form a compound.,2. Decomposition reactions,1. Redox Reaction,A decomposition rea

12、ction (分解反應(yīng)) is a reaction in which a single compound reacts to give two or more substances.,3. Displacement reactions,1. Redox Reaction,A displacement reaction (置換反應(yīng)) is a reaction in which elements reacts with a compound, displacing an element from it.,4. Combustion reactions,A combustion reaction

13、 (燃燒反應(yīng)) is a reaction in which a substance reacts with oxygen, usually with the rapid release of heat to produce a flame.,VI. Balancing Redox equations (half-reaction method),1. Identify the species being oxidized and the species being reduced from the changes in their oxidation numbers, write the s

14、keleton (unbalanced) equation of the redox reaction.,1. Redox Reaction,2. Split the the skeleton equation into two equations for the oxidation and reduction half-reactions.,3. Balance all elements in the half-reactions except O and H (mass balance, 物料平衡).,4. In acidic solution, balance O by using H2

15、O, then balance H by using H+. In basic solution, balance O by using H2O; then balance H by adding H2O to the side of each half-reaction that needs H and adding OH- to the other side (mass balance , 物料平衡).,1. Redox Reaction,In an acidic aqueous solution, H+(aq) and H2O(l) may be possible reactants o

16、r products in the equation; If this were a basic aqueous solution, OH- ion would replace the H+ ion.,5. Balance electric charge by adding electrons (e-) to the more positive side (charge balance , 電荷平衡).,1. Redox Reaction,7. Simplify the balanced equation by canceling species that occur on both side

17、s. Check the equation to make sure that both numbers of atoms and charge are balanced.,6. Multiply each half-reaction by a factor such that when the half-reactions are added, the electrons cancel (electrons cannot appear in the final equation) and add them.,Example 8-2: MnO4- reacts with H2C2O4 in a

18、cidic aqueous solution, producing Mn(II) and CO2.,Solution:,1. MnO4- + H2C2O4 Mn2+ + CO2,2. Reduction: MnO4- Mn2+ Oxidation: H2C2O4 CO2,3. Balance all elements except H and O: MnO4- Mn2+ H2C2O4 2CO2,4. Balance the O atoms by using H2O: MnO4- Mn2+ + 4H2O H2C2O4 2CO2,Balance the H atoms on either side

19、 of the half-reaction with H+: MnO4- + 8H+ Mn2+ + 4H2O H2C2O4 2CO2 + 2H+,1. Redox Reaction,5. Balance the electric charge: MnO4- + 8H+ + 5e- Mn2+ + 4H2O H2C2O4 2CO2 + 2H+ + 2e-,6. Equalize electrons transferred in the two half-reactions: 2MnO4- + 16H+ + 10e- 2Mn2+ + 8H2O 5H2C2O4 10CO2 + 10H+ + 10e-,

20、Add the two equations and cancel the electrons: 2MnO4- + 5H2C2O4 + 16H+ 2Mn2+ + 8H2O + 10CO2 + 10H+,7. Ten of H+ ions on the left are canceled by the 10 H+ on the right: 2MnO4- + 5H2C2O4 + 6H+ 2Mn2+ + 8H2O + 10CO2,Mass and charge are both balanced; so this is the fully balanced net ionic equation.,1

21、. Redox Reaction,Example 8-3: MnO4- reacts with Br- in basic aqueous solution, producing MnO2 and BrO3-.,Solution:,1. MnO4- + Br- MnO2 + BrO3-,2. Reduction: MnO4- MnO2 Oxidation: Br- BrO3-,3. Both equations are already balanced with respect to Mn and Br.,4. Balance the O atoms by using H2O: MnO4- Mn

22、O2 + 2H2O Br- + 3H2O BrO3-,Balance H by adding H2O to the side of each equations that needs H and OH- to the opposite side of each arrow: MnO4- +4H2O MnO2 + 2H2O + 4OH- Br- + 3H2O +6OH- BrO3- + 6H2O,1. Redox Reaction,Simplify each half-reaction by canceling like species on opposite sides of the arro

23、w: MnO4- +2H2O MnO2 + 4OH- Br- + 6OH- BrO3- + 3H2O,5. Balance the electric charge: MnO4- + 2H2O + 3e- MnO2 + 4OH- Br- + 6OH- BrO3- + 3H2O + 6e-,6. Equalize electrons transferred in the two half-reactions: 2MnO4- + 4H2O + 6e- 2MnO2 + 8OH- Br- + 6OH- BrO3- + 3H2O + 6e-,7. Add the two equations and can

24、cel the electrons: 2MnO4- + 4H2O + Br- + 6OH- 2MnO2 + 8OH- + BrO3- + 3H2O,2MnO4- + Br- + H2O 2MnO2 + 2OH- + BrO3-,1. Redox Reaction,As2S3 + ClO3- + H+ Cl- + H3AsO4 + SO42-,解：,1. As2S3 + ClO3- + H+ Cl- + H3AsO4 + SO42-,2. 將離子方程式拆成氧化和還原兩個半反應(yīng)： 還原反應(yīng)： ClO3- Cl- 氧化反應(yīng)：As2S3 H3AsO4 + SO42-,3. 配平除H、O外的所有元素：

25、ClO3- Cl- As2S3 2H3AsO4 + 3SO42-,4. 用H2O配平O：ClO3- Cl- + 3H2O As2S3 + 20H2O 2H3AsO4 + 3SO42-,用H+配平H:ClO3- + 6H+ Cl- + 3H2O As2S3 + 20H2O 2H3AsO4 + 3SO42- + 34H+,Exercise 2 (3),5. 配平電荷：ClO3- + 6H+ + 6e- Cl- + 3H2O As2S3 + 20H2O 2H3AsO4 + 3SO42- + 34H+ + 28e-,6. 使氧化還原兩個半反應(yīng)中的得失電子數(shù)相等： 14ClO3- + 84H+ + 84

26、e- 14Cl- + 42H2O 3As2S3 + 60H2O 6H3AsO4 + 9SO42- + 102H+ + 84e-,兩式相加，合并成離子反應(yīng)方程式： 14ClO3- + 84H+ + 3As2S3 + 60H2O 14Cl- + 42H2O + 6H3AsO4 + 9SO42- + 102H+,7. 消除等式兩邊相等數(shù)目的H+和H2O： 14ClO3- + 3As2S3 + 18H2O 14Cl- + 6H3AsO4 + 9SO42- + 18H+,2. Primary Cell and Electrode Potential,I. Primary cell (galvanic c

27、ell, voltaic cell),Zn + Cu2+ Cu + Zn2+,Where is the free energy ?,2. Primary Cell and Electrode Potential,Chemical energy heat (near the surface of Zn rod),Zn + Cu2+ Cu + Zn2+,2. Primary Cell and Electrode Potential,Each half-reaction takes place in separated container (half-cell). Two half-cells ar

28、e connected in such a way that electrons flow from one metal electrode to another through an external circuit, and ions flow from one half-cell to another through an internal cell connection.,Zn Zn2+ + 2e-,Cu2+ + 2e- Cu, Zn + Cu2+ Cu + Zn2+, electric current (i),2. Primary Cell and Electrode Potenti

29、al,A salt bridge (鹽橋) is a tube of an electrolyte in a gel(凝膠) that is connected to the two half-cells; the salt bridge allows the flow of ions (?) but prevents the mixing of the different solutions that would allow direct reaction of the cell reaction.,Salt bridge: Saturated KCl, KNO3 or NH4NO3 aqu

30、eous solution ?,2. Primary Cell and Electrode Potential,The electrode with lower electrical potential is called the negative electrode (負極).,The electrode with higher electrical potential is called the positive electrode (正極).,2. Primary Cell and Electrode Potential,A primary cell (原電池) is an electr

31、ochemical cell in which a spontaneous chemical reaction is used to generate an electric current.,Primary cells are also known as galvanic cells (伽伐尼電池) or voltaic cells (伏打電池).,An electrochemical cell (電化電池) is a system consisting of electrodes that dip into an electrolyte and in which a chemical re

32、action either uses or generates an electric current.,An electrolytic cell (電解池) is an electrochemical cell in which an electric current drives an otherwise nonspontaneous reaction.,Zn half-cell,Cu half-cell,Anode reaction: Zn Zn2+ + 2e- (oxidation reaction),Cathode reaction: Cu2+ + 2e- Cu (reduction

33、 reaction),Cell reaction (電池反應(yīng)): The net reaction that occurs in the primary cell. Zn + Cu2+ Cu + Zn2+,2. Primary Cell and Electrode Potential,The electrode at which reduction occurs is called the cathode (陰極).,The electrode at which oxidation occurs is called the anode (陽極).,Zn Zn2+ + 2e-,Cu2+ + 2e

34、- Cu,2. Primary Cell and Electrode Potential,Zinc-carbon dry cell (鋅-碳干電池),II. Some commercial primary cell,This cell has a zinc can as the anode; a graphite rod in the center, surrounded by a paste of manganese dioxide, ammonium and zinc chlorides, and carbon black, is the cathode.,The voltage of t

35、his dry cell is initially about 1.5 V.,2. Primary Cell and Electrode Potential,2. Alkaline “dry” cell （堿性干電池）,This cell is similar to the a zinc-carbon dry cell, but it has potassium hydroxide in place of ammonium chloride. This cell performs better under current drain and in cold weather.,2. Primar

36、y Cell and Electrode Potential,3. Lithium-iodine cell (鋰-碘電池),This is a voltaic cell in which the anode is lithium metal and the cathode is an I2 complex, separated by a thin crystalline layer of lithium iodide (LiI).,This battery is very reliable and is used to power heart pacemakers. The battery i

37、n implanted within the patients chest and lasts about ten years before it has to be replaced.,anode,anode,Zinc-carbon dry cell (鋅-碳干電池),Lithium-iodine cell (鋰-碘電池),7g/mol,1mol,7g/0.108mol,0.216mol,2. Primary Cell and Electrode Potential,2. Primary Cell and Electrode Potential,4. Lead storage cell (鉛

38、蓄電池),This voltaic cell consists of electrodes of lead alloy grids; one electrode is packed with a spongy lead to form the anode, and the other electrode is packed with lead dioxide to form the cathode. Both are bathed in an aqueous solution of sulfuric acid, H2SO4.,2. Primary Cell and Electrode Pote

39、ntial,5. Nickel-cadmium cell (鎳-鎘電池),It is a voltaic cell consisting of an anode of cadmium and a cathode of hydrated nickel oxide (approximately NiOOH) on nickel, the electrolyte is potassium hydroxide.,These half-reactions are reversed when the cell is recharged. Nicad batteries can be recharged a

40、nd discharged many times.,“記憶效應(yīng)”,2. Primary Cell and Electrode Potential,6. Fuel cell (燃料電池),A fuel cell is essentially a battery, but it differs by operating with a continuous supply of energetic reactants, or fuel. Such cells are used in the space shuttle orbiters to supply electric energy.,Overal

41、l reaction:,2. Primary Cell and Electrode Potential,A fuel cell,O2(g) + 4H+(aq) +4e- 2H2O(l),III. Notation for primary cells,(-) Zn Zn2+ (c1) Cu2+ (c2) Cu (+),(-) represents anode (negative) (left); (+) represents cathode (positive) (right); represents salt bridge; represents a phase boundary; c1 an

42、d c2 represent the concentrations;,2. Primary Cell and Electrode Potential,It is convenient to have a shorthand way of designating particular primary cells.,The cell terminals are at the extreme ends in the cell notation.,(1) If a gas is involved in the electrode reaction, the partial pressure must

43、be noted. (2) If a gas or two “valence ions” are involved in the electrode reaction, an “inert electrode”(Pt) (惰性電極) is demanded, and the inert electrode must be noted. (3) The different species in the same phase should be separated by a comma.,2. Primary Cell and Electrode Potential,(-)Pt H2 (pH2)

44、H+ (c1) Fe3+ (c2), Fe2+ (c3) Pt (+),Notes:,The notation for a cell that has a hydrogen electrode on the left, an Fe2+-Fe3+ electrode on the right is:,2. Primary Cell and Electrode Potential,Hydrogen electrode (氫電極): hydrogen bubbles over a platinum plate that is immersed in an acidic solution.,( Ox

45、+ ne- Red ),The cell terminals are at the extreme ends in the cell notation.,IV. Types of electrodes,(1) Metal-metal ion electrode: Consist of metal flake and the metal-ion-containing solution. e.g., Zn2+/Zn Electrode reaction: Zn2+ + 2e- Zn Notation for electrode (cathode): Zn2+ (c) Zn,2. Primary C

46、ell and Electrode Potential,(2) Gas electrode: Consist of a gas, a solution containing the corresponding ions and inert electrode. Electrode reaction: Cl2 + 2e- 2Cl- Notation for cathode: Cl- (c) Cl2(p) Pt,(3) Metal-insoluble salt anionic electrode: Consist of metal and its insoluble salt, immersed

47、into solution containing the same anion. e.g, Ag-AgCl electrode Electrode reaction: AgCl + e- Ag + Cl- Notation for cathode: Cl-(c) AgCl(s) Ag,2. Primary Cell and Electrode Potential,(4) Redox electrode: Consist of inert electrode, immersed into solution of ionic redox electric couples. Electrode re

48、action: Fe3+ + e- Fe2+ Notation for cathode: Fe3+(c2), Fe2+(c1) Pt,Example 8-4: Design a primary cell according to the following redox reaction: 5 Fe2+ + MnO4- + 8 H+ 5 Fe3+ + Mn2+ + 4 H2O and write the notation of the primary cell.,Pt was chosen as inert electrodes. Notation for anode: Pt Fe2+(c1),

49、 Fe3+(c2) Notation for cathode: MnO4-(c3), Mn2+(c4), H+(c5) Pt,2. Primary Cell and Electrode Potential,Solution:,Divide the redox reaction into two half-reactions: Oxidation: Fe2+ Fe3+ + e- Reduction: MnO4- + 8 H+ + 5e- Mn2+ + 4H2O,The notation for the cell: (-) Pt Fe2+(c1), Fe3+(c2) MnO4-(c3), Mn2+

50、(c4), H+(c5) Pt (+),V. Electrode potential (電極電位),2. Primary Cell and Electrode Potential,Potential difference,When a metal plate is inserted into a solution containing the corresponding metal ions, some metal ions enter the solution, leaving electrons in metal plate; some metal ions accept electron

51、 to deposit on the metal plate.,2. Primary Cell and Electrode Potential,2. Primary Cell and Electrode Potential,Electrode potential is determined by the tendency of electrode to give off or take on electrons.,VI. Standard electrode potential (標準電極電位),1. Standard Hydrogen Electrode (SHE): IUPAC regul

52、ation: SHE is selected as reference electrode, and its electrode potential is assumed to be zero.,Conditions: c(H+) = 1 mol L-1, p(H2) = 100 kPa Electrode reaction: 2H+(aq) + 2e- H2 (g) Notation of electrode: Anode: Pt H2 (100 kPa) H+ (a=1) Cathode: H+ (a=1) H2 (100 kPa) Pt SHE = 0.00000 V (298.15K)

53、,2. Primary Cell and Electrode Potential,Electromotive force (E, 電動勢): The (maximum) potential difference between the electrodes of a primary cell.,2. Determination of electrode potential,2. Primary Cell and Electrode Potential,E = + - -,The electromotive force of a cell is sometimes called the cell

54、 potential, or more colloquially, its voltage (電壓).,Example: Determination of electrode potential of copper,Pt H2 (100 kPa) H+ (a=1) Cu2+ (a) Cu (s),E = (Cu2+/Cu) - SHE = (Cu2+/Cu),2. Primary Cell and Electrode Potential,E = right - left = right - SHE = right,IUPAC regulation: Electrode potential is

55、 the electromotive force of the following cell.,Pt H2 (100 kPa) H+ (a=1) Mn+ (a) M,The digital voltmeter gives the reading with a sign attached.,Pt H2 (100 kPa) H+ (a=1) Zn2+ (a=1) Zn (s),E = - 0.7618V,(Zn2+/Zn) = E = - 0.7618V,2. Primary Cell and Electrode Potential,Pt H2 (100 kPa) H+ (a=1) Cu2+ (a

56、=1) Cu (s),E = + 0.3419 V,(Cu2+/Cu) = E = + 0.3419 V,3. Standard electrode potential (),The electrode potential determined under standard states is called standard electrode potential (標準電極電位).,Standard state:,2. Primary Cell and Electrode Potential,The concentrations of solute are 1 molL-1, the gas

57、 pressures are 100 kPa, and temperature has a specified value (usually 25 C).,More precisely, all the solutes should be at unit activity, not unit molarity. However, we ignore this complication here.,Standard electrode Potentials at 298.15K,(1). applies to standard states in aqueous solution. The te

58、mperature for the listed data is 298.15 K.,Notes:,(2). in the table is reduction potentials (A reduction potential is a measure of the tendency for an oxidized species to gain electrons in the reduction half-reaction).,(3). is an intensive property. Zn2+ +2e- Zn = - 0.7618 V (1/2)Zn2+ + e- (1/2)Zn =

59、 - 0.7618 V,2. Primary Cell and Electrode Potential,O + ne- R,R O + ne-,(1) The more positive the , the stronger oxidizing agent the oxidized form is; the more negative the , the stronger reducing agent the reduced form is.,(2) In a redox electric couple, stronger oxidizing form corresponds to weaker reducing form; weaker oxidizing form corresponds to stronger reducing form.,4. The significance of standard electrode potential,2. Primary Cell and Electrode Potential,MnO4-/Mn2+, Cr2O72-/Cr3+, Na+/Na,Strongest oxidizing agent in the table: MnO4- Strongest reducing agent