山大《醫(yī)學(xué)基礎(chǔ)化學(xué)》雙語課件06Electrochemistry

1、Chapter six ElectrochemistryCombination reactionDecomposition reaction Single-replacement reactionDouble -replacement reactionChemical reaction4Fe(s) + 3O2 2Fe2O3(s)CaCO3(s) CaO +CO2(g)Zn(s) + 2HCl(aq) ZnCl2 +H2(g)AgNO3 + NaCl NaNO3+AgCl(s) 6-1 Oxidation-reduction Concepts6-2 Voltaic Cells6-3 Electr

2、ode Potentials6-4 Electrode Potentials for Nonstandard Conditions6-5 Determination of pH 6-1 Oxidation-reduction ConceptsOxidation-reduction reaction Oxidizing agentReducing agentOxidation NumbersThe rules of Assigning Oxidation Numbers Oxidation-reduction reaction (Redox) is a reaction in which ele

3、ctrons are transferred from one reactant to another. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)Oxidation and reduction take place at same time Zn(s) - 2 e Zn2+(aq) lost two electronsCu2+(aq) + 2 e Cu(s) gained two electrons half-reactionZn (s) + CuSO4(aq) ZnSO4(aq) + Cu (s) Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)O

4、XIDATIONloss of electron(s) by a species; increase in oxidation number.REDUCTIONgain of electron(s); decrease in oxidation number; OXIDIZING AGENTelectron acceptor; species is reduced.REDUCING AGENTelectron donor; species is oxidized.Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Oxidation number the number of c

5、harges an atom would have in a molecule if electrons were transferred completely in the direction indicated by the difference in electronegativity. H2 (g) + F2(g) 2 HF(g)00+1-1 Remember : oxidation number has no physical reality. eg. In SO3, the oxidation number of S is +6 Rules of oxidation number

6、(1970)all the atoms in H2, F2, Be, Li, Na, O2, P4, and S8 have the same oxidation number zero. For an ion composed of only one atom, the oxidation number is equal to the charge on the ion. Thus K+ ion : + 1; Mg2+ ion: +2; Al3+ ion: +3; F- ion: - l; O2- ion: -2In a polyatomic ion, the sum of the oxid

7、ation numbers of all the elements in the ion must be equal to the net charge of the ion. NH4+ , S2O32- The oxidation number of oxygen in most compounds (for example, H2O and CaO) is -2; but In OF2, it has the oxidation number + 2; In hydrogen peroxide (H2O2) and in peroxide ion (O22-), : -1. HOOHThe

8、 oxidation number of hydrogen is + 1, except when it is bonded to a metal as in LiH, NaH, and BaH2, where its oxidation number is -1 Fluorine has the oxidation number -1 in all of its compoundsIn a neutral molecule, the sum of the oxidation numbers of all the atoms must be zero.Fe3O4 中，F(xiàn)e：+8/3； S4O6

9、2- 中，S：+5/2。 increase in oxidation number, loss of electron(s) - OXIDATIONdecrease in oxidation number, electron acceptor; - REDUCTIONMnO4 + C2O42- Mn2+ + CO2+7+3+4+2Reducing agentOxidizing agentREDUCING AGENTOXIDIZING AGENT3 H2O+ 3 Cl2 ClO3- + 6 H+ + 5Cl-self oxidation-reductionOxidizing AgentReduc

10、ing agent歧化反應(yīng): 又稱之為自身氧化還原反應(yīng)。 在歧化反應(yīng)中，化合物內(nèi)的同一種元素的一部分原子(或離子)被氧化，另一部分原子(或離子)被還原。 Balancing EquationsStep 1:Divide the reaction into half-reactions, one for oxidation and the other for reduction. Ox Cu - Cu2+ Red Ag+ - AgStep 2: Balance each element for mass. Already done in this case.Step 3: Balance eac

11、h half-reaction for charge by adding electrons. Ox Cu - Cu2+ + 2e Red Ag+ + e - AgCu + Ag+ - Cu2+ + AgBalancing EquationsStep 4:Multiply each half-reaction by a factor so that the reducing agent supplies as many electrons as the oxidizing agent requires.Reducing agent Cu - Cu2+ + 2eOxidizing agent 2

12、 Ag+ + 2 e - 2 AgStep 5:Add half-reactions to give the overall equation. Cu + 2 Ag+ - Cu2+ + 2AgThe equation is now balanced for both charge and mass.Balancing Equations Balance the following in acid solution VO2+ + Zn - VO2+ + Zn2+Step 1:Write the half-reactionsOxZn - Zn2+RedVO2+ - VO2+Step 2:Balan

13、ce each half-reaction for mass.OxZn - Zn2+RedVO2+ - VO2+ + H2O2 H+ +Add H2O on O-deficient side and add H+ on other side for H-balance.Balancing EquationsStep 3: Balance half-reactions for charge. Ox Zn - Zn2+ + 2e Rede + 2 H+ + VO2+ - VO2+ + H2OStep 4: Multiply by an appropriate factor. Ox Zn - Zn2

14、+ + 2e Red 2e + 4 H+ + 2 VO2+ - 2 VO2+ + 2 H2OStep 5: Add balanced half-reactionsZn + 4 H+ + 2 VO2+ - Zn2+ + 2 VO2+ + 2 H2OE.g. BrO3-(aq) + I-(aq) Br-(aq) + I2 (aq)Divide the skeleton reaction into two half reactions BrO3-(aq) Br-(aq) I-(aq) I2 (aq)Balance each half-reaction for mass.6H+ + BrO3-(aq)

15、 Br-(aq) + 3H2O2I-(aq) I2 (aq)Balance the charges by adding e- +5 -1 6H+ + BrO3-(aq) Br-(aq) + 3H2O 6e- + 6H+ + BrO3-(aq) Br-(aq) + 3H2O -2 0 2I-(aq) I2 (aq) 2I-(aq) I2 (aq) + 2e-Multiply the equation (s) by the appropriate factor so that the no. of e- in each half reaction is the same. 6e- + 6H+ +

16、BrO3-(aq) Br-(aq) + 3H2O 1 2I-(aq) I2 (aq) + 2e- 3Add the two half reactions (canceling the electrons lost)6e- + 6H+ + BrO3-(aq) Br-(aq) + 3H2O 6I-(aq) 3I2 (aq) + 6e-6H+ + 6I-(aq) + BrO3-(aq) Br-(aq) + 3H2O + 3I2 (aq)6-2 Voltaic CellsIn spontaneous oxidation-reduction (redox) reactions, electrons ar

17、e transferred and energy is released.The net result is that zinc metal reacts with copper ions to produce zinc ions and copper metal. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)Voltaic CellsWe can use that energy to do work if we make the electrons flow through an external device.We call such a setup a voltai

18、c cell.6-2 Voltaic Cells (Primary Cell) Voltaic cell: a setup in which a spontaneous chemical reaction generates an electric current. Spontaneous process: a physical or chemical change that occurs by itself. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)a negative electrode -anodea positive electrode - cathodean

19、 electrolyteA daniell cell uses a spontaneous chemical reaction to generate an electric current. Zn2+voltaic cell a simple device with which chemical energy is converted into electrical energyHalf-cell is that portion of an electrochemical cell in which a half-reaction takes place. Half-reactions Th

20、e oxidation and reduction reactions at the electrodes are called Zn(s) Zn2+(aq) + 2 e (Oxidation) Cu2+(aq) + 2 e Cu(s) (Reduction)Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)the pair of electrodeIn every half-reaction, there are two states with different oxidation number of common element, the state with highe

21、r oxidation number is called oxidation state and the state with lower oxidation number is called reduction state. The oxidation state and reduction state are called the pair of electrode.Zn2+/Zn, Cu2 + /CuSalt bridge: A salt bridge can be a U-tube device filled with an electrolyte, such as potassium

22、 chloride. exists to provide the electrical connection between the two reaction vessels while keeping the two reactions separate. The salt bridge allows the electron transfer between the two vessels.Notation for a Voltaic Cell for, Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Zn(s)Zn2+( aq ) Cu2+ ( aq )Cu(s )A

23、node Salt bridgeCathode Phase boundary(- )( + )Zn(s)Zn2+( aq ); Cu(s ) Cu2+ ( aq ) Mn+ + ne = M pair of electrode: Mn+/ M Ox + ne = Red Ox / Redpair of electrode:( electrode couple)Zn2+/Zn ; Cu2+/CuElectrode: MMn+ Fe2+ + Ag+ = Fe3+ + AgPt Fe3+, Fe2+ Ag+ Ag Fe3+/ Fe2+ Ag+ / Ag Electromotive ForceThe

24、maximum potential difference between the electrodes of a voltaic cell is referred to as the electromotive force (emf) of the cell, denoted E E = cathode anode = + - - E is a positive number. The types of electrode1. Metal-metal ion electrode 2. Metal-insoluble salt electrodes3. Gas electrode 4. Oxid

25、ation-reduction electrodesThe types of electrode1. Metal-metal ion electrode MMn+ : Mn+(aq) + ne = M(s) Zn(s) Zn2+( aq ); Cu(s ) Cu2+ ( aq )Ag-AgCl electrode :Ag , AgCl(s) | Cl-(c)AgCl + e- = Ag + Cl- 2. Metal-insoluble salt electrodesThe glass electrodePtHg(l)Hg2Cl2 (s) Cl- (c)Hg2Cl2 (s)+ 2e = 2Hg

26、+ 2Cl-Complex electrodeCalomel electrodeA hydrogen electrode : PtH2(p)H+(c) 2H+(aq) + 2e = H2 (g) 3. Gas electrode :4. Oxidation-reduction electrodesPt Fe3+ (c1),Fe2+(c2)Fe3+ + e = Fe2+6-3 Electrode PotentialsThe producing of Electrode PotentialsStandard electrode potential Standard hydrogen electro

27、de (SHE)Strength of an Oxidizing or Reducing AgentDisproportionation (歧化反應(yīng)) Equilibrium Constants from Electrochemistry 電 極 電 勢 的 產(chǎn) 生在金屬晶體中，存在著金屬正離子，金屬原子和自由電子。當把金屬板插入它的鹽溶液中時，有兩種反應(yīng)傾向: 在某一給定濃度的溶液中，若失去電子的傾向大于獲得電子的傾向，到達平衡時的最后結(jié)果將是金屬離子Mn進入溶液，使金屬棒上帶負電，靠近金屬棒附近的溶液帶正電。如圖：這時在金屬和鹽溶液 之間產(chǎn)生電位差。_金屬的電極電勢 金屬的電極電勢與金屬本

28、身的活潑性和金屬離子在溶液中的濃度及溫度有關(guān)。 金屬越活潑、溶液濃度越稀、溫度越高、溶解的傾向越大，金屬表面所帶負電荷越多；平衡時電極的電極電勢愈低。反之，金屬表面帶的正電荷越多，電極的電極電勢越高. Zn2/Zn電對的電極電勢比Cu2/Cu電對要負一些。由于兩極電勢不同，連以導(dǎo)線，電子流(或電流)得以通過。在銅鋅原電池中，實驗告訴我們，如將兩電極連以導(dǎo)線，電子流將由鋅電極流向銅電極，這說明Zn片上留下的電子要比Cu片上多，或 Zn2/Zn電對與Cu2/Cu電對兩者具有不同的電極電勢，綜上所述：原電池中電流的產(chǎn)生，是由于組成原電池的兩電極存在著電勢差所致，因此，可以說電極電勢的大小標志著金屬原

29、子或離子得失電子能力的大小，常用來衡量物質(zhì)氧化還原能力的強弱。The standard electrode potentialA standard electrode refers to an electrode in which the concentration of ions and the pressure of gases (in atmospheres) are equal to 1 ，temperature at 25.The potential of standard electrode called the standard electrode potential, it i

30、s designated by a superscript degree sign: Standard Hydrogen Electrode (SHE)IUPAC ：the standard hydrogen electrode (SHE) is taken to have an electrode potential of zero.H+(l.0mol/L)|H2(1 atm)|Pt ; = 0.0000 V 100 kPa (-) Pt| H2(g, 100KPa) | H+(1mol/L) | Cu2+(1mol/L) | Cu(s)( + )Eocell = o(Cu2+/Cu) o(

31、H+/H2)o(Cu2+/Cu) 0.000V = + 0.34 V0(Cu2+/Cu) = + 0.340 VCu2+ ions are more readilyreduced to Cu (s) than H+ ions are reduced to H2.Measuring the Standard Potentialof the Cu2+ / Cu Electrode(- ) Zn(s)Zn2+( 1mol/L ) H+ ( 1mol/L )H2(g)Pt(s ) ( + )Eocell =o (H+/H2) o (Zn2+/Zn) 0.000V o (Zn2+/Zn) = - 0.7

32、63 V o (Zn2+/Zn) = - 0.763 VZn2+ ions are less readilyreduced to Zn(s) than H+ ions are reduced to H2.Measuring the Standard Potentialof the Zn2+ / Zn Electrode Tab. the standard electrode potential Tab. the standard electrode potentialThe values for the table entries are reduction potentials, so li

33、thium at the top of the list has the most negative number, indicating that it is the strongest reducing agent. The strongest oxidizing agent is fluorine with the largest positive number for standard electrode potential. Applications of the standard electrode potentials Strength of oxidizing and redu

34、cing agentsDecision of the direction of redox reactionRelationship to equilibrium constantsElectrode potentials under non-standard conditions potentials for voltaic cells1. Strength of an Oxidizing or Reducing AgentThe strengths of oxidizing and reducing agents are indicated by their standard electr

35、ode potentials. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)Cu + Fe3+(aq) Fe2+(aq) + Cu2+2Li + F2 2Li+(aq) + 2 F - Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)Fe3+(aq) + Cu Fe2+(aq) + Cu2+F2 + 2Li 2 Li+(aq) +2 F -對角線法則：高電極電勢的氧化型一定能氧化低電極電勢的還原型低電極電勢的還原型一定能還原高電極電勢的氧化型Strength of an Oxidizing or Reducing AgentStrength of an

36、Oxidizing or Reducing Agent:strong oxidizing agents appear as reactants in half-equations toward the bottom best reducing agents appear on the right-hand side of half equations toward the top Example 6-1: Which of the following is the most powerful oxidizing agent? Cr3+(aq) ,Br2(l) , Cu2+ (s) Soluti

37、on: Look at the values for for the reduction of each of the above species: the largest value is the most powerful oxidizing agent Cr3+(aq) +3 e Cr(s) = - 0.74 V Cu2+ + 2e Cu(s) = + 0.34 V Br2(l) + 2e 2Br -(aq) = +1.07 V Example 6-2 Find a chemical species that will convert Ag+(aq) to Ag(s) without c

38、onverting Cu+(aq) to Cu(s). Solution: First determine what type of reaction occurs. The conversion of Ag+(aq) to Ag(s) is a reduction (the silver oxidation state goes from +1 to 0). Ag+(aq) + e Ag(s) = 0.80 V All half reactions with values more negative than 0.80 V will reduce Ag+(aq) to Ag(s). But

39、we dont want to reduce Cu+(aq) to Cu(s). Cu+(aq) + e Cu(s) = 0.52 V 例6-7 滴加氯水于含有Br-、I-離子的混合溶液中， 判斷在標準狀態(tài)下哪種離子先被氧化。 解：首先計算Cl2水氧化Br-、I-反應(yīng)的E值 （1） Cl2 + 2Br - = Br2 + 2Cl- E10 0Cl2/Cl- - 0Br2/Br- 1.36 1.087 0.273(V) （2） Cl2 + 2I- = I2 + 2Cl- E200Cl2/Cl-0I2/I- 1.36 0.54 0.82V） 因兩個反應(yīng)類型相同（轉(zhuǎn)移電子數(shù)相同），且 E20E10。

40、 故在標準狀態(tài)下，氯水首先氧化I-，后氧化Br-，這與實驗結(jié)果是一致的。 the stronger oxidizing agent will react with the stronger reduction agentRelative Oxidizing and Reducing PowerWhich of the following is most easily oxidized? Al Sn Cd Which of the following is most easily reduced? Mg2+ Na+ Sn2+ 2. Decision of the spontaneous dire

41、ction of redox reactionDecision of the spontaneous direction of redox reaction: E0 the direction of spontaneity of a reactionE0 the direction of spontaneity of a reaction Example (1) Sn2+ + 2Fe3+ = Sn4+ + 2Fe2+ (2) 2Br - + 2Fe3+ = Br2 + 2Fe2+ the direction of spontaneity of a reaction ?Sol. (1) spon

42、taneity of a reaction (2) non-spontaneity of a reaction , occurs in opposite direction Example 6-3 : MnO4- /Mn2+ half-cell is = l.49 V. Suppose this half-cell is combined with a Zn2+/Zn half-cell in a voltaic cell, with Zn2+ = MnO4- = Mn2+ = H+ =1mol/L Write equations for the half-reactions at the a

43、node and the cathode.(b) Write a balanced equation for the overall cell reaction.(c) Calculate the standard electromotive force(emf) of the cell.(d) write the notation of this cell. Solution:(a): (MnO4- /Mn2+) = 1.49 V (Zn2+/Zn) = - 0.76V . Permanganate ions will be reduced at the cathode. Cathode:

44、MnO4- Mn2+ MnO4- + 8H+ Mn2+ + 4 H2O MnO4-(aq)+8H+5eMn2+(aq)+4H2O(l) Anode: Zn(s) Zn2+ (aq) +2e(b) :2MnO4-(aq)+5Zn(s)+16H+(aq)=2Mn2+(aq)+5Zn2+(aq)+8H2O(l) (c) : E =(MnO4-/ Mn2+) - (Zn2+/Zn) = 1.49 (- 0.76) = 2.25 V(d): (-) Zn(s)Zn2+(aq)MnO4-(aq), Mn2+(aq )H+ (aq)Pt(s) (+) 3. Equilibrium Constants fro

45、m Electrochemistry T = 298.2 K 時， R = 8.315 Jmol-1 K-1; F = 9.647104 (c/mol) (1) Equilibrium Constants上式表明，在一定溫度下，氧化還原反應(yīng)的平衡常數(shù)與標準電池電動勢和反應(yīng)得失電子數(shù)有關(guān)，與反應(yīng)物的濃度無關(guān)。E0越大，平衡常數(shù)就越大，反應(yīng)進行越完全。因此，可以用E0值的大小來估計反應(yīng)進行的程度。一般說，E00.20.4V的氧化還原反應(yīng),其平衡常數(shù)均大于106（K106），表明反應(yīng)進行的程度已相當完全了。Example 6-5Calculate the equilibrium constant o

46、f the redox reaction at 25C 2 MnO4- (aq) + 5Zn(s) + 16 H+(aq) 2 Mn2+(aq) + 5 Zn2+(aq) + 8 H2O(l)SolutionLooking up two electrode potentials in table (6-1): MnO4- / Mn2+ : MnO4- +8 H+ +5 e Mn2+ + 4 H2O Zn 2+/Zn : Zn2+ + 2e Zn(s) in over reaction, 10 electrons were transferred , so n =10例6-8 計算下列氧化還原反

47、應(yīng)的平衡常數(shù)。 （1）Fe2+Ag+Fe3+Ag（s） （2）Pb2+CuPb（s）+Cu2+ 解: （1） E0 0Ag+/Ag -0Fe3+/Fe2+ 0.7996 - 0.771 0.0286 K 3.042 （2） E20 0Pb2+/Pb - 0Cu2+/Cu -0.1263 - 0.3402 -0.4665 K 1.7410-16(2).求溶度積常數(shù)許多難溶電解質(zhì)飽和溶液的離子濃度極低，用一般的化學(xué)分析方法很難準確測定其濃度,通常采取選擇適當電極組成原電池，測定其電池電動勢，進而方便、準確地確定其值。Example Calculate the equilibrium constan

48、t at 25for the reaction Ag+ + Cl- AgCl（s）and the solubility product Ksp for silver chloride.解 Ag+ + Cl- AgCl（s）+ Ag+ Ag由該式中尋找出兩個氧化還原電對，即 Ag+/Ag和AgCl（s）/Ag，Cl-電對， 其電極電勢值查表結(jié)果如下： AgCl（s）+ e Ag + Cl- 0 0.2223V Ag+ + e Ag 0 0.7996V將反應(yīng)式兩側(cè)各加一項金屬Ag，從電極電勢數(shù)值可以看出， 負極: AgCl（s）/Ag、Cl-， 正極: Ag+/Ag Ag+ + Cl- AgCl（

49、s）的平衡常數(shù)為： K = 5.62109AgCl（s） Ag+ + Cl- 例6-10應(yīng)用有關(guān)電極電勢數(shù)值確定Ag（CN）2-的 穩(wěn)定常數(shù)。解 Ag（CN）2-穩(wěn)定常數(shù)表達式為 Ag+ + 2CN- Ag（CN）2-將反應(yīng)式兩側(cè)各加一項金屬Ag，則反應(yīng)方程式為： Ag+ + 2CN- + Ag = Ag（CN）2- + Ag根據(jù)方程式: 電對組成 Ag+/Ag Ag（CN）2-/Ag，CN- K K穩(wěn) 5.5 1018 Ag（CN）2- +e Ag + 2CN- 0 - 0.31VAg+ +e Ag 0 0.7996V查表： Disproportionation _ is a process

50、 in which a single chemical species is both oxidized and reduced. Cu2+(aq) + e Cu+(aq) = 0.l6 V Cu+(aq) + e Cu(s) = 0.52 V 2 Cu+(aq) Cu2+(aq) + Cu(s) O2/H2O2: O2 +2H+2e H2O2 = 0.682 VH2O2/H2O: H2O2+2H+2e 2H2O = 1.77 V 2 H2O2 + 2H+ O2 + 2H+ + 2H2O 2 H2O2 O2 + 2H2O Example 6-4Use data in table (6-1) t

51、o decide whether the iron(II) ion is unstable with respect to disproportionation under standard conditions.Solution: Look up the table (6-1), we got two relevant half-cell potentials: Fe3+ + e Fe2+ = 0.77 V Fe2+ + 2 e Fe(s) = - 0.41 V Fe3+ + Fe(s) 2 Fe2+ Fe2O3nH2OFe2+ + 2 e Fe(s) = - 0.41 V Fe 3+ +

52、e Fe2+ = 0.77 V= 1.23 V元素電勢圖及其應(yīng)用 如鐵的元素電勢圖： -0.0365 圖中每一電對以橫線相連，并將0值標于橫線上方。1. 元素電勢圖當一種元素具有多種氧化態(tài)時，為了直觀的了解各氧化態(tài)之間的關(guān)系，將各電對的標準電極電勢按氧化數(shù)從高到低順序以圖解方式表示，這種表示元素各氧化態(tài)之間電勢變化的關(guān)系圖稱為元素電勢圖。 有些電對中含有H+或OH離子，由于溶液的pH值不同，物質(zhì)的存在形式及電極電勢值不同。因此，元素電勢圖分為酸性介質(zhì)和堿性介質(zhì)兩大類。酸性介質(zhì)電勢圖用A0 表示，堿性介質(zhì)電勢圖用B0 表示。如 2. 元素電勢圖的應(yīng)用（1）未知標準電極電勢的求算 若已知兩個或兩個

53、以上相鄰電對的標準電極電勢，即可 求算另一個電對的未知標準電極電勢。設(shè) 10、20是元素電勢圖中相鄰電對的標準電極電勢: 10、20、30 分別表示相鄰電對的已知標準電勢 n1 、n2 、n3 分別表示相鄰電對的轉(zhuǎn)移電子數(shù) 末0表示欲求電對的未知標準電勢例6-12 應(yīng)用下列元素電勢圖，求算1.50解： n1=2 n2=1 n3=3 將有關(guān)數(shù)值代入上式，則：0.3394V(2). 判斷歧化反應(yīng)能否發(fā)生：=例題：已知Br的元素電勢圖如下0.6126解：(1)0.61260.51960.7665(2)標準電極電勢應(yīng)用說明 1. 0值應(yīng)用的條件 0值是在標準狀態(tài)下水溶液中測出的，非水溶液、高溫、固相反

54、應(yīng)的情況不適用。另外，溶液中離子濃度為非標準狀態(tài)，且偏離標準狀態(tài)較大時，不宜用0值直接判斷反應(yīng)進行的方向和限度，應(yīng)經(jīng)過計算處理，才能得到正確結(jié)果。燃燒反應(yīng)能否發(fā)生E0無法判斷2. 0值與反應(yīng)速度無關(guān) 0值僅從熱力學(xué)角度衡量反應(yīng)的可能性和進行的程度。它是電極處于平衡狀態(tài)時表現(xiàn)出的特征值，與反應(yīng)平衡到達的快慢、即反應(yīng)速度的大小無關(guān)。與+7+6+7+2氧化劑還原劑實驗：應(yīng)見MnO4- 特有的顏色，（未見） 加熱,亦不見有顏色變化； 加Ag+催化，才見有MnO4- 出現(xiàn)。3.0值與反應(yīng)中物質(zhì)的計量系數(shù)無關(guān) 因為它是體系的強度性質(zhì)，取決于物質(zhì)的本性而與物質(zhì)的多少無關(guān)。Nonstandard electr

55、ode potentials are affected by the nature of electrode、the concentrations of reactants taking part in the reaction and the temperature . 6-4 Electrode Potentials for Nonstandard ConditionsFor a reduction half-reaction ox. + n e red. state R is the gas constant 8.314 J/(mol.K); T is the Kelvin temper

56、ature; n is the number of electrons transferred F is the Faraday constant, equal to 96500 C/mol. If we substitute 298K (25) for the temperature in the Nernst equation and put in values for R and F, we get: Nernst equation This equation allows us to compute the electrode potential at any concentratio

57、n of oxidation state and reduction states and at any temperature. Increasing the concentration of oxidation state, the potential should increase; In other words it should decrease the potential of the half cell that increasing concentration of reduction state;When ox.=red.=1mol.L-1 ,About Nernst equ

58、ation:（一）離子濃度的改變對電極電勢的影響 例6-15計算25時，電極Fe3+（l mol L1）， Fe2+（0.0lmolL1） Pt 的電極電勢。1. The Nernst Equation & C effect 0.771 + 0.1184 0.8894(V)解：查表知 Fe3+ + e Fe2+ 0 0.771V 根據(jù)能斯特方程式計算其電極電勢：例6-16 計算25時，Cd2+（0.1molL-1） Cd電極的電極電勢 解：查表知 Cd2+ + 2e Cd 0 - 0.4026V 例6-15和例6-16計算結(jié)果表明：當還原型物質(zhì)濃度減小時，電極電勢增大，即氧化型物質(zhì)的 氧化能力

59、增強；當氧化型物質(zhì)濃度減小時，電極電勢降低，即還原型物質(zhì)的 還原能力增強。 -0.4322(V)則Example 6-6: Calculate the electrode potential of following electrode in pH =5 solution ? Pt|MnO4- (1.0), Mn2+(1.0), H+ (10-5)Sol: Electrode half-reaction is MnO4- +8 H+ +5 e Mn2+ + 4 H2O生成沉淀對電極電勢的影響若有沉淀劑參加反應(yīng)時，由于生成難溶性沉淀改變了反應(yīng)物的離子濃度，電極電勢發(fā)生變化。例6-17 在電極反應(yīng)

60、Ag+ + e = Ag 體系中如入NaCl后， 設(shè)反應(yīng)達平衡時，Cl-1molL-1，計算其電極電勢。解：體系中加入Cl-后，發(fā)生如下反應(yīng)： Ag+ + Cl- AgCl 該反應(yīng)使溶液中Ag+離子濃度降低。 = 0.2227（V） 計算結(jié)果表明: 由于沉淀劑Cl-的加入，使氧化型物質(zhì)Ag+降低，電極電勢顯著降低，Ag+的氧化能力變?nèi)酢Ｒ话阏f，難溶性化合物的溶解度越小，對電極電勢的影響越大。 關(guān)于Ag+離子氧化能力變化數(shù)據(jù)： Ksp 電極反應(yīng) 0(V)降低 Ag+ + e Ag 0.7996 AgCl + e Ag + Cl- 0.2223 AgBr + e Ag + Br- 0.0713 A