山大《医学基础化学》双语课件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，Fe：+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歧化反应: 又称之为自身氧化还原反应。 在歧化反应中，化合物内的同一种元素的一部分原子(或离子)被氧化，另一部分原子(或离子)被还原。 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 (歧化反应) Equilibrium Constants from Electrochemistry 电 极 电 势 的 产 生在金属晶体中，存在着金属正离子，金属原子和自由电子。当把金属板插入它的盐溶液中时，有两种反应倾向: 在某一给定浓度的溶液中，若失去电子的倾向大于获得电子的倾向，到达平衡时的最后结果将是金属离子Mn进入溶液，使金属棒上带负电，靠近金属棒附近的溶液带正电。如图：这时在金属和盐溶液 之间产生电位差。_金属的电极电势 金属的电极电势与金属本

28、身的活泼性和金属离子在溶液中的浓度及温度有关。 金属越活泼、溶液浓度越稀、温度越高、溶解的倾向越大，金属表面所带负电荷越多；平衡时电极的电极电势愈低。反之，金属表面带的正电荷越多，电极的电极电势越高. Zn2/Zn电对的电极电势比Cu2/Cu电对要负一些。由于两极电势不同，连以导线，电子流(或电流)得以通过。在铜锌原电池中，实验告诉我们，如将两电极连以导线，电子流将由锌电极流向铜电极，这说明Zn片上留下的电子要比Cu片上多，或 Zn2/Zn电对与Cu2/Cu电对两者具有不同的电极电势，综上所述：原电池中电流的产生，是由于组成原电池的两电极存在着电势差所致，因此，可以说电极电势的大小标志着金属原

29、子或离子得失电子能力的大小，常用来衡量物质氧化还原能力的强弱。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 -对角线法则：高电极电势的氧化型一定能氧化低电极电势的还原型低电极电势的还原型一定能还原高电极电势的氧化型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-离子的混合溶液中， 判断在标准状态下哪种离子先被氧化。 解：首先计算Cl2水氧化Br-、I-反应的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） 因两个反应类型相同（转移电子数相同），且 E20E10。

40、 故在标准状态下，氯水首先氧化I-，后氧化Br-，这与实验结果是一致的。 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 时， R = 8.315 Jmol-1 K-1; F = 9.647104 (c/mol) (1) Equilibrium Constants上式表明，在一定温度下，氧化还原反应的平衡常数与标准电池电动势和反应得失电子数有关，与反应物的浓度无关。E0越大，平衡常数就越大，反应进行越完全。因此，可以用E0值的大小来估计反应进行的程度。一般说，E00.20.4V的氧化还原反应,其平衡常数均大于106（K106），表明反应进行的程度已相当完全了。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 计算下列氧化还原反

47、应的平衡常数。 （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).求溶度积常数许多难溶电解质饱和溶液的离子浓度极低，用一般的化学分析方法很难准确测定其浓度,通常采取选择适当电极组成原电池，测定其电池电动势，进而方便、准确地确定其值。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由该式中寻找出两个氧化还原电对，即 Ag+/Ag和AgCl（s）/Ag，Cl-电对， 其电极电势值查表结果如下： AgCl（s）+ e Ag + Cl- 0 0.2223V Ag+ + e Ag 0 0.7996V将反应式两侧各加一项金属Ag，从电极电势数值可以看出， 负极: AgCl（s）/Ag、Cl-， 正极: Ag+/Ag Ag+ + Cl- AgCl（

49、s）的平衡常数为： K = 5.62109AgCl（s） Ag+ + Cl- 例6-10应用有关电极电势数值确定Ag（CN）2-的 稳定常数。解 Ag（CN）2-稳定常数表达式为 Ag+ + 2CN- Ag（CN）2-将反应式两侧各加一项金属Ag，则反应方程式为： Ag+ + 2CN- + Ag = Ag（CN）2- + Ag根据方程式: 电对组成 Ag+/Ag Ag（CN）2-/Ag，CN- K K稳 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元素电势图及其应用 如铁的元素电势图： -0.0365 图中每一电对以横线相连，并将0值标于横线上方。1. 元素电势图当一种元素具有多种氧化态时，为了直观的了解各氧化态之间的关系，将各电对的标准电极电势按氧化数从高到低顺序以图解方式表示，这种表示元素各氧化态之间电势变化的关系图称为元素电势图。 有些电对中含有H+或OH离子，由于溶液的pH值不同，物质的存在形式及电极电势值不同。因此，元素电势图分为酸性介质和碱性介质两大类。酸性介质电势图用A0 表示，碱性介质电势图用B0 表示。如 2. 元素电势图的应用（1）未知标准电极电势的求算 若已知两个或两个

53、以上相邻电对的标准电极电势，即可 求算另一个电对的未知标准电极电势。设 10、20是元素电势图中相邻电对的标准电极电势: 10、20、30 分别表示相邻电对的已知标准电势 n1 、n2 、n3 分别表示相邻电对的转移电子数 末0表示欲求电对的未知标准电势例6-12 应用下列元素电势图，求算1.50解： n1=2 n2=1 n3=3 将有关数值代入上式，则：0.3394V(2). 判断歧化反应能否发生：=例题：已知Br的元素电势图如下0.6126解：(1)0.61260.51960.7665(2)标准电极电势应用说明 1. 0值应用的条件 0值是在标准状态下水溶液中测出的，非水溶液、高温、固相反

54、应的情况不适用。另外，溶液中离子浓度为非标准状态，且偏离标准状态较大时，不宜用0值直接判断反应进行的方向和限度，应经过计算处理，才能得到正确结果。燃烧反应能否发生E0无法判断2. 0值与反应速度无关 0值仅从热力学角度衡量反应的可能性和进行的程度。它是电极处于平衡状态时表现出的特征值，与反应平衡到达的快慢、即反应速度的大小无关。与+7+6+7+2氧化剂还原剂实验：应见MnO4- 特有的颜色，（未见） 加热,亦不见有颜色变化； 加Ag+催化，才见有MnO4- 出现。3.0值与反应中物质的计量系数无关 因为它是体系的强度性质，取决于物质的本性而与物质的多少无关。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）离子浓度的改变对电极电势的影响 例6-15计算25时，电极Fe3+（l mol L1）， Fe2+（0.0lmolL1） Pt 的电极电势。1. The Nernst Equation & C effect 0.771 + 0.1184 0.8894(V)解：查表知 Fe3+ + e Fe2+ 0 0.771V 根据能斯特方程式计算其电极电势：例6-16 计算25时，Cd2+（0.1molL-1） Cd电极的电极电势 解：查表知 Cd2+ + 2e Cd 0 - 0.4026V 例6-15和例6-16计算结果表明：当还原型物质浓度减小时，电极电势增大，即氧化型物质的 氧化能力

59、增强；当氧化型物质浓度减小时，电极电势降低，即还原型物质的 还原能力增强。 -0.4322(V)则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生成沉淀对电极电势的影响若有沉淀剂参加反应时，由于生成难溶性沉淀改变了反应物的离子浓度，电极电势发生变化。例6-17 在电极反应

60、Ag+ + e = Ag 体系中如入NaCl后， 设反应达平衡时，Cl-1molL-1，计算其电极电势。解：体系中加入Cl-后，发生如下反应： Ag+ + Cl- AgCl 该反应使溶液中Ag+离子浓度降低。 = 0.2227（V） 计算结果表明: 由于沉淀剂Cl-的加入，使氧化型物质Ag+降低，电极电势显著降低，Ag+的氧化能力变弱。一般说，难溶性化合物的溶解度越小，对电极电势的影响越大。 关于Ag+离子氧化能力变化数据： Ksp 电极反应 0(V)降低 Ag+ + e Ag 0.7996 AgCl + e Ag + Cl- 0.2223 AgBr + e Ag + Br- 0.0713 A