Chapter six Electrochemistry Combination reaction Decomposition reaction Single-replacement reaction Double -replacement reaction 4Fe(s) + 3O2 → 2Fe2O3(s) Chemical reaction CaCO3(s) → CaO +CO2(g) Zn(s) + 2HCl(aq) → ZnCl2 +H2(g) AgNO3 + NaCl → NaNO3+AgCl(s)
6-1 Oxidation-reduction Concepts 6-2 Voltaic Cells 6-3 Electrode Potentials 6-4 Electrode Potentials for Nonstandard Conditions 6-5 Determination of pH
6-1 Oxidation-reduction Concepts Oxidation-reduction reaction Oxidizing agent Reducing agent Oxidation Numbers The rules of Assigning Oxidation Numbers
Oxidation and reduction take place at same time Oxidation-reduction reaction (Redox) is a reaction in which electrons are transferred from one reactant to another. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Oxidation and reduction take place at same time
half-reaction Zn (s) + CuSO4(aq) → ZnSO4(aq) + Cu (s) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) half-reaction Zn(s) - 2 e → Zn2+(aq) lost two electrons Cu2+(aq) + 2 e → Cu(s) gained two electrons
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) OXIDATION—loss of electron(s) by a species; increase in oxidation number. REDUCTION—gain of electron(s); decrease in oxidation number; OXIDIZING AGENT—electron acceptor; species is reduced. REDUCING AGENT—electron donor; species is oxidized.
Remember : +1 -1 oxidation number has no physical reality. the number of charges 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) +1 -1 Remember : oxidation number has no physical reality. eg. In SO3, the oxidation number of S is +6
Rules of oxidation number (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: -2 In a polyatomic ion, the sum of the oxidation 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. H—O—O—H The 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 compounds In a neutral molecule, the sum of the oxidation numbers of all the atoms must be zero. Fe3O4 中,Fe:+8/3; S4O62- 中,S:+5/2。
MnO4− + C2O42- Mn2+ + CO2 - OXIDATION increase in oxidation number, loss of electron(s) - OXIDATION decrease in oxidation number, electron acceptor; - REDUCTION REDUCING AGENT OXIDIZING AGENT MnO4− + C2O42- Mn2+ + CO2 +7 +3 +4 +2 Oxidizing agent Reducing agent
self oxidation-reduction 3 H2O+ 3 Cl2 → ClO3- + 6 H+ + 5Cl- self oxidation-reduction Oxidizing Agent Reducing 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 Equations Cu + Ag+ ---> Cu2+ + Ag Step 1: Divide the reaction into half-reactions, one for oxidation and the other for reduction. Ox Cu ---> Cu2+ Red Ag+ ---> Ag Step 2: Balance each element for mass. Already done in this case. Step 3: Balance each half-reaction for charge by adding electrons. Ox Cu ---> Cu2+ + 2e Red Ag+ + e ---> Ag
The equation is now balanced for both charge and mass. Balancing Equations Step 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+ + 2e Oxidizing agent 2 Ag+ + 2 e ---> 2 Ag Step 5: Add half-reactions to give the overall equation. Cu + 2 Ag+ ---> Cu2+ + 2Ag The equation is now balanced for both charge and mass.
Add H2O on O-deficient side and add H+ on other side for H-balance. Balancing Equations Balance the following in acid solution— VO2+ + Zn ---> VO2+ + Zn2+ Step 1: Write the half-reactions Ox Zn ---> Zn2+ Red VO2+ ---> VO2+ Step 2: Balance each half-reaction for mass. Red 2 H+ + VO2+ ---> VO2+ + H2O Add H2O on O-deficient side and add H+ on other side for H-balance.
Balancing Equations Step 3: Balance half-reactions for charge. Ox Zn ---> Zn2+ + 2e Red e + 2 H+ + VO2+ ---> VO2+ + H2O Step 4: Multiply by an appropriate factor. Ox Zn ---> Zn2+ + 2e Red 2e + 4 H+ + 2 VO2+ --> 2 VO2+ + 2 H2O Step 5: Add balanced half-reactions Zn + 4 H+ + 2 VO2+ ---> Zn2+ + 2 VO2+ + 2 H2O
E.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) Br-(aq) + 3H2O 2I-(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-
[ 6e- + 6H+ + BrO3-(aq) Br-(aq) + 3H2O ] × 1 Multiply the equation (s) by the appropriate factor so that the no. of e- in each half reaction is the same. [ 6e- + 6H+ + BrO3-(aq) Br-(aq) + 3H2O ] × 1 [ 2I-(aq) I2 (aq) + 2e- ] × 3
Add 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 Cells 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) In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.
Voltaic Cells We can use that energy to do work if we make the electrons flow through an external device. We call such a setup a voltaic cell.
6-2 Voltaic Cells (Primary 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 -anode a positive electrode - cathode an electrolyte a positive electrode - cathode A 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 energy
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Half-cell is that portion of an electrochemical cell in which a half-reaction takes place. Half-reactions The oxidation and reduction reactions at the electrodes are called Zn(s) → Zn2+(aq) + 2 e (Oxidation) Cu2+(aq) + 2 e → Cu(s) (Reduction)
the pair of electrode In every half-reaction, there are two states with different oxidation number of common element, the state with higher 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 + /Cu
Salt bridge: 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. A salt bridge can be a U-tube device filled with an electrolyte, such as potassium chloride.
Notation for a Voltaic Cell for, Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Zn(s)∣Zn2+( aq ) ‖Cu2+ ( aq )∣Cu(s ) (- ) ( + ) Cathode Anode Salt bridge Phase boundary
Zn(s)∣Zn2+( aq ); Cu(s ) ∣Cu2+ ( aq ) pair of electrode:( electrode couple) Zn2+/Zn ; Cu2+/Cu Mn+ + ne = M pair of electrode: Mn+/ M Ox + ne = Red Ox / Red Electrode: M∣Mn+
Fe2+ + Ag+ = Fe3+ + Ag Fe3+/ Fe2+ Ag+ / Ag Pt∣ Fe3+, Fe2+ ‖ Ag+ ∣ Ag
Electromotive Force The maximum potential difference between the electrodes of a voltaic cell is referred to as the electromotive force (emf) of the cell, denoted E E = φ cathode – φ anode = φ+ - φ - E is a positive number.
The types of electrode 1. Metal-metal ion electrode 2. Metal-insoluble salt electrodes 3. Gas electrode 4. Oxidation-reduction electrodes
The types of electrode 1. Metal-metal ion electrode M∣Mn+ : Mn+(aq) + ne = M(s) Zn(s) ∣ Zn2+( aq ); Cu(s ) ∣Cu2+ ( aq )
2. Metal-insoluble salt electrodes Ag-AgCl electrode : Ag , AgCl(s) | Cl-(c) AgCl + e- = Ag + Cl- The glass electrode
Calomel electrode Complex electrode Pt∣Hg(l)∣Hg2Cl2 (s) ∣Cl- (c) Hg2Cl2 (s)+ 2e = 2Hg + 2Cl- Complex electrode
3. Gas electrode : A hydrogen electrode : Pt∣H2(p)∣H+(c) 2H+(aq) + 2e = H2 (g)
4. Oxidation-reduction electrodes Pt∣ Fe3+ (c1),Fe2+(c2) Fe3+ + e = Fe2+
6-3 Electrode Potentials The producing of Electrode Potentials Standard electrode potential Standard hydrogen electrode (SHE) Strength of an Oxidizing or Reducing Agent Disproportionation (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: 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.
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 potential A 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 is designated by a superscript degree sign: φ°
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 ; φ° = 0.0000 V
100 kPa
Measuring the Standard Potential of the Cu2+ / Cu Electrode (-) Pt| H2(g, 100KPa) | H+(1mol/L) || Cu2+(1mol/L) | Cu(s)( + ) Eocell = φo(Cu2+/Cu) –φo(H+/H2) φo(Cu2+/Cu) – 0.000V = + 0.34 V φ0(Cu2+/Cu) = + 0.340 V Cu2+ ions are more readily reduced to Cu (s) than H+ ions are reduced to H2.
Measuring the Standard Potential of the Zn2+ / Zn Electrode (- ) Zn(s)∣Zn2+( 1mol/L ) ‖H+ ( 1mol/L )∣H2(g)∣Pt(s ) ( + ) Eocell =φo (H+/H2)– φo (Zn2+/Zn) 0.000V – φo (Zn2+/Zn) = - 0.763 V φ o (Zn2+/Zn) = - 0.763 V Zn2+ ions are less readily reduced to Zn(s) than H+ ions are reduced to H2.
Tab. the standard electrode potential The values for the table entries are reduction potentials, so lithium 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 reducing agents Decision of the direction of redox reaction Relationship to equilibrium constants Electrode potentials under non-standard conditions potentials for voltaic cells
1. Strength of an Oxidizing or Reducing Agent The strengths of oxidizing and reducing agents are indicated by their standard electrode potentials. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Cu + Fe3+(aq) → Fe2+(aq) + Cu2+ 2Li + F2 → 2Li+(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 Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Fe3+(aq) + Cu → Fe2+(aq) + Cu2+ F2 + 2Li →2 Li+(aq) +2 F -
Strength of an 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) Solution: Look at the values for φ° for the reduction of each of the above species: the largest value is the most powerful oxidizing agent Cr3+(aq) +3 e → Cr(s) φ° = - 0.74 V Cu2+ + 2e → Cu(s) φ° = + 0.34 V Br2(l) + 2e → 2Br -(aq) φ° = +1.07 V
Example 6-2 Find a chemical species that will convert Ag+(aq) to Ag(s) without converting 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) φ° = 0.80 V All half reactions with φ° values more negative than 0.80 V will reduce Ag+(aq) to Ag(s). But we don't want to reduce Cu+(aq) to Cu(s). Cu+(aq) + e → Cu(s) φ° = 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Eφ值 (1) Cl2 + 2Br - = Br2 + 2Cl- E10 = φ0Cl2/Cl- - φ0Br2/Br- =1.36 – 1.087 =0.273(V) (2) Cl2 + 2I- = I2 + 2Cl- E20=φ0Cl2/Cl--φ0I2/I- =1.36 – 0.54 =0.82V) 因两个反应类型相同(转移电子数相同),且 E20>E10。 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 agent
Relative Oxidizing and Reducing Power Which 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 direction of redox reaction E>0 the direction of spontaneity of a reaction E<0 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) spontaneity of a reaction (2) non-spontaneity of a reaction , occurs in opposite direction
Example 6-3 : MnO4- /Mn2+ half-cell is φ°= 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 anode 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.
Permanganate ions will be reduced at the cathode. Solution: (a): φ°(MnO4- /Mn2+) = 1.49 V φ°(Zn2+/Zn) = - 0.76V . Permanganate ions will be reduced at the cathode. Cathode: MnO4- → Mn2+ MnO4- + 8H+ → Mn2+ + 4 H2O MnO4-(aq)+8H++5e→Mn2+(aq)+4H2O(l) Anode: Zn(s) → Zn2+ (aq) +2e (b) : 2MnO4-(aq)+5Zn(s)+16H+(aq)=2Mn2+(aq)+5Zn2+(aq)+8H2O(l)
(c) : E° =φ°(MnO4-/ Mn2+) - φ°(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 from Electrochemistry T = 298.2 K 时, R = 8.315 J·mol-1 ·K-1; F = 9.647104 (c/mol)
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,E0≥0.2~0.4V的氧化还原反应,其平衡常数均大于106(K>106),表明反应进行的程度已相当完全了。
Example 6-5 Calculate the equilibrium constant of the redox reaction at 25°C 2 MnO4- (aq) + 5Zn(s) + 16 H+(aq) → 2 Mn2+(aq) + 5 Zn2+(aq) + 8 H2O(l)
Solution Looking 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
解: (1) E0 = φ0Ag+/Ag -φ0Fe3+/Fe2+ [例6-8] 计算下列氧化还原反应的平衡常数。 (1)Fe2++Ag+=Fe3++Ag(s) (2)Pb2++Cu=Pb(s)+Cu2+ 解: (1) E0 = φ0Ag+/Ag -φ0Fe3+/Fe2+ = 0.7996 - 0.771 = 0.0286 K = 3.042
(2) E20 = φ0Pb2+/Pb - φ0Cu2+/Cu = -0.1263 - 0.3402 = -0.4665 K= 1.74×10-16
(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 constant at 25℃for the reaction Ag+ + Cl- AgCl(s)and the solubility product Ksp for silver chloride.
+ Ag Ag+ + Cl- AgCl(s) + Ag 解 将反应式两侧各加一项金属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- φ0 = 0.2223V Ag+ + e = Ag φ0 = 0.7996V
AgCl(s) Ag+ + Cl- 从电极电势数值可以看出, 负极: AgCl(s)/Ag、Cl-, 正极: Ag+/Ag K = 5.62×109 AgCl(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-
[Ag(CN)2]- +e Ag + 2CN- φ0= - 0.31V 查表: [Ag(CN)2]- +e Ag + 2CN- φ0= - 0.31V Ag+ +e Ag φ0= 0.7996V K = K稳 =5.5 ×1018
2 Cu+(aq) Cu2+(aq) + Cu(s) Disproportionation __ is a process in which a single chemical species is both oxidized and reduced. Cu2+(aq) + e Cu+(aq) φ° = 0.l6 V Cu+(aq) + e Cu(s) φ° = 0.52 V 2 Cu+(aq) Cu2+(aq) + Cu(s)
O2/H2O2: O2 +2H++2e H2O2 φ° = 0.682 V H2O2/H2O: H2O2+2H++2e 2H2O φ° = 1.77 V 2 H2O2 + 2H+ O2 + 2H+ + 2H2O 2 H2O2 O2 + 2H2O
Example 6-4 Use data in table (6-1) to 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+ φ°= 0.77 V Fe2+ + 2 e→ Fe(s) φ°= - 0.41 V Fe3+ + Fe(s) →2 Fe2+
Fe2O3·nH2O Fe2+ + 2 e→ Fe(s) φ°= - 0.41 V Fe 3+ + e → Fe2+ φ°= 1.23 V
元素电势图及其应用 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。 如铁的元素电势图: -0.0365 图中每一电对以横线相连,并将φ0值标于横线上方。
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φA0 表示,碱性介质电势图用φB0 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 φ10、φ20是元素电势图中相邻电对的标准电极电势:
φ10、φ20、φ30…… 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 φ末0表示欲求电对的未知标准电势
解: n1=2 n2=1 n3=3 将有关数值代入上式,则: [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:
(2). 判断歧化反应能否发生: = 0.3394V
例题:已知Br的元素电势图如下 0.6126
解:(1) 0.6126
(2) 0.7665 0.5196
标准电极电势应用说明 反应能否发生E0无法判断 1. φ0值应用的条件 φ0值是在标准状态下水溶液中测出的, 非水溶液、高温、固相反应的情况不适用。 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φ0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
与 实验:应见MnO4- 特有的颜色,(未见) 2. φ0值与反应速度无关 φ0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。 +7 +6 +7 +2 与 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.φ0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。
6-4 Electrode Potentials for Nonstandard Conditions Nonstandard electrode potentials are affected by the nature of electrode、the concentrations of reactants taking part in the reaction and the temperature .
For a reduction half-reaction ox. + n e red. state R is the gas constant 8.314 J/(mol.K); T is the Kelvin temperature; 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 concentration of oxidation state and reduction states and at any temperature.
About Nernst equation: 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 ,
1. The Nernst Equation & C effect (一)离子浓度的改变对电极电势的影响 1. The Nernst Equation & C effect [例6-15]计算25℃时,电极Fe3+(l mol ·L-1), Fe2+(0.0lmol·L-1) Pt 的电极电势。 解:查表知 Fe3+ + e Fe2+ φ0 = 0.771V 根据能斯特方程式计算其电极电势: = 0.771 + 0.1184= 0.8894(V)
则 例6-15和例6-16计算结果表明: [例6-16] 计算25℃时,Cd2+(0.1mol·L-1) Cd电极的电极电势 解:查表知 Cd2+ + 2e Cd φ0 = - 0.4026V 则 = -0.4322(V) 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。
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Ag+ + e = Ag 体系中如入NaCl后, 设反应达平衡时,[Cl-]=1mol·L-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 φ0(V) Ag+ + e Ag 0.7996 AgCl + e Ag + Cl- 0.2223 降低 降 低 AgBr + e Ag + Br- 0.0713 AgI + e Ag + I- -0.1519
3. Dependence of potential on complexation: [例6-18] 在Fe3+,Fe2+都为1.0mol·L-1的溶液中,加入NaF, 使反应后溶液中F-的浓度为1.0mol·L-1,计算此时电对 Fe3+/Fe2+ 的电极电势。 解:若Fe3+与F-形成FeF3配合物(Fe3+的配位数一般为6,FeF3是化合物Fe(H2O)33F的简写式)则 Fe3+ + 3F- Fe3+ + 3F- FeF 3 因为FeF3的β3很大,而且溶液中F-是过量的,所以可认为Fe3+完全转化为FeF3。
Fe3+ + 3F- FeF 3 从配位平衡分析,FeF3配位物存在着如下平衡 平衡时各组分的浓度为 x 1.0 1.0-x 则 计算结果表明:NaF的存在,使Fe3+/Fe2+电对的电极电势降低,Fe3+的氧化能力减弱。 根据能斯特方程式
(二)离子浓度改变对氧化还原反应方向的影响 非标准状态下,对于两个电势比较接近的电对,判断反应进行的方向时,应考虑离子浓度变化的影响。 [例6-19] 判断 2Fe3++2I-=2Fe2++I2 (1)在标准状态下、 (2)[Fe3+]=0.001mol·L-1 [I-]=0.001mol·L-1 [Fe2+]=1mol·L-1时反应进行的方向。 解:(1)在标准状态 I2 + 2e 2I- φ0 =0.535V Fe3+ + e Fe2+ φ0 = 0.771V 判断反应正向进行: 2Fe3++2I- Fe2+ +I2
(2)在非标准状态: 氧化剂 还原剂
电池电动势 判断反应逆向进行:2Fe2++I2=2Fe3++2I-
[例6-20] 对于Cu2+ + Cd = Cd2+ + Cu的反应, (1)在标准状态下、(2)[Cu2+]=1×10-6mol·L-1 [Cd2+]=1mol·L-1时,是否都能发生金属镉置换Cu2+离子的反应。 解:(1)标准状态 则 能 ,标准状态下上述反应能够自发进行。
(2) 当[Cu2+]=l×10-6mol·L-1 、[Cd2+]=1mol·L-1 根据能斯特方程式计算: 电池电动势 计算结果表明,反应正向自发进行
例6-19 结果告诉我们,对于电池电动势比较小的反应,离子浓度的改变有可能引起反应方向的改变; 而例6-20结果表明:电池电动势较大的反应,离子浓度的改变一般不影响反应方向。 通常认为,电池电动势大于0.5V以上,离子浓度改变一般不致于引起反应逆转,所以常用标准电极电势判断反应的方向。此时,离子浓度的改变对电极电势的影响可以忽略。
(三)介质的酸度对电极电势的影响 在许多电极反应中包含着H+和OH-,说明溶液酸度将会对电极电势产生影响。 [例6-21] 在下列电极反应中,Cr2O72- + 14H+ + 6e = 2Cr3++7H2O φ0 = 1.33V。若将[Cr2O72-]和[Cr3+]都定为1mol·L-1,只改变H+离子浓度,对电极电势有何影响? 解:根据能斯特方程式,则有:
当[H+]=1mol·L-1时 当[H+]=10-3mol·L-1时 计算结果表明: Cr2O72-在强酸性溶液中的氧化性比 在弱酸性溶液中强。 在实验室和工业生产中,总是在较强酸性溶液中使用K2Cr207作为氧化剂。 当[H+]=10-3mol·L-1时
E=0 φ φ φ φ
Calculating Redox Equilibrium Constants Ex . : Calculate the equilibrium constant for the reaction 2Fe3+ + 3I- 2Fe2+ + I3- Sol: 2Fe3+ + 2e- 2Fe2+ φ0 = 0.771 V I3- + 2e- 3I- φ0 = 0.536 V
2Fe3+ + 3I- 2Fe2+ + I3-
电势法测定溶液的pH 电池电动势测定法(简称电势法):利用测定电池电动势以求溶液中待测离子浓度及某物质相对含量的分析方法。 参比电极(reference electrode) pH值测定 原电池 指示电极(indicator electrode) 参比电极:电位恒定,准确已知,常用甘汞电极 指示电极:电极电势能够正确地反映出待测离子的浓度
Pt∣Hg(l)∣Hg2Cl2 (s) ∣Cl- (c) 甘汞电极 Pt∣Hg(l)∣Hg2Cl2 (s) ∣Cl- (c) (参比电极) Hg2Cl2 (s)+ 2e 2Hg + 2Cl- 由上式可以看出,当温度一定时甘汞电极的电位主要决定于aCl-。当 aCl-一定时,其电极电位是个定值。 不同浓度的KCl溶液,使甘汞电极的电位具有不同的恒定值,如表所列。25℃时甘汞电极的电极电势:
Hydrogen Electrode 指示电极 H+(c)|H2(p)|Pt ; 2H+ + 2e H2(g)
Ag , AgCl(s) | Cl- (c) , H+(c) 玻璃电极 Ag , AgCl(s) | Cl- (c) , H+(c)
玻璃电极 应该指出,膜电位的产生不是由于电子的得失或转移,而是由于H+在溶液和硅胶层界面间进行迁移,改变界面上电荷的分布产生了相界电位,膜内外的相界电位差就是膜电位。 玻璃膜电极具有内参比电极,如Ag,AgCl 电极,因此整个玻璃膜电极的电位,应是内参比电极电位与膜电位之和,即
测定溶液的pH,常用玻璃电极作pH指示电极,饱和甘汞电极作参比电极,组成原电池。 (-) Ag , AgCl(s) | H+(c)| 待测pH溶液 || SCE (+) 电池电动势为: 与电极材质有关, 使用时需用标准溶液标定
在一定温度下,SCE为常数, 0玻未知, 将电极插入pHs的标准缓冲溶液中,测定电池电动势为 (1) 再将电极插入pHx的溶液中,测定电池电动势Ex (2) (2) -(1) 式: 整理得:
一、生物氧化概念、意义 H2O+CO2+能量 营养物 (糖、脂、蛋白质) 意义 用于生命活动 用于维持体温 生物体 营养物 (糖、脂、蛋白质) [O] 意义 用于生命活动 用于维持体温 实质是需氧细胞在呼吸代谢过程中所进行的一系列氧化还原反应过程。又称细胞氧化或组织呼吸。
分别进入两条呼吸链的底物 苹果酸 异柠檬酸 琥珀酸 β-羟丁酸 谷氨酸 FAD(Fe-S) 异柠檬酸 琥珀酸 β-羟丁酸 谷氨酸 FAD(Fe-S) NAD+ FMN CoQ b c1 c aa3 O2 丙酮酸 FAD α-酮戊二酸 脂肪酰CoA α-磷酸甘油
NADH呼吸链电子传递和水的生成 FADH2呼吸链电子传递和水的生成 O2 MH2 Fe S O2- M 2H+ H2O H2O Fe S 1 2 O2 MH2 还原型代 谢底物 FMNH2 NAD+ CoQ 2e 2Fe2+ 细胞色素 b- c- c1 -aa3 Fe S O2- M 氧化型代 谢底物 CoQH2 FMN 2Fe3+ NADH+H+ 2H+ H2O FADH2呼吸链电子传递和水的生成 H2O FAD FADH2 琥珀酸 Fe S 2Fe2+ 2Fe3+ 细胞色素 b- c1 - c-aa3 CoQH2 CoQ 1 2 O2 O2- 2H+ 延胡索酸 2e
电子传递体的排列顺序证据: 1.拆开和重组 2.氧化还原电位:从低电位向高电位传递 ⊿Gº'=-nF⊿Eº' ⑴NADH脱氢酶抑制剂:阻断复合物Ⅰ(Fe-S中心)→Q ,包括鱼藤酮,安密妥和杀粉蝶菌素 ⑵细胞色素b抑制剂:阻断Cytb→Cytc1,包括抗霉素A ⑶细胞色素氧化酶抑制剂:阻断Cytaa3→O2,包括氰化物(CN-),CO,H2S,叠氮化物(N3-)。 ⊿Gº'=-nF⊿Eº'
氧化磷酸化偶联部位 (琥珀酸) 底物 FAD (Fe-S) 底物 NADH FMN CoQ Cyt b c1 c aa3 O2 ADP+Pi ATP ~p ~p ADP+Pi ATP ADP+Pi ATP
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