在半反应中，没有H+参与的电对氧化能力不变；Hg22+、Cl2、Cu2+ ； H在氧化型一边的电极电位下降，氧化能力减弱；Cr2O72-、MnO4-、H2O2；
＋
H在还原型一边的电极电位上升，氧化性增强。 14、求298.15K，下列电池的电动势，并指出正、负极：
Cu(s) │ Cu2+(1.0?10-4 mol·L-1)‖Cu2+(1.0?10-1 mol·L-1) │ Cu(s) 0.05916V-12+
解 ?右＝?(Cu2+/ Cu)+lg(1.0?10)＝?(Cu/ Cu)－0.029 6V
2θθθθ?左＝?(Cu2+/ Cu)+ 0.05916Vlg(1.0?10-4)＝?(Cu2+/ Cu)－0.118 4V
＋
2右边为正极，左边为负极。 E = 0.118 4V－0.029 6V = 0.088 8 V
15、已知298.15K下列原电池的电动势为0.388 4V：
（－）Zn(s)│Zn2+(x mol·L-1) ‖Cd2+(0.20 mol·L-1) │Cd (s)（＋）
则Zn2+离子的浓度应该是多少？ 解 查表知
?θθ(Cd2+/Cd) = - 0.403V；?(Zn2+/Zn) = - 0.762V
θθE?＝?(Cd2+/Cd) -?(Zn2+/Zn) = -0.403V - (-0.762V) = 0.359V
0.05916V0.05916V[Zn]由E＝E- lgQ ＝0.388 4V = 0.359V -lg0.222?2?得 [Zn2+] = 0.021 mol·L-1 16、298.15K， Hg2SO4(s) + 2e-
Hg22+ (aq) + 2e-
试求Hg2SO4的溶度积常数。
解 将两个电极组成原电池；Hg22+ （aq）+ SO42（aq）＝ Hg2SO4(s)
－
2Hg(l) + SO42- (aq) ?θ=0.6125 V 2Hg(l) ?θ= 0.7973 V
Eθ＝0.7973V-0.6125V＝0.185v , n=2
lgK= 2×E/0.059 16v，K= 1.8×106 Ksp =
???1= 5.6×10-7 ?K17、已知298.15K下列电极的标准电极电位
Hg2Cl2(s) + 2e-
2Hg（l） + 2Cl- (aq) ?θ= 0.268V
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问当KCl的浓度为多大时，该电极的?= 0.327V。 解 ?(Hg2Cl2/Hg) =
?θ(Hg2Cl2/Hg) +
0.05916V1lg?2 2[Cl]0.327V＝0.268V＋
0.05916V1lg?2 2[Cl]解得 [Cl-] = 0.1 mol·L-1
18、在298.15K，以玻璃电极为负极，以饱和甘汞电极为正极，用pH值为6.0的标准缓冲溶液组成电池，测得电池电动势为0.350V；然后用活度为0.01 mol·L-1某弱酸（HA）代替标准缓冲溶液组成电池，测得电池电动势为0.231V。计算此弱酸溶液的pH值，并计算弱酸的解离常数Ka。
解 根据 pH?pHs?(E?Es)F
2.303RT(0.231V?0.350V)pH?6.0?＝4.0
0.05916VH+]= 1.0×10-4 mol·L-1
又 c = 0.01 mol·L-1
[H?]2Ka = = 1.0×10-6
cExercises
1. What is the value of the equilibrium constant at 25℃ for the reaction (refer to the table of standard electrode potential): I2(s) + 2Br(aq)
－
2I-(aq) + Br2(l)？
Solution ?θ( I2/ I-)=0.5355V； ?θ( Br2/ Br-)=1.066V
lgK = nFE/RT =
?θ2?(0.5355V?1.066V)＝-17.94
0.05916VK? = 1.15×10-18
?2. What is ?Gand E at 25℃ of a redox reaction for which n=1 and equilibrium constant K = 5
??×103 ？
Solution ?G= -RTln K = - 8.314J·K-1·mol-1×298K×ln(5×103)= -21100J·mol-1
?-21100J?mol-1 ?G=-nFE, n=1, E= - = 0.219V -196500C?mol???3. Balance the following aqueous skeleton reactions and identify the oxidizing and reducing agents: (1) Fe(OH)2(s) + MnO4-(aq) → MnO2(s) + Fe(OH)3(s) (basic)
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(2) Zn(s) + NO3-(aq) → Zn2+(aq) + N2(g) (acidic) Solution (1) 3 Fe(OH)2(s)+ MnO4-(aq) + 2H2O＝MnO2(s)+3Fe(OH)3(s)＋OH-(aq) MnO4-(aq) is the oxidizing agent
(2) 5Zn(s) + 2NO3-(aq)＋12H+ ＝ 5Zn2+(aq) + N2(g) +6H2O NO3-(aq) is the oxidizing agent
4. Write the cell notation for the voltaic cells that incorporate each of the following redox reactions: (1) Al(s) + Cr3+(aq) → Cr(s) +Al 3+(aq )
(2) Cu2+(aq) + SO2(g) + 2H2O(l) → Cu(s) +SO42-(aq) +4H+(aq) Solution (1) (－) Al(s) │Al3+(c1) ‖Cr3+(c2) │Cr(s) (+)
(2) (－) Pt(s)│SO2(g) │SO42-( c1), H+( c2) ‖Cu2+(c3) │Cu(s) (+)
5. A primary cell consists of SHE (as an anode) and a Cu2+/Cu electrode. Calculate [Cu2+] when Ecell = 0.25V.
Solution ?θ(Cu2+/Cu)=0.342V
?(Cu2+/Cu)＝?θ(Cu2+/Cu)+
0.05916Vlg[Cu2?] 2E =?( Cu2+/Cu)－?（SHE） 0.25 V＝0.342V＋
－
0.05916Vlg[Cu2?] - 0.000V 2 [Cu2+]＝7.8×104 mol·L-1
6. A primary cell consists of Ni2+/Ni and Co2+/Co half cells with the following initial concentrations: [Ni2+]=0.8mol·L-1; [Co2+]=0.2 mol·L-1. (If the volume of solution is the same)
(1) What is the initial E? (2) What is E when [Co2+] reaches 0.4 mol·L-1? (3) What is the equilibrium constant K?
(4) What is the value of [Ni2+]/[Co2+] when E=0.025V Solution (1) ?θ(Co2+/Co)= - 0.28V, ?θ(Ni2+/Ni) = - 0.257V
?(Co2+/Co)=?θ(Co2+/Co)+
?0.05916Vlg0.2＝－0.300V 20.05916V?(Ni2+/Ni)=?θ(Ni2+/Ni)+ lg0.8＝－0.260V
2The initial E = - 0.260V – (- 0.300V) = 0.040V Cell reaction: Ni2+ (aq)+ Co → Co2+ (aq) + Ni , n=2 (2) When [Co2+] = 0.4mol·L-1，[Ni2+] = 0.6mol·L-1;
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?(Co2+/Co)=?θ(Co2+/Co)+
0.05916Vlg0.4＝－0.292V 20.05916V?(Ni2+/Ni)=?θ(Ni2+/Ni)+ lg0.6＝－0.264V
2E = - 0.264V – (- 0.292V) = 0.028V
(3) E＝－0.257V– (- 0.28V)＝0.023V , and n=2
lgK = nE/0.05916v =
???2?0.023V= 0.778, K= 6.0
0.05916V0.05916v0.05916V[Ni2?]θ(4) E?E? lgQ?E?lg2?n2[Co]0.05916V[Ni2?]lg0.025V=0.023V+ 2[Co2?][Ni2?]＝1.17 2?[Co]7. A concentration cell consists of two hydrogen electrodes. Electrode A has H2 at 0.9 atm bubbling into 0.1 mol·L-1 HCl， Electrode B has H2 at 0.5atm bubbling into 2.0 mol·L-1 HCl. Which electrode is the anode? What is the E? What is the equilibrium constant K?
Solution hydrogen electrode half-reaction 2H+(aq) + 2e
θ－? H2 ， n=2
0.05916V0.12lgElectrode A： ?A=?A + ＝-0.058V 20.90.05916V2.02lgElectrode B： ?B=?B + ＝0.0267V 20.5θElectrode A is the anode.
E = 0.0267V – (- 0.058V) = 0.0847V
∵ E＝0.0V， lgK = nE/0.05916V ∴ K=1.
8. In a test of a new reference electrode, a chemist constructs a primary cell consisting of a Zn2+/Zn electrode and the hydrogen electrode under the following conditions: [Zn2+]=0.01 mol·L-1; [H+]=2.5 mol·L-1; pH2 =30 kPa, Calculate the E at 25℃.
Solution ?θ( Zn2+/Zn)= - 0.762V
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?(Zn2+/Zn)=?θ(Zn2+/Zn)+
+
θ0.05916Vlg0.01＝－0.821V 20.05916V2.52?(H/H2)=?(H/H2) +lg＝0.039V
20.3+
E ＝0.039V－(－0.821V)＝0.860V
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