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1 Chapter 4 Reactions in Aqueous Solution The Important types of aqueous reaction, These include- 1.Precipiatation reactions 2.acid-base reactions 3.Oxidation-reduction reactions

2 4-1: Solute Concentration: 1. Molarity (M): 容積莫耳濃度 2. Normality (N): 當量濃度 4-2: Precipitation reaction 1. Net Ionic Equations fig Stoichiometry 4-3: Acid-base reactions 1. Strong and Weak Acids and Bases 2. Equations for Acid Base Reactions 3. Acid Base Titrations 4-4: Oxidation-reduction reactions 1. Oxidation number 氧化數 2. Balancing half-equations (oxidation or reduction) 3. Balancing Redox Equations

Stoichiometry 4-3: Acid-base reactions 1. Strong and Weak Acids and Bases 2.

3 溶液 (solution)- 由二種或二種以上的純物質均勻混合形成一個單相 (single phase), 其中含溶質與溶劑溶質 (solute)- 被溶解的物質, 溶液中較少量的物質溶劑 (solvent)- 溶解其他物質者, 溶液中較大量的物質溶液溶劑溶質碳酸飲料 (l) H 2 O 糖, CO 2 空氣 (g) 合金 (s) N 2 Pb O 2, Ar, CH 4 Sn 4.1

4 溶質的濃度定量的溶劑或溶液中, 所含溶質之量的多寡, 稱為該溶質的濃度共有八種表示法 Cl 1. 重量百分率 (weight percentage),% 2. 克式量濃度 (formality) 3. 體積莫耳濃度 (molarity), M 4. 重量莫耳濃度 (molality), m 5. 當量濃度 (normality),n 6. 莫耳分率 (mole fraction),x i 7. 百萬分數 (parts per million),ppm 8. 十億萬分數 (parts per billion),ppb 4.1

$莫耳分率 (mole fraction),x i 7. 百萬分數 (parts per million),ppm 8.$

5 範例 4.1 鹽酸的瓶子上標示著濃鹽酸, 表示 1 公升的這種溶液含有 12.0mol 的 HCl, 亦即 HCl = 12.0 M (a) 在 25.0 ml 的這種溶液中, 含有多少 moles 的 HCl? n HCl = 25.0 ml 1L 1000 ml 12.0 mol 1L HCl = 0.300mol HCl (b) 需多少體積的濃鹽酸才能含有 1.00 mol 的 HCl? V HCl = 1.00mol HCl 1L 12.0mol HCl 1000mL 1L = 83.3mL HCl 4.1

6 4-1 Solute Concentrations: Molarity (M) 1.The concentration of a solute in solution can be expressed in terms of its Molarity Molarity (M)= moles. of. solute liters. of. solution The symbol [ ] is commonly used to represent the molarity of a species in a solution. 表示在溶液中的莫耳濃度 Ex a solution containing 1.20mol of substance A in 2.50L of solution 1.20mol [ A] = = 0.480mol = M 2.50L L 2.The molarity of a solution can be used to Calculate: (1) The number of moles of solute in a given volume of solution. (2) The volume of solution containing a given number of moles of solute.

表示在溶液中的莫耳濃度 Ex a solution containing 1.20mol of substance A in 2.50L of solution 1.20mol [ A] = = 0.480mol = 0. 480M 2.50L L 2.

7 4.5

8 Ex 4-1: The bottle labeled dilute nitric acid in the lab contains 6.0 mol of per liter of solution 6.0 M a) How many moles of are in 75 ml of this solution? b) What volume of dilute nitric acid must be taken to contain one mole of? Sol: 1L 60mol a) n HNO = 75mL = 0.45moleHNO mL 1L b) 1L V = 1.00molHNO3 = 0. 17L 6.0molHNO 3

9 3.An ionic solid dissolves in water, the cations and anions separate from each other NaCl ( S ) + MgCl H O Mg ( aq ) + 2Cl ( aq 2( S) ) Cl 2+ molarity = molarity MgCl Mg MgCl 2 = molarity = 2 molarity Cl Cl MgCl2 = 2 ( NH4) 3 PO4( S) 3NH ( aq+ PO ) 4 ( aq) molarity = 3 molarity molarity + NH 4 PO 4 H 2 O (aq ) 3 = 1 molarity Na + Cl ( aq ) 2 ( NH PO 4 ) 3 ( NH4) 3 PO4 4

molarity = 2 molarity Cl Cl MgCl2 = 2 ( NH4) 3 PO4( S) 3NH + 3 4 ( aq+ PO ) 4 ( aq) molarity =

10 Ex 4-2: Give the concentration, in moles per liter, of each ion in a) 0.08 M b) 0.40 M a) K SO + + K = = 0.16 M 2 = = 0.08 M b) FeCl 3 ( aq 3( S ) + ) Cl 3+ = = 0.40 M Fe Cl 2 4( S) SO 4 2K + ( aq) SO 4 = = 1.2 M 2 3 aq Fe + ( )

08 = 0.08 M b) FeCl 3 ( aq 3( S ) + ) Cl 3+ = 1 0.40 = 0.

11 4-2 Precipitation Reactions 沈澱反應 Sometimes when water solutions of two different ionic compounds are mixed, an insoluble solid separate out of solution. precipitate Ksp solubility product 溶解度積沈澱由兩種不同溶液的陽離子與陰離子相互作, 不可溶的固體將會被分離出來

12 Figure 4.3 The precipitation diagram

13 沈澱反應規則所有鹼金屬 (IA 族 ) 及銨 (NH 4+ ) 化合物均具可溶性所有含 NO 3- ( 硝酸根 ) ClO 3- ( 氯酸根 ) ClO 4- ( 過氯酸根 ) 的化合物均可溶除 IA 族之氫氧化物 Ba(OH) 2 為可溶 Ca(OH) 2 為微溶外, 其餘之氫氧化物 (OH - ) 均不可溶多數氯化物 (Cl - ) 溴化物 (Br - ) 碘化物 (I - ) 是可溶, 但含 Ag + Hg 2 +2 Pb +2 例外所有碳酸根 (CO 3-2 ) 磷酸根 (PO 4-3 ) 及硫化物 (S -2 ) 均為不溶性大多數 SO 4-2 ( 硫酸根 ) 為可溶性, 但 CaSO 4 ( 硫酸鈣 ) 及 Ag 2 SO 4 ( 硫酸銀 ) 為微溶 ;BaSO 4 ( 硫酸鋇 ) HgSO 4 ( 硫酸汞 ) 及 PbSO 4 ( 硫酸鉛 ) 為不溶性 4.2

物 (I - ) 是可溶, 但含 Ag + Hg 2 +2 Pb +2 例外所有碳酸根 (CO 3-2 ) 磷酸根 (PO 4-3 ) 及硫化物 (S -2 ) 均為不溶性大多數 SO 4-2 ( 硫酸

14 Ex 4-3: Using the precipitation diagram (fig 4.3), predict what will happen when the following pairs of aqueous solutions are mixed. (a) CuSO 4 和 NaNO 3 CuSO 4(aq) NaNO 3(aq) Cu +2 (aq) + Na + (aq) + SO 4-2 (aq) NO 3 - (aq) Possible precipitates:cu(no) 3 及 Na 2 SO 4,form figure, both of these compounds are soluble So no precipitate forms. (b) Na 2 CO 3 和 CaCl 2 Na 2 CO 3(aq) 2Na + (aq) + CO 3-2 (aq) CaCl 2(aq) Ca +2 (aq) + 2Cl - (aq) Possible precipitates: NaCl andcaco 3,form figure NaCl is soluble, CaCO 3 is not. So CaCO 3 precipitates. 4.1

4,form figure, both of these compounds are soluble So no precipitate forms.

15 Precipitation Reactions 沈澱 Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) Molecule reaction equation 分子反應方程式 Pb NO Na + + 2I - PbI 2 (s) + 2Na + + 2NO 3 - Ionic reaction equation 離子反應方程式 Pb I - PbI 2 (s) PbI 2 Net ionic equation 淨離子反應方程式 Na + and NO 3- is 旁觀 (no part in the reaction)ion 4.2

2Na + + 2NO 3 - Ionic reaction equation 離子反應方程式 Pb 2+ + 2I - PbI 2 (s) PbI 2

16 1.Net ionic equation 1. 寫出平衡之分子反應方程式 2. 寫出其全部之離子反應方程式 3. 刪除反應式兩邊之旁觀離子 Write the net ionic equation for the reaction of silver nitrate with sodium chloride. AgNO 3 (aq) + NaCl (aq) AgCl (s) + NaNO 3 (aq) Ag + + NO 3- + Na + + Cl - AgCl (s) + Na + + NO 3 - Ag + + Cl - AgCl (s) 4.2

17 Net ionic equation must show atom balance There must be the same number of atoms of each element on both sides. 方程式兩邊原子數相同 charge balance There must be the same total charge on both sides. 方程式兩邊總電荷必須相等 In particular all of the chemical equations written throughout this chapter are net ionic equations. 4.2

方程式兩邊原子數相同 charge balance There must be the same total charge on both

18 Ex 4-4: Write a net ionic equation of the followings: (a) NaOH andcu(no 3 ) 2 Ions present:na +, OH -, Cu +2, NO 3 - Possible precipitates:nano 3,Cu(OH) 2, NaNO 3 is soluble, Cu(OH) 2 is not Cu +2 (aq) + 2OH - (aq) Cu(OH) 2(s) (b) BaCl 2 和 Ag 2 SO 4 Ions present:ba +2, Cl -, Ag +, SO 4-2 Possible precipitates:baso 4,AgCl,Both compounds are insoluble. Ba +2 (aq) + Ag + (aq) + SO 4-2 (aq) Cl - (aq) BaSO 4(s) AgCl (s) (c) (NH 4 ) 3 PO 4 和 K 2 CO 3 Ions present:nh 4+, PO 4-3, K +, CO 3-2 there is no precipitation reaction. 4.2

present:ba +2, Cl -, Ag +, SO 4-2 Possible precipitates:baso 4,AgCl,Both compounds are insoluble.

19 2.Stoichiometry 化學計量 Calculate mole-mass relations in reactions is readily applied to solution reaction represented by net ionic equations. 計算反應中莫耳 - 質量的關係. 以淨離子反應方程式來表達溶液的反應以一般的方法來解決限量反應物在淨離子反應方程式中的變化 4.2

20 Ex 4-5: When aqueous solutions of sodium hydroxide and iron nitrate are mixed, a red precipitate forms. (a) Write a net ionic equation for the reaction. (b) Determine the mass of precipitate formed when 30.0mL of 0.125M Fe(NO 3 ) 3 reacts. 3 a) ( aq) + (b) = 0.125mol/L m Fe(OH )3 = Fe + 3 OH ( aq ) n Fe 3+ = = n MM Fe( OH ) 3 ( s) 3 = mol 3 3 ( ( )) = 0.401g

0mL of 0.125M Fe(NO 3 ) 3 reacts. 3 a) ( aq) + (b) = 0.

21 Exam4.5 (c) Whe aqueous solutions of sodium (NaOH) and iron (III) nitrate Fe(NO 3 ) 3 are mixed,a red precipitate forms. Calculate the mass of precipitate formed when50.00ml,0.200m NaOH and 30.00mL,0.125MFe(NO 3 ) 3? Fe +3 (aq) + 3OH - (aq) Fe(OH) 3(s) 0.125molFe(NO 1molFe n + Fe 3 = LFe(NO3) 3 = molfe 1LFe(NO3) 3 1molFe(NO3) 3 ) n - OH 1L mol = 50.0mL NaOH = molOH 1000 ml L If Fe +3 is limiting reactant: W 1molFe(NO gFe(OH) Fe(OH) = molfe = 0.401gFe(OH) molFe 1molFe(NO3) 3 If OH - is limiting reactant W Fe(OH) 3 = mol OH - 1 mol Fe(NO 3 mol OH 3 - ) 3 ) g Fe(NO 1mol Fe(NO ) ) 3 = g Fe(OH) 3 4.2

22 範例 4.5 考慮範例 4.4 的淨離子反應方程式, 需要多少體積的 0.106MCu(NO 3 ) 2 溶液, 用以形成 6.52g 的 Cu(OH) 2? Cu +2 (aq) + 2OH - (aq) Cu(OH) 2(s) V Cu(OH) = gCu(OH) molCu 1molCu(OH) 2 1LCu(NO3 ) molCu(NO 1molCu(NO 1molCu 3 ) 1molCu(OH) gCu(O H) ) 2 = 0.630LCu(NO 3 )

23 4-3 Acid-base reactions Acid Acidic solution: 1. sour taste 2. Litmus( 石蕊 ) turns from blue to red. 3.It is a species that produces H + ion in water solution. Base Basic solution: 1. slippery feeling 2. Litmus turns from red to blue 3.It is a species that produces OH - ion in water solution. 2HCl (aq) + Mg (s) 2HCl (aq) + CaCO 3 (s) MgCl 2 (aq) + H 2 (g) CaCl 2 (aq) + CO 2 (g) + H 2 O (l) 4.3

24 1. Strong and Weak Acids and Bases. There are two types of acids, strong and weak, Strong acid:ionize completely, forming H+ ions and anions. Table 4.1 Ex + HCl( aq) H ( aq) + Cl ( aq) Start 0.10 mol - - balance 0.00 mol 0.10 mol 0.10 mol ionize completely Weak acid:only partially ionized to H + ions in water. Double arrow implies that this reaction does not go to completion. HB H + B + ( aq) aq ( aq) ) ( ) 開始 0.10 mol - - 平衡 0.09 mol 0.01 mol 0.01 mol partially ionized

25 Table4.1 Common Strong Acids and Bases Acid Name of Acid Base Name of Base HCl Hydrochloric acid 氫氯酸 LiOH Lithium hydroxide 氫氧化鋰 HBr Hydrobromic acid 氫溴酸 NaOH Sodium hydroxide 氫氧化鈉 HI Hydriodic acid 氫碘酸 KOH Potassium hydroxide 氫氧化鉀 HNO 3 Nitric acid 硝酸 Ca(OH) 2 Calcium hydroxide 氫氧化鈣 HClO 4 Perchloric acid 過氯酸 Sr(OH) 2 Strontium hydroxide 氫氧化鍶 H 2 SO 4 Sulfuric acid Ba(OH) 2 Barium htdroxide 硫酸氫氧化鋇 4.3

26 Strong base:in water solution is completely ionized to OH - ions and cations. Ex. NaOH Ca ( OH ) 2 ( S Na + ( s) ( aq) + OH ( aq) ) ( ) 2OH aq 2 Ca + aq ( ) Weak base: They react with molecules, acquiring H + ions and leaving OH - ions behind. Ex NH 3( aq) H 2 O NH + OH 4 ( aq (aq) ) + + 開始 0.10 mol - - 平衡 mol mol mol partially ionized. +

27 2. Equations for Acid-Base reactions Table 4.2 Types of Acid-Base Reactions 1. Strong acid-strong base 中和 (neutralization) H + (aq) + OH- (aq) H 2 O (l) net ionic Equation 2. Weak acid-strong base HB (aq) H + (aq) + B- (aq) H + (aq) + OH- (aq) H 2 O (l) HB (aq) + OH - (aq) H 2 O (l) + B- (aq) net ionic Equation 3. Strong acid-weak base NH 3(aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) H + (aq) + OH- (aq) H 2 O (l) H + (aq) + NH 3(aq) NH 4 + (l) 總反應 H + (aq) + B (aq) BH (aq) + net ionic Equation 4.3

28 Neutralization Reaction( 中和反應 ) acid + base salt + water HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O H + + Cl - + Na + + OH - H + + OH - Na + + Cl - + H 2 O H 2 O 4.3

29 Ex 4-6:Write a net ionic equation for each of the following reactions in dilute water solution. a) Hyopchlorous acid 次氯酸 and calcium hydroxide b) Ammonia with perchloric acid 過氯酸 c) Hydriodic acid () with sodium hydroxide Ans: a) HClO weak acid strong base HClO (aq ) + OH (aq + ) ClO (aq ) b) weak base strong acid 3 ( aq) + NH (aq) H + c) strong acid strong base H + (aq) OH + (aq (l) ) NH + 4 ( aq) O H 2 O H 2

30 Write a net ionic equation for each of the following reactions in dilute water solution (a) 亞硝酸 (HNO 2 ) 與氫氧化鈉 (NaOH) 的反應 HNO 2(aq) H + (aq) + OH - (aq) H 2 O (l) H + (aq) + NO - 2 (aq aq) HNO 2(aq) + OH - (aq) H 2 O (l) + NO - 2 (aq aq) 總反應 (b) 乙胺 (CH 3 CH 2 NH 2 ) 與過氯酸 (HClO 4 ) 的反應 C 2 H 5 NH 2(aq) + H 2 O (l) C 2 H 5 NH + 3 (aq) + OH - (aq) H + (aq) + OH - (aq) H 2 O (l) C 2 H 5 NH 2(aq) + H + (aq) C 2 H 5 NH + 3 (l) 總反應 (c) 溴酸 (HBr) 與氫氧化鉀 (KOH) 的反應 H + (aq) + OH - (aq) H 2 O (l) 4.3

31 3.Acid-Base Titration titration 滴定 Measuring the volume of a standard solution (a solution of known concentration) required to react with a measured amount of sample. 用已知濃度之標準溶液, 滴定未知溶液, 而測定出其樣品的含量 equivalence point 當量點 The number of equivalent mole of base equal to the number of equivalent mole of acid. 酸的莫耳數等於鹼的莫耳數 Neutralization point 中和點酸鹼滴定後, 其滴定終點恰為 ph7 強酸強鹼 End point 滴定終點 The point of the indicator change color 酸鹼滴定, 指示劑變色點 4.3

32 Ex4.7 In titration, it is found that 25.0mL of 0.500M NaOH is required to react with (a) a 15.00mL sample of HCl. What is the molarity of HCl? (b) A 15.0mL sample of a weak acid, H 2 A.What is the molarity of H 2 A, assuming the reaction to be An aspirin tablet weighing 2.50g. What is the percentage of acetylsalicylic acid,, in the aspirin tablet? HC 9 H 7 O 4(aq) + OH - (aq) C 9 H 7 O 4 - (aq) + H 2 O (l) Sol: (a) n HCl = n NaOH = ( = mol mol [HCl] = = 0.833M L 10 L)(0.500mo l/l) mol ( b). nh 2A = n = = 0.01mol OH M H = = 0.417mol / L 2 A L 2 H A + 2OH 2H O + A 2 2 ( c) nhc H 9 9 massnhc nhc H 7 9 O 4 7 O H 7 4 = noh O g = 2.50g = mol = mol g / mol 100 = 90.0 = 2.25g 4.3

33 4-4 Oxidation-Reduction Reactions Common type of reaction in aqueous solution involves a transfer of electrons between two species. Such r reaction is called an oxidationreduction or redox reaction. () Zn (s) + 2H + (g) Zn +2 (aq) + H 2(g) Zn Zn e - 2H + + 2e - H 2 Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) Zn + 2H + + 2e - Zn 2+ + H 2 + 2e - Zn (s) + 2H + (g) Zn +2 (aq) + H 2(g) 4.4

34 4-4 Oxidation-reduction Reactions Redox reactions The type of reaction in aqueous solution involves a transfer of electron between two species.the reactions is called an oxidation-reduction Loses electron Donates electron Increase in oxidation numbers Is oxidized 被氧化 Reducing agent Gains electron Accept( Receives) electrons Decrease in oxidation numbers Is reduced Oxidizing agent 1.Oxidation and reduction occur together, 2.There is no net change in the number of electrons in a redox reaction. Zn + H Zn + H + 2+ ( s) 2 ( aq) ( aq) 2( g) 4.4

35 4.4

36 1.Oxidation number 在氧化還原反應中其原子的淨電荷數, 陰電性較大的原子傾向於獲得電子而得到負的氧化數 1. The oxidation number of an element in an elementary substance is O 元素 ( 未化合狀態 ) 其氧化數為 0 Na, Be, K, Pb, H 2, O 2, P 4 = 0 2. The oxidation number of an element in a monatomic ion is equal to the charge of that ion. 單原子離子其氧化數恰為其電荷數 Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = -2 3.Certain elements have the same oxidation number in all or almost all their compounds. (1) group metals1 as +1 ; group metals 2 as +2 (2) O is an oxidation number of 2. But is H 2 O 2 及 O 2 2- is 1. 1/2 (3) Hydrogen is an Oxidation number of +1,the in metal hydrides, Where hydrogen is present the H is 1. H 2 O, H = +1; LiH, H =

37 4.he sum of the oxidation numbers in a neutral species is 0 ;in a polyatomic ion,it is equal to the charge of that ion. 中性分子中, 所有原子之氧化數總和為零在多原子離子中所有元素的氧化數總和必定與其電荷數相同 4.4

38 Oxidation numbers of all the elements in the following? IF 7 F = -1 7x(-1) +? = 0 I = +7 NaIO 3 Na = +1 O = -2 3x(-2) + 1 +? = 0 I = +5 K 2 Cr 2 O 7 O = -2 K = +1 7x(-2) + 2x(+1) + 2x(?) = 0 Cr =

39 Ex 4-8 :What is the oxidation number o f phosphorus in sodium phosphate,? Na 3 PO 4 In the dihydrogen phosphate ion H? 2 PO 4 Sol: Na PO 3 4 H 2 PO 4 ( + 1) 3+ x = + 5 x + ( 2) 4 = 0 ( + 1) 2+ x+ ( 2) 4= 1 x =+ 5

40 Zn (s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu (s) Zn Zn e - Zn is oxidized Zn is reducing agent Cu e - Cu Cu 2+ is reduced Cu 2+ is oxidizing agent 銅與硝酸銀反應生成硝酸銀與銀? 在本反應中何者為氧化劑? Cu (s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag (s) Cu Cu e - Ag + + 1e - Ag Ag + is reduced Ag + is oxidizing agent 4.4

41 Types of Oxidation-Reduction Reactions Combination Reaction A + B C S + O 2 SO 2 Decomposition Reaction C A + B KClO 3 2KCl + 3O 2 4.4

42 Types of Oxidation-Reduction Reactions Displacement Reaction A + BC Sr + 2H 2 O Sr(OH) 2 + H TiCl 4 + 2Mg Ti + 2MgCl Cl 2 + 2KBr 2KCl + Br 2 AC + B Hydrogen Displacement Metal Displacement Halogen Displacement 4.4

43 2. Balancing half-equation (Oxidation or Reduction) To balance half equation : 1.Balance the atoms of the element being oxidized or reduced. 2.Balance the oxidation number by adding electrons. 平衡氧化數 : 氧化數多的一邊加電子 3.Balance charge by adding ions in acidic solution, ions in basic solution. 平衡電荷數 : 酸溶液加 ( 電荷數較少之一邊 ), 鹼溶液加 ( 電荷數較多之一邊 ) 4. Balance hydrogen by adding molecules. 平衡 H: 5. Balance Oxygen number. 平衡 O

44 Balancing half-equation Oxidation reaction 氧化反應 Fe +2 (aq) Fe +3 (aq) ( 鐵的氧化數 : +2 +3) Fe +2 (aq) Fe +3 (aq) + e - Reduction reaction 還原反應 Cl 2 (aq) 2Cl - (aq) ( 氯的氧化數 : 0-1) Cl 2 (aq) + 2e - 2Cl - (aq) 4.4

45 Ex 4-9: Balance the following half-equations A) MnO - 4 (aq) Mn +2 (aq) (acidic solution) B) Cr(OH) 3(s) CrO -2 4 (aq) (basic solution) (A) MnO 4 - (aq) Mn +2 (aq) (acid solution) (a) There is one atom of Mn on both sides,mn MnO 4 - (aq) Mn +2 (aq) (b) Mn an oxidation number of +7 +2,is reduced reaction MnO 4 - (aq) + 5e - Mn +2 (aq) (c) The total charge of balance,left -6,right +2, 酸性溶液中, 以 H + 平衡電荷 MnO 4 - (aq) + 5e - + 8H + (aq) Mn+2 (aq) (d) Balance the eight H ions on the left,add four 4 H 2 O to the right MnO 4 - (aq) + 5e - + 8H + (aq) Mn+2 (aq) + 4H 2 O (l) (e) 方程式左右各有 4 個 O, 為己平衡之還原半反應 MnO 4 - (aq) + 5e - + 8H + (aq) Mn +2( aq) + 4H 2 O(l 4.4

46 (B) Cr(OH) 3(s) CrO 4-2 (aq) basic solution (a) 兩邊均含一個 Cr, 所以不必平衡 Cr Cr(OH) 3(aq) CrO 4-2 (aq) (b) Cr 的氧化數 +3 +6, 為氧化半反應 Cr(OH) 3(aq) CrI -2 4 (aq) + 3e - (c) 平衡總電荷, 左邊 0, 右邊 -5, 鹼性溶液中, 以 OH - 平衡電荷 Cr(OH) 3(aq) + 5OH - (aq) Mn+2 (aq) + 3e- (d) 平衡 H, 左邊 8 個 H, 所以右邊加入 4 H 2 O Cr(OH) 3(aq) + 5OH - CrO -2 4 (aq) + 3e - + 4H 2 O (l) (e) 方程式左右各有 8 個 O, 為己平衡之還原半反應 Cr(OH) 3(aq) + 5OH - CrO -2 4 (aq) + 3e - + 4H 2 O (l) 4.4

47 3. Balancing Redox equation 1. Split the equation into tow half-equations, one for reduction, the other for oxidation. 將反應方程式分別以兩個半反應方程式表示, 一個氧化, 一個還原 2. Balance one of the half-equations with respect to both atoms and charge as described above 平衡其中一個半反應方程式中的原子數及電荷數 3. Balance the other half-equation 平衡另一個半反應方程式中的原子數及電荷數 4. Combine the two half-equations in such a way as to eliminate electrons(2 + 3) 消去電子藉著消去方程式兩邊的電子數, 將兩個半反應方程式結合起來 4.4

48 Ex 4-10:Balance the following redox equations (a) Fe +2 (aq) + MnO 4 - (aq) Fe +3 (aq) + Mn +2 acid solution (1) 氧化 : Fe +2 (aq) Fe+3 (aq) Fe +2 (aq) Fe+3 (aq ) + e- 還原 : MnO 4 - (aq) Mn +2 (aq) MnO 4 - (aq) + 5e - + 8H + (aq) Mn+2 (aq) + 4H 2 O (l) 為使電子可消去, 故在氧化反應方程式乘以 5 5Fe +2 (aq) 5Fe+3 (aq) + 5e- MnO - 4 (aq) + 5e - + 8H + (aq) Mn+2 (aq) + 4H 2 O (l) MnO 4 - (aq) + 5Fe H + (aq) Mn+2 (aq) + 5Fe+3 (aq) + 4H 2 O (l) 4.4

49 (b) Cl 2(aq) + Cr(OH) 3(aq) Cl - (aq) + CrO 4-2 (aq) 鹼性溶液 (1) 氧化 : Cr(OH) 3(aq) CrO 4-2 (aq) Cr(OH) 3(s) + 5OH - (aq) CrO 4-2 (aq ) + 3e- + 4H 2 O (l) 還原 : Cl 2(aq) 2Cl - (aq) Cl 2(aq) + 2e - 2Cl - (aq) 為使電子可消去, 故還原反應方程式 3 為使電子可消去, 故在氧化反應方程式 2 3Cl 2(aq) + 6e - 6Cl - (aq) 2Cr(OH) 3(s) + 10OH - (aq) 2CrO 4-2 (aq ) + 6e- + 8H 2 O (l) 3Cl 2(aq) + 2Cr(OH) 3(s) + 10OH - (aq) 2CrO 4-2 (aq ) +6Cl- (aq) + 8H 2 O (l) 4.4

50 Ex 4-11:As you found in Example 4.10, the balanced equation for the reaction between and in acidic solution is What volume of M KMnO4 solution is required to react completely with 27.5 ml M Fe(NO3) 2? MnO 4 - (aq) + 5Fe H + (aq) Mn+2 (aq) + 5Fe+3 (aq) + 4H 2 O (l) n + 2 Fe n +2 Fe = L n KMnO molFe 1L (NO 3 1molMnO 5molFe V = ) 2 KMnO 4 n M 1molFe 1molFe(NO n - = mol = MnO KMnO 4 KMnO ) 2 mol 4 = 6.88 MnO mol Fe + 2 V MnO - 4 = 1.38 = mol L MnO - 4 1mol KMnO 1mol MnO L 0.684mol KMnO 4 4.4

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