Tuyen tap BT Dien Hoa

7/30/2019 Tuyen tap BT Dien Hoa

1/26

I. OLYMPIC HA HC VIT NAM:OLYMPIC HA HC SINH VIN TON QUC 2003 (Bng A):

Brom lng tc dng c vi H3PO3theo phn ng:H3PO3 + Br2 + H2O H3PO4 + 2H

+ + 2Br-1) Tnh hng s cn bng ca phn ng 298K2)

Tnh th in cc chun E

o

(H3PO4/H3PO3) nu bit Eo

(Br2/2Br-

) = 1,087V3) Tnh th in cc chun Eo(H3PO3/H3PO2) nu bit Eo(H3PO4/H3PO2) = 1,087VCho bit cc s liu sau 298K:

H+(dd) H3PO4(dd) Br-(dd) H3PO3(dd) Br2(l) H2O(l)

Hott(kJ/mol) 0 -1308 -141 -965 0 -286So(J/mol.K) 0 -108 83 167 152 70BI GII:

1) Hop = -339kJSop = -331JK-1.Gop = -240,362kJ lgK 42,125 K = 1,33.1042.

2) Gop = -nFEop Eop = 1,245VE

o

(Br2/2Br-

) - Eo

(H3PO4/H3PO3) = Eop = 1,245V

Eo(H3PO4/H3PO3) = -0,158V - 0,16V3) H3PO4 + 4H

+ + 4e H3PO2 + 2H2O Eo1 = - 0,39V (1)

H3PO4 + 2H+ + 2e H3PO3 + H2O E

o1 = - 0,16V (2)

Ly phng trnh (1) (2) ta c: H3PO2 + 2H+ + 2e H3PO2 + H2O Eo3 = ?Go3= Go1 - Go2 -2FEo3 = -4FEo1(-2FEo2) Eo3 = -0,62V

OLYMPIC HA HC SINH VIN TON QUC 2003 (Bng A):Cho bit cc th in cc chun: Eo(Cu2+/Cu) = 0,34V; Eo(Cu2+/Cu+) = 0,15V; Eo(I2/2I-) =

0,54V.1) Hi ti sao ngi ta c th nh lng Cu2+trong dung dch nc thng qua dung dch KI?

Cho bit thm rng dung dch bo ho ca CuI trong nc nhit thng (25oC) c

nng l 10-6M2) S dng tnh ton xc nh xem Cu c tc dng c vi HI gii phng kh H2 hay

khng?3) Mui Cu2SO4 c bn trong nc hay khng? Gii thch.

BI GII:1) Cu2++ e Cu+ Eo1 = 0,15V

Cu2+ + I- + e CuI Eo2 = ?

Cu

ICuEE

oo2

12 lg059,0

[Cu2+] = [I-] = 1M

M

I

KCu s 1210

Eo2 = 0,15 + 0,059lg1012 = 0,86 > Eo(I2/I

-)Vy c phn ng: Cu2+ + 3I- CuI + I2.nh lng I2theo phn ng: I2 + Na2S2O3 Na2S4O6 + 2NaI

2) Cu2++ 2e Cu Eo1 = 0,34VCu2++ e Cu+ Eo2 = 0,15V Cu++ e Cu Eo3 = 0,34.20,15 = 0,53VCuI + e Cu + I- Eo4 = Eo3 + 0,059lg10-12 = -0,17V


2/26

Vy c phn ng: 2Cu +2HI 2CuI + H23) Cu++ e Cu Eo1 = 0,53V

Cu2++ e Cu+ Eo2 = 0,15V2Cu+ Cu + Cu2+ Eo = 0,530,15 = 0,38VVy Cu2SO4l mui tan trong nc, khng bn trong dung dch:

Cu2SO4 Cu + CuSO4OLYMPIC HA HC SINH VIN TONQUC 2005 (Bng A): loi tr cc ion NO3-trong nc (cc ion NO3-c mt trong nc xut pht t phn

bn) c th kh n thnh NO2-bng cch cho i qua li c cha bt Cd.1) Vit na phn ng ca hai cp NO3-/HNO2 v HNO2/NO trong mi trng axit. Chng

minh rng HNO2b phn hy trong mi trng pH = 0 n 6.2) pH = 7, nng NO3- l 10-2M. Vit phn ng gia Cd v NO3-. Hi NO3-c b kh

hon ton 25oC trong iu kin ny khng? Tnh nng NO3-cn li trong nc khicn bng.

3) Tnh th kh (th oxy ha - kh) chun ca cp NO3-/NO2- pH = 14 v 25oCCho bit cc s liu sau 25oC: Eo(NO3-/HNO2) = 0,94V; Eo(HNO2/NO) = 0,98V;

E

o

(Cd

2+

/Cd) = -0,40V; Ka(HNO2) = 5.10

-4

; Ks(Cd(OH)2) = 1,2.10

-14

.BI GII:1) NO3

- + 3H+ + 2e HNO2 + H2O; Eo = 0,94V

HNO2 + H+ + e NO + H2O; E

o = 0,98V pH = 0 th Eo(HNO2/NO) > Eo(NO3-/HNO2) nn HNO2b phn hy theo phn ng:3HNO2 NO3

- + 2NO + H+ + H2O pH = 6 th:Eo(NO3

-/HNO2) = 0,94 + 0,059/2(lg10-6) =

Eo(HNO2/NO) = 0,98 + 0,059lg10-6 = 0,626V

Eo(HNO2/NO) vn ln hn Eo(NO3-/HNO2) nn HNO2vn khng bn

2) Cd + NO3- + H2O Cd2++ NO2- + 2OH-Gi thit phn ng l hon ton th [Cd2+] = [NO3-]b = 10-2M pH = 7 th [Cd2+] = Ks/[OH-]2= 1,2M. Nng Cd2+sau phn ng nh hn nhiu so vi

1,2M nn khng c kt ta Cd(OH)2. tnh [NO3-] khi cn bng cn tnh hng s cn bng K ca phn ng trn:Cd + NO3

- + H2O + 3H+ K Cd2+ + NO2

- + 2OH- + 3H+

Cd2+ + HNO2 + 2H2O 2K Cd2+ + H+ + NO2

- + 2H2OK = K1.K2.K3.

14214445

45

11

10.325,1)10.(10.5.10.65,2

10.65,242,45

059,0

)40,094,0(2lg

K

KK

Hng s K rt ln nn phn ng gn nh hon ton. pH = 7 ta c:

Cd + NO3- + H2O Cd

2+ + NO2- + 2OH-

Ncb: (10-2x) = x = 10-2 x = 10-2 10-7Nh vy ta c:

K1 K1


3/26

MNO 3332722

14 10.55,7)10.(10.10

10.325,1

3) VEE

K oNONO

o

NONO017,0

059,0

)40,0(2lg

23

23

/

/

1

OLYMPIC HA HC SINH VIN TON QUC 2005 (Bng A):C th ho tan hon ton 100mg bc kim loi trong 100ml dung dch amoniac nng 0,1M khi tip xc vikhng kh c khng?

Cho bit nguyn t khi ca Ag = 107,88; hng s in li baz ca amoniac l Kb =1,74.10-5; cc hng s bn ca phc [Ag(NH3)i]+tng ng l: lg1 = 3,32(i = 1) v lg2 = 6,23(i =2).

Cc th kh (th oxy ha - kh) chun 25oC: Eo(Ag+/Ag) = 0,799V; Eo(O2/OH-) =0,401V. p sut ring phn ca oxy trong khng kh l 0,2095atm. Phn ng c thc hin 25oC.BI GII:

NAg = 0,100 : 107,88 = 9,27.10-4mol

S mol cc i ca NH3cn to phc l: 9,27.10-4

. 2 = 1,854.10-3

M ngha lnh hnnhiu so vi s mol NH3c trong dung dch (10-2M). Vy NH3rt d ho tan lng Ag nuxy ra phn ng.

Chng ta s kim tra kh nng ho tan theo quan im in ha v nhit ng: Ag+ + e Ag E1 = E

o1 + 0,059lg[Ag

+]

O2 + 4e + H2O 4OH-

422

2lg4

059,0

OH

PEE

Oo

Khi cn bng E1 = E2. Trong dung dch NH3= 0,1M (lng NH3 phn ng khng ngk) ta c: [OH-] = (Kb.C)1/2 = 1,32.10-3M

E2 = 0,5607V.

V E2 = E1nn t tnh ton ta c th suy ra c [Ag

+

] = 9,12.10

-5

MNng tng cng ca Ag+trong dung dch:[Ag+]o = [Ag

+] + [Ag(NH3)+] + [Ag(NH3)2

+]= [Ag+](1 + 1[NH3] + 12[NH3]

2) = 15,5MGi tr ny ln hn nhiu so vi lng Ag dng cho phn ng. V vy cc iu kin in

ha v nhit ng thun li cho vic ho tan 0,100g AgOLYMPIC HA HC SINH VIN TON QUC 2005 (Bng B):

1) Trn hai th tch bng nhau ca hai dung dch SnCl2 0,100M v FeCl30,100M. Xc nhnng cc ion thic v st khi cn bng 25oC. Tnh th ca cc cp oxy ha - kh khicn bng.

2) Nhng mt si Ag vo dung dch Fe2(SO4)3 2,5.10-2M. Xc nh nng ca Fe3+; Fe2+ vAg

+

khi cn bng 25

o

C. Tnh th ca cc cp oxy ha - kh khi cn bng.Cho bit Eo(Sn4+/Sn2+) = 0,15V; Eo(Fe3+/Fe2+) = 0,77V; Eo(Ag+/Ag) = 0,80VBI GII:

1) Sn2+ + 2Fe3+ Sn4+ + 2Fe2+Ncb: 0,05- x 0,052x x 2x

lgK = 2(0,77015)/0,059 = 21 K = 1021.K rt ln v nng Fe3+cho phn ng nh hn nhiu so vi Sn2+phn ng gn nh

hon ton: 2x 0,05


4/26

[Fe2+] = 0,05M; [Sn4+] = 0,025M; [Sn2+] = 0,025M; [Fe3+] = M

K = MFe 1232

21

2

2

10.58,10025,0

10.1.025,0

)05,0.(025,0

Khi cn bng Ecb = 0,77 + 0,059lg M15,0025,0

025,0lg

2

059,015,0

05,0

10.58,1 12

2) Ag + Fe3+ Ag+ + Fe2+ncb: 0,05x x x

lgK = (0,770,80)/0,059 = -0,51 K = 0,31Ta c:

MFe

MFeAgxx

x

33

222

10.6

10.38,431,005,0

Ecb = V72,010.38,4lg059,080,010.38,4

10.6lg059,077,0 2

2

3

K THI CHN HC SINH GII QUC GIA VIT NAM 2002 (BNG A)1. Bit th oxi ha-kh tiu chun :

EoCu

2+/Cu

+= +0,16V, E

oCu

+/Cu = +0,52V, E

oFe

3+/Fe

2+= +0,77V, E

oFe

2+/Fe

= -0,44V

Hy cho bit hin tng g xy ra trong cc trng hp sau:(a) Cho bt st vo dung dch Fe2(SO4)3 0,5M.(b) Cho bt ng vo dung dch CuSO4 1M.

2. Dung dch X gm Na2S 0,010M, KI 0,060M, Na2SO4 0,050M.Axit ho chm dung dch Xn pH = 0. Thm FeCl3cho n nng 0,10M.

i Tnh th ca cc platin nhng trong dung dch thu c so vi cc calomen boho (Hg2Cl2/2Hg,2Cl-).

ii Biu din s pin, vit phng trnh phn ng xy ra ti cc in cc v phnng tng qut khi pin hot ng.

Cho : axit c H2S pK1 = 7,00, pK2 = 12,90; HSO4- c pK = 2,00; Tch s tan

ca PbS = 10-26; PbSO4 = 10-7,8

; PbI2 = 10-7,6

.E

oFe

3+/Fe

2+= 0,77 V ; E

oS/H2S = 0,14V ; E

oI2/2I

-= 0,54V ; Ecal bo ho = 0,244V

BI GII:1. a) E

oFe

3+/Fe

2+= +0,77 V > Eo Fe2+/Fe = -0,44 V nn:

Tnh oxi ho: Fe3+

mnh hn Fe2+Tnh kh: Fe mnh hn Fe2+Do phn ng t pht xy ra gia 2 cp l: 2 Fe3+ + Fe 3 Fe2+Nh vy Fe tan trong dung dch Fe(SO4)3 to thnh mui FeSO4, lm nht mu vng ( hoc nu) ca ion Fe3+v cui cng lm mt mu (hoc to mu xanh nht) dung dch. b) E

oCu

+/Cu = + 0,52 V > Eo Cu2+/Cu+ = + 0,16 V nn:

Tnh oxi ho: Cu+ mnh hn Cu2+

Tnh kh: Cu+ mnh hn CuDo phn ng t pht xy ra giaa 2 cp l: Cu+ + Cu+ Cu2+ + Cu


5/26

Phn ng nghch (Cu2+phn ng vi Cu to thnh ion Cu+) khng xy ra. Do khi b bt ngvo dung dch CuSO4khng xy ra phn ng v quan st khng thy hin tng g. 2. Axit ho dung dch X:S

2-+ 2H

+ H2S (C H2S = 0,010 < S H2S nn H2S cha bo ho, khng thot ra

khi dung dch)

Phn ng: 2 Fe3+ + H2S 2 Fe2+ + S + 2 H+ K=10210,1 0,01

0,08 0,02 0,022 Fe

3++ 2I

- 2 Fe2+ + I2 K=10

7,8

0,08 0,06 0,02

0,02 0,08 0,030Thnh phn trong dung dch: Fe3+ 0,020 ; Fe2+ 0,080 ;I2 0,030M ;H

+0,02M

E Fe3+

/Fe2+

= 0,77 + 0,059 lg 0,02/0,08 = 0,743V (cc dng)Ecal= 0,244V ( cc m)

Epin = E+ E= 0,743 0,244 = 0,499VS pin:Hg | Hg2Cl2 | KCl bh || Fe

3+, Fe

2+| Pt

Phn ng: 2 Hg + 2 Cl- = Hg2Cl2 + 2 e+ 2x Fe

3++ e = Fe

2+

2 Hg + 2 Fe3+

+ 2 Cl-

= Hg2Cl2K THI CHN HC SINH GII QUC GIA VIT NAM 2002 (BNG A)

Cho dng in 0,5A i qua dung dch mui ca mt axit hu c trong 2 gi. Kt qu sauqu trnh in phn l trn catot to ra 3,865 gam mt kim loi v trn anot c kh etan v kh

cacbonic thot ra.1. Cho bit mui ca kim loi no b in phn? Bit rng 5,18 gam ca kim loi y c 1,59gam Cu t dung dch ng sunfat.2. Cho bit mui ca axit hu c no b in phn?3. Vit cc phng trnh phn ng xy ra trn cc in cc. BI GII:1. in lng Q = It = 0,5 x 2 x 3600 = 3600 coulomb dng to ra 3,865 g kimloi. T nh lut Faraday, ng lng

Khi lng mol ca kim loi: A = n. . V kim loi ny y ng ra khi dung dch nnng lng ca Cu:

Cu= A/2 = 63,6/2 = 31,8 v t phn ng:2 + Cu2+ = Cu+ 2 +

ta c: : 31,8 = 5,18 : 1,59, suy ra = 103,6Trong phn ng y Cu, kim loi ch c th c mc oxi ho t 1 n 3, do s chn khi

lng mol nguyn t t 3 kh nng sau:

6,1033600

9650865.3

n

A

+


6/26

A1 = 103,6 x 1 = 103,6

A2 = 103,6 x 2 = 207,2

A3 = 103,6 x 3 = 310,8V khng c nguyn t vi A > 240 v bng 104 c tnh kim loi v c mc oxi ho l +1.

Do kim loi phi tm ch c th l Pb (A = 207,6).

2. Ti anot khi in phn c C2H6v CO2thot ra l sn phm ca s oxi ho anionhu c, mui ny c cng thc Pb(RCOO)2. S to ra etan.(CH3 - CH3) v CO2 t nhm COO

- chng t mui in phn l

Pb(CH3COO)2 .

R R

3. Cc phn ng xy ra trn cc in cc:Ti catot: Pb2+ + 2 e = PbTi anot: CH3COO

-- e = CH3COO

CH3COO

= CH3

+ CO2

2 CH3

= C2H6Tng qut: 2 CH3COO

-2e = C2H6+ CO2.

K THI CHN HC SINH GII QUC GIA VIT NAM 2004 (BNG A)

Dung dch A gm AgNO3 0,050 M v Pb(NO3)2 0,100 M.

1. Tnh pH ca dung dch A.2. Thm 10,00 ml KI 0,250 M v HNO3 0,200 M vo 10,00 ml dung dch A. Sau phn ng ngi tanhng mt in cc Ag vo dung dch B va thu c v ghp thnh pin (c cu mui tip xc haidung dch) vi mt in cc c Ag nhng vo dung dch X gm AgNO3 0,010 M v KSCN 0,040 M.

a) Vit s pin .b) Tnh sc in ng Epinti 25

0C .

c) Vit phng trnh phn ng xy ra khi pin hot ng. d) Tnh hng s cn bng ca phn ng .

Cho bit : Ag+ + H2O AgOH + H+

(1) ; K1= 1011,70

Pb2+

+ H2O PbOH+

+ H+

(2) ; K2= 107,80

Ch s tch s tan pKs : AgI l 16,0 ; PbI2 l 7,86 ; AgSCN l 12,0 .

;RT

ln = 0,0592 lgF= 0 ,799 V

E0Ag+/Ag

2. Epins thay i ra sao nu: a) thm mt lng nh NaOH vo dung dch B ; b) thm mt lngnh Fe(NO3)3vo dung dch X?

BI GII:

1. Ag+

+ H2O AgOH + H+

; K1 = 10-11,7

(1)Pb

2++ H2O PbOH

++ H

+; K2 = 10

-7,8 (2)

Do K2 >> K1nn cn bng 2 quyt nh pH ca dung dchPb

2++ H2O PbOH + H

+; K2 = 10

-7,8 (2)

C 0,10

0,10 x x x


7/26

8,72

10x1,0

x

x = 10-4,4

= H+ ; pH = 4,40

2.a) Dung dch B: Thm KI : CAg+ = 0,025 M; CPb2+ = 0,050CI- = 0,125M ; CH+ = 0,10M

Ag+

+ I

AgI0,025 0,125

- 0,10

Pb2+

+ 2 I

PbI20,05 0,10

- -

Trong dung dch c ng thi hai kt ta AgIv PbI2

AgI Ag+ + I ; Ks1 = 1.10-16

(3)

PbI2 Pb2+

+ 2 I

; Ks2 = 1.10-7,86

(4)

Ks1


8/26

V179,0E

10.33,3lg0592,0799,0Aglg0,05920,799E

2

11

2

V E2 > E1, ta c pin gm cc Ag trong X l cc + , cc Ag trong B l cc

S pin:AgI AgSCNPbI2 SCN

0,03 M

b) Epin = 0,1790,001 = 0,178V

c)Phng trnh phn ng: Ag + I AgI + e

AgSCN + e Ag + SCN

AgSCN + I

Ag + SCN

d) K = = = 104

3. a) Khi thm lng nh NaOH vo dung dch B , c th xy ra 3 trng hp: - Lng NaOH qa t khng trung ho HNO3: S to phc hidroxo ca Pb

2+vn

khng ng k, do Epinkhng thay i.- Lng NaOH trung ho HNO3: C s to phc hidroxo ca Pb

2+ do Pb2+

gim, Nng I-s tng ln, do nng Ag+gim xung, E1gim ; vyEpintng.-Lng NaOH d trung ho htHNO3 v ho tan PbI2to thnh PbO2

, do Pb2+gim vEpintng. PbI2 + 4 OH

PbO2

+ 2 H2O + 2 I

b) Thm t Fe3+ vo dung dch X: Fe3+ + SCN FeSCN2+

Nng ion SCNgim, do nng ion Ag+tng, E2tng EpintngK THI CHN HC SINH GII QUC GIA VIT NAM 2005 (BNG A)

pH = 0 v 25oC th in cc tiu chun Eoca mt s cp oxi ha-kh c cho nhsau:

2IO4/I2(r) = 1,31V; 2IO3

/I2(r) = 1,19V; 2HIO/I2(r) = 1,45V ; I2(r)/2I

= 0,54V

(r): ch cht trng thi rn.1. Vit phng trnh na phn ng oxi ha - kh ca cc cp cho.2. Tnh Eoca cc cp IO4

/ IO3

v IO3

/ HIO

3. V phng din nhit ng hc th cc dng oxi ha-kh no l bn, cc dng no l khngbn? Ti sao?4. Thm 0,40 mol KI vo 1 lt dung dch KMnO40,24 M pH = 0

a) Tnh thnh phn ca hn hp sau phn ng.b) Tnh th ca in cc platin nhng trong hn hp thu c so vi in cc calomen bo

ho.

5. Tnh Eo ca cp IO3/ I2(H2O).

I2(H2O) ch iot tan trong nc.Cho bit: Eo = 1,51 V ; E ca in cc calomen bo ha bng 0,244 V ;

Ag Ag

KsAgSCNKsAgI10

12

1016

MnO4/Mn

2+RTF


9/26

25oC, ln = 0,0592 lg ; tan ca iot trong nc bng 5,0.104M.

BIGII:1. 2 IO4

+ 16 H

++ 14 e I2(r) + 8 H2O ; E

o= 1,31 V = E

o1

2 IO3

+ 12 H+

+ 10 e I2(r) + 6 H2O ; Eo

= 1,19 V = Eo2

2 HIO + 2 H+

+ 2 e I2(r) + 2 H2O ; Eo

= 1,45 V = Eo3

I2 (r) + 2 e 2 I

; Eo

= 0,54 V = Eo

4

2. 2 IO4

+ 16 H+

+ 14 e I2(r) + 8 H2O ; K1= 10

I2 (r) + 6 H2O 2 IO3 + 12 H+ + 10 e ; K21 = 10

2 IO4

+ 4 H+

+ 4 e 2 IO3

+ 2 H2O ; K5

= 10

K5 = K1. K21 Eo5 = E

o=

4

1014 21OO

EE = 1,61 V

2 IO3

+ 12 H+

+ 10 e I2(r) + 6 H2O ; K2

= 10

I2(r) + 2 H2O 2 HIO + 2 H+

+ 2 e ; K31

= 10

2 IO3

+ 10 H+

+ 8 e 2 HIO + 4 H2O ; K6=K2. K3

1

K6= 10 = K2. K31

Eo

= Eo6=

8

210 32OO

EE = 1,125 (V)

3. V E < Eo

nn HIO s t oxi ho - kh

4 2 HIO + 2 H+ + 2 e I2(r) + 2 H2O

2 HIO + 4 H2O 2 IO3

+ 10 H+

+ 8 e

10 HIO 4 I2(r) + 2 IO3

+ 2 H+ + 4 H2O

Vy dng km bn nht v mt nhit ng hc l HIO, cc dng khc: IO4, IO3

,

I2, I u bn pH = 0.

4. a)Eo = 1,51V >>Eo (Eo4nh nht) nn u tin s xy ra phn ng:

2 MnO4

+ 8 H+

+ 5 e Mn2+ + 4 H2O

IO4-/IO3

-

14Eo1/0,0592

-10Eo2/0,0592

4Eo5/0,0592

10Eo2/0,0592

-2Eo3/0,0592

8Eo6/0,0592

IO3-/HIO

IO3-/HIO HIO/I2

MnO4-/Mn2+ I2/2I

-


10/26

5 2 I I2(r) + 2 e

2 MnO4

+ 10 I

+ 16 H+ 2 Mn2+ + 5 I2(r) + 8 H2O ; K =

10

163

CO 0,24 0,4

C 0,04 2 0,04 10 0,04 2 0,04 5

C 0,16 0 1 0,08 0, 2MnO4

cn d s oxi ho tip I2 thnh IO3

.

Eo

= 1,51 V > Eo

= 1,19 V

2 MnO4

+ 8 H+

+ 5 e Mn2+

+ 4 H2O

I2(r) + 6 H2O 2 IO3

+ 12 H+

+ 10 e

2 MnO4

+ I2(r) + 4 H+

2 IO3

+ 2 Mn2+

+ 2 H2O ; K =10

176

CO 0,16 0,2 0,08

C 0,082 0,08 0,082 0,082C 0 0,12 1 0,16 0, 24

Thnh phn hn hp sau phn ng: IO3

= 0,16 M; Mn2+

=0,24 M; I2(H2O)=5.104M; )

I2(r) = 0,12M; pH = 0.

b) Trong hn hp c cp IO3-/I2 (r) nn: VHIOEE o

rIIO18,1lg

10

0592,0 223)(/ 23

E so vi in cc calomen bo ho: 1,18 0,244 = 0,936V

5. 2IO3-+ 12H

++ 10e I2(r) + 6H2O : K2 = 10

10.1,19/0,0592

I2(r) I2(H2O) S = 5.10-4M

2IO3- + 12H+ + 10e I2(H2O) + 6 H2O; K7=10 =1010.1,19/0,0592.SSuy ra E7

o= 1,17V

K THI CHN HC SINH GII QUC GIA VIT NAM 2005 (BNG A)Mt bnh in phn cha dung dch NaOH (pH=14) v mt bnh in phn khc cha

dung dch H2SO4(pH = 0) 298K. Khi tng hiu in th t t hai cc mi bnh ngii ta thyc kh ging nhau thot ra c hai bnh ti cng in th.1. Gii thch hin tng trn. Vit cc phng trnh phn ng xy ra mi bnh (khng xt s tothnh H2O2 v H2S2O8).

2. Tnh hiu in th ti thiu phi t vo hai cc mi bnh cho qa trnh in phn xy ra. 3. Ngi ta mun gim pH ca dung dch NaOH xung cn 11. C th dng NH4Cl c khng?Nu c, hy gii thch v tnh khi lng NH4Cl phi dng gim pH ca 1 lt dung dch

NaOH t 14 xung cn 11.4.Khi pH ca dung dch NaOHbng 11, thhiu in th ti thiu phi t vo hai cc ca bnhin phn cho qa trnh in phn xy ra l bao nhiu?Cho bit: Eo = 0,4V; Eo =1,23 V; pKb(NH3)= 4,75.

BI GII:1.Trong th nghim ny, nc b in phn cng mt in th.

E7o

/0,0592

H2O, 1/2 O2/2OH 2H+, 1/2 O2/H2O


11/26

a) Dung dch NaOH: anot: 2 OH H2O + 1/2 O2 + 2 e catot: 2 H2O + 2 e H2 + 2 OH

H2O H2 + 1/2 O2

b) Dung dch H2SO4: anot: H2O 1/2 O2 + 2 H+ + 2 e catot: 2 H+ + 2 e H2

H2O H2 + 1/2 O2

Kh thot ra 2 bnh u l hidro v oxi2.

a) Dung dch NaOH:Eant= 0,4 V

Ecatt= 0 + lg (1014

)2

= 0,83 V

Umin = Eant Ecatt = 0,4 + 0,83 = 1,23 V

b)Dung dch H2SO4:Eant= 1,23 V

Ecatt= 0 V

Umin = Eant Ecatt = 1,23 V

(khi tnh Uminkhng xt n qu th).

3. C th dng NH4Cl gim pH ca dung dch NaOH t 14 xung 11.

NH4+

+ OH

NH3 + H2O

pOH ca dung dch NaOH thmNH4Cl gim pH ca dung dch NaOH t 14

xung 11 c tnh theo cng thc:

pOH = pKb + lg

3 = 4,75 + lg

Suy ra [NH4+] = 0,0178[NH3]

Khi pH ca dung dch NaOH gim t 14 xung 11 th [OH] ca dung dch gim i: 1 10

3= 0,999 mol. y chnh l s mol NH3hnh thnh. Vy [NH3] = 0,999 mol/L v:

[NH4+] = 0,0178 0,999 0,0178 (mol/L)

S mol NH4Cl phi thm vo 1 lt dung dch:

n = n + n = 0,0178 + 0,999 = 1,0168 (mol)

Khi lng NH4Cl phi thm vo 1 lt dung dch: 1,0168 53,5 = 54,4 (gam)

2

0,0592

[NH3]

[NH4+]

[NH3]

[NH4+]

NH4+ NH3


12/26

4. Khi pH = 11, dung dch NaOH:

Eanot= 0,4 V + lg

Ecatt= 0 + lg (1011

)2

Umin = Eanot Ecatot = 0,4 + 3 0,0592 + 0,0592 11 1,23 V

II. OLYMPIC HA HC QUC T:K THI OLYMPIC HA HC QUC T LN TH 30:

Vng kim loi thng c pht hin trong cc loi aluminosilicat v b phn tnnhuyn trong cc khong cht khc. Vng c th c tch bng cch cho nghin vn tc dngvi dung dch natri xianua sc khng kh. Trong qa trnh ny vng km loi c chuyn chmthnh [Au(CN)2]

-tan c trong nc (phn ng (1)).Sau khi t n cn bng, phn dung dch (pha dung dch) c bm ra v vng kim loi

c thu hi bng cch cho phc vng tc dng vi km, km c chuyn thnh [Zn(CN)4]2-(phn ng 2).

1) Vit v cn bng cc phng trnh ion ca phn ng (1) v (2).Vng trong t nhin thng dng hp kim vi bc v bc cng b oxy ho bi dung dchnatri xianua sc khng kh.

2) 500L dung dch cha [Au(CN)2]- 0,0100M v [Ag(CN)2]-0,0030M c cho bay hi nch cn mt phn ba th tch ban u v c x l bng km (40g). Gi thit rng s sailch so vi iu kin tiu chun l khng quan trng v cng gi thit l cc phn ng oxyha kh xy rahon ton. Hy tnh cc nng ca [Au(CN)2]- v [Ag(CN)2]- sau khiphn ng kt thc.Cho bit:[Zn(CN)4]

2- + 2e- Zn + 4CN- Eo = -1,26V[Au(CN)2]

- + e- Au + 2CN- Eo = -0,60V[Ag(CN)2]

- + e- Ag + 2CN- Eo = -0,31V

3) [Au(CN)2]-l mt phc rt bntrong mt s iu kin nht nh. Nng ca natri xianual bao nhiu gi c 99% theo s mol ca vng trong dung dch dng phc xianua?Bit Kb([Au(CN)2]-) = 4.1028.

4) c mt s c gng pht trin cc qa trnh tch chit vng khc thay th cch trn.Ti sao?. Hy chn phng n ng:

a) Dung dch natri xianua n mn cc dng c khai thc m.b) Natri xianua thot vo nc ngm trong t v to ra hydroxianua rt c vi nhiu

ng vt.c) Vng thu c t phng php ny khng tinh khit.

BI GII:1) Phn ng (1): 4Au + 8CN- + O2 + 2H2O 4[Au(CN)2]- + 4OH-

Phn ng (2): Zn + 2[Au(CN)2]-

[Zn(CN)4]2-

+ 2Au2) EoAg/Zn = -0,31(-1,26) = 0,95VEoAu/Zn = -0,60(-1,26) = 0,66VEoAg/Zn > E

oAu/Zn. V vy phc Ag(I) s b kh trc.

nAg(I) = 1,5molnAu(I) = 5,0molnZn = 0,61mol1mol Zn s phn ng vi 2 mol Ag(I) hoc Au(I).

0,05922

1(103)20,0592

2


13/26

Nn 0,61 mol Zn s tiu th 1,2mol [Ag(CN)2]-[Ag(CN)2]

-cn li = 1,5 1,2 = 0,3mol[Au(CN)2]

-khng b kh.Nng [Au(CN)2]-khi phn ng kt thc = 0,010.3 = 0,030MNng [Ag(CN)2]-khi phn ng kt thc = 0,3.3/500 = 0,002M

3) Au+

+ 2CN- [Au(CN)2]

-

Kb = 4.1028

.

10099

)(

)(

)(

2

2

2

2

CNAuAu

CNAu

CNAu

CNAuKb

Nn 100[Au(CN)2]- = 99[Au+] + 99{[Au(CN)2]

-}Do [Au+] = [Au(CN)2]-/99Thay vo Kbta tnh c [CN-] = 5.10-14M

4) b)

K THI OLYMPIC HA HC QUC T LN TH 32:

n mn kim loi thng i km vi cc phn ng in ha. Vic n mn r st trn b mtcng theo cch ny. Phn ng in cc ban u thng l:

(1) Fe(r) Fe2+(aq) + 2e(2) O2 + 2H2O + 4e 4OH-(aq)T bo in ha ng vi cc phn ng trn c biu din nh sau (t=25oC):Fe(r)Fe2+(aq)OH-(aq), O2(k)Pt(r).Th chun 25oC:Fe2+(aq)+ 2e Fe(r) Eo = 0,44V.O2 + 2H2O + 4e 4OH-(aq) Eo = 0,40V.Chobit:

RTln10/F = 0,05916V ( 25

oC).F = 96485C.mol-1.1. Tnh Eoca phn ng 25oC.2. Vit phn ng xy ra hai na pin v ton b phn ng.3. Tnh K ca phn ng.4. Phn ng xy ra trong 24 gi v I = 0,12A. Tnh khi lng Fe chuyn thnh Fe2+ sau

24 gi. Bit oxy d.

5. Tnh E ca phn ng bit:[Fe2+] = 0,015M; pHna pin phi = 9,00, p(O2) = 0,700bar.

BI GII:1. Eo

(pin)= Eo

phi

- Eotri

= 0,40

(-0,44) = 0,84V2. Phn ng xy ra hai na pin:

Tri: 2Fe 2Fe2+ + 4e (nhn 2)Phi: O2 + 2H2O + 4e 4OH-Ton b phn ng: 2Fe + O2 + 2H2O 2Fe2+ + 4OH-

3. K = [Fe2+][OH-]4/p(O2)G = -nFEo(pin) = -RTlnK K = 6,2.1056 (M6bar-1)

4. Q = It = 10368C.


14/26

n(e) = Q/F = 0,1075mol m(Fe) = 3,00g.

5.

)(

Felog

05916,0

2

422

)()(Op

OH

nEE opinpin

pH = 9,00 [H+] = 10-9M v [OH-] = 10-5MK THI OLYMPIC HA HC QUC T LN TH 37:

Chiufen, th trn m nm trn i ca min Bc i Loan, l mt ni m bn c thkhm ph ra lch s i Loan. cng l mt trong nhng ni c m vng ln nht chu .Chnh v vy Chiufen thng c gi l th vng ca chu . KCN thng c dng chit vng t qung. Vng tan trong dung dch xianua trong s c mt ca khng kh to thnhAu(CN)2

-bn vng trong dung dch nc.4Au(r) + 8CN-(aq) + O2(k) + 2H2O(l) 4Au(CN)2

-(aq) + 4OH-(aq) 1. Vit cng thc cu to ca Au(CN)2-, ch ra v tr lp th ca tng nguyn t.2. Cn bao nhiu gam KCN chit vng t qung?

Nc cng thy, l mt hn hp gm HCl v HNO3ly theo t l 3:1 v th tch, c

tm ra v pht trin bi cc nh gi kim thut ho tan vng. Qa trnh ny l mt phn ng oxyha - kh xy ra theo phng trnh:Au(r) + NO3

-(aq) + Cl-(aq) AuCl4-(aq) + NO2(k)

3. Vit hai na phn ng v s dng n cn bng phng trnh trn.4. Ch ra qa trnh no l oxy ha, qa trnh no l kh.

Vng khng h phn ng vi axit nitric. Tuy nhin vng c th phn ng vi nc cngthy v to thnh ion phc AuCl4-. Cho bit cc th sau:

Au3+(aq) + 3e- Au(r) Eo = +1,50VAuCl4

-(aq) + 3e- Au(r) + 4Cl-(aq) Eo = +1,00V5. Tnh hng s cn bng K = [AuCl4-]/[Au3+][Cl-]4.6. Vai tr ca HCl l sinh ra Cl-. i vi phn ng trn th Cl- c vai tr g?

a) Cl- l tc nhn oxy ha.b) Cl-l tc nhn kh.c) Cl-l tc nhn to phc.d) Cl-l cht xc tc.

BI GII:1. Cu trc ng thng:

AuC CN N

-

2.

4Au + 8KCN- + O2 + 2H2O 4KAu(CN)2 + 4KOH mKCN = (20/197).(8/4).65,12 = 13,024g

3. Oxy ha: eAuClClAuaqaqr

34)(4)()(

Kh ha: 3NO3-(aq) + 6H+(aq) + 3e- 3NO2(k)+ 3H2O(l)Au(r) + 3NO3

-(aq) + 6H

+(aq) + 4Cl

-(aq) AuCl4

-(aq) + 3NO2(k) + 3H2O(l).


15/26

4. Tc nhn oxy ha: HNO3Tc nhn kh: Au

5. Au3+(aq) + 3e Au(r) Eo = +1,50VAu(r) + 4Cl-(aq) AuCl4-(aq) + 3e -Eo = -1,00V Au(r) + Au3+(aq) + 4Cl-(aq) AuCl4

-(aq) + Au(r) Eo = 0,50V

Cch 1: E = Eo

(0,059/n)lgQLc t cn bng: Q = K, E = 0; K = [AuCl4-]/[Au3+][Cl-]4Eo = (0,059/n)lgK K = 1025,42 = 2,6.1025.Cch 2: Go1+ Go2 = Go3(-nFEo1) + (-nFE

o2) = -RTlnK

E = (RT/nF)lnK = (0,059/n)lgK K = 1025,42 = 2,6.1025.6. a)III. BI TP CHUN B CHO OLYMPIC HA HC QUC T:OLYMPIC HA HC QUC T LN TH 32:

K thut in ha hc thng c dng xc nh tnh tan ca cc mui kh tan. Do scin ng l hm bc nht theo logarit ca nng cho nn c th xc nh c cc nng d

rt nh.Bi tp ny s dng mt pin in ha gm hai phn, c ni bng cu mui. Phn bn trica s pin l mt thanh Zn(r) nhng trong dung dch Zn(NO3)2(aq) 0,200M; cn phn bn phil mt thanh Ag(r) nhng trong dung dch AgNO3(aq) 0,100M. Mi dung dch c th tch 1,00L ti25oC.

a) V gin pin v vit phng trnh phn ng tng ng ca pin.b) Hy tnh sc in ng ca pin v vit phng trinh phn ng khi pin phng in.

Gi s pin phng in hon ton v lng Zn c dc) Hy tnh in lng phng thch trong qa trnh phng in.

Trong mt th nghim khc, KCl(r) c thm vo dung dch AgNO3 pha bn phi capin ban u. Xy ra s kt ta AgCl(r) v lm thay i sc in ng. Sau khi thm xong, sc in

ng bng 1,04V v [K+

] = 0,300M.d) Hy tnh [Ag+] ti cn bng.e) Hy tnh [Cl-] ti cn bng v tch s tan ca AgCl.

Th in cc chun ti 25oC nh sau:Zn2+(aq) + 2e Zn(r) E

o = -0,76VAg+(aq) + e Ag(r) E

o = +0,80V

BI GII:a) Gin pin: Zn(r)Zn2+(aq) Ag+(aq)Ag(r).

Gin pin ny tho quy c ca IUPAC vi qa trnh oxy ha (cho electron) pha bn

tri: Tri, oxy ha: Zn(r) Zn2+

(aq) + 2ePhi, kh: Ag+(aq) + e Ag(r) (nhn cho 2)Phn ng ca pin Zn(r) + Ag+ Zn2+(aq) + 2Ag(r).

b) Eopin = Eo(phi)Eo(tri) = 1,56VPhng trnh Nernst tng ng vi pin nu trn (Cng c th trnh by theo na pin v cc

bn phn ng):


16/26

Hot a ca mt cht tan tnh gn ng t nng ca cht y (chia cho nng tiuchun) v hot ca mt cht rn bng 1

Vaa

aa

nEE

Zn

Zno

pinpin 52,1)100,0(

200,0lg

2

05916,056,1

).(

).(lg

05916,022

Ag

2

Ag2

Tr s dng ca Epincho thy rng phn ng pin vit nh trn l phn ng c th t xy ratrong qa trnh phng in.Phn ng c th t xy ra: Zn(r) + Ag+ Zn2+(aq) + 2Ag(r).

c) Khiphng in hon ton, Epin= 0 v phn ng trong pin t cn bng0 = 1,560,05916/2.lgK K = 5,5.1052; ngha l cn bng ca phn ng pin dch chuyn

hn v bn phi, nn thc t khng cn ion Ag+trong dung dchLng Ag+v electron vn chuyn:n(Ag+) = [Ag+].V = 0,100mol v n(e-) = n(Ag+) = 0,100molHng s Faraday F l s in lng ng vi 1 mol electronQ = n(e-).F = 9648,5C

d) Gi x l nng Ag+cui ([Ag+])in cc bn tri khng i, ngha l nng [Zn

2+

] duy tr ti 0,200M MAgx 1010.3,7

2

200,0lg

2

05916,056,104,1

e) [Cl-] = nng thm - nng gim do AgCl kt ta = 0,300 (0,100 - 7,3.10-10)= 0,200M

Ks(AgCl) = 7,3.10-10.0,200 = 1,5.10-10M2.

OLYMPIC HA HC QUC T LN TH 31:Ga tr Eocho cc bn phn ng ca Fe v Ce nh sau:

Fe3+ + e- Fe2+ Eo = 0,77V

Ce4+ + e Ce3+ Eo = 1,61V

Th ti im tng ng ca s chun Fe

2+

v Ce

4+

l 1,19V. Hai cht ch th mic th s dng xc nh im tng ng:

DiBolane(dip): InOx + 2e Inred Eodip = 0,76VTm khng mu

p-nitro-di-bolane(pn): InOx + 2e Inred Eopn = 1,01V

Tm khng muC hai cht ch th u i mu khi [InOx]/[Inred] = 10. Vy cht ch th no l thch hp cho

s chun Fe2+- Ce4+?BI GII:

Vi di-bolane: redoxh

EE odipdd lg2

059,0

Khi [InOx]/[Inred] = 10 th:

Vred

oxhEE odipdd 79,0lg

2

059,0

Ti 0,79V, tnh [Fe3+]/[Fe2+]


17/26

2,2lg059,077,079,0

lg1

059,0

2

3

2

3

2

3

Fe

Fe

Fe

Fe

Fe

FeEE

o

Fedd

Di-bolane khng phi l cht ch th thch hp v [Fe3+

] gp 2,2 ln [Fe2+

]Vi p-nitro-di-bolane:

redoxh

EE opndd lg2

059,0

Khi [InOx]/[Inred] = 10 th:

Vred

oxhEE opndd 04,1lg

1

059,0

Ti 0,79V, tnh [Fe3+]/[Fe2+]

4

2

3

2

3

2

3

10.80,3lg059,077,004,1

lg1

059,0

Fe

Fe

Fe

Fe

Fe

FeEE

o

Fedd

Vy cht ch th thch hp l p-nitro-di-bolaneOLYMPIC HA HC QUC T LN TH 30:

Cc bn phn ng sau c lin quan n s hnh thnh urani trong dung dch nc:U3++ 3e U Eo = -1,798VU4++ e U3+ Eo = -0,607VUO2

2++ e UO2+ Eo = +0,062VUO2

2+ + 4H++ 2e U4+ + 2H2O Eo = +0,327VUO2

2+ + 4H++ 6e U + 2H2O Eo = -1,444VUO2

+ + 4H++ e U4+ + 2H2O Eo = +0,620Va) Xc nh mc oxy ha ca cc tiu phn khc nhau c cha urani xut hin trong cc bn

phn ng trn.b) Bng cch phn tch cc bn phn ng trn, hy xc nh din tin ha hc ti u ca mtmu nh urani rn tip xc vi dung dch 1M ca mt axit mnh n chc HX, c mthydro di p sut 1atm, tt c ti 25oC. Vit cc phng trnh phn ng, cn bng v thin cc ca tt c cc phn ng (Gi thit rng baz lin hp X-khng phn ng ng kvi urani hoc hp cht ca urani).

c) Tiu phn bn nht ca urani ti pH = 6 l g? (v cc iu kin khc u coi nh chun). d) Xc nh khong pH ca dung dch hoc axit trung ho m dung dch 1M ca UO2+l bn:i) Vi cc iu kin khc u chun (nh P(H2) = 1, nng ca cc tiu phn c cha urani

= 1.ii) Vi P(H2) = 1,0.10-6atm v cc iu kin khc u coi nh chun.

iu kin no l thch hp hn cho s hnh thnh urani trong cc lung nc thin nhin?BI GII:a) Urani kim loi c s oxi ha bng khng theo nh ngha. S oxi ha cc tiu phn khc

ca U: U(III) [U3+]; U(IV) [U4+]; U(V) [UO2+]; U(VI) [UO22+].b) Cc iu kin m t l iu kin chun, nn c th dng th kh chun xc nh cc

phn ng t xy ra theo chiu no. Cng cn phi xt cc bc kh:2H++ 2e H2 Eo = 0,000VC hai bn phn ng to ra U kim loi:


18/26

U3++ 3e UUO2

2+ + 4H++ 6e U + 2H2OT Urani b oxi ha k tip:2U + 6H+ 2U3+ + 3H2 Eopin = +1,798VU + 2H+ + 2H2O UO22+ + 3H2 Eopin = +1,444V

Bt k qa trnh no trn y chim u th, khng c sn phm s cp no U3+

hocUO22+l sn phm chnh.U(III) t oxy ha chuyn thnh U(IV):2U3+ + 2H+ 2U4+ + H2 Eopin = +0,607VTrong khi U(VI) li t kh to li U(IV) hoc U(V):UO2

2+ + 2H+ + H2 U4+ + 2H2O Eo = +0,327VUO2

2+ + H2 2UO2+ + 2H+ Eo = +0,062Vv U(V) cng t kh thnh U(IV):2UO2

+ + 6H+ + H2 2U4+ + 4H2O Eo = +0,620VV U4+l tiu phn duy nht khng th t phn ng vi H+ hoc H2, nn sc nhiu nht

trong dung dch nc iu kin ny.

(B sung: khi vn cn U kim loi, qa trnh: U

4+

+ 3U 4U

3+

E

o

pin = +1,191Vl thun li, nhng ch c th xy ra n khi ht U, sau U(III) s b oxi ha thnh U(IV) nh ni trn).

c) Do cc iu kin l tiu chun tr [H+] = 1,0.10-6M, c th dng dng n gin ca phngtrnh Nernst nh sau cho cc phng trnh c lin quan:2U + 6H+ 2U3+ + 3H2 Eopin = +1,798V

Epin = Eopin(RT/6F)ln([H+]-6) = +1,444V

U + 2H+ + 2H2O UO22+ + 3H2 Eopin = +1,444VEpin = E

opin(RT/6F)ln([H+]-2) = +1,326V

2U3+ + 2H+ 2U4+ + H2 Eopin = +0,607VEpin = E

opin(RT/2F)ln([H+]-2) = +0,253V

UO2

2+ + 2H+ + H2 U

4+ + 2H2O Eo

pin= +0,327V

Epin = Eopin(RT/2F)ln([H+]-2) = -0,027VUO2

2+ + H2 2UO2+ + 2H+ Eopin = +0,062VEpin = E

opin(RT/2F)ln([H+]2) = +0,293V

2UO2+ + 6H+ + H2 2U4+ + 4H2O Eopin = +0,620V

Epin = Eopin(RT/2F)ln([H+]-6) = -0,444V

Ton b cc qa trnh trn c tr s Eopindng khi xt theo chiu thun, nn tt c u xyra c theo chiu ny ti pH = 0. Tri li, U(VI) U(IV) v U(V) U(IV) t xy ra theo chiunghch ti pH = 6 (nh cho thy qua tr s m ca Epin); nay U(IV) li b oxy ha thnh U(V) hocU(VI). Do U(VI) l mc oxy ha duy nht khng t phn ng c vi H+ hay H2trong iu kinny nn tiu phn chim s lng ln s l UO22+.

d) gii cu hi ny, cn xem xt cc phn ng c lin quan n UO2+.UO22+ + H2 2UO2+ + 2H+ Eo = +0,062V

Epin = Eopin(RT/2F)ln([H+]2.P(H2)-2)

2UO2+ + 6H+ + H2 2U4+ + 4H2O Eo = +0,620V

Epin = Eopin(RT/2F)ln([H+]-6.P(H2)-2)

Ti cc iu kin lc u, Epin= 0, tng ng vi cc trng hp sau:


19/26

2/1)/2(2 .2

RTFE

H

opinePH cho qa trnh U(VI) U(V) v 6/1)/2(2 .

2

RTFE

H

opinePH

cho

qa trnh U(IV) U(V)Cc biu thc trn cho gi tr ban u ca:

i) [H+] < 11,2M hay pH > -1 U(V) bn hn U(IV) v [H+] < 3,19.10-4M hay pH > 3,50 U(V) bn hn U(IV).ii) [H+] < 1,12.10-5M hay pH > 4,95 U(V) bn hn U(IV) v [H+] < 3,19.10-2M hay pH >1,50 U(V) bn hn U(IV).Nh vy UO2+bn hn cc mc oxy ha khc trn khong pH 3,5 7 (ta ch ang xt cc

dung dch axit v trung ha) di kh quyn hydro tiu chun, nhng ch bn v khng b oxy hathnh U(VI) ti pH c ga tr ln hn 4,95.

p sut ring phn ca H2trong kh quyn rt thp, nn iu kin ca cu sau (cu ii) gnhn vi iu kin thc ca mi trng qa t. UO2+khng thc s bn nh kt qa ca pin chothy.OLYMPIC HA HC QUC T LN TH 32:

Mangan v St trong hu ht t trng an Mch c ngun gc ch yu t cc enzym

trong cc vt liu hu c cht. Trong iu kin axit v kh, cc nguyn t ny c mt didng MnIIv FeII.Ti nhng ni m nc ngm trn ra b mt theo cc con sui, cc ion b oxy ha bi oxy

khng kh.

a) Vit phng trnh phn ng oxy ha mangan (II) thnh mangan (IV) oxit vi dioxi (O 2).b) Hy tnh G1oti 25oC ca phn ng cu a vi cc s liu cho di y:2MnO2(r) + 4H

+(aq) + 2e

-Mn2+ + 2 H2O(l) E

o2 = 1,21V

3O2(k) + 4H+

(aq) + 4e-2H2O(l) E

o3 = 1,23V

c) Hy tnh hng s cn bng K1ti 25oC ca phn ng cu aGi thit rng mangan (II) to phc vi vt liu hu c trong mn ca t. Cng gi thit

rng hng s to phc mangan (II) mn l 105M-1v nng ca ligand bng 10-4M

d)Hy tnh hng s cn bng ca phn ng:

2Mn(mn)2+

(aq) + O2(k) + 2H2O(l) 2MnO2(r) + 2mn(aq) + 4H+

(aq)

t trng axit c th c pH bng 5 v t kim c th c pH bng 8e) T cc s liu l thuyt cho trn hy tnh nng ca Mn(mn)2+ln lt ti pH bng 5 v

pH = 8. Bit p(O2) = 0,2 bar v t trng c cha lng MnO2d.f) Mangan c cy trng hp th qua trung gian phc mn. Loi t canh tc no gp vn

thiu mangan d c lng ln mangan trong t?BI GII:

a) 2Mn2+(aq) + O2(k) + 2H2O(l) 2MnO2(r) + 4H+(aq)b) G2o = -2FE2o = -233,5kJ.mol-1

G3o

= -2FE3o

= -474,7kJ.mol-1

G1o

= -2G2o

+ G3o

= -7,7kJ.mol-1

c) G1o = -RTlnK1vi K1 = 22,4M2bar-1d) Mn2+(aq) + mn(aq) Mn(mn)2+ (aq) = 105M-12Mn(mn)

2+(aq) + O2(k) + 2H2O(l) 2MnO2(r) + 2mn(aq) + 4H

+(aq)

149

2

1

2

2

42

10.24,2)(

barMK

OpmunMn

HmunK

(*)

e) pH = 5: Thay [H+] = 10-5M vo biu thc (*) trn ta tnh c [Mn(mn)2+] = 4,7.10-10M


20/26

pH = 8: Thay [H+] = 10

-8M vo biu thc (*) trn ta tnh c [Mn(mn)2+] = 4,7.10-16Mf) t kim

OLYMPIC HA HC QUC T LN TH 33:a) Mt dung dch cha ion Sn2+c chun in ha bng dung dch Fe3+. Th kh tiu chun

ca cp Sn4+/2+ v Fe3+/2+ c cho di y:

Sn

4+

+ 2e = Sn

2+

E

o

= 0,154VFe3+ + e = Fe2+ Eo = 0,771Vi) Vit phng trnh phn ng tng qut v tinh nng lng t do ca phn ng.ii) Xc nh hng s cn bng ca phn ng.

b) Nu 20ml dung dch Sn2+0,10M c chun bng dung dch Fe3+0,20M. Tnh sc inng ca pin trong cc iu kin sau:i) Khi thm 5mL Fe3+.ii) Ti im tng ng.iii) Khi thm 30mL dung dch Fe3+

in cc calomel (EoS.C.E= 0,242V) c s dng lm in cc chun.c) Mt trong s nhng phng php phn tch quan trng nh lng ion Cu2+l php chun

iot. Trong phn ng ny Cu2+

s b oxy ha v Cu

+

bng I

-

v I2t do sinh ra s phn ngvi dung dch chun Na2S2O3. Phn ng chun Cu2+xy ra theo phng trnh:2Cu2+ + 4I- 2CuI(r) + I2(aq)Th chun ca cc qa trnh oxy ha - kh s l:Cu2+ + e = Cu+ Eo= 0,153VI2 + 2e = 2I

- Eo= 0,535VDa vo th chun th phn ng trn khng th xy ra c. Tuy nhin phn ng ny vn

xy ra mt cch nh lng. Chng ta hy tm hiu l do thng qua cc d kin sau:i) trong dung dch nc th CuI c tch s tan b vi Ksp = 1,1.10-12. Tnh th kh chun

biu kin ca qa trnh CuI(r) = Cu+ + I-.ii) S dng kt qa cu i), tnh th kh chun biu kin ca phn ng v hy da vo cho

bit chiu phn ng.

iii) Tnh hng s cn bng ca phn ng t cc d kin cho cu ii).BI GII:

a) i) Phn ng xy ra theo phng trnh:Sn2+ + 2Fe3+ = 2Fe2+ + Sn4+ Eo = 0,617V

G = -nFE = -119kJii) 0,0592/n(lgK) K = 6,92.1020.b) Trc im tng ng, E ca phn ng c tnh bi phng trnh:

4

2

/..lg

2

0592,024

Sn

SnEEE

o

SnSnkh

o

ECSoxpin

i) Khi thm 5,00mL Fe3+th lng Sn2+chuyn sang Sn4+s l 5,00/20,00.

Nh vy:

30,20/0,5

0,20/0,154

2

Sn

Sn

Thay vo biu thc tnh E ta thu c kt qa E = -0,102Vii) Ti im tng ng ta tnh E theo cng thc:

VEE

E

o

Fe

o

Sn 118,03

2 2324 /Fe/Sn


21/26

iii) Khi thm 30,00mL Fe3+vo th trong dung dch cn d 10mL Fe3+:

20,10

0,203

2

Fe

Fe

Thay vo biu thc tnh E ta c: E = 0,511Vc) i) G = -RTlnKsp = 68,72kJ

Mt khc ta c: G = -nFE Eo = -0,707Vii) Cu+ + I- = CuI Eo= 0,707V

Cu2+ + e = Cu+ Eo= 0,153VPhn ng kh Cu2+bng I-xy ra theo phng trnh:Cu2+ + I- + e = CuI(r) E

o = 0,86VT y ta tnh c ga tr Eoca phn ng:2Cu2+ + 4I- 2CuI(r) + I2(aq) Eo = 0,325V.Eoc ga tr dng cho bit rng phn ng xy ra theo chiu thun. iu ny c th c

gii thch do I-va ng vai tr cht kh va ng vai tr tc nhn to kt ta. S to thnh kt taCuI l bc chnh ca phn ng, s to kt ta CuI lm cho ion Cu+b tch ra khi dung dch va phn ng din ra theo chiu thun.iii) Go = -nFEo = -31,3kJ

Mt khc Go = -RTlnK K = 2,9.105.OLYMPIC HA HC QUC T LN TH 34:

ng c l mt phn quan trng trong cuc sng hin i. Cc xe hichy in ang cpht trin bo m s vn chuyn trong tng lai. Mt trong s nhng phng php chnh chocc phng tin chy bng in l s dng cc ngun nng lng in thch hp. Acquy th cnphi sc li khi ht in nn n khng th s dng lin tc. S dng phng php in situ tng qut to ra in trong cc pin nhin liu l mt phng php rt hp dn. Pin nhin liu l mt t boin ha m trong cc phn ng ha hc din ra lin tc. Pin nhin liu s dng cc phn ngchy sinh ra in. Cc bn phn ng cng din ra cc in cc v electron c chuyn hathng qua mt mch in kn. Cc electron c phn lp bi mi trng ion cha dung dch hoccht rn trng thi nng chy (u c tnh dn in). Cc phn ng din ra cc in cc ca pinnhin liu hydro oxy vi cht in ly l dung dch KOH l:

O2(k) + 2H2O + 4e = 4OH-(aq) (1)

H2(k) + 2OH-(aq) = 2H2O + 2e (2)

Phn ng tng qut s l: 2H2 + O2 = 2H2O (3)Sn phm ca phn ng l nc v hiu sut l t 5060%

1) Vit phn ng xy ra catot2) Vit phn ng xy ra anot3) Nu cht in ly l axit photphoric th phn ng s xy ra nh th no?

S thay i ca nng lng Gibbs Gol thc o s chuyn dch chiu ca phn ng. Schuyn i nng lng c cho bi h thc Go = -nFEo. n l s electron vn chuyn trong phn

ng v F l hng s Faraday (F = 96487C). Th in cc chun ca O2(k) 25oC l +1,23V.4) Hy tnh Goca pin nhin liu trong mi trng axit (cu 3)

Qa trnh sinh ra nng lng do s t chy nhin liu l kh kim sot. H Lan, khthin nhin l mt ngun nng lng hp dn do n lun c sn. Cc nh my in hin i c thchuyn ha c 35 40% ngun nng lng sinh ra do t chy kh thin nhin. Phn ng chyta nhit ca kh thin nhin (metan) c biu din bi phng trnh:

CH4(k) + 2O2(k) CO2(k) + 2H2O(k)+ nng lng


22/26

Nng lng sinh ra thng c s dng mt cch gin tip si m cn nh hay chy cc my mc. Tuy nhin, trong nhng loi ximng chu nhit cha cc oxit kim loi trngthi rn (tn ti trng thi ion) lm nhng cht dn in th kh thin nhin c th c s dngmt cch trc tip, khng cn xc tc v hiu sut chuyn ha cao hn hn (75%). Phn ng trongloi pin nhin liu ny l:

CH4(k) + 2O2(k) CO2(k) + 2H2O(k)5) Vit cc phn ng catot v anot.Mt loi pin chu nhit khc tn dng cacbonat nng chy nh l cht dn in. Hydro

c s dng lm nhin liu, oxy c trn vi CO26) Vit phn ng catot v anot v phn ng xy ra.

BI GII:1) catot oxy b kh v ion hydroxit (phn ng 1)2) anot hydro b oxy ha thnh nc (phn ng 2)3) Anot: 2H2 4e + 4H+

Catot: 4e + 4H+ + O2(k) 2H2O(k)Phn ng chung: 2H2(k) + O2(k) 2H2O(k)

4)

Th kh chun ca phn ng xy ra catot: 0VTh kh chun ca phn ng xy ra anot: +1,23VS electron vn chuyn: 4eGo = -nFEo = -474716J

5) CH4(k) + 2(O2-, cht dn in) CO2(k) + 2H2O(k)+ 4eO2(k) + 4e

2O2-(cht dn in)CH4(k) + 2O2(k) CO2(k) + 2H2O(k)

6) Anot: 2H2(k) + 2CO32-(l) 2CO2(k) + 2H2O(k)+ 4eCatot: O2(k) + 2CO2(k)+ 4e 2CO32-(l)Phn ng chung: 2H2 + O2 = 2H2O

OLYMPIC HA HC QUC T LN TH 35:Pin Ni

Cd (Nicad) c s dng rng ri trong cc loi thit b b tn nh in thoi ding, my quay phim xch tay, laptop, v.v Pin Ni Cd c ga va phi v c chu trnh sng cao

ng thi c th hot ng c nhit rt thp hay rt cao. N khng cn phi c bodng v c th c np in 2000 ln.. Mt t bo ca pin Ni Cd thc hin hai na phn ngsau:

Cd(OH)2(r)+ 2e Cd(r) + 2OH- Eo1 = -0,809V2NiO(OH) + 2H2O + 2e 2Ni(OH)2(r) + 2OH- Eo2 = -0,490VEo1; E

o2l th kh chun 25o-C.

1) Phn ng no xy ra catot? Vit phng trnh Nernst.2) Phn ng no xy ra anot? Vit phng trnh Nernst.3) Vit phn ng chung.4) Tnh E ca phn ng 25oC.5) Tnh khi lng Cd cha trong 1 chic in thoi di ng c s dng pin Ni Cd. Bit

cng sut thng thng ca pin l 700mAh.BI GII:

2NiO(OH) + 2H2O + 2e 2Ni(OH)2(r) + 2OH- Eoc = -0,490V

2ln2

OHF

RTEE occ

Cd(r) + 2OH- Cd(OH)2(r) + 2e Eoa = -0,809V


23/26

21

ln2

OHF

RTEE oaa

Cd(r) + 2NiO(OH)(r) + 2H2O 2Ni(OH)2(r) + Cd(OH)2(r)nap

phong

E = E

o

aEo

c = 1,299V700mAh = 0,700A . 3600s = 2520CnCd = 2520/2.96485 = 0,013mol mCd = 0,013.112,4 = 1,47g

OLYMPIC HA HC QUC T LN TH37:Mt s hp cht v c c s bin i s oxy ha, chng hn nhiu hp cht ca Mn c

phm vi s oxy ha t 0 n +7. Th kh tiu chun ca mt na phn ng c o da vo incc hydro. Trong bi tp ny, s kh Mn2++ 2e Mn c th c vit l: Mn2+(1,5)Mn. Mntrong dung dch axit xy ra mt lot s kh: Mn3+ Mn2+ Mn c th c vit nh sau:Mn3+(1,5)Mn2+(-1,18)Mn.

S kh t xy ra nu nh th kh l dng. S Frost v nEo(n l s electron themgia vo na phn ng) ca cp X(N)/X(O) theo s oxy ho N ca nguyn t, c dng ch ra

cc vi ht bn nht ca cc hp cht c cc s oxy ha khc nhau. Hnh di l s Frost caMn3+/Mn2+/Mn.

1) Th kh ph thuc nng ca vi ht trong mt dung dch. MnCO3c tch s tanKsp=1,8.10

-11. Hy dng phng trnh Nernst tnh th ti 25oC ca mch in ha gm:Mn(r)Mn2+(aq)Mn2+(aq)/MnCO3Mn(r)nu nng Mn2+ in cc bn phi l1,0.10-8M.

2) i vi oxy, th kh tiu chun trong dung dch axit nh sau: O2(0,7)H2O2(1,76)H2O. Tnhth kh ca na phn ng kh O2 thnh H2O. H2O2c th t phn hy c khng?Xenon diflorua c th c to ra khi t mt bnh sy kh cha kh xenon v kh flo

di nh sng mt tri. Na phn ng kh XeF2xy ra nh sau:XeF2(aq) + 2H

+(aq)+ 2e Xe(k) + 2HF(aq) Eo = 2,32V

3) Hy dng m hnh VSEPR d on s cp electron v hnh dng phn t ca XeF2. Hychng t rng trong dung dch nc, XeF2

phn hy to ra O2

, tnh Eoca phn ng ny.Liu s phn hy c xy ra thun li trong mt dung dch ctnh axit hay tnh baz?

Hy gii thch bit 2H2O O2 + 4H+ + 4e c Eo = -1,23V.BI GII:

1) Epin = Eo(0,0592/2)lg([Mn2+]phi/[Mn2+]tri)Ksp = [Mn

2+][CO32-] = 1,8.10-11

[Mn2+]phi = 1,0.10-8M v [Mn2+]tri = 1,0M; E

o= 0,00V (do tt c u ca Mn)Vy Epin = 0,237V


24/26

2) Phn ng kh oxy thnh nc c Eo = (0,70 + 1,76)/2 = 1,23VQa trnh O2 + 4H

++ 4e 2H2O Eo = 1,23VNh vy qa trnh 2H2O2 2H2O + O2 c Eo = 1,06V > 0,00VVy phn ng xy ra theo chiu thun.

3) S cp electron l 5 (lng thp tam gic) nhng c 3 cp electron cc mt phng binnn cu to hnh hc ca XeF2l ng thng.2H2O O2 + 4H+ + 4e Eo = -1,23VXeF2(aq) + 2H

+(aq)+ 2e Xe(k) + 2HF(aq) Eo = 2,32V

2XeF2(aq) + 2H+

(aq) + 2H2O Xe(k) + O2(k) + 4HF(aq) Eo = 1,09VVy phn ng ny din ra d dng trong mi trng axit.

IV . OLYMPIC HA HC CC NC TRN TH GII:OLYMPIC HA HC C 2004:

Th in cc chun ca bc v niken ln lt l Eo(Ag+/Ag) = +0,80V v Eo(Ni2+/Ni) = -0,23V. Mnh no sau y l ng?

a) Ag+ l tc nhn oxy ha cn Ni2+l tc nhn kh.b) Ag+l tc nhn oxy ha mnh hn Ni2+v Ag l cht kh mnh hn Ni.c)

Ni

2+

c th b kh bi bc kim loi.d) Ag+l tc nhn oxy ha mnh hn Ni2+v Ni l cht kh mnh hn Ag.e) Ni2+l tc nhn oxy ha mnh hn Ag+v Ag l cht kh mnh hn Ni.

BI GII: Cu dOLYMPIC HA HC ITALY 1999:

Cch biu din ng ca pin Daniell l:a) Cu2+(aq)Cu(r)Zn(r)Zn2+(aq).b) Zn2+(aq)Zn(r)Cu(r)Cu2+(aq).c) Zn(r)Zn2+(aq)Cu2+(aq)Cu(r).d) Cu(r)Cu2+(aq)Zn2+(aq)Zn(r).

BI GII: Cu cOLYMPIC HA HC ITALY 1999:

in phn mt dung dch CuSO4(100mL; 0,150M) vi cng 1,25A cho ti khi ton bCu thot ra th thi gian cn thit l:

a) 77,2 phtb) 60,3 phtc) 19,3 phtd) 38,6 pht

BI GII: Cu dOLYMPIC HA HC O 1999:

Trong php phn tch ny tc nhn oxy ha l ion Ce4+, n d b kh thnh ion Ce3+. Ga trca th oxy ha - kh ca nhng ion ny ph thuc vo nhng anion c mt. Cht chnh oxyha dung dch Ce4+ l As2O3. Cho As2O3tc dng vi NaOH ri axit ha thu c asenit (AsO33-),ion ny b Ce4+ oxy ha thnh asenat (AsO43-), xc tc l mt lng nh OsO4, cht ch th oxy ha- kh l feroin.

Vit phng trnh ion ca phn ng chun asenit bng Ce4+v tnh th im tngng (Eeq) khi lm vic vi pH = 1.

Eo1(AsO43-/AsO3

3-) = 0,56VEo2(Ce

4+/Ce3+; HClO4) = 1,70VBI GII:


25/26

AsO33- + 2Ce4+ + H2O 2Ce3+ + AsO43- + 2H+

3

3

3

42

13

3

23

411 lg

2

059,0lg

2

059,0lg

2

059,0

AsO

AsOHE

AsO

HAsOEE oo

Khi pH = 1 th E1 =

3

3

3

41 lg

2

059,0059,0

AsO

AsOEo

3

4

3

3

3

4

3

3

3

4

3

4

21

21

3

4

22

lg2

059,0lg059,0lg

2

059,0lg059,0059,0

lg2

059,0

AsO

AsO

Ce

Ce

AsO

AsO

Ce

CeEE

EEE

Ce

CeEE

oo

eq

o

im tng ng:

90,03

4

3

4

3

3

eqECe

Ce

AsO

AsOV

OLYMPIC HA HC C 1999 (Vng 2):

Cho ba t bo in ha vi cc hiu in th tng ng 298K:1) Hg/HgCl2, KCl (bo ho)//Ag

+ ( a = 0,0100mol/L)/Ag E = 0,439V2) Hg/HgCl2, KCl (bo ho) = AgI (bo ho)/Ag E = 0,089V3) Ag/AgI (bo ho), PbI2 (bo ho) // HgCl2, KCl (bo ho)/Hg E = 0,230V

a) Hy trnh by khi nim hot ngha l g v n c ng dng u trong in hahc?

b) Tnh TAgI.c) Tnh T ca PbI2Bit Eo(Ag+/Ag) = 0,799V; R = 8,314J/K.mol; F = 96487C.

BI GII:a) Do tc ng tng h ca cc ion vi nhau m nhiu khi cc dung dch vi nhng nng

C 0,01M khng cn x s nh nhng dung dch l tng na. Khi ngi ta tnh tonbng nng hu hiu v gi l hot . nhng nng thp th hot bng nng.T in p ca t bo1 ngi ta c th tnh c in th ca in cc calomen[Hg/Hg2Cl2, KCl (bo ho)].Sau c th tnh c nng ion bc trong t bo 2. V rng y nng ion ioduabng nng ion bc, cho nn kt qa l tch s ho tan ca bc iodua T = [Ag+]2.Cng t hiu in th ca t bo 3 ngi ta c th tnh c nng ion bc ri t sdng tch s ho tan va tnh c trn c th tnh c nng ion iodua trong t bony. Nng iodua ny ln hn nhiu so vi nng ion bc. Lng d ion iodua phihnh thnh do s ho tan ch iodua, mt na lng d cho bit nng ion ch. Bng

cch c th tnh c tch s ho tan ca PbI2.b) Tnh in th ca in cc calomen: E1 = E(Ag+/Ag)Ecalomen

Vi VAgaF

RTEAgAgE

o 681,0)(ln)/(

Tnh nng ionbc:E2 = E(AgI (bo ha)/Ag//Ag)Ecalomen E(AgI (bo ha)/Ag//Ag) = 0,331VMt khc ta c:


26/26

E(AgI (bo ha)/Ag//Ag) = MAgaVAgaF

RTEo 810.22,1)(331,0)(ln

Vi i lng ny th [Ag+] = a(Ag+) TAgI = 1,48.10

-16.c) E3 = EcalomenE(AgI (bo ho), PbI2 (bo ho)/Ag)

E(AgI (bo ho), PbI2 (bo ho)/Ag) = 0,012VTng t nh phn b, kt qa thu c y l [Ag+] = 4,89.10-14MPhn nng ion iodua sinh ra t AgI cng bng nh vy.S dng tch s ho tan i vi AgI tnh c phn b ta c:[I-] = 3,02.10-3MPhn nng in iodua sinh ra t AgI l khng ng k so vi ga tr ny. Nh vy nng

ion ch s l 0,5[I-].T(PbI2) = 0,5[I

-]3 = 1,38.10-8M3.OLYMPIC HA HC C 1999 (Vng 3):

Cho cc th chun sau y:AgBr(r)+ e Ag + Br- Eo1 = 0,0713V

Ag+

+ e Ag Eo2 = 0,7996Va) Trn c s hy tnh TAgBr 25oC.

b) T hy tnh Goi vi qa trnh: AgCl(r) Ag+(aq) + Cl-(aq). Bit SAgCl(25oC) =1,274.10-5M.

BI GII:a) Eo1c lin quan n mt bn t bo, trong trn mt lp kt ta bc bromua nng

ion bromua l 1,00M. Nh c Eo2ngi ta c th theo phng trnh Nernst tnh c nng ion bc trong t bo v t nh c [Br-] = 1,00M c th tnh c tch s ho tan.

21313

121 10.81,410.81,4ln MTMCF

RTEE AgBr

oo

b)T ho tan ta c: [Ag

+] = [Cl-] = 1,274.10-5

M v nh vy l tch s ho tan T =(1,274.10-5)2. T cng l hng s cn bng KCi vi phn ng: AgCl(r) Ag+(aq) + Cl-(aq)KCngoi ra cng l hng s cn bng nhit ng lc hc. iu ny c ngha l: Go = -RTlnT = 55,8kJ/mol.

Tuyen tap BT Dien Hoa

Documents

Transcript of Tuyen tap BT Dien Hoa