S GD-T TNH BC NINH

TRNG THPT CHUYN BC NINH

TI Phn loi, xy dng cu trc cc bi tp v pin

in ha phc v cho vic bi dng hc sinh gii Quc gia

Ngi thc hin: Th.s Vng B Huy

T: Ha hc
Trng: THPT Chuyn Bc Ninh

Nm hc: 2012 - 2013

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PIN IN HA
Phn I: TNG QUAN
I. TM QUAN TRNG, VAI TR CA PHN NG TRONG PIN IN:
Trong chng trnh ho hc ph thng c cp ti phn ng oxi ha - kh, trong
loi phn ng ny c th chia lm hai loi:
Loi th nht l phn ng trong c s trao i electron trn b mt ca cc cht tip,
y l loi phn ng oxi ha kh ph bin.
Loi th hai l phn ng trao i electron trn b mt ca hai in cc, loi ny c
gp trong in phn v trong pin in.
Phn ng oxi ha kh xy ra trong pin in chng trnh hc sinh cc trng ph
thng gp khng nhiu. Nhng trong chng trnh chuyn ha v c bit trong cc bi thi
hc sinh gii cp Quc gia v Quc t c cp ng k.
II. TNH HNH THC T NI DUNG KIN THC V PIN IN TRONG
CC TI LIU HIN HNH:
Trong cc ti liu hin hnh th ch yu l cc ti liu cho hc sinh n luyn thi i
hc v cao ng, cc ti liu ny qu nhiu m khng c s thng nht mt l thuyt chun
mc m nng v bi tp tnh ton gy cho hc sinh li hc th ng, khng c tnh sng
to v km t duy v ho hc. Bn cnh nhng ti liu ny thy cc ti liu dnh cho hc
sinh gii, hc sinh chuyn cn t v c bit ti liu dnh cho thi HSG Quc gia, Quc t
thc s l qu t.
V ni dung kin thc cc loi phn ng trong cc ti liu hin hnh ch yu cp
nhiu v phn ng trao i ion to ra cht kt ta, phn ng axit - baz v phn ng oxi
ho kh do s trao i electron trn b mt ca cc cht. Cn phn ng oxi ha kh xy ra
trong pin in trong sch gio khoa vit rt s ng v c bit mi lin h v nh hng
ca cc loi phn ng ha hc trong pin in sch gio khoa ph thng l khng c. Mt
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khc trong cc ti liu sch dng cho sinh vin cng ch vit theo mt h thng c bn,
cha c h thng su rng v a dng cc loi pin in. iu c bit l cha xy dng
mi lin kt cc loi phn ng trong pin in.
Chnh v nhng kh khn trn y, ti xy dng ti: Phn loi, xy dng
cu trc cc bi tp v pin in ha phc v cho vic bi dng hc sinh gii Quc gia
III. MC TIU CA TI:
Mc tiu ca ti ny l xy dng cu trc cc bi tp v pin in, qua phn
loi, nh gi tc dng ca cc bi tp phn ng oxi ha - kh trong pin in v cc phn
ng lin quan phc v cho vic bi dng hc sinh gii Quc gia.
IV. PHNG PHP THC HIN:
- Nghin cu l thuyt v cc qu trnh oxi ha, qu trnh kh cc in cc, s vn
dng l thuyt phn ng oxi ha kh trong ging dy ho hc trng ph thng v cc
trng chuyn. Xt cc mi quan h gia cc loi phn ng khc nhau c lin quan n
phn ng oxi ha kh.
- iu tra thu thp cc ti liu c lin quan n phn ng oxi ha kh qua h
thng sch gio khoa, sch bi tp ph thng, cc thi tuyn sinh vo cc trng i hc
v cao ng, cc thi hc sinh gii cc tnh, thi hc sinh gii Quc gia vng mt v
vng hai, ti liu Olympic Vit Nam Quc t, ti liu chun b cho thi Olympic Quc t
v mt s ti liu khc.
- Phn tch c chn lc cc bi tp trong sch gio khoa, cc thi hc sinh gii v
mt s bi xut thm phn loi nh gi cc bi tp v phn ng trong pin in.
- T nh ra cc cu trc cc bi tp v phn ng xy ra trong pin in.
V. VAI TR CA BI TP TRONG VIC BI DNG HC SINH GII
QUC GIA:
t ra c cc yu cu cho hc sinh trong qu trnh ging dy th vic la chn,
xy dng cc bi tp l mt vic lm ht sc quan trng v cn thit i vi mi gio vin.
Thng qua bi tp, gio vin s nh gi c kh nng nhn thc, kh nng vn dng
kin thc ca hc sinh. Bi tp l phng tin c bn nht dy hc sinh tp vn dng
kin thc vo thc hnh. Qu trnh vn dng kin thc thng qua cc bi tp c rt nhiu
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hnh thc phong ph v a dng. Thng qua vic gii cc bi tp m kin thc c cng
c khc su, chnh xc ho, m rng v nng cao. Cho nn bi tp va l ni dung, va l
phng php, va l phng tin thc y vic dy tt v hc tt.
c bit, bi tp ho hc l phng tin c bn dy hc sinh vn dng cc kin
thc ho hc vo thc t i sng, sn xut v tp nghin cu khoa hc. Bi tp ho hc
c tc dng gio dc tr dc v c dc rt ln. l:
- Rn luyn cho hc sinh kh nng vn dng c cc kin thc hc.
- o su v m rng kin thc mt cch phong ph, hp dn.
- n tp, cng c v h thng ho kin thc mt cch d hiu nht.
- Rn luyn c k nng cn thit v ho hc, nh k nng cn bng phng trnh, k
nng tnh ton, k nng t duy c th ring ca mn ho hc,.v.v...
- Pht trin nng lc nhn thc, tr thng minh cho hc sinh.
- Gio dc t tng, o c, tc phong nh rn luyn tnh kin nhn, trung thc, sng
to, chnh xc, khoa hc. Nng cao lng yu thch hc tp b mn. Qua , pht trin mt
cch ton din nhn cch cho hc sinh.

Phn II: NI DUNG

A. C S L THUYT:
I. IN CC V TH IN CC:
1. Quy c v th in cc:
+ Th kh (xy ra qu trnh kh)<Eox/kh>: ox + ne kh
+ Th oxi ho (xy ra qu trnh oxi ho)< Ekh/ox>: kh ox + ne
2. Phng trnh tnh Eox/kh:
Qu trnh kh: a.ox + ne + bB + .... l.kh + mM + ...
Theo phng trnh Nec (Nernst):
[kh]l.[M]m. ...
0
E = E ox/kh - (0,0592/n)lg[ox]a.[B]b. ...
II. CC LOI IN CC V TH IN CC:
Phn tch: Phn ny gip cho hc sinh phn loi cc loi in cc, hiu c cc
qu trnh xy ra mi in cc. T gip hc sinh bit cch xy dng mt pin in
thng qua cc loi in cc, tnh th ca in cc khi c nhiu qu trnh xy ra, tnh sut
in ng ca pin,...
1. in cc loi I:
L mt h in ho gm mt kim loi tip xc vi mt dung dch cha ion kim loi
(Mn+) hoc ion phc ca kim loi.
y l h in ha dng kh ng vai tr l in cc c nhng trong dung dch cha
ion kim loi hoc ion kim loi tn ti bi ion phc.
- in cc kim loi M nhng trong dung dch chc ion Mn+:
M M+ (CM)
- in cc kim loi M nhng trong dung dch chc ion [M(L)m] n+:
M [M(L)m] n+ (CM)
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2. in cc loi II:
L mt h in ha gm mt kim loi b bao ph mt hp cht kh tan (mui hoc
hiroxit hoc oxit) nhng vo mt dung dch cha anion ca hp cht kh tan :
M, MA A(C) .
V d 1: in cc Ag, AgCl KCl(C). Cho Ks = 10-10, EoAg+/Ag = 0,799V. Tnh EoAgCl/Ag v
EAgCl/Ag khi C = 2M.
Phn tch: y l mt h in ha c qu trnh xy ra ca phn ng oxi ha kh
v cn bng ca hp cht t tan. Hc sinh c th vn dng a dng vi cc hp cht t tan
khc.
Hng dn:
* Tnh EoAgCl/Ag:
+ Tnh theo t hp cn bng:
Ag+ + e Ag

K1 = 100,799/0,0592

AgCl Ag+ + Cl-

Ks = 10-10

AgCl + e Ag + Cl- K = 10E/0,0592

=> K = K1.Ks
=> EoAgCl/Ag = EoAg+/Ag + 0,0592lgKs = 0,799 + 0,0592lg10-10 = 0,207(V)
+ Hoc tnh theo biu thc:
EAgCl/Ag = EoAg+/Ag + 0,0592lg[Ag+]
= EoAg+/Ag + 0,0592lgKs/[Cl-]
- Khi th kh chun ly [Cl-] = 1M
=> EoAgCl/Ag = EoAg+/Ag + 0,0592lgKs = 0,799 + 0,0592lg10-10 = 0,207(V)
- Khi nng khng chun CCl- = 2M, ta c:
EAgCl/Ag = EoAg+/Ag + 0,0592lgKs/CCl- = 0,799 + 0,0592lg10-10/2 = 0,189(V)
Hoc EAgCl/Ag = EoAgCl/Ag + 0,0592lg1/CCl- = 0,207 + 0,0592lg1/2 = 0,189(V)
V d 2: Vit in cc calomen, bn phn ng, biu thc EoHg2Cl2/Hg v EHg2Cl2/Hg:
in cc: Hg, Hg2Cl2 KCl (C); c bn phn ng:
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Hg2Cl2 + 2e 2Hg + 2ClC cc biu thc:

- EHg2Cl2/Hg = EoHg22+/Hg + (0,0592/2)lg[Hg22+]
= EoHg22+/Hg + (0,0592/2)lgKs/(CCl-)2
=> Khi CCl-= 1M, ta c:
EoHg2Cl2/Hg = EoHg22+/Hg + (0,0592/2)lgKs
=> Khi CCl- 1M, ta c:
EHg2Cl2/Hg = EoHg22+/Hg + (0,0592/2)lgKs/(CCl-)2
= EoHg2Cl2/Hg - 0,0592lgCClNhn xt: Khi cho KCl bo ho, nng Cl- khng i, nn th kh EHg2Cl2/Hg
khng i, v vy thng dng in cc calomen bo ho lp pin in o sut in
ng ca pin ri tnh cc hng s Ka, Ks, hng s to phc, o pH, ...
V d 3: Cho in cc thu ngn oxit Hg, HgO OH -. Vit bn phn ng in cc v biu
thc th kh lin quan ti pH.
Hng dn:
Bn phn ng: HgO + 2e + H2O Hg + 2OH=> EHgO/Hg = EoHgO/Hg - (0,0592/2)lg[OH-]2
= EoHgO/Hg - 0,0592lgKw/[H+]
= EoHgO/Hg - 0,0592lgKw + 0,0592lg[H+]
= 0,924 - 0,0592lg10-14 - 0,0592pH
= 0,0952 - 0,0592pH
3. in cc oxi ho - kh (in cc Redox):
L mt in cc tr (Pt, Cgr, ...) nhng vo dung dch c hai dng ox(a M) v kh (b M):
V d 4: Vit cc qu trnh v phng trnh th kh ca cc in cc.
Pt Fe2+(C1 mol/l); Fe3+( C2 mol/l).
Pt Mn2+( C1 mol/l); MnO4-( C2 mol/l); H+ (C3 mol/l).
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Pt Cr3+( C1 mol/l); Cr2O72-( C2 mol/l) H+ (C3 mol/l).

Pt Br-( C1 mol/l); Br2(C2 mol/l).
(Cc gi tr C1, C2, C3 c th ging nhau hoc khc nhau)
...
Hng dn:
Vit cc qu trnh di dng tng qut:
a.ox + ne + bB + .... l.kh + mM + ...
Theo phng trnh Nec (Nernst):
E = E0ox/kh - (0,0592/n)lg(([kh]l.[M]m. .../[ox]a.[B]b...)
Nhn xt:
y l cc qu trnh ox/kh ph bin, thng gp. Th kh ca mi cp ox/kh c th lin
quan n pH hoc khng lin quan n pH.
4. in cc kh:
L mt h in ho gm in cc tr (Pt) tip xc ng thi vi kh v dung dch cht
in li:
a) in cc kh hiro:
V d 4: Vit in cc, na phn ng v biu thc lin quan ti th kh ca in cc hiro
vi axit mnh v axit yu HA c hng s cn bng Ka:
Phn tch: Bi ton c bn l hc sinh thit lp mt h in ha ca mt axit mnh
vi mt in cc chun. T sut in ng ca pin o c, tnh ra th in cc hiro ri
xc nh c pH ca dung dch. Cn nu o c pH ca dung dch v sut in ng
ca pin th xc nh c th kh ca mt cp ox/kh cn nghin cu. Bi ton mc
nng cao hn l cho h in ha vi mt axit yu. Qua v d ny gip hc sinh c th tm
c hng s cn bng ca mt axit yu thng qua thit lp mt pin in gia mt in
cc Pt nhng trong dung dch axit HaA (bit nng ) vi mt in cc chun.
Hng dn:
+ Axit mnh (H+):
Pt, H2(x atm) H+(C).
Na phn ng: 2H+ + 2e H2
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=> E2H+/H2 = Eo2H+/H2 + (0,0592/2)lg[H+]2/PH2

= 0 - 0,0592pH - (0,0592/2)lgPH2
Nu PH2 = 1 atm; => E2H+/H2 = - 0,0592pH
+ Nu l axit yu:
HA H+ + A-

Ka

=> E2HA/H2 = Eo2H+/H2 + (0,0592/2)lg[H+]2/PH2

= 0 + 0,0592lg[H+]

(*)

Vi PH2=1 atm; [H+] = (Ka.[HA])1/2 (Ka.CHA)1/2 (khi gi thit HA l axit yu hay rt yu)
thay vo (*), ta c:
E2HA/H2 = (0,0592/2)lgKa + (0,0592/2)lgCHA => Khi bit CHA, o c E2HA/H2 l
tnh c Ka.
b) in cc kh clo:
V d 5: Vit in cc, na phn ng v biu thc lin quan ti th kh ca in cc kh
clo.
Hng dn:
in cc Pt, Cl2( x atm) Cl- (C)
Na phn ng: Cl2 + 2e 2Cl=> ECl2/2Cl- = EoCl2/2Cl- + (0,0592/2)lgPCl2/(CCl-)2
Khi x = 1 ; => ECl2/2Cl- = EoCl2/2Cl- - 0,0592lgCClc) in cc kh oxi:
V d 6: Vit in cc, na phn ng v biu thc lin quan ti th kh ca in cc kh
oxi.
Phn tch: Phn ny gip cho hc sinh hiu qu trnh kh ca oxi trong cc mi
trng khc nhau v kh nng oxi ha ca oxi ph thuc vo pH ca mi trng. Qua biu
thc thy c nu pH cng nh th kh nng oxi ha ca oxi cng mnh.
Hng dn:
+ Vi in cc:
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Pt, O2( P), H2O 4OH- (C)

Na phn ng:
O2 + 4e + 2H2O 4OHE(O2,H2O)/OH- = Eo(O2,H2O)/OH- + (0,0592/4)lg(PO2/[OH-]4
Nu PO2= x =1 atm, th E(O2,H2O)/OH- = Eo(O2,H2O)/OH- - 0,0592lg[OH-]
= Eo(O2,H2O)/OH- + 0,0592pOH
= Eo(O2,H2O)/OH- + 0,0592(14 - pH) (*)
+ Vi in cc:
Pt, O2(P), H+(C) 2H2O
Na phn ng: O2 + 4e + 4H+ 2H2O
=> E(O2,H+ )/H2O = Eo(O2,H+ )/H2O + (0,0592/4)lgPO2.[H+]4
Nu PO2= x = 1atm; => E(O2,H+ )/H2O = Eo(O2,H+ )/H2O + 0,0592lg[H+]
= Eo(O2,H+ )/H2O - 0,0592pH
= Eo(O2,H+ )/H2O - 0,0592(14 - pOH)

(**)

(T Eo(O2,H2O)/OH- <=> Eo(O2,H+ )/H2O thng qua t hp cn bng ca H2O).

III. CC LOI PIN IN:

III.1- Pin khng ni lng:

Pin khng ni lng l mt loi pin c hai in cc cng nhng vo mt dung dch in li
V d 7: Vit s pin, na phn ng v phng trnh phn ng khi pin hot ng trong
cc trng hp sau:
1. Pin gm hai in cc Pt nhng trong dung dch HCl, kh clo hai in cc c P
khc nhau. Hoc mt in cc bm kh H2, cn in cc kia bm kh clo. Hoc in cc
Ag, AgCl c nhng trong dung dch HCl vi in cc kh clo.
2. Pin Zn - Hg c nhng trong dung dch KOH C M.

EoZnO22-/Zn=-1,22V;

EoHgO/Hg=0,12V
3. Pin Zn - PbO2 c nhng trong dung dch H2SO4 38%. EoPbO2/Pb = 1,455V
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4. Pin Zn - O2 c nhng trong dung dch NH4Cl CM.

5. a) Vit s ca c quy ch, cc bn phn ng v phng trnh phn ng khi c
quy ch phng in v np in.
b) Khi np in vi I = 19,3A, t = 1,5 gi. Hi c bao nhiu gam PbSO4 b phn tch?
6. C mt pin in (gi l pin nhin liu, dng cung cp in nng v nc tinh
khit cho cc chuyn gia bay trong v tr) gm in cc anot (C-Ni), in cc catot c (CNi-NiO) nhng vo Na2CO3 nng chy v np H2 vo in cc anot, O2 vo in cc catot.
Vit cc bn phn ng, phng trnh phn ng khi pin hot ng v s pin.
Phn tch: Cc v d trn c h thng mt s dng h in ha ca cc cp oxi
ha kh nhng trong cng mt dung dch in li: c th l dung dch axit, dung dch
baz, dung dch mui hoc dng nng chy. xy dng c pin theo dng bi ny,
hc sinh phi hiu cc qu trnh xy ra mi in cc, mun vit ng dng kh, dng oxi
ha ca mi qu trnh l phi hiu s tn ti ca mi dng trong mi trng .
Hng dn:
1. Pin gm hai in cc Pt nhng trong dung dch HCl, kh clo hai in cc c P khc
nhau. Hoc mt in cc bm kh H2, cn in cc kia bm kh clo. Hoc in cc Ag,
AgCl c nhng trong dung dch HCl vi in cc kh clo.
S pin: (-) Pt Cl2 (P2) HCl(aq) Cl2(P2), Pt (+) (Vi P2 < P1)
H in ha ca loi pin ny ch l do s chnh lch v p sut ca cng mt dng kh,
cng to cho th khc nhau v c hnh thnh pin in.
Na phn ng anot (-): 2Cl- Cl2 + 2e
Na phn ng catot (+): Cl2 + 2e 2ClHoc: S pin: (-) Pt H2 (P2) HCl(aq) Cl2(P2), Pt (+)
Na phn ng anot (-): H2 2H+ + 2e
Na phn ng catot (+): Cl2 + 2e 2Cl=> Phn ng xy ra trong pin: H2 + Cl2 2HCl
Hoc: S pin: (-) Ag,AgCl HCl Cl2(P atm), Pt (+)
Na phn ng anot (-): Ag + Cl- + 1e AgCl
Na phn ng catot (+): Cl2 + 2e 2Cl11

=> Phn ng khi pin hot ng:

2Ag + Cl2 2AgCl
2. Pin Zn - Hg c nhng trong dung dch KOH CM. EoZnO22-/Zn=-1,22V; EoHgO/Hg= 0,12V
Hng dn: y l pin in gm hai in cc l dng kh c nhng cng trong
dung dch KOH. Trong loi pin ny cn nm c dng oxi ha tn ti l g ? vit cho
ng hc sinh cn hiu Zn (II) trong mi trng kim v mi qu trnh khi vit phi ng
cho mi trng ca pin hot ng.
S pin in:
(-) Zn KOH (C) HgO, Hg (+)
Na phn ng anot (-): Zn + 4OH- Zn(OH)42- + 2e
Na phn ng catot (+): HgO + 2e + H2O Hg + 2OH=> Phn ng khi pin hot ng:
Zn + HgO + 2OH- + H2O Zn(OH)42- + Hg
3. Pin Zn - PbO2 c nhng trong dung dch H2SO4 38%. EoPbO2/Pb = 1,455V
Hng dn: y l mt pin in gm hai in cc dng kh v in cc tr (c
cha dng ox/kh) cng nhng trong dung dch H2SO4 long.
S pin in:
(-) Zn Zn2+(C), H2SO4 (C%) PbSO4, PbO2, Pt (+)
Na phn ng anot (-): Zn Zn2+ + 2e
Na phn ng catot (+): PbO2 + 2e + 4H+ + SO42- PbSO4 + 2H2O
=> Phn ng khi pin hot ng:
Zn + PbO2 + 4H+ + SO42- Zn2+ + PbSO4 + 2H2O
4. Pin Zn - O2 c nhng trong dung dch NH4Cl CM.
Hng dn: y l mt pin in gm hai in cc c hai dng ox/kh cng nhng
trong dung dch NH4Cl. vit ng cho loi pin ny hc sinh phi xc nh c in
cc, qu trnh xy ra in cc v s tn ti dng oxi ha ca ion Zn2+.
S pin in:
(-) Zn Zn(NH3)42+, NH4Cl (aq) O2, Pt (+)
Na phn ng anot (-): Zn + 4NH4+ [Zn(NH3)4]2+ + 2e + 4H+
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Na phn ng catot (+): O2 + 4e + 4H+ 2H2O

=> Phn ng khi pin hot ng:
2Zn + O2 + 8NH4+ 2[Zn(NH3)4]2+ + 2H2O + 4H+
5. a) Vit s ca c quy ch, cc bn phn ng v phng trnh phn ng khi c
quy ch phng in v np in.
b) Khi np in vi I = 19,3A, t = 1,5 gi. Hi c bao nhiu gam PbSO4 b phn tch?
Hng dn: y l mt loi pin c c ch thun nghch. vit ng cho mi qu
trnh, hc sinh bit kt hp qu trnh oxi ha kh v cn bng ca hp cht t tan.
a) + Khi pin phng in, c s pin in:
(-) Pb H2SO4 38% PbO2 (+)
Na phn ng anot (-): Pb + SO42- PbSO4 + 2e
Na phn ng catot (+): PbO2 + 2e + SO42- + 4H+ PbSO4 + 2H2O
=> Phn ng khi pin hot ng:
Pb + PbO2 + 2H2SO4 2PbSO4 + 2H2O
+ Khi pin np in (nh mt bnh in phn):
Na phn ng anot (+): PbSO4 + 2H2O PbO2 + 4H+ + 2e + SO42Na phn ng catot (-): PbSO4 + 2e Pb + SO42=> Phn ng khi pin np in:
2PbSO4 + 2H2O PbO2 + Pb + 2H2SO4
b) Theo phn ng khi np in ta c:
mPbSO4 = M.nPbSO4 = 303(It/2F).2 = 303(10.1,5.3600/96500) = 169,55 (g)
6. C mt pin in (gi l pin nhin liu, dng cung cp in nng v nc tinh khit
cho cc chuyn gia bay trong v tr) gm in cc anot (C-Ni), in cc catot c (C-NiNiO) nhng vo Na2CO3 nng chy v np H2 vo in cc anot, O2 vo in cc catot.
Vit cc bn phn ng, phng trnh phn ng khi pin hot ng v s pin.
Hng dn: y l mt loi pin gm hai in cc l cc cht khng than gia qu
trnh oxi ha kh c nhng cng trong cht in li nng chy. Loi pin ny ban u
hc sinh kh hnh dung cc phn ng xy ra mi in cc. hiu v vit c th hc
13

sinh suy lun dng oxi ha v dng kh to ra khi pin lm vic s tham gia phn ng
ion CO32- khng i.
+ Na phn ng anot: H2 + CO32- CO2 + H2O + 2e
Na phn ng catot: 1/2 O2 + 2e + CO2 CO32=> Phng trnh phn ng khi pin hot ng:
H2 + 1/2 O2 H2O
=> S pin:
(-) C-Ni, H2 Na2CO3(n/c) O2, C-Ni-Ni (+)

III.2- Pin ni lng:

Pin ni lng l loi pin gm hai in cc c nhng trong hai dung dch in li v c
to ni gia hai dung in li thng bng cu mui cha dung dch KCl bo ha.
Tng qut: (-) kh1/ox1 ox2/kh2 (+)
1. Pin nng :
V d 8: Cho qu trnh xy ra trong pin nh sau:
1)

Ag+ (C1) Ag+ (C2)

2)

HCl (C1) HCl (C2)

Thit lp s pin v na phn ng khi pin hot ng.

Hng dn: y l mt loi c cng mt dng oxi ha kh, nhng do s chnh
lch v nng , nn c gi tr th kh khc nhau v hnh thnh c pin in. Hc sinh
cn xc nh th kh ca cp no ln hn.
1. S pin in c xy ra qu trnh Ag+ (C1) Ag+ (C2).
(-) Ag Ag+(C2) Ag+(C1) Cu (+)
Na phn ng anot: Cu Cu2+ + 2e
Na phn ng catot: Cu2+ + 2e Cu
2. S pin in c xy ra qu trnh HCl (C1) HCl (C2):
(-)Pt, H2 (P) HCl(C1) HCl(C2) H2(P) Pt(+) (Vi C2 > C1)
Na phn ng anot: H2 2H+ + 2e
Na phn ng catot: 2H+ + 2e H2
14

Hoc: S pin in:

(-)Pt, Cl2 (P) HCl(C1) HCl(C2) Cl2(P) Pt(+) (Vi C2 < C1)
Na phn ng anot: 2Cl- Cl2 + 2e
Na phn ng catot: Cl2 + 2e 2Cl2. Pin in ch c cc phn ng oxi ha kh:
V d 9: Vit cc s pin, cc na phn ng v phng trnh phn ng khi pin hot ng
ca cc cp oxi ha kh cho sau:
1) Zn2+/Zn vi Cu2+/Cu.
2) Fe3+/Fe2+ vi Cr2O72-(H+)/Cr3+/
3) Br2/2Br- vi MnO4-(H+)/Mn2+.
Hng dn: Dy l loi pin in ph bin v thng gp. Nu theo nh tnh hc
sinh xc nh cp no c dng oxi ha mnh hn th bn phi (c th kh chun ln l
in cc dng), cp cn li bn tri (c th kh chun nh l in cc m)
1) Do tnh oxi ha ca ion Cu2+ > Zn2+ (hoc E(Cu2+/Cu) > E(Zn2+/Zn), nn c s pin:
(-) Zn ZnSO4C1 CuSO4C2 Cu (+)
Na phn ng anot: Zn Zn2+ + 2e
Na phn ng catot: Cu2+ + 2e Cu
=> Phn ng khi pin hot ng: Zn + Cu2+ Zn2+ + Cu
2) Do tnh oxi ha ca ion Cr2O72- > Fe3+ (hoc E(Cr2O72-/Cr3+) > E(Fe3+/Fe2+), nn c s pin:
(-) Pt Fe2+; Fe3+ (aq) Cr2O72-; Cr3+(aq) Pt (+)
Na phn ng anot: Fe2+ Fe3+ + e
Na phn ng catot: Cr2O72- + 6e + 14H+ 2Cr3+ + 7H2O
=> Phn ng khi pin hot ng: 6Fe2+ + Cr2O72- + 14H+ 6Fe3+ + 2Cr3+ + 7H2O
3) Do tnh oxi ha ca ion MnO4- > Br2 (hoc E(MnO4-/Mn2+) > E(Br2/2Br-), nn c s pin:
(-) Pt Br2(C1)Br-(C2) MnO4-; Mn2+(aq) Pt (+)
Na phn ng anot: 2Br- Br2 + e
Na phn ng catot: MnO4- + 5e + 8H+ Mn2+ + 4H2O
=> Phn ng khi pin hot ng: 10Br- + MnO4- + 8H+ Mn2+ + 5Br2 + 4H2O
15

3. Pin in c cc phn ng ph:

Trong loi pin in c bn m hc sinh thng l cc qu trnh xy ra trong pin khi pin
hot ng ch l cc qu trnh oxi ha kh. nng cao bi ton v c tnh ng dng,
trong pin in cn c cc phn ng axit baz, phn ng to hp cht t tan, phn ng to
phc.
V d 10: Vit s pin in, cc qu trnh xy ra mi in cc khi pin hot c phn
ng:
1)

H+ + RCOO- RCOOH

2)

2Ag+ + SO42- Ag2SO4

3)

Ag2SO4 + 2Cl- 2AgCl + SO42-

4)

Ni2+ + 4CN- Ni(CN)4 2-

5).

[Cu(NH3)4]2+ + 4CN- [Cu(CN)4]2- + 4NH3

6)

AgCl + 2CN- [Ag(CN)2]- + ClPhn tch: y l mt loi pin in m phng trnh phn ng ca pin in li

khng phi l phn ng oxi ha kh (phn ng axit baz, phn ng to hp cht t tan,
phn ng chuyn t cht t tan sang cht t tan hn, phn ng to phc, phn ng chuyn
t phc km bn sang phc bn, phn ng chuyn t hp cht t tan sang dng phc,...).
vit c s ca loi pin in ny, hc sinh phi xc nh:
- Hai in cc u c cng dng ox/kh, ch c iu s tn ti ca mi dng oxi ha hoc
dng kh (thng dng oxi ha) l khc nhau.
- Gi tr th ca dng ox/kh no ln hn.
Hng dn: T cc cp ox/kh v nhn thy gi tr th ca cc cp,... c cc s
pin in v cc na phn ng trn in cc l:
1) S pin in c phn ng: H+ + RCOO- RCOOH
(-) Pt, H2(P) RCOO-(C1) H+(C2) H2,(P) Pt (+)
Na phn ng anot: H2 + 2RCOO- 2RCOOH + 2e
Na phn ng catot: 2H+ + 2e H2
2) S pin in c phn ng: 2Ag+ + SO42- Ag2SO4
(-) Ag, Ag2SO4 K2SO4(C1) Ag+(C2) Ag (+)
16

Na phn ng anot: Ag + SO42- Ag2SO4 + 2e

Na phn ng catot: Ag+ + 1e Ag
3) S pin in c phn ng: Ag2SO4 + 2Cl- 2AgCl + SO42(-) Ag, AgCl KCl(C1) K2SO4(C2) Ag2SO4, Ag (+)
Na phn ng anot: Ag + Br- AgBr + e
Na phn ng catot: AgCl + e Ag + Cl4) S pin in c phn ng: Ni2+ + 4CN- Ni(CN)4 2(-) Ni Ni(CN)42-, KCN(C1) Ni2+(C2) Ni (+)
Na phn ng anot: Ni + 4CN- Ni(CN)42- + 2e
Na phn ng catot: Ni2+ + 2e Ni
5) S pin in c phn ng: [Cu(NH3)4]2+ + 4CN- [Cu(CN)4]2- + 4NH3
(-) Cu Cu(CN)42-; KCN(C1) NH3(C2); Cu(NH3)42+ Cu (+)
Na phn ng anot: Cu + 4CN- Cu(CN)42- + 2e
Na phn ng catot: Cu(NH3)42+ + 2e Cu + 4NH3
6) S pin in c phn ng: AgCl + 2CN- [Ag(CN)2]- + Cl(-) Ag Ag(CN)2-; KCN(C) KCl(C) AgCl, Ag (+)
Na phn ng anot: Ag + 2CN- Ag(CN)2- + e
Na phn ng catot: AgCl + e Ag + Cl-

B. BI TP VN DNG:
I. Vit s pin, cc qu trnh, phng trnh phn ng khi
pin hot ng, tnh sut in ng ca pin:
V d 11: Thm 0,40 mol KI vo 1 lt dung dch KMnO4 0,24 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
Cho pH = 0 v 25oC th in cc tiu chun E o ca mt s cp oxi
ho - kh c cho nh sau: 2IO4/ I2 (r) = 1,31V; 2IO3/ I2 (r) = 1,19V;
17

2HIO/ I2 (r) = 1,45 V; I2 (r)/ 2I = 0,54V ; MnO4-/Mn2+ = 1,51V; E ca

in cc calomen bo ho bng 0,244 V; tan ca it trong nc bng
5,0.10 4 M.
Phn tch: y l bi ton c bn tnh Epin to bi in cc chun v mt in cc
ch c dng oxi ha, dng kh v mi trng. im nng cao ca bi ny l tnh thnh
phn gii hn ca phn ng oxi ha kh.
Do Eo(MnO4-/Mn2+) = 1,51V >> Eo(I2/2I-) = 0,53V; nn u tin s xy ra phn ng:
2 MnO4
CO

0,24

2 Mn2+ + 5 I2(r) + 8 H2O ; K = 10 165,54

0,4

C 0,08
C

+ 10 I + 16 H+
0,4

0,16

0,08

0, 2

Do Eo MnO4-/Mn2+ = 1,51V > Eo IO3-/I2 = 1,19V; nn MnO4 cn d s oxi ho tip I2 thnh

IO3 theo phn ng:
2 MnO4 + I2(r)

+ 4 H+ 2 IO3 + 2 Mn2+ + 2 H2O ; K

= 10 176
CO

0,16

0,16

0,2

0,08

0,08
0,12

0,16

0, 24

Thnh phn hn hp sau phn ng: IO 3 0,16 M; Mn2+0,24 M; I2 (H2O) 5.

104M; I2(r) 0,12 M; pH = 0.
b)

Trong hn hp c cp IO3/ I2 (r) nn:

E = Eo (IO3-/I2(r) + (0,0592/10)lg [IO3]2 [H+]12
= 1,19 + (0,0592/10)lg (0,16)2 = 1,18(V)
E so vi in cc calomen bo ho: Epin = 1,18 0,244 = 0,936(V)

18

Nhn xt: Cc bi ton pin in dng trn rt ph bin. T cc cp ox/kh phn ng

vi nhau v tha nhn dng kh hoc dng oxi ha ht. Lc tnh th kh ca cp m c
dng oxi ha hoc dng kh cn d.
V d 12: Dung dch A gm AgNO3 0,050 M v Pb(NO3)2 0,100 M v HNO3 0,200M.
Thm 10,00 ml KI 0,250 M vo 10,00 ml dung dch A, thu c dung dch B. Ngi ta
nhng mt in cc Ag vo dung dch B v ghp thnh pin (c cu mui tip xc hai dung
dch) vi mt in cc c Ag nhng vo dung dch X gm AgNO 3 0,010 M v KSCN
0,040 M.
a) Vit s pin.
b) Tnh sc in ng Epin ti 250C.
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= 10 11,70

Pb2+ + H2O

PbOH+ + H+

(2) ; K2= 10 7,80

Ch s tch s tan pKs: AgI l 16,0 ; PbI2 l 7,86 ; AgSCN l 12,0; E0Ag+/Ag = 0,799V.
Phn tch: y l bi ton h in ha gm hai in cc u l in cc kim loi
c bao ph bi hp cht t tan, cc qu trnh xy ra mi in cc c nh hng ca
hp cht t tan, nn th kh ca mi cp s thay i. hiu v lm c dng bi ny,
hc sinh phi nm trc cn bng ca hp cht t tan, phn nng cao ca bi ton l cho h
in ha c nhiu hp cht t tan, nn hc sinh phi xt n cn bng chnh. Khi hc sinh
tnh th kh ca cp ox/kh, thng hc sinh kh hiu gi tr ca dng oxi ha v dng kh
trong biu thc tnh theo phng trnh Nec (hoc hiu l gi tr ban u ca dng oxi ha,
dng kh) l bao nhiu ? iu ny gio vin nhn mnh cho hc sinh thy c gi tr
dng oxi ha, dng kh trong biu thc tnh, chnh l gi tr tn ti trng thi cn bng
ca cn bng chnh.
Hng dn:
1a) Dung dch B: Thm KI : CAg+ = 0,025 M; CPb2+ = 0,050
CI- = 0,125M ; CH+ = 0,10M
19

Ag+

AgI

I-

0,025

0,125

0,10

Pb2+

PbI2

2 I-

0,05

0,10

Trong dung dch c ng thi hai kt ta AgI v PbI2

AgI

Ag+ + I-

Ks1 = 1.10-16 (3)

PbI2

Pb2+ + 2I-

Ks2 = 1.10-7,86 (4)

Ks1 << Ks2, vy trong dung dch cn bng (4) l ch yu. S to phc
hiroxo ca Pb2+ l khng ng k v c H+ d:
Pb2+ + H2O

PbOH + H+ ;

PbOH 10
Pb 10

7 , 8

10 6,8 PbOH Pb 2

PbI2

Trong dung dch

Pb2+ +
x

(2x)2x = 10-7,86

x = 1,51 . 10-3M

.10
Ag KI 31,02
.10

16

s1

K2 = 10-7,8

2I-

Ks2 = 1.10-7,86

2x
2x = [I-] = 2,302 . 10-3M

3,31.10 14 M .

E ca cc Ag trong dung dch A:

Ag+ + e
E 1 E 0Ag

Ag

0,0592 lg Ag 0,799 0,0592 lg 3,31.10 14

Ag

E 1 0,001V

Dung dch X
20

Ag+ + SCN-

AgSCN

0,01

0,04

(0,03-x)

KS-1 = 1012,0

(0,01-x)

x(0,03-x) = 10-12
=> Ag x

10 12
3,33.10 11
2
3x10

E 2 0,799 0,0592 lg Ag 0,799 0,0592 lg 3,33.10 11

E 2 0,179V

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

S pin
AgI
PbI2

Ag

b)

AgSCN
SCN - 0,03M

Ag

Epin = 0,179 0,001 = 0,178V

c) Phng trnh phn ng

Ag +
AgSCN
AgSCN

I-

AgI
+

Ag

I-

AgI

d)

K s ( AgSCN )
K s ( AgI)

+
+

SNC SNC -

10 12
10 4
16
10

V d 13:
1. Thit lp s pin v vit na phn ng khi pin hot ng xy ra phn ng:
CH3COO- + HSO4- CH3COOH + SO4221

2. Tnh Gpin
3. Tnh nng mol cc ion trong dung dch khi Ipin = 0.
4. Ghp pin xung i:
(-) Pt H2 CH3COO- (0,080M) HSO4- (0,050M) H2 Pt (+)
vi pin: (-) Ag,AgCl HCl (1,50M) KCl(bo ho) Hg2Cl2, Hg (+)
Cho EoAgCl/Ag = 0,222V; EHg2Cl2/Hg = 0,244V; KaCH3COOH = 10-4,76 ; KaHSO4- = 10-2,00
Vit cc bn phn ng ng xy ra mi in cc v cc phng trnh phn ng ?
Hng dn:
1. y l mt dng pin in m phn ng xy ra khi pin hot ng khng phi l
phn ng oxi ha kh (phn ng axit baz). xy dng c s pin in hc sinh
phi xc c dng oxi ha v dng kh mi in cc v th kh no ln hn xc
nh in cc catot (cc dng).
Do ion H+ t HSO4- nhiu hn ion H+ t CH3COO- nn c s pin l:
(-) Pt H2 (1 atm) CH3COO- (0,08M) HSO4- (0,05M) H2 (1 atm) Pt (+)
Na phn ng antot:
H2 + 2CH3COO- 2CH3COOH + 2e
Na phn ng catot:
2HSO4- + 2e H2 + SO42=> Phn ng xy ra khi pin hot ng:
CH3COO- + HSO4- CH3COOH + SO422. Tnh Gpin
Hng dn: tnh c Gpin ca loi pin in m phn ng xy ra ca pin khng
phi phn ng oxi ha kh (phn ng axit baz hoc phn ng to hp cht t tan hoc
phn ng to phc), hc sinh tnh theo th kh ca qu trnh kh mi in cc, sau
tnh ra sut in ca pin v tnh nng lng Gipx: Gpin = - nFEpin
+ Tnh E(-):
Theo cn bng:
CH3COO- + H2O CH3COOH + OHCo

1
22

Kb = Ka-1.Kw = 10-9,24

[]

(1-x)

=> K = ....= x2/(1-x) = 10-9,24 (vi 0 < x < 1) => x = [OH-] = 10-4,62
=> [H+] = 10-14/10-4,62 = 10-9,38
=> E(-) = 0 + (0,0592/2)lg[H+]2/PH2 = 0,0592lg10-9,38 = - 0,56(V)
+ Tnh E(+):
Theo cn bng:
HSO4Co

[]

(1-y)

H+ + SO42y

K = 10-2

=> K = .... = y2/(1-y) = 10-2 (vi 0 < y < 1) => y = [H+] = 0,095
=> E(+) = 0,0592lg[H+] = 0,0592lg0,095 = - 0,061(V)
Vy Epin = E(+) - E(-) = - 0,061 - (-0,56) = 0,499(V)
=> G = - nFEpin = - 2.96500.0,499 = - 96307(J) = - 96,307kJ
3. Tnh nng mol cc ion trong dung dch khi Ipin = 0.
Hng dn : Khi Ipin = 0, tc l pin ngng hot ng, lc ny hc sinh hiu rng
E(+) = E(-), nn Epin = 0 => G = 0, c ngha phn ng phn ng xy ra trong pin t ti
trng thi cn bng. V vy tnh nng ca cc ion cn bng .
Khi I = 0 c ngha l pin ngng phng in, tc l phn ng trong pin t n trng thi
cn bng:
CH3COO- + HSO4- CH3COOH +
Co

0,08

0,05

[]

(0,03+x)

(0,05-x)

SO42- K = 10-2.(10-4,76)-1 = 102,76

(0,05-x)

=> K = .... = (0,05-x)2/(0,03+x).x = 102,76 => x = 1,43.10-4(M)

=> [CH3COOH] = [SO42-] = 0,05-1,43.10-4 = 0,049857(M); [HSO4-] = 1,43.10-4M;
[CH3COO-] = 0,03 + 1,43.10-4 = 0,030143(M).
4. Ghp pin xung i:
(-) Pt H2 CH3COO- (0,080M) HSO4- (0,050M) H2 Pt (+)
vi pin: (-) Ag, AgCl HCl (1,50M) KCl(bo ho) Hg2Cl2, Hg (+)
Cho EoAgCl/Ag = 0,222V; EHg2Cl2/Hg = 0,244V; KaCH3COOH = 10-4,76 ; KaHSO4- = 10-2,00
23

Vit cc bn phn ng ng xy ra mi in cc v phng trnh phn ng?

Phn tch: ghp xung i hai pin in (ni hai cc dng vi nhau v hai cc
m vi nhau). Ch c iu hc sinh phi xc nh c u l pin, u l in phn. Mun
vy hc sinh phi tnh th kh mi in cc v tnh sut in ng ca pin. Nu sut
in ca pin no ln hn thi ng vai tr l pin, cn li l bnh in phn. iu nng cao
ca bi ton ny l tnh th kh mi in cc.
Hng dn;
+ Xt pin 1: (-) Pt H2 CH3COO- (0,080M) HSO4- (0,050M) H2 Pt (+)
- in cc anot:
CH3COO- + H2O CH3COOH + OHCo

0,08

[]

(0,08-x)

K = 10-9,24

=> K = ... = x2/(0,08-x) = 10-9,24 (vi 0 < x < 0,08) => x = 6,78.10-6
=> E(-) = 0,0592lg[H+] = 0,0592lg(10-14/6,78.10-6) = -0,52(V)
- in cc catot:
HSO4- H+ + SO42Co

0,05

[]

(0,05-x)

K = 10-2

=> K = ... = x2/(0,05-x) = 10-2 (vi 0 < x < 0,05) => x = 0,018
=> E(+) = 0,0592lg[H+] = 0,0592lg0,018 = - 0,103(V)
=> Epin(1) = E(+) - E(-) = -0,103 -(-0,52) = 0,417 (V)
+ Xt pin 2: (-) Ag, AgCl HCl (1,50M) KCl(bo ho) Hg2Cl2, Hg (+)
- Bn phn ng anot:
Ag + Cl- AgCl + e
=> EAgCl/Ag = EoAgCl/Ag + 0,0592lg(1/CCl-) = 0,222 + 0,0592lg(1/1,5) = 0,212(V)
=> Epin(2) = E(+) - E(-) = 0,244 - 0,212 = 0,032(V)
V Epin(1) = 0,417 > Epin(2) = 0,032(V), nn pin (1) c vai tr cung cp in cho pin (2) (pin
c np in). Do vy s pin c ni nh sau v cc bn phn ng xy ra:
Pin:

(-) Pt H2 CH3COO- (0,080M) HSO4- (0,050M) H2 Pt (+)

24

H2 + 2CH3COO-2CH3COOH +2e
I
/p:

2HSO4- +2e SO42- + H2

I

(-) Ag, AgCl HCl (1,50M) KCl(bo ho) Hg2Cl2, Hg (+)

2AgCl + 2e 2Ag + 2Cl-

2Hg + 2Cl- Hg2Cl2 + 2e

- Phn ng xy ra trong pin (phng in):

HSO4- + CH3COO- CH3COOH + SO4- (Phn ng t xy ra)
- Phn ng xy ra khi np in:
2Hg + 2AgCl Hg2Cl2 + 2Ag (phn ng khng t xy ra)

II. Tnh hng s v s thay i sut in ng ca pin.

1. V d minh ha:
V d 14: Cho s pin:
(-) Zn Zn2+(0,10M) KCl (0,50M) AgCl, Ag (+)
E0Zn

2+

/Zn

= - 0,763 V ; E0Ag /Ag = 0,799 V ; Epin = 1,017 V

+

Tnh tch s tan ca mui AgCl.

+ Phn tch:
y l dng bi ton t thit lp mt loi pin vi nng cc cht ly tu . Thng qua
cc gi tr th in cc tiu chun ca cc cp oxi ho - kh cho bit v da vo thc
nghim o c sc in ng ca pin t tnh ra c tch s tan ca hp cht t tan
tip xc vi in cc kim loi.
Tu theo cch trnh by, y hc sinh c th vit ngay theo h thc tnh sc in ng
ca pin:
Epin = E(+) E(-)
0

= ( E Ag /Ag + 0,0592.lgKs(AgCl) 0,0592.lg0,5 ) ( - 0,763 +

+

0,0592
2+
lg[Zn ]
2

Thay cc gi tr th in cc tiu chun, nng cc cht v sc in ng ca

25

pin tnh ra c: Ks(AgCl) = 10-10,00

* Tm li:
+ Cc dng bi tp tnh tch s tan (hoc tnh hng s axit, hoc s baz, hoc s to phc)
t th in cc, sut ng pin in o c cng a dng, phong ph. Nhng gii quyt
c dng bi tp ny i hi hc sinh phi nm vng v phn ng oxi ho-kh, phn ng
to hp cht t tan (phn ng axit baz, phn ng to phc) v hc sinh phi c trang b
kin thc v pin in mt cch vng vng.
+ Cc bi ton ngc ca dng bi ton trn thng gp nhiu hn. V d cho cc phn
ng to ra hp t tan (phn ng axit baz, phn ng to phc) da vo lng cht v gi tr
hng s bi cho tnh c nng cn bng ca cc ion, tnh ra th in cc cho tng cp
oxh/kh v t xy dng s pin, tnh sc in ng ca pin.

2. Bi tp vn dng c tnh cht tng hp:

V d 15: C mt dung dch axit yu HA C(mol/l), c nng khng qu b li l mt
axit yu v mt in cc calomen bo ha (c Ecal (bo ho) = 0,244V).
1. Vit s pin in hnh thnh pin in. Gii thch ngn gn.
2. Nu cch lm xc nh c pH ca dung dch axit HA (v d: Tnh pH ca
dung dch HA khi Epin = 0,303V).
3. Nu cch lm thc nghim xc nh hng s cn bng ca HA trn c s pin
in.
4. Nu thay dung dch HA bng dung dch NaA, in cc kh oxi, th s pin nh
th no ? Gii thch ? Cho Eo(O2, H+/H2O) = 1,24V
5. Nh t t axit H2SO4 (long) vo in cc cha HA th Epin thay i nh th no?
Phn tch:
y l mt dng bi ton pin in gia mt in cc c lin quan ti cn bng axit
baz ca mt axit yu vi mt in cc so snh (in cc kim loi b bao ph mt hp
cht t tan). Ni dung ca bi ton mang tnh cht thc nghim: T cch o sut in ng
ca pin tnh c pH ca dung dch. Da vo cn bng ca axit khi tnh c pH v da
vo nng ban u cho ca axit, ta tnh c hng s cn bng axit.
26

1. V axit HA l axit yu nn: Eo(2HA/H2) < Eo(2H+/H2) = 0V < Ecal (bo ho) = 0,244V, v vy
c s pin in l
(-) Pt, H2(1 atm) HA (C) KCl (bo ho) Hg2Cl2, Hg (+)
2. Ta lp dng c theo pin in trn, dng vn k o c sut in ng ca pin v
da biu thc:
E(-) = Eo2H+/H2 + (0,0592/2)lg[H+]2/PH2
= 0,0592lg[H+] = - 0,0592pH
=> Epin = E(+) - E(-) = 0,244 - (-0,0592pH); y o c Epin th s tnh c pH.
V d: Tnh pH ca dung dch HA khi Epin = 0,303V).
Ta c Epin = E(+) - E(-) = 0,244 - (-0,0592pH) = 0,303
=> pH = 1
3. Khi xc nh hng s cn bng ca mt axit yu HA (c nng mol/l l C). Ta
lp dng c s pin nh phn 1, o c sut in ng ca pin. T E pin, tnh c pH,
t pH tm c: [H+] = [A-] = 10-pH v [HA] = C 10-pH; => Ka = ([H+].[A-])/[HA]. V d:
Tnh Ka(HA) khi C = 10-2; Epin = 0,444V. Cho Ecal (bo ho) = 0,244V.
+ Tng t phn (2), ta c:
Epin = E(+) - E(-) = 0,244 - (-0,0592pH) = 0,444
=> pH = 3,38. Hay [H+] = 10-3,38 (M)
Xt cn bng:

HA

H+ + A-

K=?

Co 10-2
[ ] (10-2-10-3,38)

10-3,38 10-3,38

=> K = (10-3,38)2/(10-2-10-3,38) = 10-4,76

4. Xt qu trnh:
O2 + 4e + 4H+ 2H2O
4 H2O H+ + OH-

K1 = 104.1,23/0,0592
Kw = 10-14

O2 + 4e + 2H2O 4OH-

K2 = 104E/0,0592

=> K2 = K1.Kw4
Eo = (1,23 14.0,0592) = 0,40V
M Eo(O2, H2O/OH-) = Eo + (0,0592/4)lg(1/[OH-)4
27

Do EoO2/OH-(A-) > Eo(O2/OH-) = 0,40 > Eo(calomen) = 0,244V, nn c s pin l:

(-) Hg2, Hg2Cl2 KCl(bo ha) NaA O2(1atm), Pt (+)
5. Khi nh t t dung dch H2SO4 vo in cc c A- lm nng OH- gim, nn
E(+) tng, v vy Epin tng.
* Nhn xt:
+ y l mt bi ton thit lp pin dng c bn l lp in cc nghin cu vi mt
in cc tiu chun. nng cao bi ton, ngi ta cho in cc nghin cu vi mt in
cc bt k: c th l in cc bao bc bi hp cht ion tan hoc in cc kim loi nhng
trong dung dch c ion phc ca kim loi ,...
+ Bi ton trn tnh hng s cn bng tng qut cho mt axit n axit. C th p
dng cho c bi axit a axit c Kai >> Ka(i+1).
V d 16: 1. Trnh by cch lm th nghim thng qua pin in tnh c hng s K s
ca mui AgI.
2. C mt in cc Ag c bao ph bi hp cht t tan AgI, dung dch KI 1,000.10 1

M lp vi in cc calomen bo ha v o c sut in ng ca pin l 0,333V. Tnh

tch s tan ca AgI. Bit EoAg+/Ag = 0,799V; Ecalomen(bo ho) = 0,244V.

3. Sut in ng ca pin s thay i nh th no khi:
a) Thm NaI 0,1M.
b) Thm NaCl 0,1M.
c) Thm dung dch NH3 0,2M.
d) Thm dung dch KCN 0,2M.
e) Thm dung dch HNO3 0,2M.
(u c thm vo in cc nghin cu)
Cho pKs(AgCl: 10,00; AgI: 16,00): Ag(NH3)2+ = 107,24; Ag(CN)2- = 1020,48.
Phn tch v hng dn:
1. Dng bi ton xy dng mt pin in t mt in cc ca hp cht t tan vi mt
in cc chun. hc sinh chun b v lm c th nghim dng ny, cn hiu v lm
theo tng bc sau:
28

+ Ly mt in bc c bao ph bi hp cht t tan (l hp cht ang cn xc nh tch s

tan, v d: AgI) v nhng trong dung dch mui cha anion ca hp cht t tan (v d
dung dch KI bit nng ).
+ Chn mt in cc th hai thng l in cc chun (v d in cc calomen bo ha
hoc in chun ca Ag nhng trong mui AgNO3 1M).
+ Ni hai in cc c mc vn k xc nh chiu ca dng in (xc nh in cc) v o
sut ng ca pin. T tnh c Ks.
V d: (-) Ag, AgI KI 0,1M KCl(bo ha) Hg2Cl2, Hg (+)
Hoc:
(-) Ag, AgI KI 0,1M AgNO3 1M Ag (+)
2. Khi lp pin, o c Epin, xc nh c chiu ca dng in v xc nh du ca
in cc. T vit cc qu trnh xy ra ca in cc nghin cu. Lp biu thc tnh th
kh iu kin nghin cu theo phng trnh Nec v xy dng biu thc E pin lin quan ti
hng s KS. iu quan trng l hc sinh hiu c nng bt k ca ion kim loi trong
phng trnh Nec, l nng cn bng ca ion kim loi trong cn bng to ra hp cht t
tan.
Gi s qua thc nghim xc nh c s pin nh sau:
(-) Ag, AgI KI 0,1M KCl(bo ha) Hg2Cl2, Hg (+)
- cc (-) c:
E(-) = Eo(Ag+/Ag) + 0,0592lg[Ag+] = Eo(Ag+/Ag) + 0,0592lgKs/[I-]
= 0,799 + 0,0592lg10 + 0,0592lgKs = 0,7398 + 0,0592lgKs
Epin = 0,244 0,8582 0,0592lgKs = 0,333
K = 10-16.
Nhn xt: Tng t cho cc bi ton tnh tch s tan ca mt s hp cht t tan nh:
AgCl, AgBr, Ag2SO4, ... Thng l mui t tan ca kim loi hot ng trung bnh v yu.
3. nh gi c Epin tng hay gim, hc sinh cn nm c:
+ Cht cho vo c phn ng c vi ion no to ra cht t tan khng (Phn ng mi
xy ra c th l phn to ra hp cht t tan khc, phn ng to phc, phn ng oxi ha kh,
phn ng axit baz).
29

+ Phn ng c nh hng ti cn bng ca cht t tan khng.

+ Khi c nh hng n cn bng ca cht t tan th xt xem ion m tham gia qu trnh kh
c tng hay gim ri xc nh th kh, t mi xc nh c s thay i Epin.
Hng dn:
a) C s pin:
(-) Ag, AgI KI 0,1M KCl(bo ha) Hg2Cl2, Hg (+)
Khi thm mui NaI vo in cc anot lm cho cn bng AgI Ag+ + I- chuyn dch
theo chiu nghch, nng d ion Ag+, do E(-) gim. V vy sut in ng ca pin tng.
b) Khi thm NaCl 0,1M vo in cc anot, c phn ng:
Khi thm CH3COONa vo in cc anot c phn ng:
AgI + Cl- AgCl + I-

K = 10-16.(10-10)-1 = 10-6 (nh)

Theo phn ng trn thy K rt nh, mt khc nng NaCl li long, nn qu trnh
chuyn sang AgCl l rt t. V vy sut in ng ca pin coi khng i.
c) Khi thm NH3 0,2M vo in cc anot, c phn ng:
AgI + 2NH3 Ag(NH3)2+ + I-

K = 10-16.(10-7,24)-1 = 10-9,76 (rt nh)

Theo phn ng trn thy K rt nh, mt khc nng NH 3 li long, nn qu trnh

chuyn sang Ag(NH3)2+ l rt t. V vy sut in ng ca pin coi khng i.
d) Khi thm dung dch KCN 0,2M vo in cc anot, c phn ng:
AgI + 2CN- Ag(CN)2+ + I-

K = 10-16.(10-20,48)-1 = 104,48 (ln)

Theo phn ng trn thy K ln, nn qu trnh chuyn sang phc bn gn nh hon ton,
do ion Ag+ gim i, E(-) gim. V vy sut in ng ca pin tng.
e) Khi thm dung dch HNO3 0,2M vo in cc anot, c phn ng I - b oxi ha bi HNO3
lm nng ion I-, dn n nng ion Ag+ tng (trong mi trng axit ion Ag + khng
tham gia qu trnh to phc hiroxo), nn E (-) tng. V vy sut in ng ca pin gim,
n lc no c th i chiu dng in.
V d 17: Cho s pin:
(-) Ag AgNO31,000.10-1M; NH3 1M Ag2SO4(bo ho) Ag (+)
1. Tnh hng s to phc Ag(NH3)2+ bit EoAg+/Ag = 0,800V; KsAg2SO4 = 1,100.10-5;
Epin = 0,390V
30

2. Sut in ng ca pin thay i nh th no nu ta thay i nh sau:

a) Thm mui NaCN 1M vo in cc anot.
b) Thm HCl vo in cc anot.
c) Thm CH3COONa 0,01M vo in cc anot.
d) Thm mui BaCl2 bo ho vo in cc catot.
e) Thm mui NaCN bo ho vo in cc catot.
Cho Ag(CN)2- = 1020,48; KsCH3COOAg = 10-2,7 ; KsBaSO4 = 10-9,96.
Phn tch: y l bi ton xc nh hng s to phc ca mt phc cht, da trn
pin in khi bit th kh ca mt in cc v o c E pin. hiu v lm c hc sinh
cn tnh thnh tho cn bng to phc. Theo bi ny cho bit nng ban u ca ion
kim loi, ca phi t, t tm nng mi ion trng thi cn bng, ri lp biu thc
lin quan ti hng s to phc ca ion kim loi thay vo phng trnh Nec. iu quan
trng l hc sinh hiu c nng bt k ca ion kim loi l nng cn bng ca ion
kim loi trong cn bng to phc.
1. Tnh Ag(NH3)2 = ?
+ Tnh E(+):
Theo cn bng: Ag2SO4 2Ag+ + SO422S

Ks = 1,10.10-5

=> [Ag+]2.[SO42-] = (2S)2.S = Ks

=> [Ag+] = 2S = 2.(Ks/4)1/3
=> E(+) = EoAg+/Ag + 0,0592lg[Ag+]
= EoAg+/Ag + 0,0592lg2.(Ks/4)1/3 = 0,8 + 0,0592lg2.(1,1.10-5/4)1/3 = 0,708(V).
+ Tnh E(-):
Theo cn bng:
Ag+ + 2NH3 Ag(NH3)2+
Co

0,1

1
(0,8+2x)

=?

0
(0,1-x)

=> = (0,1-x)/x.(0,8+2x) 0,1/0,8x

31

=> [Ag+] = x = 1/8

=> E(-) = EoAg+/Ag + 0,0592lg[Ag+] = 0,8 + 0,0592lg1/8
= 0,747 - 0,0592lg
=> Epin = 0,708 - 0,747 + 0,0592lg = 0,390
=> = 107,247
2. Xt cc cn bng:
a) Cho s pin:
(-) Ag Ag(NH3)2+ (0,01M) Ag2SO4 (bo ho) Ag (+)
Khi thm mui NaCN vo in cc anot c phn ng:
Ag(NH3)2+ + 2CN- Ag(CN)2- + 2NH3

K = 10-7,24.1020,48 = 1013,24(rt ln),

nn khi cho mui NaCN vo lm nng ion Ag + gim do E(-) gim. V vy sut in
ng ca pin tng.
b) Thm HCl vo in cc anot.
Khi thm HCl vo in c anot c phn ng:
Ag(NH3)2+ + 2H+ Ag+ + NH4+
Do phn ng trn lm nng ion Ag + tng, nn E(-) tng ln. V vy sut in ng ca
pin gim v n lc no pin ngng hot ng v i chiu dng in trong pin.
c) Thm CH3COONa 0,01M vo in cc anot.
Khi thm CH3COONa vo in cc anot c phn ng:
Ag(NH3)2+ + CH3COO- CH3COOAg + 2NH3

K = 10-7,24.(10-2,7)-1 = 10-4,54

(nh)
Theo phn ng trn thy K nh, mt khc nng CH3COONa li long v vy lng
phc mt i rt t. V vy sut in ng ca pin coi khng i.
d) Thm mui BaCl2 bo ho vo in cc catot.
Khi thm BaCl2 vo in cc catot c phn ng:
Ba2+ + 2Cl- + Ag2SO4 BaSO4 + 2AgCl

K = 10-4,83.(10-9,96)-1.1010.2 = 1025,13 (rt ln)

Theo cn bng trn cho thy nng ion Ag + in cc catot gim, nn lm E (+) gim =>
lm sut in in ca pin gim v n lc no c th pin i chiu dng in.
e) Thm mui NaCN bo ho vo in cc catot.
32

Khi thm mui NaCN bo ho vo c phn ng:

Ag2SO4 + 4CN- 2Ag(CN)2- + SO42-

K = 10-4,83.(1020,48)2 = 1036,13 (rt ln)

T phn ng trn thy K rt ln, nn phn ng xy ra nhanh v hon ton v vy E (+)

gim nhanh, nn Epin gim nhanh v n lc no c th pin i chiu dng in.

C. BI TP THAM KHO C TNH CHT TNG HP:

V d 19:
1. Thm H2SO4 vo dung dch gm Pb(NO3)2 0,010 M v Ba(NO3)2 0,020 M cho n nng
0,130 M (coi th tch dung dich khng i khi thm axit).
Hy tnh pH v nng cc ion kim loi trong dung dch A thu c.
H = 1 atm) c nhng trong dung dch
2. a) Hy biu din s pin gm in cc hiro (p
2

CH3COOH 0,010 M ghp (qua cu mui) vi in cc Pb nhng trong dung dch A. Hy

ch r anot, catot.
b) Thm 0,0050 mol Ba(OH)2 vo 1 lt dung dch pha cc hiro (coi th tch khng thay
i). Tnh Epin v vit phng trnh phn ng xy ra khi pin hot ng.
Cho: pKa (HSO4-) 2,00 ; pKa (CH3 COOH) 4,76;
cpKs (BaSO4) 9,93 ; pKs (PbSO4) 7,66 .
o 2+
/Pb= - 0,123 V.
(RT/F) ln = 0,0592lg ; EPb

Hng dn gii:
1.

Pb(NO3)2
0,010
----Ba(NO3)2
0,020
----H2SO4
0,130
----HSO4 +
0,130
0,110
HSO4
0,110

Pb2+

2NO3

0,010
Ba2+

2NO3

0,020
H+

HSO4

0,130
2+

Ba
0,020
----Pb2+
0,010

0,130
BaSO4 +

PbSO4 +
33

H+ ;
0,130
0,150
H+ ;
0,150

107,93

105,66

C
[ ]

0,100
----0,160

Thnh phn ca h: HSO4

0,100 M
+
H
0,160 M
BaSO4 , PbSO4

HSO4
H+
+
SO42 ;
10-2
0,100
x
x
(0,100 - x)
(0,160 + x)
x
-2
x (0,160 + x)/(0,100 - x) = 10
x = [SO42] = 5,69.10-3 (M)
[ H+] = (0,160 + x) = 0,1657 (M)
pH = 0,78
2+
2
-9,93
-3
[Ba ] = K(BaSO
) /[SO4 ] = 10
/5,69.10 = 2,0.10-8 (M)
S
2
-7,66
[Pb2+] = K((PbSO
) /[SO4 ] = 10
/5,69.10-3 = 3,84.10-6 (M)
S
4

2. a) Cc Hiro: 2 H+

CH3COOH
C
0,01
[ ]
0,01 - x
2
x /(0,01 - x) = 10-4,76
E2H

/H2

+ 2e
H+

H2
+ CH3COO

; K a = 10-4,76

x
x
+
x = [H ] = 4,08.10-4 M

pH = 3,39

= - 0,0592 pH = - 0,0592.3,39 = - 0,2006 (V)

Cc Pb/PbSO4:
PbSO4 +

Pb

2e

SO42

E = EPbSO /Pb + (0,0592/2) lg(1/[SO42]).

4

o
o
/Pb = EPb2+/Pb+ (0,0592/2) lg KS(PbSO )
M EPbSO
4

= - 0,123 + (0,0592/2) lg10-7,66 = - 0,350 V

Vy E = - 0,350 + (0,0592 / 2)lg(5,69.10-3)-1 = - 0,284 V < E2H+/H2

(Cng c th tnh theo cp Pb2+/Pb:

E = - 0,123 + (0,0592/2) lg [Pb2+] = -0,123 + (0,0592/2) lg3,84.10-6
= - 0,283 V < E2H /H
.
+

Vy cc Pb l anot; cc hiro l catot.

()
b)

PbSO4, SO42BaSO4 , H SO4-

Pb

0,005
-----

CH3COO + H2O
0,010
0,010 - x
2
x /( 0,010 - x) = 10-9,24

H2 (Pt)

(CH3COO)2Ba +

2 CH3COOH + Ba(OH)2
0,010
-----

C
[ ]

CH3COOH

0,005
CH3COOH + OHx

2H2O

K b = 10-9,24

x = 10

-5,62

34

pH = 8,38

(+)

E2H

/H2

= - 0,0592 pH = - 0,0592.8,38 = - 0,496 V

(anot)
(catot)

EPbSO /Pb = - 0,284 V

Vy Epin = - 0,284 - (- 0,496) = 0,212 V.
Phn ng trong pin: anot
H2
2 H+ + 2e
2 CH3COO- + 2 H+
2 CH3COOH
2 CH3COO + H2
2 CH3COOH + 2e
catot PbSO4 + 2 e
Pb + SO42
Phn ng xy ra trong pin:
PbSO4 + H2 + 2 CH3COOPb + SO42 + 2 CH3COOH
4

3. a) Phng trnh cc phn ng xy ra trn b mt cc in cc ca b m:

Anot (cc +):
2 H2O - 4 e
O2 + 4 H+ (s oxi ho)
Catot (cc -): 2 Ni2+ + 4 e
2 Ni
(s kh)
Phng trnh ca phn ng tng cng l:
2 Ni2+ + 2 SO42- + 2H2O
2 Ni + O2 + 2 H2SO4
V d 20:
Dung dch A gm Fe(NO3)3 0,05 M; Pb(NO3)2 0,10 M; Zn(NO3)2 0,01 M.
1. Tnh pH ca dung dch A.
2. Sc kh H2S vo dung dch A n bo ho ([H2S] = 0,10 M), thu c hn hp B.
Nhng kt ta no tch ra t hn hp B?
3. Thit lp s pin bao gm in cc ch nhng trong hn hp B v in cc platin
nhng trong dung dch CH3COONH4 1 M c bo ho bi kh hiro nguyn cht p
sut 1,03 atm. Vit phn ng xy ra trn tng in cc v phn ng trong pin khi pin lm
vic.
Cho: Fe3+ + H2O FeOH2+ +

H+

lg*1 = -2,17

Pb2+ + H2O PbOH+

H+

lg*2 = -7,80

Zn2+ + H2O ZnOH+

H+

lg*3 = -8,96

E0

=
Fe3+ /Fe2+

0
0,771 V; ES/H
= 0,141 V; E 0
2S

=
Pb2+ /Pb

-0,126 V ; 25 oC: 2,303

RT
ln = 0,0592lg
F

pKS(PbS) = 26,6; pKS(ZnS) = 21,6; pKS(FeS) = 17,2. (pKS = - lgKS, vi KS l tch s tan).
pK a1(H2S) = 7,02; pK a2(H 2S) = 12,90; pK

Hng dn gii:
1.
Fe3+ + H2O FeOH2+ +

a(NH +
4)

H+
35

= 9,24; pK a(CH3COOH) = 4,76

*1 = 10-2,17

(1)

Pb2+ + H2O PbOH+ + H+ *2 = 10-7,80

(2)
2+
+
+
-8,96
Zn + H2O ZnOH
+ H
*3 = 10
(3)
+
-14
H2O OH + H
Kw = 10
(4)
So snh (1) (4): *1. CFe3+ >> *2. CPb2+ >> *3. C Zn2+ >> Kw tnh pHA theo (1):
Fe3+ + H2O FeOH2+ + H+
*1 = 10-2,17
(1)
C
0,05
[]
0,05 - x
x
x
+
[H ] = x = 0,0153 M pHA = 1,82.
0
0
2. Do E Fe3+ /Fe2+ = 0,771 V > ES/H2S = 0,141 V nn:

1/ 2Fe3+ + H2S
2Fe2+ + S + 2H+
K1 = 1021,28
0,05
0,05
0,05
2+
+

2/ Pb + H2S
PbS + 2H
K2 = 106,68
0,10
0,05
0,25
2+
+
3/ Zn + H2S ZnS + 2H
K3 = 101,68
4/ Fe2+ + H2S FeS + 2H+
K4 = 10-2,72
K3 v K4 nh, do cn phi kim tra iu kin kt ta ca ZnS v FeS:
'
'
V mi trng axit C Zn2+ = C Zn 2+ = 0,010 M; CFe2+ = C Fe2+ = CFe3+ = 0,050 M.
i vi H2S, do Ka2 << Ka1 = 10-7,02 nh kh nng phn li ca H2S trong mi
'
trng axit khng ng k, do chp nhn [H +] = CH+ = 0,25 M tnh CS2- theo cn
bng:
H2S S2- + 2H+
Ka1.Ka2 = 10-19,92
CS' 2-

= Ka1.Ka2

0,1
[H 2S]
-19,92
(0,25) 2 = 10-19,72.
[H ]2 = 10

'
'
'
'
Ta c: C Zn2+ . CS2- < KS(ZnS) ZnS khng xut hin; CFe2+ . CS2- < KS(FeS) FeS khng
tch ra.
Nh vy trong hn hp B, ngoi S, ch c PbS kt ta.

3. E PbS/Pb = E Pb2+ /Pb = E 0Pb2+ /Pb +

EPt = E 2H + /H 2 =

0,0592 K S(PbS)
0,0592
lg
lg [Pb2+] = - 0,126 +
= - 0,33 V
2
[S2- ]
2

0,0592 [H + ]2
lg
, trong [H+] c tnh nh sau:
2
p H2

CH3COONH4 NH 4 + CH3COO1
1

+
NH 4
NH3 + H
Ka = 10-9,24 (5)
CH3COO- + H2 O CH3COOH + OH- Kb = 10-9,24 (6)
Do Ka = Kb v C NH +4 CCH3COO- pH = 7,00 [H+] = 10-7
(c th tnh [H+] theo iu kin proton hoc t hp 2 cn bng (5) v (6))
36

Vy: E 2H+ /H =
2

0,0592 [H + ]2
0,0592 10-14
lg
=
lg
= -0,415 V < EPbS/Pb = - 0,33 V
2
p H2
2
1,03

in cc ch l catot, in cc platin l anot. S pin:

+
2+
2+
(-)Pt(H2)CH3COO- 1M; NH
4 1M S; PbS; H2S 1M; H 0,25M; Fe 0,05M; Zn 0,01M Pb (+)
(p = 1,03 atm)

PbS + 2H+ + 2e Pb + H2S

2H+ + 2e
H2
2xH+ + CH3COO- CH3COOH

Trn catot:
Trn anot :

H2 + 2CH3COO- 2CH3COOH + 2e
Phn ng trong pin: PbS + H2 + 2H+ + 2CH3COO- Pb + H2S +
2CH3COOH
V d 21:(Chn QT 2006).
Dung dch A c to thnh bi CoCl2 0,0100 M, NH3 0,3600 M v H2O2
3,00.103 M.
1. Tnh pH v nng ion Co2+ trong dung dch A.
2. Vit s pin v tnh sc in ng E ca pin c hnh thnh khi
ghp (qua cu mui) in cc Pt nhng trong dung dch A vi in
cc Ag nhng trong dung dch K 2CrO4 8,0.103 M c cha kt ta
Ag2CrO4.
Cho: pKa: NH4+ 9,24; HCrO4 6,50; pKs (ch s tch s tan)
Ag2CrO4: 11,89.
EO: Co3+/Co2+ 1,84V; H2O2/2OH 0,94V; Ag+/Ag 0,799V.
Log hng s to phc: Co 3+ + 6NH3

Co(NH 3)63+ ;

lg1

= 35,16
Co 2+ + 6NH3

RT
4,39 F

Co(NH 3)62+ ;

ln = 0,0592 lg
Hng dn gii:
1.

CoCl2
0,0100
-----

Co2+

0,0100
37

2Cl

lg2 =

To phc ca ion coban vi NH3

Co2+
+ 6 NH3
0,0100
0,3600
----0,3000

Co(NH3)62+

2 = 104,39

0,0100

Oxi ho Co(NH3)62+ bi H2O2.

2 Co(NH3)62+
Co(NH3)63+ + e
H2O2 + 2e
2OH

2 (0,94 E
o2 )

2 Co(NH 3)63+ + 0,0592

2OH ; K = 10

2 Co(NH3)62+ + H2O2
(1)
o

Tnh th chun E2 ca cp Co(NH3)63+/Co(NH3)62+ :

Co(NH3)63+
Co3+ + 6 NH3
Co3+ + e
Co2+
;
2+
2+
Co
+ 6 NH3 Co(NH3)6
Co(NH3)63+ + e

E1

0,0592

Co(NH3)62+

2
o
E2 = E1 +10,0592 lg

K2 = K1 11 2

;
11
0,0592
K1 = 10
;o E2 2

K2 = 10

E2 = 1,84 + 0,0592 (4,39 35,16) = 0,0184 (V)

2 (0,94 E
o2 )

2 (0,94

K = 100,0592

= 0,0184
K = 10
)

= 10 31

0,0592

2 Co(NH3)62+ + H2O2
2 Co(NH3)63+ + 2OH ;
(1)
0,0100
0,0030
0,0040
----0,0060
0,0060
Thnh phn gii hn ca h:
Co(NH3)62+
0,0040 M

Co(NH3)63+
0,0060 M

NH3
0,3000 M

K = 1031

OH
0,0060 M

Tnh pH ca dung dch:

S phn li ca cc phc cht trong dung dch khng ln v ln v
c NH3 d. Tnh pH theo cn bng:
NH3

H2O

NH4+

OH

(2)
C
[ ]
0,3000 M

0,3000
(0,3000 - x)

x (0,0060 +
x)
0,3000 - x

x
= 104,76
38

6.10-3
(6.10-3 + x)

x = 7,682.104 <<

[OH] = 6,768.103

pH = 11,83

Tnh nng ca Co2+ trong dung dch:

Kt qu tnh theo (2) cho thy [NH3] 0,3000.
Co(NH3)62+
Co2+ + 6 NH3
C
0,0040
[ ]
(0,0040 - x)
x
0,3000
x (0,3)6
= 10-4,39
0,0040 x

0,3000

10-4,39

x = [Co2+] = 2,117.10-4 <<

V vy vic coi [NH3] 0,3000 l ng.

2. Tnh Epin
E ca in cc Pt:
[Co(NH3)63+] Co(NH
= C ) 3+
= 0,0060 M (v 1 >> ; c d NH3)
3 6

[Co(NH3)62+] = 4.10-3 2,117.10-4 = 3,788.10-3 (M)

6.10-3+ 0,0592 lg
E Co(NH
= E) +/ Co(NH ) 2+
= 0,0184
3
3 6

Pt

3 6

3,788.10-

0,0320 (V)

Tnh E ca in cc Ag: Th ca in cc Ag do cp
Ag2CrO4/2Ag quyt nh (hoc Ag+/Ag).
Ag2CrO4 + 2e
2Ag
+
CrO42
o
Ag
EAg = E2CrO4/2Ag
o

Tnh E4 :

+
lg
CrO42
0,0592
2

Ag2CrO4
2

2Ag+

Ag+ + 2e

CrO422Eo3 ; Ks = 10-11,89
0,0592

Ag

2Eo

K 32 = 10

0,0592

(E3 = 0,799 V)
Ag2CrO4 + 2e
K4 = K32. Ks

2Ag

0,0592

E4 = E3 2+

CrO42 ; K4 = 10

lg Ks = 0,447 (V)

Tnh nng CrO42:

Co
C

CrO42
+
-3
8.10
8.10-3 - x
x

H2O

HCrO4
x

-3

8.10 - x

Ag2CrO4

= 10-7,5

OH ;

Kb = 10-7,5

x
x = 1,6.10-5 << 8.10-3

2Ag+

C
39

CrO42
8.10 -3

Ks = 10-11,89

[]

2x
8.10-3 + x
(2x)2 (8.10-3 + x) = 10-11,89
x = 6,3.10-6 << 8.10-3
[CrO42] = 8.10-3 M
1

EAg = 0,447
+
lg
2
0,0592 CrO4
o
2

/Ag= E
(C th tnh theoAgEAg
+

= 0,5090 (V)

+ 0,0592 lg [Ag+])

EAg > EPt

Ag l catot, Pt l anot.

S pin:
(a)
Pt Co(NH3)62+ , Co(NH3)63+ , NH3

Ag2CrO4 , CrO42

Ag

(c)
Epin = Ec Ea = 0,5090 0,0302 = 0,479 (V)
V d 22: (Chn QT 2007) Dung dch A gm FeSO4 0,020 M; Fe2(SO4)3 v H2SO4.
a) Ly chnh xc 25,00 ml dung dch A, kh Fe3+ thnh Fe2+; chun Fe2+ trong hn hp
( iu kin thch hp) ht 11,78 ml K2Cr2O7 0,0180 M. Hy vit phng trnh ion ca
phn ng chun . Tnh nng M ca Fe2(SO4)3 trong dung dch A.
b) Tnh nng M ca H2SO4 trong dung dch A, bit dung dch ny c pH = 1,07.
c) Ghp cc Pt nhng trong dung dch A (qua cu mui) vi cc Ag nhng trong dung
dch AgNO3 0,0190 M c thm K2CrO4 cho n nng 0,0100 M (coi th tch c gi
nguyn).
Hy cho bit anot, catot v vit phng trnh phn ng xy ra khi pin hot ng. Tnh
sc in ng ca pin.
Cho pKa: HSO4- 1,99; Fe3+( Fe3+ + H2O FeOH2+ + H+) 2,17;
Fe2+( Fe2+ + H2O FeOH+ + H+) 5,69.
Ch s tch s tan pKs ca Ag2CrO4 11,89.
Eo : Fe3+/ Fe2+ 0,771 V; Ag+/Ag 0,799 V; (RT/F)ln = 0,0592 lg.
Hng dn gii:
2

3. a) Phn ng c/ Cr2O 7 + 6 Fe2+ + 14 H+ 2 Cr3+ + 6 Fe3+ + 7 H2O

CFe

= CFeSO 4 + 2 CFe 2 (SO 4 ) 3 = 0,02 + 2C1

2

CFe . 25,00 = 6 (CCr 2 O 7 . VCr 2 O 7 ) 25,00(0,020 + 2C1) = 6(0,0180 . 11,78)

40

C1 = 0,01544 M

hay

CFe2(SO4)3 = 0,01544 M.

b) Trong dd A c: Fe2+ 0,020 M; Fe3+ 2C1; H+ (C, M); HSO 4 (C, M); cc cn bng:
H3O+ + OH-

2 H2O
Fe2+ + 2 H2O

Kw = 10-14

FeOH+ + H3O+

Ka1 = 10-5,96

(1)
(2)

Fe3+ + 2 H2O

FeOH2+ + H3O+

Ka2 = 10-2,17

(3)

HSO 4 + H2O

SO 24

Ka = 10-1,99

(4).

+ H3O+

So snh ta thy (3) v (4) l ch yu v tng ng nhau. p dng /lut bo ton

2

proton, ta c [H3O+] = CH + [FeOH2+] + [SO 4 ]

(a)

T (3) c [FeOH2+] / [Fe3+] = Ka2 / [H3O+] [FeOH2+] / CFe

= Ka2 / Ka2 + [H3O+]

= 10-2,17 / (10-2,17 + 10-1,07) [FeOH2+] = 0,0736 CFe = 0,0736 . 0,015445 . 2.

T/ t, t (4) c [SO 24 ] / [HSO 4 ] = Ka / [H3O+]
[SO 24 ] / CHSO 4 = 10-1,99/ (10-1,99 + 10-1,07) [SO 24 ] = 0,107 C;
3

P/ trnh (a) tr thnh [H3O+] = C + 0,0736 CFe + 0,107 C

(b).

T (b) CH 2 SO 4 = C = (10-1,07 0,0736 . 0,03089) / 1,107 CH 2 SO 4 = C = 0,07483 M.

c) EPt = E Fe 3 / Fe 2 = E 0

Fe

/ Fe

+ 0,0592 lg([Fe3+]/[Fe2+])

Fe3+ + 2 H2O
C
[

FeOH2+ + H3O+

10-2,17

0,03089
]

0,03089 x

x .10-1,07 / (0,03089 x) = 10-1,07

= 0,02862 M [Fe2+] = CFe

10-1,07

x = 0,002273 [Fe3+] = 0,03089 0,002273

= 0,020 M (v Ka1 rt b).

Vy: EPt = 0,771 + 0,0592 lg ( 0,02862 / 0,020) = 0,780 V.

2 Ag+

CrO 24

0,019

0,010

5. 10-4
Ag2CrO4

Ag2CrO4

2 Ag+

CrO 24
5.10-4

C
[ ]

2x
41

5.10-4 + x

Ks = 10-11,89

( 2x )2 (5.10-4 + x) = 10-11,89 4x3 + 2,0.10-3x2 - 10-11,89 = 0 x = 2,08.10-5

[Ag+] = 2x = 4,96.10-5 M.

C:
o

EAg = E Ag

/ Ag

+ 0,0592 lg [Ag+] = 0,799 + 0,0592 lg4,96.10-5 = 0,544 V.

V EAg < EPt nn cc Ag l anot; cc Pt catot.

2 Ag

CrO 24 Ag2CrO4 + 2e

catot 2x Fe3+

Phn ng trong pin: anot

2 Ag

+ CrO 24 + 2Fe3+

Fe2+

Ag2CrO4 + 2 Fe2+

Epin = EPt - EAg = 0,780 0544 = 0,236 V.

V d 23: (Chn QT 2008)
1. Tnh Eo . Thit lp s pin v vit phng trnh phn ng xy ra trong pin c ghp bi
cp CrO42-/CrO2 v NO3-/NO iu kin tiu chun.
Cho: Cr3+ + H2O CrOH2+ +

H+

Fe2+ + H2O FeOH+ +

Cr(OH)3 Cr3+

H+

3 OH

1= 10-3,8

2 = 10-5,92

KS = 10-29,8

Cr(OH)3 H+ + CrO2- + H2O

K = 10-14

H2O H+ + OH- Kw =10-14

E0CrO42-/Cr(OH)3, OH- = -0,13 V ; E0NO3-, H+/NO = 0,96 V.
2. xc nh hng s to phc (hay hng s bn) ca ion phc [Zn(CN) 4]2-, ngi ta lm
nh sau:
Thm 99,9 ml dung dch KCN 1M vo 0,1 ml dung dch ZnCl 2 0,1 M thu c
100ml dung dch ion phc [Zn(CN)4]2- (dung dch A). Nhng vo A hai in cc: in cc
km tinh khit v in cc so snh l in cc calomen bo ho c th khng i l 0,247
V (in cc calomen trong trng hp ny l cc dng). Ni hai in cc vi mt
in th k, o hiu in th gia chng c gi tr 1,6883 V.
Hy xc nh hng s to phc ca ion phc [Zn(CN) 4]2-. Bit th oxi ho - kh tiu
chun ca cp Zn2+/Zn bng -0,7628 V.
Hng dn gii:
c. Tnh E0

(E0x)
42

CrO42- + 4 H2O + 3 e Cr(OH)3 +

Cr(OH)3 H+ + CrO2 + H2O
H+ + OH H2O

5 OH

K1 = 10
K2 = 10-14
Kw-1 =1014

CrO42- + 2 H2O + 3 e CrO2- + 4 OHK3 = K1. K2. Kw-1

E0

= -0,13 V < E0

K3 =10
= 0,96 V

Vy cp CrO42-/CrO2- l anot v cp NO3-/NO l catot. S pin:

(-) Pt CrO42- 1M ; CrO2- 1M ; OH- 1M

NO3- 1M ; H+ 1M (Pt) NO, pNO = 1atm (+)

Phn ng xy ra trong pin:

Ti catot:

NO3- + 4 H+ + 3 e NO + 2 H2O

Ti anot:

CrO2- + 4 OH4

CrO42- + 2 H2O + 3e

HOH

NO3- + CrO2-

H+ + OHCrO42- + NO

2. Phn ng to phc:
Zn2+ + 4 CN [[Zn(CN)4]2-]
Theo bi, rt d CN- nn s to phc xy ra hon ton, tnh nng ion Zn2+ cn bng
to phc theo nng ion phc, phi t v lin quan vi hng s to phc. T biu thc:
E (o c) = 1,6883 = ECal EZn = 0,247 EZn
EZn = 0,247 1,6883 = 1,4413 V
EZn = 0,7628 + (0,0592/2)lg [Zn2+] = 1,4413 (V)
=> tnh c = 1018,92
CCr3
*
+1

V d 24: (Chn QT 2009).

1 + hCho: E 0 = 0,80V; E 0
= -0,15V; E 0Au
Ag Ag
AgI/Ag,I
+

3+

/Ag

= 1,26V; E 0Fe3+ /Fe = -0,037V; E 0Fe2+ /Fe = -0,440V

Eo(MnO4-/Mn2+) = 1,51V; Eo(MnO4-/MnO2) = 1,70V

Hy:
1. a. Thit lp mt s pin xc nh tch s tan ca AgI. Vit cc phng trnh
phn ng xy ra trn mi in cc v trong pin.
b. Tnh tan (s) ti 25oC ca AgI trong nc.
2. a. Lp pin in trong xy ra s oxi ho ion Fe 2+ thnh ion Fe3+ v ion Au3+ b
kh thnh ion Au+. Vit cc phng trnh phn ng xy ra trn mi in cc v trong pin.
b. Tnh sc in ng chun ca pin v hng s cn bng ca phn ng xy ra trong pin ny.
43

3. Lp pin in trong c in cc Pt nhng vo dung dch thu c t 2,24 lt kh H 2S

( ktc) li vo 100 ml dung dch FeCl 3 2M in cc Pt nhng vo dung dch c MnO 4- 1M
(pH = 0). Vit cc phng trnh trn cc in cc v phn ng xy ra trong pin. Tnh Go
Hng dn gii:
1. a. xc nh tch s tan K S ca AgI, cn thit lp s pin c cc in cc Ag
lm vic thun nghch vi Ag+. in cc Ag nhng trong dung dch no c [Ag +] ln hn
s ng vai tr catot. Vy s pin nh sau:
(-) Ag I-(aq), AgI(r) Ag+(aq) Ag(r) (+)
Hoc:

(-) Ag, AgI(r) I-(aq) Ag+(aq) Ag(r) (+)

Phn ng cc m:

Ag(r) + I(aq)

AgI(r) + e

K 11

Phn ng cc dng:

Ag+(aq) + e

Ag(r)

K2

Phn ng xy ra trong pin: Ag+(aq) + I-(aq)

AgI(r)

Trong K S-1 = K 11 .K2 = 10( EAg+ /Ag -EAgI/Ag,I- ) / 0,059 1,0.1016

0

K S-1

(1)

KS = 1,0.1016.

b. Gi S l tan ca AgI trong nc nguyn cht, ta c:

AgI

Ag+ + IS

KS = 10-16

V qu trnh to phc hidroxo ca Ag+ khng ng k, I- l anion ca axit mnh HI, nn

S=

KS =1,0.10-8 M

2. Theo qui c: qu trnh oxi ha Fe2+ xy ra trn anot, qu trnh kh Au3+ xy ra trn catot,
do in cc Pt nhng trong dung dch Fe3+, Fe2+ l anot, in cc Pt nhng trong dung dch
Au3+, Au+ l catot:
(-)

Pt Fe3+(aq), Fe2+(aq) Au3+(aq), Au+(aq) Pt

Phn ng cc m:

2x

Phn ng cc dng:
Phn ng trong pin:

(+)

Fe2+(aq)

Fe3+(aq) + e

K 11

Au3+(aq) + 2e

Au+(aq)

K2

Au+(aq) + 2Fe3+(aq)

Au3+(aq) + 2Fe2+(aq)

(2)

K = (K 11 )2.K2 = 102( EAu3+ /Au -EFe3+ /Fe2+ ) / 0,059

0

Trong th kh chun ca cp Fe3+/Fe2+ c tnh (hoc tnh theo hng s cn bng) nh

sau:
44

Fe3+ + 3e

Fe

E0(1) = -0,037 V,

G0(1) = -3FE0(1)

Fe2+ + 2e

Fe

E0(2) = -0,440 V,

G0(2) = - 2F E0(1)

Fe3+ + e

Fe2+ E0(3) =

0
0
-G 0 (3)
= G (1) - G (2) = 3E0(1)- 2E0(2) = 0,77V
F
F

K = (K 11 )2.K2 = 102(1,260,77) / 0,059 = 1016,61

iu kin tiu chun, sc in ng chun ca pin trn s l:
0
0
E0pin = E Au3+ /Ag+ - E Fe3+ /Fe2+ = 0,49 V

3. Lp pin in trong c in cc Pt nhng vo dung dch thu c t 2,24 lt kh H 2S (

ktc) li vo 100 ml dung dch FeCl3 2M in cc Pt nhng vo dung dch c MnO4- 1M (pH =
0). Vit cc phng trnh trn cc in cc v phn ng xy ra trong pin. Tnh Go
Hng dn gii:
pH = 0, th kh chun Eo(MnO4-/Mn2+) =1,51V> Eo(Fe3+/Fe2+) = 0,771V; nn c s pin l
(-) Pt Fe3+ 1M; Fe2+ 1M MnO4- 1M; H+ 1M Pt (+)
Bn phn ng cc m:

Fe2+ Fe3+ + e

Bn phn ng cc dng:

MnO4- + 5e + 8H+ Mn2+ + 4H2O

=> Phn ng xy ra khi pin hot ng:

5Fe2+ + MnO4- + 8H+ 3Fe3+ + Mn2+ + 4H2O

K = 105E = 105(1,51-0,771)/0,0592 = 1062,42

=> Go = -nFEo = -5.96500.(1,51-0,771) = -356,568kJ

45

mc lc
Tr

Phn I: Tng quan

I- Tm quan trng, vai tr ca phn ng trong pin

in. .............
II. Tnh hnh thc t ni dung kin thc v pin in trong cc ti liu hin
hnh.

III-

Mc

tiu

thc

hin. ................................................................
VI- Vai tr ca bi tp trong vic bi dng HSG Quc

ti. ....................................................................
IVPhng
php

ca

gia. ........

Phn II: NI DUNG

A. C S L THUYT:
I. IN CC V TH IN CC:
1. Quy c v th in cc:
2. Phng trnh tnh Eox/kh:
II. CC LOI IN CC V TH IN CC:
1. in cc loi I:
2. in cc loi II:
3. in cc oxi ho - kh (in cc Redox):
4. in cc kh:
III. CC LOI PIN IN:
III.1- Pin khng ni lng:
III.2- Pin ni lng:
1. Pin nng :
2. Pin in ch c cc phn ng oxi ha kh:
3. Pin in c cc phn ng ph:
B. BI TP VN DNG:
I. Vit s pin, cc qu trnh, phng trnh phn ng khi pin hot
46

4
5
5
5
5
5
5
5
7
8
10
10
14
14
15
15
17
17

ng, tnh sut in ng ca pin:

II. Tnh hng s v s thay i sut in ng ca pin.

25

1. V d minh ha:

25

2. Bi tp vn dng c tnh cht tng hp:

C. BI TP THAM KHO C TNH CHT TNG HP:

26
33

Bc Ninh, ngy 10 thng 09 nm 2012

Ni tip nhn kin
/c: Vng B Huy, t trng t Ha trng THPT Chuyn Bc Ninh
Email: huyvuongba@gmail.com

47