PP-Giai-BT-DienPhan-HoChiTuan-DHY-HN

8/6/2019 PP-Giai-BT-DienPhan-HoChiTuan-DHY-HN

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http://ebook.here.vn - Th vin Bi ging, thi trc nghim

Phng php gii bi tp v in phn

Bin son H Ch Tun - H Y H Ni

I NHC LI L THUYT

1) in phn cht in li nng chy: p dng i vi MCln, M(OH)n v Al2O3 (M l kim loinhm IA v IIA)2) in phn dung dch cht in li trong nc:- Vai tr ca nc: trc ht l dung mi ha tan cc cht in phn, sau c th tham gia trctip vo qu trnh in phn:+ Ti catot (cc m) H2O b kh: 2H2O + 2e H2 + 2OH

+ Ti anot (cc dng) H2O b oxi ha: 2H2O O2 + 4H

+ + 4e- Ti catot (cc m) xy ra qu trnh kh M+, H+ (axit), H2O theo quy tc:+ Cc cation nhm IA, IIA, Al3+ khng b kh (khi H2O b kh)+ Cc ion H+ (axit) v cation kim loi khc b kh theo th t trong dy th in cc chun (ion ctnh oxi ha mnh hn b kh trc): Mn+ + ne M+ Cc ion H+ (axit) d b kh hn cc ion H+ (H2O)

+ V d khi in phn dung dch hn hp cha FeCl3, CuCl2 v HCl th th t cc ion b kh l:Fe3+ + 1e Fe2+ ; Cu2+ + 2e Cu ; 2H+ + 2e H2 ; Fe

2+ + 2e Fe- Ti anot (cc dng) xy ra qu trnh oxi ha anion gc axit, OH(baz kim), H2O theo quy tc:+ Cc anion gc axit c oxi nh NO3, SO4

2, PO43, CO3

2, ClO4khng b oxi ha+ Cc trng hp khc b oxi ha theo th t: S2> I> Br> Cl> RCOO> OH> H2O3) nh lut Faraday

m =

Trong :+ m: khi lng cht gii phng in cc (gam)+ A: khi lng mol ca cht thu c in cc+ n: s electron trao i in cc+ I: cng dng in (A)+ t: thi gian in phn (s)+ F: hng s Faraday l in tch ca 1 mol electron hay in lng cn thit 1 mol electron

chuyn di trong mch catot hoc anot (F = 1,602.10-19

.6,022.1023

96500 C.mol-1

)

II MT S C S GII BI TP V IN PHN

- Khi lng catot tng chnh l khi lng kim loi to thnh sau in phn bm vo- m (dung dch sau in phn) = m (dung dch trc in phn) (m kt ta + m kh)- gim khi lng ca dung dch: m = (m kt ta + m kh)- Khi in phn cc dung dch:

+ Hiroxit ca kim loi hot ng ha hc mnh (KOH, NaOH, Ba(OH)2,)+ Axit c oxi (HNO3, H2SO4, HClO4,)


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+ Mui to bi axit c oxi v baz kim (KNO3, Na2SO4,) Thc t l in phn H2O cho H2 ( catot) v O2 ( anot)- Khi in phn dung dch vi anot l mt kim loi khng tr (khng phi Pt hay in cc than ch)th ti anot ch xy ra qu trnh oxi ha in cc- C th c cc phn ng ph xy ra gia tng cp: cht to thnh in cc, cht tan trong dungdch, cht dng lm in cc. V d:

+ in phn nng chy Al2O3 (c Na3AlF6) vi anot lm bng than ch th in cc b n mn

dn do chng chy trong oxi mi sinh+ in phn dung dch NaCl khng mng ngn to ra nc Giaven v c kh H 2 thot ra catot

+ Phn ng gia axit trong dung dch vi kim loi bm trn catot- Vit phn ng (thu hoc nhng electron) xy ra cc in cc theo ng th t, khng cn vit

phng trnh in phn tng qut- Vit phng trnh in phn tng qut (nh nhng phng trnh ha hc thng thng) tnhton khi cn thit

- T cng thc Faraday s mol cht thu c in cc- Nu bi cho I v t th trc ht tnh s mol electron trao i tng in cc (ne) theo cng

thc: ne = (*) (vi F = 96500 khi t = giy v F = 26,8 khi t = gi). Sau da vo th t inphn, so snh tng s mol electron nhng hoc nhn vi ne bit mc in phn xy ra. V d d on xem cation kim loi c b kh ht khng hay nc c b in phn khng v H 2O c bin phn th in cc no- Nu bi cho lng kh thot ra in cc hoc s thay i v khi lng dung dch, khilng in cc, pH,th da vo cc bn phn ng tnh s mol electron thu hoc nhng miin cc ri thay vo cng thc (*) tnh I hoc t- Nu bi yu cu tnh in lng cn cho qu trnh in phn th p dng cng thc: Q = I.t =ne.F- C th tnh thi gian t cn in phn ht mt lng ion m bi cho ri so snh vi thi giant trong bi. Nu t < t th lng ion b in phn ht cn nu t > t th lng ion cha b

in phn ht- Khi in phn cc dung dch trong cc bnh in phn mc ni tip th cng dng in v thigian in phn mi bnh l nh nhau s thu hoc nhng electron cc in cc cng tn phinh nhau v cc cht sinh ra cc in cc cng tn t l mol vi nhau- Trong nhiu trng hp c th dng nh lut bo ton mol electron (s mol electron thu c catot = s mol electron nhng anot) gii cho nhanh

III MT S V D MINH HA

V d 1: in phn ha ton 2,22 gam mui clorua kim loi trng thi nng chy thu c 448 mlkh ( ktc) anot. Kim loi trong mui l:A. Na B. Ca C. K D. Mg

Hng dn: nCl2 = 0,02Ti catot: Mn+ + ne M Theo lbt khi lng mM = m(mui) m(Cl2) = 2,22 0,02.71 = 0,8 gam

Ti anot: 2Cl Cl2 + 2e Theo lbt mol electron ta c nM = M = 20.n n = 2 v M l Ca

(hoc c th vit phng trnh in phn MCln M + n/2Cl2 tnh) p n B

V d 2: Tin hnh in phn (vi in cc Pt) 200 gam dung dch NaOH 10 % n khi dung dchNaOH trong bnh c nng 25 % th ngng in phn. Th tch kh ( ktc) thot ra anot v


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catot ln lt l:A. 149,3 lt v 74,7 lt B. 156,8 lt v 78,4 ltC. 78,4 lt v 156,8 lt D. 74,7 lt v 149,3 lt

Hng dn: mNaOH (trc in phn) = 20 gamin phn dung dch NaOH thc cht l in phn nc: H2O 1/2 O2 (anot) + H2 (catot)

NaOH khng i m (dung dch sau in phn) = 80 gam m (H2O b in phn) = 200 80 =

120 gam nH2O = 20/3 mol VO = 74,7 lt v VH = 149,3 lt p n D

V d 3: Sau mt thi gian in phn 200 ml dung dch CuSO 4 ( d = 1,25 g/ml) vi in cc graphit(than ch) thy khi lng dung dch gim 8 gam. lm kt ta ht ion Cu 2+ cn li trong dungdch sau in phn cn dng 100 ml dung dch H2S 0,5 M. Nng phn trm ca dung dchCuSO4 ban u l:A. 12,8 % B. 9,6 % C. 10,6 % D. 11,8 %

Hng dn: nH2S = 0,05 mol- Gi x l s mol CuSO4 tham gia qu trnh in phn: CuSO4 + H2O Cu + 1/2O2 + H2SO4 (1) m (dung dch gim) = m Cu(catot) + m O2(anot) = 64x + 16x = 8 x = 0,1 mol - CuSO4 + H2S

CuS + H2SO4 (2) nH2S = nCuSO4 = 0,05 mol

- T (1) v (2) nCuSO4 (ban u) = 0,1 + 0,05 = 0,15 (mol) C% = p n B

V d 4: in phn 100 ml dung dch CuSO4 0,2 M vi cng dng in 9,65A. Tnh khilng Cu bm vo catot khi thi gian in phn t1 = 200 s v t2 = 500 s. Bit hiu sut in phn l100 %A. 0,32 gam v 0,64 gam B. 0,64 gam v 1,28 gamC. 0,64 gam v 1,60 gam D. 0,64 gam v 1,32 gam

Hng dn: nCuSO4 = 0,02 = nCu2+

Thi gian cn thit in phn ht Cu2+ l t = s t1 < t < t2 Ti t1 c 1/2s mol Cu2+ b in phn m1 = 0,01.64 = 0,64 gam v ti t2 Cu

2+ b in phn ht m2 =1,28 gam p n B

V d 5: in phn 200 ml dung dch CuSO4 vi in cc tr v cng dng in 1A. Khi thy catot bt u c bt kh thot ra th dng in phn. trung ha dung dch thu c sau khi in

phn cn dng 100 ml dung dch NaOH 0,1M. Thi gian in phn v nng mol ca dung dchCuSO4ban u l:A. 965 s v 0,025 M B. 1930 s v 0,05 M

C. 965 s v 0,05 M D. 1930 s v 0,025 M

Hng dn: nNaOH = 0,01 mol- Khi catot bt u c bt kh (H2) thot ra chng t CuSO4 b in phn ht theo phng trnh:CuSO4 + H2O Cu + 1/2O2 + H2SO4- nNaOH = nOH= 0,01 mol nH2SO4 = 0,5.nH

+ = 0,5.nOH= 0,005 (mol) nCu = nCuSO4 =

0,005 (mol) = 0,005 t = 965 s v CM(CuSO ) = M (hoc c th davo cc phn ng thu hoc nhng electron in cc tnh) p n A


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V d 6: in phn 200 ml dung dch hn hp AgNO3 0,1 M v Cu(NO3)2 0,2 M vi in cc trv cng dng in bng 5A. Sau 19 pht 18 giy dng in phn, ly catot sy kh thy tng mgam. Gi tr ca m l:A. 5,16 gam B. 1,72 gam C. 2,58 gam D. 3,44 gam

Hng dn: nAg+ = 0,02 mol ; nCu2+ = 0,04 mol

- Ta c ne = mol- Th t cc ion b kh ti catot:Ag+ + 1e Ag (1) sau (1) cn 0,06 0,02 = 0,04 mol electron0,02 0,02 0,02Cu2+ + 2e Cu (2) sau (2) cn d 0,02 mol Cu2+0,02 0,04 0,02m (catot tng) = m (kim loi bm vo) = 0,02.(108 + 64) = 3,44 gam p n D

V d 7: Ha tan 50 gam tinh th CuSO4.5H2O vo 200 ml dung dch HCl 0,6 M thu c dungdch X. em in phn dung dch X (cc in cc tr) vi cng dng in 1,34A trong 4 gi.Khi lng kim loi thot ra catot v th tch kh thot ra anot ( ktc) ln lt l (Bit hiusut in phn l 100 %):A. 6,4 gam v 1,792 lt B. 10,8 gam v 1,344 ltC. 6,4 gam v 2,016 lt D. 9,6 gam v 1,792 lt

Hng dn: nCuSO4.5H2O = nCuSO4 = 0,2 mol ; nHCl = 0,12 mol

- Ta c ne = mol- Th t in phn ti catot v anot l:Ti catot: Cu2+ + 2e Cu Cu2+ cha b in phn ht m (kim loi catot) = 0,1.64 = 6,4gam

0,1 0,2 0,1Ti anot:

2Cl Cl2 + 2e ne (do Clnhng) = 0,12 < 0,2 mol ti anot Cl b in phn ht v0,12 0,06 0,12 n nc b in phn ne (do H2O nhng) = 0,2 0,12 = 0,08 mol2H2O O2 + 4H

+ + 4e0,02 0,08

V (kh thot ra anot) = (0,06 + 0,02).22,4 = 1,792 lt p n A

V d 8: C 200 ml dung dch hn hp Cu(NO3)2 v AgNO3, in phn ht ion kim loi trongdung dch cn dng cng dng in 0,402A trong 4 gi. Sau khi in phn xong thy c 3,44gam kim loi bm catot. Nng mol ca Cu(NO3)2 v AgNO3 trong hn hp u ln lt l:A. 0,2 M v 0,1 M B. 0,1 M v 0,2 MC. 0,2 M v 0,2 M D. 0,1 M v 0,1 M

Hng dn:

- Ta c ne = mol- Ti catot: Ag+ + 1e Ag Ta c h phng trnh:

x x (mol)Cu2+ + 2e Cu CM Cu(NO3)2 = CM AgNO3 = 0,1 M p n D

y y (mol)


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V d 9: Ha tan 4,5 gam tinh th MSO4.5H2O vo nc c dung dch X. in phn dung dch Xvi in cc tr v cng dng in 1,93A. Nu thi gian in phn l t (s) th thu c kim loiM catot v 156,8 ml kh ti anot. Nu thi gian in phn l 2t (s) th thu c 537,6 ml kh . Bitth tch cc kh o ktc. Kim loi M v thi gian t ln lt l:A. Ni v 1400 s B. Cu v 2800 sC. Ni v 2800 s D. Cu v 1400 s

Hng dn: Gi nMSO4 = nM2+

= x mol

V d 10: Mc ni tip hai bnh in phn: bnh (1) cha dung dch MCl 2 v bnh (2) cha dungdch AgNO3. Sau 3 pht 13 giy th catot bnh (1) thu c 1,6 gam kim loi cn catot bnh (2)thu c 5,4 gam kim loi. C hai bnh u khng thy kh catot thot ra. Kim loi M l:

A. Zn B. Cu C. Ni D. PbHng dn: - Do hai bnh mc ni tip nn ta c:

Q = I.t = M = 64 Cu p n B

V d 11: in phn nng chy Al2O3 vi anot than ch (hiu sut in phn 100 %) thu c m kgAl catot v 67,2 m3 ( ktc) hn hp kh X c t khi so vi hiro bng 16. Ly 2,24 lt ( ktc)hn hp kh X sc vo dung dch nc vi trong (d) thu c 2 gam kt ta. Gi tr ca m l:A. 54,0 kg B. 75,6 kg C. 67,5 kg D. 108,0 kg

Hng dn: 2Al2O3 4Al + 3O2 (1) ; C + O2 CO2 (2) ; 2C + O2 2CO (3)- Do X = 32 hn hp X c CO2 ; CO (x mol) v O2 d (y mol)- 2,24 lt X + Ca(OH)2 d 0,02 mol kt ta = nCO2 trong 67,2 m

3 X c 0,6 CO2

- Ta c h phng trnh: v 0,6 + x + y = 3 x = 1,8 v y = 0,6

T (1) ; (2) ; (3) mAl = kg p n B

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PP-Giai-BT-DienPhan-HoChiTuan-DHY-HN