Sử dụng bài tập hoá học phần kim loại để bồi dưỡng năng lực tự học cho học sinh trung học phổ thông

sit DUNyJU TAP HOA HOG PHI N KIM lOlU fl( BOI Dli 6 NANG UlC 111 HOC

CHO HOC SINH TRUNO HOC PHO THONG

Trong qud trinh dgy hgc Hod hgc, vlfc si>

dyng cdc bdi tfp JBT) s8 mang Igi hlfu qud

coo, hgc sinh (HS) ttip thu kiln ihik mft

cdch nhanh chdng, hung thO hgc tdp, ghi nhd nfi

dung bdi hgc Iflu non,... Tuy nhlftn, trong thyc t l

vlfc s\t dyng bdi tgp (BT) Hod hgc d l bdi dudng

ndng Iyc ty hgc (NLTH) cho HS cdn hgn chl. Phdn

Idn gido vidn (GV) Idn kSp chO ylu vdi phuong

phdp thuylt h-lnh, rdt it ngudi si> dyng BT Hod

ngc hodc nlu cd sid dyng thi vdn chiro mang Knh

hf thdng. Do dd, mdc dO tdn rdt nhilu ihdl gion

d trdn k^ nhung hifu qud cOo vlfc hhh hfl kidn

thijc cuo HS chua dgt nhu mong mudn. Ngodi

ra, qud trinh td chuc thudng xuydn v l thilt kd,

xdy dyng cdc logl BT Hod hgc trong td chuydn

mdn Widng lidn tyc vd mang Hnh cdp nhdt. VI

v ^ , mft h-ong nhffng phirang tifn td chi>c hogt

dfng tv Iyc hgc tdp, phdt huy h'nh Kch eye cOa HS

Id sd dyng ngudn BT d l giup HS thyc hlfn tdt vlfc

tg hgc, tv dgc, tg kilm tra ddnh gid kdt qud hgc

Idp cua m)nh. Sd dgng BTnhv ihi ndo dibit

dudng NLTH cho HS9 Bdn thdn mft BT hod hgc

chua cd tdc dyng gl cd: Khdng phdi mft BT hod

hgc •ha/ thi ludn cd tdc dyng Kch cycl Vdn d l

phy thufc chu ylu Id «ngudl si) dyng nd'. Ldm

thi ndo d l bilt h-oo dOng ddi tugng, phdi bidt

cdch khoi thdc hlft d l mgi khfa cgnh cOa bdi

todn, giOp HS ty m)nh Hm ra cddi gldl, lOc dd BT

hod hgc thdt sy cd ^ nghTo. Vdl hudng Hdp cdn

trdn, ch6ng tdi xin don cO* vifc si> dyng BT Hod

hgc {Phdn Kim ^ / - Hod hgc 12) d l bdi dudng

NLTH cho HS THPT h-ong mdt sd hudng hgp sou:

1 . BT rdn luyf n m^t s^ ndng Iyc tv hgc

1) Rdn luy^ ndng Igc lu duy dde Idp. Trong

qiid trinh dgy hgc cdn phdi rdn cho HS thdi quen

suy nghT dfc Ifp, ddy Id mft trong nhOng yni td

Kin d l cho sdng Igo cua qud trinh tg Iyc hgc tdp.

Cdc BT bdi dudng NLTH phdi hudng HS vdo

nhi>ng suy nghT dfc Ifp, giOp cdc em Hm tdi,

phdt hlfn ra kiln thik. Vidvl: Hda ton 50 gam

Hnh Ihl CuSO^.SHjO vdo 200 ml dung djch HCl

# .

O niS. NGUYEN THI VAN -

0,d M thu clu9c dung dich X. Oem difn phdn dung

d|ch X (cdc difn cvc trcr) vdl cudng d f ddng difn

1,34A Irong 4 gid. Tlnh UnSl \uqng ldm logl thodt

ra d cotot vd th< Hch khi thodi ra d onot (dldc)?

|blA hlfu sueft difn phdn Id 100%|.

M ldm duqc BT ndy, HS phdt dva vdo cfng

thuc Faraday dtf tfnh Iu9ng chdt thu du^c d cotdt.

Cd th^ sd dung phuong pndp bdo todn electron

vd n6m vOng thd ti/ difn phdn d ode difn OfC di

xdc dinh lu^ng Uif thu du ^ d andt.

"C.SO, s v - "cso, - O'^nwl; n ^ - 0,12 mol.

Tacd: n.=—=i2:ii=0,2niol • F 26,8

Thd tl/ difn phdn tgi cotot vd onot Id: - Cotot:

Cu'* + 2e-»Cu

0,1 0,2 0,1

-» Cu^* chua bi difn phdn hft -> m ^tgts,^.)"

0,l.d4-d,4gam;-Anot:

2Cl- -> Ojt 2e

0,12 0,0d 0,12

-> n, (do Ch nhudngi - 0,12 < 0,2 mol tgl

onot Ch da bi difn phdn hft vd djn nudc bl difn

1*6" ".1*,»*-^-0. 2 - 0.12-0.08 ""I

2Hp-»0, + 4H* + 4e

0,02 0,08

V w*«~»™, = (0.06 + 0,02).22,4 - 1,792 lit

21 Rin luyin ndng h/c phdt hlin y^tdivd gldl

quy^ vdn di. Ddy id mft trong nhOng ndng luc

c6n thift phdt rfn luyfn cho HS vl nd cd tdc dung

rfl Idn ktch thich hogt dfng tu duy sdng tgo cuo

HS. Khi gldl <fiySl v6n di , ddi hdi HS phdi huy

dfng cdc ki ^ thdc, Idl llfu thu thf p, xsm x4t ddnh

gid thdng tin mft cdch chinh xdc, khoa hgc qua

dd giOp HS ndm ch& kiln thdc. Dfng ihdl hinh

Ihdnh cho HS Id khd ndng nhfn IhOc v ^ d l nhanh,

cd hung ihu nhdt Id trong nhDng v^n 66 mdt.

Khi xdy dl/ng hf thing BT hod hgc, GV cf n

chgn nhOng BT df t HS vdo trgng thdi phdi suy

•liiiiiiiGiio die HHMII nyti Ha S« - TM l M

Tw Chi aim due $6 286 m a. g/aom