guía Electroquímica

Vista previa del material en texto

DUOS He
de la celda
calculamos k de KCI
[K(1] =0.056 [NaCl =0.0836 molI Pero tienen #eg.:# mol, por ser electrolito 1:1
G =0.0239 r- G
=0.0285 -1
- 134.5--1eq-1cm2) (0.056 mol(-)
=7.53x10-3 cm
=134.5 leq(m k1
=
1000 (m3(1
=
1000 cm3 L-
I conduct especifica de la sol de Nall
kNal1 =I GNal1 =10.351(m-1(10.0285)=8.98+10-3imt
="* =7.53 x10-3 -cm
-
=0.315cm-1
0.0234 -1
... se obtiene In conductancia
MNal1 =1000 (m3(-1.KNAC
=
1000(m3(-18.98x18 (m-1)
=
107.4 eq-'im2
(Na(1 0.0836 molL-1
a con la ecuncion:
n+=2 =2 +Fa +
MN+=196,500 Asmol-") (7.91x10-4cm"s "X-1) =76.3315--mol- cm2
B
v =NE =7.91 x10
- 4.cms"v-20xsm" =0.01582 (ms-1
t =I = 4(M =252.84450865
0.01582 (mS-1
I =v
IOR 1 == =2 +FN +
MNx +=146,500 As Mot-' 15,79-10-4(M2 y-1s-1) =50.1AXcm=mol-1
con AV-1 =1:. MMa+=50.1 -IM2MOL-1
&Null =126.451-Mol- CM2 ANA41=144.7 --mol-cm2 NaOH247.8 --mol- (M2
I
- Null =1Qu++
2 &
+MNH411 =ANAG+ MATHOH:Mr+-is-+Mat +ir++ Mor
+ YuH= Mai
&NH40H = -NaCl +HaN + -YNnOH =(-126.45) + 149.7+247.8 [--mol- (m2]
&ONHOH=271.05 -1-Mol- cm2