Electroquímica 4

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t
=-in. 10
.......
en
su
t = s
It
⑪ ⑦
C.Amóolico C. Central C. Aniones
⑧ &
⑰ ⑦
* ↓[C] [CJeto Y[]
- ⑰↓ O ↑[A] [AJcte [A]~④
⑰④↳
to
CA
↓
C C telS
Citral
chagla de Nettorf
La disminución de la viterface de un electrodo es proporcionala la
velocidadcon la queelson se abija delelectrodo.
Anódica Católica Ag" +1é -- Agi #29=1.68x10-04
S
me=117799 12094s leq - 96500C
eq X/imca =0.10336% 0.19398% PE==107.879/mol 1.68N0
Confral 12/mol
0.14948% PE=107.879/27 X=145.442
0.16034g =MAg
104 -107.879
x -0.160349
MMia=74.459hnd ⑰ 0 +[A]
117.799 --100 IS
PE ==74.459d X
->0.14948
->a[]
1ef/md x =0.1759
PE=74.459/24 1x=0.1759-0.1219
117.7997 100 1x =0.05329
x -0.10336
X =0.1219 -Perpués 1eq-- In.759
x-0.05329
↑
x=7.19x104sq
~
10q -> 96,500C tt =#eqF 120.99=m
7.15x0*2q ⑪ X #eqX Tr =0.19398 Pequée --
Tre =0.14948 120.999 -100
x =69.022 t =7.15x10-47 ⑧a[]
x -0.19598 +=0.48
1.48 x10-424 X=0.23479 -
t =0: 1 =0.48 Yo[1] AX=0.23479-0.1819
Anter 1x=0.05369
d
X - 0.1448
↑tt =0.48 120.999--100 109--f4.45,
x =0.181, anter x -0,053g
Método de la frontera movil
Cantidad de masa depende
t=0
⑪ ⑰
0
O
del
pero equivalento yla carga
↑
S
/↑ U Wr
-
R
⑦ An
=#eq
Mc= 0.300g (
- Qu Infon1-> Cam Olag
anódico
1.43g después C*
+2 -l
2 1.210 anter
Da= 0.22g PE =MM =63.549/nd
&- 2*Frd
Ot
-SO PE=31.77g
~
/
-
~
#09 =
E =0.229 =6.92x10431.779/21#27 =
E =
0.200,
31.779/09 t=R- =HeqF=6.92x10"
#21-9.44x0eq Q#eqF 9.44x109
t =0.733
-
tt
=
1 - 0.733
++=0.267
Este t= MMAgros=1709/mo
t. =1970 Agt
+1e
-Ag
*&f28.2359 de H20
Y 0.0999 AgNOs
0
-4
28.4359 8.187 AgNOs t= 3.062x10 09 PE=MM
=
1709tnd
9.886x18424 #8 1emo
Q=#eq f H20 AgNOs t=0.512 #29
a+ =4.886x10
- 4(96,500C 28,4359 - P.187
Q=95.3999 28.235g
-X t,it. =1 *81 =0.0869
tt=1-0.512 1704/09
An =0,1859-0.099, X=0. 1859 -Popre tt =0.488 #84=5,062x109
Am =0.0869
KCl Ag+1-Ag Cl
-
--Clette It= -#19#24PE==107.879/ad
8.137gt C.A 109/md P.E. = 35,459/mol 11 =3.86x10.14
0.08979 Ag PE=1.07.879/24 109/mol 7.94N044
p.E. =35.499/09 +1 =0.48
t= #11 =m =0.08479 t= 0.52P.E. 107.879/nd #81=v
t. = R- #04:7.94x101 P.E.
Q #89=0.013759
35.499407
#89= 3.86x10
-
19