TSI1 Problème 1 : Pile électrochimique

e r
U1= 12 V
R= 21 Ω
U2= 10,5V
e U1
i= 0
e U2
U1U2
e r
U1=e= 12 V
U2=Ri =e−ri ⇐⇒ r=R(e−U2
U2
=21(12 −10,5)
10,5=3Ω
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u i
u=e−ri
u i
Pge r
i
pg=u.i = (e−ri).i =ei −ri2
ri2
V= 250 mL
n0= 1,5mmol HClO
HClO(aq)+H2O(l)−→ ClO−
(aq)+H3O+ (aq)
Qr=[H3O+][ClO−]
[HClO]
ClO−H3O+
Qr= 0
ClO−c= 1,4.10−5mol.L−1
HClO(aq)+H2O(l)−→ ClO−
(aq)+H3O+ (aq)
n0
n0−x
n0−xfxfxf
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c=xf
V⇐⇒ xf=cV
xf= 1,4.10−5.0,25 = 3,5.10−6mol
n(HClO) = 1,5.10−3n(ClO−)=3,5.10−6mol n(H3O+) = 3,5.10−6mol
Qr=(xf/V )2
(n0−xf)/V
Qr= 3,2.10−8
K0= 3,2.10−8
S
E= 520.109kW.h
η= 15% = 0,15
Φ = 1600 kW.h/m2
E= ΦSη
S=E
Φη
S=520.109
1600×0,15 = 2,2.109m2
d
d∼20 km
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