[PDF] Oxydoréduction Exercice 2 : Équilibrage de ré

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n ā= 1/n ā

ā E ā E

ā E

Ģ Ģ6= Ģ 6=()

Ģ 6= () 6=()

() ()6=() Ģ Ģ6= +

Ģ ()6= () 6=Ģ ()

ȕ ; ā 0

; ā 0

E(()/()) =;E(Ģ/()) =;E(()/) =

;E(Ģ/Ģ) =;E(Ģ/()) =; K(Ģ) =

1;2 (()) = 9;7 RT(X)/F '0;06(X)

E(/()) =;E(Ģ/) =;

K(Ģ) = 3;2

ā E(Ģ/)

ā10;0 ě 10;0

ā ā = 0

E(Ģ/()) =; E(()/) =;

j;jj;j (1) =; =; V0=

E ā ā U=

V2V1 1;5 1;5

V0=; ȕā c0=;

c=; V ā

V ; ; ; ȕ ā

ȕ E=f(V) E(V) 0;24

āě 0;52

ā 0;76

Nā 0;89

E(/) =;E(/) =; ȕ E=;

E(/());E(/());E(/());E(/());

(1)()j(;) +(;)jj(;)j()(2):

āU ȕ Ģ ā ǃ

(1)

ā U=; ȕ

E(/) =; E(/()) =; ȕ

c=; ; ě

ā () ā ā () Ģ = 107

ȕāK ā () Ģ() ā

ā E(/()) ā

E ā ā

E(/()) =;E(/()) =;E(Ģ/()) =;

2(Ģ) = 8;9 4([]) = 24;9

ā E ā E

ā E

E(Ģ/)

E(Ģ/) =; ā ā ȕ

ě c=

ā E(/()) =; (()) = ā

[]2ā

Ģ[]2 =;

āā c0=

ā A

ā24681012141;11;21;31;41;5O

2(g)H 2O H 2(g)A pHE(V)

ā E(/) = 7

ȕ ɵP=;

ā ȕā E(/) ā

ě c=

= 0;;< E <; ě ā () ā ā c [] E= ā () ě c = 1 ā = 10 ā ě ā ȕā (0)Ģ(+V II)Ģ(I)(+I)Ģ(+V)Ģ(+I) (0)(+IV)Ģ(+V I)(+III)(+III)(+III)(+II)Ģ(+V II)

ķā ()e() Ģ E=E(()/())0;06(/c)

ā ā ȕ =cpKs

E';

Ģ eĢ E=E(ĢĢ) + 0;03(Ģc/Ģ)';

Ģe() Ģ E=E(Ģ()) +

0;06c/2=;

()e

E=E(()) +

0;06P(())h

P0 p =K1;2 ȕě ā E=;

Ģ e () E=E(Ģ/())+0;03Ģh4P0

P(())c

E=;

Ģ () ķā Ģ e

e() ā

E(())E(Ģ) =0;06

2hP2(())c2

3P20 =0;06 2K;

ȕɵ K= 101;33

Ģ e= E=E1+0;06

2h3

Ģ e=Ģ E=E2+0;06

2h2

ȕ ě ā []/Ka ā

E(ĢĢ) =E10;03K=;

e Ģ=() E1=E1+0;06 3h4 c () e= E2=E2+0;06 2h4c e Ģ= E3=E3+0;06 5h8

5E3= 3E1+ 2E2 E3= 1;51

0;052x0;053x5xā

ā K= 106(E1E2)/0;06= 1047 ā ȕě

ā x=; =

;=3 r h4c5

K2=;ā()ā

20:10351;67:102'; ;

ȕā ā12

E1=E(/) + 0;06[]1

[]1'E(/) + 0;018

E2=E(/) + 0;06[]2

[]2'E(/)0;018

ǃ+ ā 1 ā

0;12(2) = 0;036 ā 2 1

1 ȕā =

ȕ ā[1][1]= [2][2]=c=; ȕā

Q=cFV0=;

ā ā E1E2 ā ā

(E1E2)Q= ɧ ȕā āā

ȕā ā ȕāE2E1 ć ā U=V2V1

U= 1;5E2E1=0;036 E2 E1

2 ā 1 āā

ā 0 ā

1+2=2+1

0;1x0;1x0;05 +x0;05 +x

ȕā 0;05+x0;1x= 101;5/0;12'3;16:1012 ā []1= []2=;[]2= []1= 0;15/101:5/0:12'

U= 1;5 E2 E1

1 ā 2 āā ā

1+2=2+1

0;1 +x0;1 +x0;05x0;05x

ȕā 0;1+x0;05x= 101;5/0;12'3;16:1012ā

[]2= []1=;[]1= []2= 0;15/101:5/0:12'

ȕā'0c0V0cV/2cV/2cV

ě ȕācV2c0V0'0c0V02c0V0

ȕā ě ā E>0;6 ā

ȕāVV= 2c0V0/c=

V=ȕȕā e=E()

E=E(/) +0;06

2x

1x x=V/V e=; ě ȕā

x >1 ā e=E(/)+0;06(x1)

V= x= 1;5 e=; V= x= 2e=; ȕā

ā 3E=E() + 2E() e=;

E=E(()/())RT

F c ā

āā ā () ā E=E(/()) +

RT F c=E(()/())RT F c ā E(()/()) =E(/()) + RT F /c2 'E(/())0;06() =; ā

ā E(()/())'E(/())0;03(()) =;

E(Ģ/())'E(/())0;03

=; E(Ģ/())'

E(/())0;03

12 ȕā

E1=E( /()) + 0;06[]1/cE2=E( /) + 0;06[]2/c:

ȕ Ģ U=E2E1= 0;06[]2

[]1=; ā

ȕā2

E1 āā ȕ

2U=E2E1= 0;06[]2/[]1>;

[]2 []1= 101:08/0;06= 1018 []1=; ; ā = [][Ģ]2/([Ģ]c) = 1020 () E1=E(/())E2=E(/)

K= 10(E1E2)/0;06'105;8 ā ā ȕ

ā āā Q=c3/([][Ģ]2) = []2/(2[]c)

ā K= 1/(K2)'108;17 ȕ

ěc(1) (2c2)22c

ȕā K=c3/(4c3(1)3) 1'1;25:102 ȕ āā

ā98;7 []';[Ģ]';[] =

c2/[Ģ]' ȕ ā ȕ e=() e= ()!()=p

2() !K(T) =e

pF RT

E(())E(/())

ā ā 1E(/())> E(/())

E(/())<0

ā ě ā 0

[]+e=() Ģ!E=E([Ģ]/()) +RT

F[[]]c

[Ģ]2 e=()!E=E(/()) +RT F[] c RT

F[][Ģ]2

[[]]c=E([Ģ]/())E(/())

0;062([]) =E([]/())E(/())

ć0;06

32([]) =E([Ģ]/())E(/()):

āE([Ģ]/()) =; E(Ģ/()) =;

ȕ Ģ K'8:1020 [Ģ]K'1050

Ģ e=() E=E(Ģ/()) +RT

3Fh4[Ģ]P

cP() e=()E=E(/()) +RT 3F[]3 c3

Ģ ()=() K'10

30;06

E(Ģ/())E(/())

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