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Unformatted text preview: A practical form at
Since:
For gases: 25oC: E ( 25o C ) = E o ( 25o C ) − Nernst Equation
0.0592V
log Q
v ΔG(T,p) = ΔGo (T)+ RT ln Q(p) →Temp. & press. dependence For solutions: ΔG(T,b) = ΔGo (T)+ RT ln Q(b) →Temp. & concen. dependence Thus: RT
RT
E (T , p, b) = E (T ) −
ln Q ( p, b) = E o (T ) −
ln
vF
vF
o ∑
∑ Products Reactants Temperature, pressure & concentration dependence of emf At equlibrium: Q = K ΔG = 0 E=0 v
aJ J
v
aJ J Concentration dependence of E
T=25oC Zn(s) + Cu2+(aq,1M) → Cu(s) + Zn2+(aq,1M) Eo(25oC)=1.10V Zn(s) + Cu2+(aq,2M) → Cu(s) + Zn2+(aq, 0.01M) E(25oC)=? [Cu2+] = 2 M, [Cu−] = 0.01 M 0.0592V
0.0592V
0.01
E ( 25 C ) = E ( 25 C ) −
= 1.17V
log Q = 1.10V −
log
v
2
2
o o o •
• Concentration Cell
Spontaneous reaction when the oxidation and reduction reactions are the same,
as long as the electrolyte concentrations are different: driven purely by mixing
the more concentrated solution has lower entropy than the less concentrated electrons will flow from the electrode in the less concentrated solution to the
electrode in the more concentrated solution
oxidation of the electrode in the less concentrated solution will increase the ion
concentration in the solution – the less concentrated solution has the anode
reduction of the solution ions at the electrode in the more concentrated solution
reduces the ion concentration – the more concentrated solution has the cathode
Cu(s)⏐ Cu2+(aq) (0.010 M) ⏐⏐ Cu2+(aq) (2.0 M)⏐ Cu(s)
2+
⎡Cuaq ,2 M ⎤
RT
E = Eo −
ln ⎣ 2+ ⎦
vF ⎡Cuaq ,0.01M ⎤
⎣
⎦ = 0V − ( −0.068V) = 0.068V Activity of ionic solutions
For compound MpXq ↔ p M+ + q X.
ideal
q
Gm = Gm + RT ln γ +p + RT ln γ −
ideal
q
ideal
= Gm + RT ln γ +pγ − = Gm + RT ln γ ±p + q define mean activity coefficient γ±,
q
γ ± = (γ +pγ − )1/( p + q ) Anions and cations are attracted to
each other lead to extra stability Ionic solution should be describe by
DebyeHuckel limiting law log γ ± = −  z+ z−  AI 1/2 A=0.509 at 25oC, 1
2 bi
I is ionic strength: I = ∑ zi
bo
2i For MpXq↔ p M+ + q X of molality b 1
1
b
2
2
o
2
2b
I = ( b+ z+ + b− z− ) / b = ( pz+ + qz− ) o = k o
2
2
b
b
For salt solution like KCl I = b / bo log γ ± ∝ I 1/2 = b1/2 15 Generalize to any electrolyte
In using the Nernst equation, we assumed ideal solution
for ions
RT
E = Eo −
ln
vF ∑
∑ v
aJ J Products
v
aJJ ∑
∑ RT
= Eo −
ln
vF Products Reactants Should be a J = γ ± E = Eo − b
bo bJvJ Reactants For simplicity, we replace b/bo with b RT
ln
vF ∑
∑ v
bJ J Products
v
bJ J Reactants RT
ln
E=E −
vF
o v
bJ J ∑
∑ Products Reactants + RT
ln
vF ∑
∑ γ Jv Products Reactants
v
bJ J RT
∑v
+
ln γ ± J
v
bJ J vF J γ Jv J Measuring Eo for Electrolytes
Pt(s)H2(g)HCl(aq)AgCl(s)Ag(s)
½H2(g)+AgCl(s) H+(aq)+Cl(aq)+Ag(s) RT
RT
RT
o
2
2
E=E −
ln aH + aCl − = E −
ln b −
ln γ ±
F
F
F
o since ln γ ± ∝ I 1/2 = b1/2 for singe charged compond 2 RT
2 RT
o
ln b = E −
ln γ ± ≈ E o + cb1/2
E+
F
F
Plot 2 RT
E+
ln b vs b½ can determine Eo precisely
F Eo+cb Temperature dependence of E • We can determine the ΔG from measurement of E
ΔG = −vFE ∂G
= −S
∂T So: dE o Δ r S o
∂E ⎞
ΔrS
⎛
→
=
Eo depends only on T
=
⎜
⎟
dT
vF
⎝ ∂T ⎠ p vF
⎛o
dE o ⎞
Δ r H o = Δ rG o + T Δ r S o = −vF ⎜ E − T
⎟
dT ⎠
⎝ This expression provides a noncalorimetric method for measuring ΔrHo
Through convention ΔfHo(H+,aq)=0, it allows calculation of ΔfHo of other
ions in solution...
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This note was uploaded on 09/25/2013 for the course CHE 301 taught by Professor Raineri,f during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Raineri,F
 Physical chemistry, Atom, pH

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