1. Suppose we have the following cell at 25˚
(a) First user standard condition determines which cation undergoes reaction by
comparing their standard reaction potential.
(b) Calculate standard EMF
(c) Suppose the following concentration for Ca2+ and K+:
[Ca2+(aq)] = 1M
[K+(aq)] = 10-6 M
Calculate EMF.
[ Ca 2+ ( aq )]1 / 2
concentration.
[ K + ( aq )]
(d) Plot the cell potential as a function of
[ Ca 2+ ( aq )]1 / 2
at which the direction of reaction would
(e) What is the ratio of
[ K + ( aq )]
change?
(f) Start with [Ca2+] = 0 and [K+] = 1M we allow a current to flow. At equilibrium
when there no potential anymore what will be the final concentration of [Ca2+]?
(g) What will be the total amount of charge transfer?
K
Ca
K2SO4
CaSO4
2. Consider a z-z electrolyte

ρ( x ) = n 0 ze0 exp

( −zF ϕ( x )) − exp ( zF ϕ( x )) 
RT
RT


(a)Derive the equation for capacity
C=
εε0
 zF ϕ 
cosh 
,
LD
 2RT 
LD =
1
κ
(b) Obtain an equation for charge density σ for small potential. Can this equation describe
the experimental data (the plot of the charge density as a function of potential is in the
lecture note).
(a) We write both reactions in reduction forms:
K+ + e- → K
EK0 = −2.92 V
Ca2+ + 2e- → Ca
0
ECa
= −2.76 V
0
Since ECa
> EK0 the reduction of Ca is more favorable, therefore Ca electrode is the
cathode side.
E 0 = Ec0 − Ea0 = −2.76 − ( −2.92 ) = 0.16 V
(b)
And the overall reaction will be:
1
1
K + Ca 2+ → K + + Ca
2
2
(c) using Nerst Eq.
+


RT  K ( aq ) 
E =E −
ln
ν F  Ca 2+ ( aq )1 / 2 
0


(1)

where K + ( aq ) = 1 M
and Ca 2+ ( aq ) = 10−6 M
1/ 2
⇔ Ca 2+ ( aq )
= 10−3 M
1 

 0.001 

E = 0.16 − 0.0258 ln 
E = −0.018 V
Note: The sign of the EMF potential (E=-0.018) shows that the direction of the
reaction is changed due to the low concentration of Ca 2+ ( aq ) and therefore Ca
will be in anode side.
(d) Using Equation (1)
+


RT  K ( aq ) 
E =E −
ln
1/ 2
 Ca 2 + ( aq )

F
0


Now we write this equation with respect to the ratio
Ca

2+
1/ 2
( aq )
K

+
1/ 2




2+

RT  Ca ( aq )
0
E =E +
ln
 K + ( aq )
F




( aq )
E
Ca

2+
K

(e) At point P: E=0 and therefore:
1/ 2
2+

RT  Ca ( aq )
0
E =E +
ln
 K + ( aq )
F




=0


( aq )
+
1/ 2
( aq )
1/ 2
 Ca 2 + ( aq )

ln  
 K + ( aq )

 
Ca

2+
K


=


1/ 2
( aq )
+
−0.16
= −6.201
0.0258
= 2.02 x10−3
( aq )
(2)
(f) The initial concentrations of K and Ca are K + ( aq ) = 1 M and
i

Ca
2+
Ca

( aq ) = 0 respectively. In this case because the ration of
i
2+
K

1/ 2
( aq )
+
( aq )
= 0 according to the graph we have back reaction. Now if we
allow the current to flow, the initial concentration of K + ( aq ) reduced and the
concentration of Ca 2+ ( aq ) increases the EMF potential become zero. In this
case the ratio of both concentrations satisfy the equation (2). Hence
Ca

1/ 2
2+
( aq )
+
( aq )
K

= 2.02 x10−3 (3)
f
f
Index f specifies the final concentration. The final concentration can be calculation
by mass balance (for each mol of K+ there will be ½ mol of Ca2+ )
Ca

2+
( aq ) − Ca 2+ ( aq ) =
f
Ca

i
2+
( aq ) − 0 =
f
1 +
+
K ( aq ) − K ( aq )



f
i
2
(
1
1 − K + ( aq )
f
2
Replacing that into equation (3) we will have:
( (
1 / 2 1 − K + ( aq )
K

+
( aq )
f
))
1/ 2
f
= 2.02 x10−3
)
K

+
( aq ) = 4.08 x10−6
f
+
Now each K produces one electron and therefore the number of electron transfer per
volume is given by:
n = N A [ K + ( aq )] = 6.02 x1023 x 4.08 x10−6 = 2.45x1018
Each electron has e0=1.602x10-19C. Therefore the amount of the charge transfer becomes:
e = e 0 xn = 1.602 x10−19 x 2.45x1018 = 0.39 C
(a) According to Poisson’s equation:
d 2ϕ −ρ( x )

 zF ϕ( x ) 
 zF ϕ( x )  
=
, ρ( x ) = n 0 ze0 exp  −
 − exp 

2
dx
εε0
RT 

 RT


replacing the charge density into Poisson’s equation we have:
d 2ϕ −n 0 ze0 
 zF ϕ( x ) 
 zF ϕ( x )  
=
 − exp 

2
exp  −
dx
εε0 
RT 

 RT

We define:
φ=
⇒
zF
ϕ
RT
d 2 φ zF d 2 ϕ
=
dx 2 RT dx 2
⇒
(2)
d 2 ϕ RT d 2 φ
=
(3)
dx 2 zF dx 2
n 0e0 = N AC 0e0 = FC 0 (4)
Applying Eq. (3), (4) into (1):
(1)
⇒
d 2 φ − z 2F 2C 0
2z 2F 2C 0
=
exp(
−φ
)
−
exp(
−φ
)
=
sinh( φ )
(
)
dx 2
RT εε0
RT εε0
Now we define:
⇒
κ2 =
−2z 2F 2C 0
RT εε0
Therefore eq. (5) becomes:
d 2φ
= κ 2 sinh( φ )
dx 2
We multiply both sides into
dϕ
dx
d ϕ d 2φ
dϕ
= κ 2 sinh( φ )
2
dx dx
dx
1 d  dφ 
2 d
cosh( φ )

 =κ
2 dx  dx 
dx
2
Note:
(6)
d
d
cosh( x ) = sinh( x ),
sinh( x ) = cosh( x )
dx
dx
By taking an integral on both sides of (6) we get:
∫
x =∞
 dφ 
d 
 dx 
x =0
= 2κ
∫ d ( cosh( φ ))
ϕ= 0
2
2
ϕ=ϕ∆
Note: at x=0 we have ϕ = ϕ∆ and in x=∞, ϕ = 0
 dφ 


 dx  x =∞
2
 dφ 
− 
 dx  x = 0
2
= 2κ 2 ( cosh( φ = 0 ) − cosh( φ∆ ))
(5)
 dφ 

 dx  x = 0
= 2κ 2 (1 − cosh( φ∆ ))
 dφ 

 dx  x = 0
= 2κ 2 (1 − cosh( φ∆ ))
2
⇒
0−
⇒
0−
2
x

2
Using 1 − cosh( x ) = 2 sinh 2 
 dφ 


 dx  x = 0
2
⇒
⇒

= 4κ 2  sinh 2 (
 dφ 


 dx  x = 0


= 2 κ  sinh(

φ∆ 
)
2 
φ∆ 
)
2 
Applying eq.(2) we have:
⇒
⇒
zF  d ϕ∆

RT  dx
 d ϕ∆ 


 dx  x = 0


 x =0
= 2κ

zF

ϕ∆  
 2RT


= 2κ  sinh 

RT 
 zF

ϕ∆  
 sinh 
zF 
 2RT

According to the definition of charge density σ:
⇒
 d ϕ∆ 

 dx  x = 0
σ = εε0 
Replacing the eq.(7) into this equation we will have:
⇒
σ=
2εε 0RT 
 zF

κ  sinh 
ϕ∆  
zF
 2RT


And finally the definition for capacity is given by:
∂σ
 ∂ϕ∆

C =



 2εε 0 RT   zF 

κ
zF   2RT 

C =
( cosh ( ϕ ) )
∆
(7)
C = εε0 κ ( cosh ( ϕ∆ ) )
C=
εε0
1
cosh ( ϕ∆ ) ) , LD =
(
LD
κ
(b) The equation we have for charge density:
σ=
2εε0RT κ
 zF ϕ 
sinh 

zF
 2RT 
For small potential:
zF ϕ  zF ϕ
=
 2RT  2RT

sinh 
Therefore
σ=
2εε0RT κ zF ϕ
= εε0 κϕ
zF
2RT
According to eq.4
κ = zF
2c 0
εε0RT
Replacing this equation into the equation for σ we will get
σ = zF ϕ
2εε0c 0
RT
This equation shows that charge density is a linear function of potential which is in
agreement with experimental plot, however this equation shows that charge density is
also a linear function of z which is in disagreement with the experiment. Overall this
equation fails to explain the experiment