Energy Storage (Supercapacitor vs.

Battery)
𝑿 𝑖𝑛 𝑿 𝑜𝑢𝑡

𝜌
Capacitors ∮ 𝑿 ∙ 𝑑 𝑨=∫ 𝑑𝑉
𝜖
𝜎
𝑿 𝑜𝑢𝑡 − 𝑿 𝑖𝑛=
𝜖
we solve
with the boundary condition;
At X=0; X =- =

Distance=- At X=H; -X= =

Energy of =- =
=

On closing the switch, charges to appear on both plates, afterwards open the switch.
: permittivity
𝜔 𝑟𝑖𝑔h𝑡 →𝑙𝑒𝑓𝑡 𝑋 𝑄 𝑡𝑒𝑠𝑡 𝐿 𝑞𝐿
Δ 𝜙= Δ 𝐸=𝜙𝑙𝑒𝑓𝑡 − 𝜙𝑟𝑖𝑔h𝑡 = = =
𝑄 𝑡𝑒𝑠𝑡 𝑄𝑡𝑒𝑠𝑡 𝜀
stored charge
Energy=Potential*Charge
dW=(
(def. of capacitance) W=
0
• Supercapacitors: very large surface area per volume: very
porous solid materials: graphene, etc.

• Supercapacitors are ideal when a quick charge is needed to fill a

short-term power need; whereas Batteries are chosen to provide
long-term energy.
3. Electrochemical Cells and Electrodes
An electrochemical cell is composed of two electrodes and an electrolyte.

the lead-acid cell

cathode
anode
− −
𝑒 𝑒

Figure 3.1 An example of a simple electrochemical cell is the lead-acid cell, here shown at
rest.
Equilibrium cell : electrochemical equilibria at both electrodes

(Ⅰ) at the lead dioxide electrode :

−
− +¿ ( aq ) +2 e ( Pb O2 ) ⇌ Pb S O4 +5 H 2 O (l ) ¿
Pb O 2 ( s )+ HS O 4 ( aq )+ 3 H 3 O

(Ⅱ) at the lead electrode :

− −
+¿ ( aq ) + 2 e ( Pb ) ⇌ Pb ( s ) +HS O4 ( aq ) + H 2 O ( l) ¿
Pb SO 4 ( s ) + H 3 O ❑
 Then the overall reaction : (Ⅰ) – (Ⅱ)

(**)
(At the both electrodes, )

this reaction can occur only when there is a passage of electrons

through an outside circuit.

−1
¿ − 387 𝑘𝐽𝑚𝑜 ℓ ¿

(big chemical energy) (small chemical energy)

(**)
(small electronic energy) (big electronic energy)

Thus, reaction(*) tends to occur in the forward direction.

For this purpose, reaction (**) must occur in the reverse direction through outside circuit.
 Although the forward reaction occurs only when move to
through outside circuit :

means that electron activity in is higher than that in .

How to find -
voltmeter
+ V –
∆E

𝑃𝑏𝑂2 𝑃𝑏
no outside circuit ( electrolyte at equilibrium)

(E   PbO2   Pb )
high E ( electrolytes high activity of electron
low E(

(Ⅰ) (Ⅱ)
(through electrolyte) :for this purpose we have to impose energy to electrons :
:this energy comes from the chemical energy released from the reaction:

전자의 포텐셜 에너지 변화 = :

)
(procedure)

(a) Write down equilibrium reaction at electrode Ⅰ with electrons on the

Write down equilibrium reaction at electrode Ⅱ left hand side.

(b) Overall reaction : Ⅰ- Ⅱ (Then, electrons transfer from Ⅱ to Ⅰ through the outside circuit)

(c) Find for overall reaction

(d) (cell voltage)

(The above formula is valid when the current is zero: i.e. at the equilibrium state.
If current flows, there appears concentration gradient in the electrolyte, and we cannot find . )

standard cell voltage

Measurement of cell voltage provides the accurate thermodynamic data !!!

𝑜 𝑜
Δ 𝐺 =−𝑛𝐹 Δ 𝐸
 change with respect to T yields and etc.

𝑜
Δ 𝑆 =𝑛𝐹
𝑑 Δ 𝐸𝑜
𝑑𝑇
𝑜 𝑜 𝑜
, Δ 𝐻 = Δ 𝐺 +𝑇 Δ𝑆 =𝑛𝐹 𝑇
2 𝑑 Δ 𝐸𝑜
𝑑𝑇 𝑇 ( )
Activity coefficients are measured by studying the effect of concentration on .
 1)If -= nF, equilibrium state;
cathode potential anode potential =

2)If -nF,
Energy -------- (𝜑 ¿ ¿ 𝑐 ¿ 𝜑 𝑎 )¿
the forward chemical reaction dominates;

𝐸𝑎−𝑛𝐹𝜑𝑎
In the forward chemical reaction, electrons
- -------- are pumped out of left electrode and pushed
)= into the right electrode.
nF Left electrode becomes cathode and right
--------
=
Energy

electrode becomes anode.

Electrolytes
- 3)If -nF,
−
𝑒 −(chemical energy re-
𝑒 electrons are pumped out of the right elec-
leased)
trode and pushed into the left electrode. Left
electrode becomes anode, the right electrode
is cathode.
Lower en- Higher en- Chemical reaction proceeds backwardly,
ergy elec- ergy elec- Gibbs free energy of electrolytes increases at
tron tron the expense of electron energy.
Cells not at equilibrium. (current flows through an outside circuit)

chemical energy converted to electric energy

(- : Galvanic cell

Figure 3.2 In its galvanic mode the lead-acid cell provides energy to a “load”. The cell
voltage is less than its equilibrium value.
Cells not at equilibrium.

( energy being stored in the “load” )

Figure 3.2 In its galvanic mode the lead-acid cell provides energy to a “load”.
The cell voltage should be less than its equilibrium value.

(Ⅰ)
Electrolyte 를 기준 : Cathode(cation+electrode) : reduction occurs

(Ⅱ)
Electrolyte 를 기준 : Anode(anion+electrode) : oxidation occurs

chemical energy converted to electric energy(- : Galvanic cell

(100% efficiency)

Δ 𝐸=1.9𝑉 <2.1𝑉
 Cells not at equilibrium.
electric energy converted to chemical energy
(- : electrolytic cell

( energy being spent in the “load” )

−
𝑒 −
𝑒

Figure 3.3 In its electrolytic mode the Gibbs energy of the lead-acid cell is increased at the expense
of external electrical energy. The cell voltage now exceeds its equilibrium value.
Electric current is being forced by a d. c. voltage source through the cell
()
Voltage source 가 전자에 해주는 일 =-nF()=nF
nF- (Ⅰ)) 에 의해 증가함 .

→ convert electric energy to chemical energy : Electrolytic cell.

(Ⅰ)
anode : oxidation

(Ⅱ)
cathode : reduction

at the expense of external electric energy.

 Figure 3.4 In this polarization curve for a lead-acid cell, the colored segments correspond to
operation in the galvanic and electrolytic modes.

Current
: equilibrium voltage, open-circuit voltage, null voltage :
: is determined by the thermodynamics of electrolyte.

In the Galvanic mode, the current should increase asE decreases to consume the same

Energy=Charge x Potential
Power=Current x Potential
 Thermodynamics of equilibrium cells

𝐀𝐠 𝐏𝐭
H2

AgCl

H O
C l −+¿2¿
H

The laws of chemical thermodynamics are obeyed by an electrochemical cell at equilibrium.

· Equilibrium is achieved by balancing the cell’s voltage against an external source.

or
· Open-circuited (above figure)

(Ⅰ)
(Ⅱ)

(overall) Ⅰ – Ⅱ

− −
e ( A 𝑔 ) ⇌ e ( Pt )
 ΔG
electron moves from Pt to Ag through the outside circuit.

at higher activity than

𝜙Ⅱ < 𝜙Ⅰ
(open-circuited)

전자의 이동까지 고려한 반응식 :

electrochemical Gibbs free energy

 electrochemical activity (

: potential 에 전하 인 입자 1 개를 넣기 위한 전기적인 일

𝑧 𝒾 𝑄𝑒 (𝜙 − 𝜙 0 ) 𝑧𝒾 𝐹
(𝜙 −𝜙 0 )
𝑘𝐵 𝑇 𝑅𝑇
¿ 𝑎𝒾 𝑒 =𝑎 𝒾 𝑒

전자의 경우
𝐹
− ( 𝜙 −𝜙 )
~
𝑎𝑒 = 𝑎 𝑒 𝑒 𝑅𝑇
− −
0

=
Thus,

(chemical equilibrium constant)

0
¿ ∆ 𝐺 +𝑅𝑇 ln 𝐾 + 𝐹 ∆ 𝐸
Since no current flows (open-circuited)
and .

Then,
𝑜
∆ 𝐺 =∆ 𝐺 +𝑅𝑇 ln 𝐾 + 𝐹 ∆ 𝐸 𝑛 +𝐹 (∆ 𝐸 − ∆ 𝐸𝑛 )
zero at equil zero by definition of

if becomes positive( i.e., backward reaction occurs :electrolytic cell )

if short-circuited,

Circumstance Cell voltage Activities

experimental cell
Experimental values
(above figure)

short-circuited 0 Adjusted ()

standard condition Unity

Thus

for the above reaction,

The null voltage

ionic conc of HCl=30.0 mM

()

We can estimate using measured data of .

(Summary)
voltmeter
V

Ⅰ Ⅱ
electrolytes

− −
oO + ⋅ ⋅ ⋅+ n e ⇄ rR +⋅ ⋅ ⋅ qQ + ⋅ ⋅ ⋅+ n e ⇄ pP +⋅ ⋅ ⋅

Ⅰ)

Ⅱ)

Ⅰ) - Ⅱ)

oO+ pP +⋅⋅⋅⇄ qQ+rR+⋅⋅⋅

 Then the null potential is given by

where

or , where is the standard half cell potential or the standard electrode potential.

and are compiled as data.

 Ⅰ Ⅱ

Ⅲ
𝑒 − Equilibrium 𝑒
−

(Pb)
)
)
(Pb)
-)=2FE

+ + +=0 at equilibrium
=2F
=0

2F=0 : =-
Cells with junctions
Daniell cell : a Galvanic cell

Figure 3.6 The Daniell cell incorporates a porous diaphragm.

Ⅰ) cathode :
Ⅱ) anode :

To prevent the parasitic reaction ,

a porous diaphragm is inserted.

Overall reaction : Ⅰ-Ⅱ :

: (parasitic reaction 이 일어나면 이 부분이 없음 )

 null voltage

Additional potential difference is caused by the liquid junction potential.

Passage of and contribute to the liquid junction potential.

(Liquid Junction Potential :caused by concentration gradient)

+
Gi  G 0 i  RT ln ai  z i F m m+=m , m-=m
At the interface, G  G +¿ 𝑧
+¿ + 𝑝
− −
𝑧 =0 ¿
¿
𝑝
G i  G 0
i  RT ln a  i  z i F 
G  i  G  i  RT ln a  i  z i F 
0

  RT a i
i  i  ln 
zi F a i
Considering both of cation and anions
t RT  
 a  t  RT a 
     ln  ln
 
a  z F

a 
z F
    
a  p  a a  p  a
   
a  p  a a  p  a
t  RT a  t  RT a 
     ln    ln 
 

z F a z F a

(I) (II)

ΔE=-+(-) (As accumulate in the

(-) becomes from positive to negative.)
Henderson equation: to be derived later:
(assuming linear variation of concentration in the interfacial area )

if (KCl 경우 )
(Liquid Junction Potential : Another Viewpoint)

𝜇 𝑖 𝜇 𝑖 + 𝑑 𝜇𝑖
high low
conc. 𝑥 𝑥+ 𝑑𝑥 conc.
of ion of ion
ⓐ
∅ 𝑖 ∅𝑖 + 𝑑 ∅ 𝑖 ⓑ

− −
𝑡 𝑡
( ∅❑ + 𝑑 ∅❑ ) + − ( 𝜇− +𝑑 𝜇− )= ( ∅ − ) + −
𝜇−
𝐹𝑧 𝐹𝑧

𝑡 +¿
𝑑 𝜙=− ¿
𝐹 𝑧 +¿ 𝑑𝜇 𝑡
− ¿
+¿ − 𝑑𝜇− ¿
𝐹 𝑧−
 Assume constant

𝑑 𝜇𝑖 =𝑅𝑇𝑑 ln 𝑎𝑖

due to the electroneutrality,

𝑡 + ¿=1 −𝑡 − ¿
(Assume )
KCl :

HCl :

(since moves faster, the lower concentration region becomes more crowded
with , resulting in high ).
Concentration cell (using KCl as a salt bridge)

concentration cell: concentration difference produces potential

Figure 3.7 This concentration cell uses a KCl salt bridge to prevent the transfer of Cu 2+ ions be-
tween the half-cells. The concentration cL of CuCl2 in the left-hand half-cell exceeds that, cR in the
right-hand half-cell.

(right side) oxidation

(anode) moves leftside
moves rightside

(salt bridge)
through right diaphragm, moves rightside

through left diaphragm, moves leftside

 (left side) reduction
(cathode) moves leftside |𝑍 𝒾|𝑢𝒾 𝐶 𝒾
𝑡𝒾 =
moves rightside ∑ | 𝑍 𝒾|𝑢 𝒾 𝐶 𝒾
𝒾

Since for KCl

there is no liquid junction potential ( electro neutrality taken care of by the transfer of excess KCl)
(I) .
(II

(I)-(II) yields:

:concentration of CuCl2 in the left-hand half-cell

- :concentration of CuCl2 in the right-hand half-cell
= and =

∆ 𝐺𝑜 1
∆ 𝐸 𝑛 =− − 𝑅𝑇 ln ¿ ¿
2𝐹 2𝐹
(Since,
Since
at the left electrode, disappear
while at the right electrode, produced according to the Le Chatelier principle

Electrons ejected from the left electrode and

electrons absorbed at the right electrode through the electrode-electrolyte interface,
which is cause by the chemical energy release from the chemical reaction.

Then left electrode becomes cathode and right electrode becomes anode:

The electrons entering to the anode is energized by the chemical energy released
and pushed to the load(lamp or motor)through an outside wire,
and eventually enter into the cathode after their energy exhausted.

-
(Concentration Cells : with and without salt bridge)

Concentration Cells : Symmetric apart from differences in concentration

(With salt bridge: we can ignore liquid junction potential)

𝑍𝑛 𝑍 𝑛 ( ) 3.5
2+¿ 𝐶 ¿𝑙

𝐾 +¿ 𝑀
¿ 𝑍𝑛
−
2+¿(𝐶 𝑟 )¿

𝐶𝑙 ( 2 𝐶 𝑙 ) 𝐶𝑙 −𝐶𝑙 ( 2 𝐶 𝑟 )
−
𝑍𝑛
(left)
(right)

Overall reaction

Assume

left barrier
𝐾 +¿(𝑎𝑞 , 3500 𝑚𝑀 )¿
−
𝐶𝑙 (𝑎𝑞 , 2𝐶 𝑙 )
 2+¿ (𝑎𝑞 , 𝐶𝑟 )¿
𝑍𝑛
right barrier
−
𝐶𝑙 (𝑎𝑞 , 3500𝑚𝑀 )

Therefore, no liquid junction potential (charge is carried mainly by )

(Without salt bridge: measured potential influenced by ionic transfer in the electrolyte)

𝑍𝑛 𝑍𝑛 ( )
−
2+¿ 𝐶 𝑙 ¿

𝐶𝑙 ( 2 𝐶 𝑙 )
𝑍𝑛 ( )
−
2+¿ 𝐶 𝑟 ¿

𝐶𝑙 ( 2 𝐶 𝑟 )
𝑍𝑛
cathode anode

Assume
2+¿ (𝑎𝑞 ,𝐶 𝑟 ) ¿
𝑍𝑛
−
𝐶𝑙 ( 𝑎𝑞, 2 𝐶 𝑙 )

−
2+¿ (𝑎𝑞 ,𝐶 𝑙 )+2 𝑒 ( 𝑍 𝑛 , 𝑙) → 𝑍 𝑛 ( 𝑠 ) ¿
𝑍 𝑛
2 +¿ ( 𝑎𝑞 , 𝐶𝑟 ) ¿
𝑍 𝑛 ( 𝑠 ) → 2 𝑒− ( 𝑍 𝑛, 𝑟 ) + 𝑍 𝑛
electrode reaction would be accompanied by the transfers

𝑡 2 +¿ (𝑎𝑞 , 𝐶 𝑟 ) → 𝑡
+ ¿𝑍
2 +¿ ( 𝑎𝑞, 𝐶
𝑙)
¿
¿
¿

+ ¿ 𝑍𝑛 𝑛
¿

− −
2 𝑡 − 𝐶𝑙 ( 𝑎𝑞 ,2 𝐶 𝑙 ) →2 𝑡 − 𝐶𝑙 ( 𝑎𝑞 ,2 𝐶 𝑟 )

this kind of cell is called a concentration cell with transference

(concentration cell with liquid junction potential : without salt bridge)

 𝜙1 𝑐 1> 𝑐 2 𝜙2
𝑐1 𝑐2
, ,

cathode anode
− −

𝑍 𝑛2+¿ +2 𝑒
→ 𝑍 𝑛( 𝑠 ) ¿
𝑍 𝑛2+¿ +2 𝑒
→ 𝑍 𝑛( 𝑠 ) ¿

∆ 𝐸=𝐸 𝑐𝑎𝑡h𝑜𝑑𝑒 − 𝐸 𝑎𝑛𝑜𝑑𝑒 + 𝐸 𝐿

𝐸 𝑐𝑎𝑡h𝑜𝑑𝑒 − 𝐸 𝑎𝑛𝑜𝑑𝑒

− ( 𝑡 − − 1 ) 𝑅𝑇
( 𝐸 ¿ ¿ 𝑎 − 𝐸 𝑏 ) ≡ 𝐸 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑙𝑦𝑡𝑒= ¿

( ) ( )
𝑏 𝑏
+¿ 𝐶 𝑡 𝑅𝑇 𝐶
𝐹𝑧 ln ⁡ 𝑎
+ − ln ¿
𝐶 𝐹𝑧 −
𝐶𝑎
the and transport not included in the above electrode reaction!
 −
+ ¿= 2, 𝑧 =− 1 , 𝑐 −=2 𝑐 +¿ = 2 𝑐 ¿ ¿
𝑧

¿ 𝐸 𝑐𝑎𝑡h𝑜𝑑𝑒 − 𝐸 𝑎𝑛𝑜𝑑𝑒 + 𝐸𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑙𝑦𝑡𝑒

Movements of and are taken care of by and .