Electrodeposition of Manganese Dioxide
Josh Farris
Shawn Martin
Department of Materials Science and Engineering
Michigan Technological University
MY 3110 Technical Report
April 28, 2004
Problem Statement
The process of manufacturing MnO2 will be discussed using electrolysis methods. This
paper will determine the mathematical aspects and processes of producing MnO2 using
thermodynamics of materials and electrochemistry.
Proposed Approach
Electrodeposition is the deposition of a conductive material from a plating solution by the
application of electric current. Electrolytic Manganese Dioxide (EMD) can be formed
from the direct electrolysis of an aqueous bath of Manganese Sulfate and Sulfuric acid.
Two graphite cathodes and a single titanium anode are placed in this aqueous bath at an
elevated temperature and superatmospheric pressure. The temperature ranges from
between 120 degrees C and 155 degrees C. By having the electrolysis take place at
higher temperatures and pressures, this allows there to be a significantly higher current
density with respect to the total anode surface. With such an increase in current density,
a higher rate of production of MnO2 is achieved. This allows there to be smaller
electrolysis units and cuts down on capital investment. With an increase in the current
density passivation may occur. Passivation is the formation of an insulating oxide film
that can build up on the titanium anode, which can disrupt the deposition of the MnO2 by
blocking the MnO2 from being deposited onto the anode. At high temperatures and
pressures this eliminates the problem of passivation.
One negative effect caused by the raise in temperature, at normal current densities, is that
it has the tendency to reduce the average specific surface area (SSA) of the MnO2
product. This has a negative impact on the overall performance of the material when it is
used as cathode active material in an alkaline cell. Some ways of solving this problem
are by increasing the current density, or by adding a soluble titanium dopant to the
aqueous bath. By adding a soluble titanium dopant, such as TiSO4, this incrementally
increases the SSA of the MnO2 product. The desired SSA of the MnO2 is between 18
and 45 m^2/g for good performance in an alkaline cell. This is because if you have more
than 45 m^2/g, the individual crystallites become smaller and the number of crystallites
per unit volume becomes larger. This leads to smaller pores between the crystallites and
makes it harder for water molecules to enter and hydroxyl ions to leave when the MnO2
is used in an alkaline cell. This decreases the reaction rate of the battery which results in
poor performance. If you have a SSA of less than 18 m^2/g, there is an insufficient
surface for the electrochemical reaction to take place.
The soluble titanium dopant, TiSO4, makes beneficial changes on the primarily γ-MnO2
crystalline structure. This enhances the performance characteristics of the product by
having a higher open circuit voltage (OCV) and a higher capacity when used for an
alkaline cell.
The resulting MnO2 can have a slightly lower real density of MnO2 (4.2-4.38 g/cm^3).
This is because of higher level of crystal imperfections, such as combined water, cation
vacancies, twinning faults, and [SO4]- anions. These also contribute to higher OCV and
better performance.
Advantages of this procedure are that it produces a high quality MnO2, the rate of
production can be increased by several fold due to the fact that the current density is
increased, less needed investment in plating equipment and size of the plant needed in
order to perform the operation, and the labor can be decreased because of the shorter
plating cycles and time spent monitoring the process.
Process
The electrolysis bath has an MnSO4 concentration between 0.2 and 2.0 mol/liter and the
H2SO4 concentration is between 0.1 and 1 mol/liter. Usually this is a 2 to 1 ratio
MnSO4 and the H2SO4, respectively. An anode and a cathode are inserted in the
aqueous bath and are enclosed in a casing forming the cell. The electrodes are connected
by a direct current source, outside the casing. The anode is connected to the positive
terminal of the direct current source, where the oxide is formed, while the cathode is
connected to the negative terminal. The bath is then heated to a temperature of about 120
degrees C with a current density of between 12.5 and 37 A/ft^2 of the anode surface. The
electrolysis bath is replenished continuously as the electrolytes are spent so the
concentration can remain constant at all times. The MnO2 becomes deposited on the
anode and accumulates for approximately 1 ½ to 3 weeks until a thickness of about 1-3
cm is achieved. The anodes are removed and the MnO2 can be removed by mechanical
shock. The MnO2 that is recovered can be crushed, ground, washed, neutralized and
dried conventionally. Then this is used as cathode material along with Zinc and
Potassium Hydroxide.
A comparison to the conventional EMD process is:
- the operating temperature is 80-98 ° C instead of 120-155 ° C
- normal atmospheric pressure
- current density is 5-8 A/ft^2, rarely exceeds 10 A/ft^2 because of passivation
Calculations and Results
The Mn2+ cation that is floating around in the aqueous solution has some H2O molecules
attached to them. In order for the cation to move into the anode, it has to discharge the
molecules and move across the double layer and enter the anode. The double layer
consists of a plane of charges on the anode and a plane of charges in the solution. A
model that we used for this is called the Helmholtz model. The rate of discharge of
cations across the double layer without a potential difference applied is given by the
Eyring equation
k+(→M)=(kT/h)exp{-ΔGm‡(→M)/RT}
[1]
where k is boltzmann’s constant, T is temperature (K), h is planck’s constant, ΔGm‡ is the
activation Gibbs function at the point where it reaches its maximum, and R is the gas
constant. (→M) means the flow is from the solution to the metal.
As some electrical potential is introduced, the Gibbs function of activation changes
across the double layer and gives us a new equation for the Gibbs function as
‡
m (→M)=
‡
m is
ΔGm‡(→M)+z+FαΔφ(M,S) [2]
the Gibbs function of activation for the motion towards the electrode
with an applied potential, z+ is the cation charge, F is the faraday, α is the fraction of the
distance between the inner helmholtz plate and the outer helmholtz plate where the Gibbs
function is the largest, and the Δφ(M,S) is the applied voltage plus the equilibrium
potential from the reaction.
In the presence of a potential difference , the rate coefficient changes a bit and is
calculated by the equation
+
(→M)=(kT/h)exp{-
‡
m (→M)/RT}
[3]
This is just substituting the newly found Gibbs function with the potential difference for
the equilibrium value.
Once the rate coefficient is found, it can be used to find the flow of current at the anode.
At a region right outside the double layer, there will be an average number of cations per
unit area (σ+ (outside)). The number of cations that are passing through unit area of the
double layer per unit time gives the current density
j+(→M)=z+e σ+
+
(→M) [4]
e is the charge on an electron.
On the inner side of the double layer there is a number of cations per unit area, σ+(inside).
There will be a flow of current away from the electrode as well
j+(←M)=z+e σ+
+
(←M)
[5]
There will be a net flow of current towards the electrode and it will be the difference of
equations [4] and [5]. This gives the dependence of the net current density on the
potential difference.
When the electrode is at equilibrium with the solution, there is a potential difference,
Δφe(M,S). This value can be referenced in a handbook for the reaction
Mn4++2e-→Mn2+
Δφe=1.21
The difference between the actual potential difference and the equilibrium value is known
as the overpotential, η. This leads to the Butler-Volmer equation
j+(→M,net)=je+{exp[-z+α ηF/RT]-exp[z+(1-α) ηF/RT]}
The overpotential is the voltage that is needed in order to overcome the reversal of Mn4+
back to solution as Mn2+. This voltage makes the reaction happen in only one direction.
The Butler-Volmer equation gives a final result of the net flux of cations deposited onto
the electrode.
* Mathcad document attached of calculations used in the experiment
Conclusion
This process is definitely better than the conventional method for preparing MnO2. It
gives a higher quality product that has a lower discharge rate when it is placed in the
alkaline cells. It is more cost efficient for companies to produce the material as well.
References
US Patent No: 6585881 B2 Date of Patent: July 1, 2003
Handout we received from you
Ruetschi, Paul. Cation-Vacancy Model for MnO2, Journal of the Electrochemical
Society, Dec. 1984
F  96487
J
Ion rate transfer from Mn2+(aq) to Mn4+(aq)
z  2
1

mol V
R  8.3144
GfMn4  o F z
J
o  1.21V
T  393 K
I
5 1
GfMn4  2.335  10
J
mol
mol K
 34
h  6.626
 J s
kb   1.38 10


K
 23 J


number of charge
H2SO4  .009
MnSO4  .016
mol
L
Mn2  1
U  .1 .101 .8
Eyring Equation:
  GfM n4 

kb  T 
R T

k1 
 e
h
 24
k1  4.191  10
mol
L
mol
L
at equilibrium at 393 K rate of formation of EMD
Mn2  .016
Hz
5 J
Gtot  2.299  10 
mol
z  2
H2( U)  U
mol
L
H2  .009
non equilibrium G
This is the charge on the Mn ion (+2)
With potential (non-equilibrium):
  .5
U  1.2 1.23 10
Vo( U)  U V
Gnon( U)  GfMn4  z F   Vo( U)
Gnon( U)  GfMn4  z F   Vo( U)
o  1
Gtot
z F
o  1.191V
  o 
R T
2 F
  1.201V
 ln


 Mn2 
H2
When a potential difference is added to the system the Gibbs energy changes as we move
across the double layer.
5 10
5
0
Gnon( U)
5 10
5
1 10
6
10
5
0
U
Gibbs Energy of Mn2+ going to Mn4+ under a potential
of 1.41 volts.
4 1
Gnon( 1.41)  9.745  10
J
mol
The rate coefficient equation under a potential difference:
  Gnon( 1.41) 

kb  T 
R T

k2 
 e
h
 42
k2  3.469  10
k2
k1
rate coefficient under a potential difference of -1.41 volts.
Hz
 19
The flux of charges depositing onto the
Anode is given by the following equation:
 8.278  10
F  96487
for reduction
J
mol V
GfMn4 

T
Q  
10
 19
jright( Q)  kb   exp  

z

1.062

10

C

7.13


exp

2

0.5



h
2
R T 
  R T  

m
17
solving for Q
The overpotential in this equation is determined by subtracting
the equilibrium potential from the applied voltage.
Given
  exp .25 V
 
jright( Q)   exp .25 V


 
R T
R T
F


F


 
.068
2
.4 cm

Find ( Q)  3267.5592000000000000 ln  1.4217413562637086053 10
-42
 A s 



3267.5592000000000000 ln  1.4217413562637086053 10


-42
2
5 kg m
Q  2.848  10
2
s mol
Used
A
 A s 
 J

2  mol
C cm 
2
m
2
 J
5 kg m

 2.848  10
2  mol
2
C cm 
s mol
2
m
  2 ( Q)  J  
  2 ( Q)  J  
Given
17





G
GfMn4 



T
mol
10
fMn4
T
mol  
 19
 19
kb   exp  
 exp 2 0.5
 kb  exp 

   z 1.062 10  C 7.13
   z 1.062 10  C inside_concent  exp 2 0.5
h
2
R T 
h
R T
R T 
  R T  






m
In order to find the net flux across the double layer, we need to have the average number
of cations/unit area on the outside of the OHP and the average number of cations/unit
area on the inside of the IHP. Then you can plug these into the butler-volmer equation
and find a net flux across going back and forth from the anode.
79
Find ( inside_concent) 
8.3584769549344291729 10
17 1
outside  7.13 10 
2
m
2
m
79
( butler  volmer)equation
inside 
thx mathcad...
( funny)
8.3584769549344291729 10
2
m
17
 19
je_forward  z 1.062 10
 C 7.13
10
2
 k1 exp  2 0.5

m
 C k1 inside exp 2 0.5
 19
je_backward  z 1.062 10

JenetM  je_forward exp 2 0.5

1.21 V F 


R T
1.21 V F 
R T


.050 V F 
R T
.050 V F 

  exp 2 0.5 R T   je_backward



54 C
JenetM  1.0716467040505287894 10 
2
m s
basically the reaction will not occur at equilibrium.
 56
53
2.4481756061374865327 10
10
U  .05 .1 6
 ( U)  U V
  exp .5 2 5 V
 56
10

jNetB_V

F
  exp .5 2 5 V F    1.321  108



( R T)  

( R T) 
Composition ratios of the solvents in equilibrium
R  8.3144
GfH2SO4  745000
T  393
GfMnO2  465100
GfH2O  237100
GfMnSO4  1080000


GfH2SO4  GfMnO2  GfH2O  GfMnSO4
ln( A )
R T


GfH2SO4 GfMnO2 GfH2O GfMnSO4
R T
e
CH2SO4
CMnSO4
14
 1.665  10
14
1.665  10
Composition ratios in equilibrium.
Gibbs Free energy of formation for the entire reaction (non-equilibrium):
CbH2SO4  .009
CbH2SO4
CbMnSO4  .016
CbMnSO4
 0.563
GfH2  0




  2.299  105
GfMnSO4  R T ln CbH2SO4  GfH2  GfMnO2  GfH2O  GfMnSO4  R T ln CbMnSO4
Gtot
5
2.299  10