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REDUCTION ON ELECTROLYSIS

Introduction

Electrochemistry is the branch of chemistry that deals with the interconversion of

electrical energy and chemical energy. Electrochemical processes are redox reaction in

which the energy released by a spontaneous reaction is converted to electricity or in

which electrical energy is used to occurring a no spontaneous reaction. Redox reactions

involve the transfer of electrons from a reducing agent to an oxidizing agent.

Electrochemical cells are used in many of electrical tools like radio, automobile

machine, toys, space equipment and so on. The examples of electrochemical cell that

usually used in those equipments are the dry cell battery and lead storage battery. When

they are in using, there are chemical reactions that result electric current. In contrast,

there are the usages of electrical energy to result the chemical reaction

According to those phenomena, we know that electrochemical cells are divided

into two groups beside the reaction type. They are galvanic cell and electrolytic cells.

Galvanic cell (Voltaic cell) is an electrochemical cell that generates electricity by means

of a spontaneous redox reaction. When electric current from an external source is passed

through an electrochemical cell, a no spontaneous reaction called electrolysis occurs,

and the cell is called electrolytic cell.

Electrolyte is a solution that if it is diluted on solvent (water) produce electricity.

The classification of electrolyte:

Strong electrolyte is the electrolyte that has ability to produce electric current

perfectly, because this ion is dissociated perfectly.

Weak electrolyte is the electrolyte that has characteristic produce electric current

badly because the molecule is not dissociated perfectly.Their Power to become strong or weak electrolyte affect by certain aspect as will be

explain below. Strong electrolyte its mean good conductor of electricity and weak

electrolyte mean bad conductor.

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The Relationship between electrolyte and chemical bonding

a. The electrolyte properties in ionic compound.

The compound that on the atom is combined by ionic bonding is called ionic

compound. On this compound, there are particles that called cations (positive charge)

and anion (negative charge).

If the ionic compound was dissolved in water, so the ions will free moving. The

free moving ion will produce electric conductivity.

b. The electrolyte properties in covalent compound

The compound that the atom is combined by covalent bonding is called covalent

compound. The covalent compound is polar, because the difference of electro negativity

that higher enough between two atoms to produce double pole that polarized so produce

polarity molecule. But, when its dissolved by water, so produce electricity.

It caused by the solvent (water) is the double pole molecule that is the polar

covalent compound that help to decompose covalent compound to produce negative or

positive ions.

example:

1. HCl(l) + H2O(l) H3O+

(aq) + Cl(aq)

2. HBr + H2O(l) H3O+

(aq) + Br(aq)

The covalent compounds that produce electricity badly are ammonia and vinegar

solution that involved in weak electrolyte. This compound will be react with water to

produce ion, but only a half will decompose to becoming ion so the total ion in solution

in small amount.

Beside electrolyte solution, metal also can to produce electricity because metal

has electron that can free moving on conduction wire. So, the electricity can be taken by

ions or electron. This fact is called conductivity. The electric conductivity is thetransport phenomenon that is the charge moving (electron or ion) through the system.

The conductivity (L) is the opposite from resistor (R). The conductivity formula are

R

lL=

:soR

=l

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LKorKorKLorl

L kLl

ll=

=

=

=

Note :

K= conductivity (mho.m-1), l/ = cell constant conductivity (m-1).

Based on formula above, the electrolyte conductivity is proportional with the electrode

surrounding but opposite with the distance of electrode.

Electric Conductivity Mechanism

The power of a conductor involves the transfer of electrons from a high negative

potential to the lower negative potential. The mechanism of this transfer is not the same

for various conductors. In an electronic conductor, such as solids and molten metal,delivery takes place via direct electron transfer through conduction with the influence of

the applied potential. In this case the constituent atoms are electrically conductive not

involved in the process. However, the electrolytic conductor, which includes an

electrolyte solution and salts, conductivity takes place through the displacement of ions,

both positive and negative toward the electrodes. This migration involves not only the

flow of electricity from one electrode to another electrode, but also involves the

transfort material from one part to another conductor. Flow of electricity in electronics

conductivity is always accompanied by chemical changes at the electrodes and the

reaction is specific and will depend on the substances making up the conductor and also

its electrode.

The mechanism of the flow of electricity through an electrolytic conductors will

be more easily understood through the following example:

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Figure 1. electrolysis cell

Electrolytic cell in the image above, consists of two electrodes of copper (Cu)

associated with the direct current source B and the second electrode is dipped into a

solution of two copper sulphate, CuSO4. Electrode D is connected to the negative pole

(-) of B and A electrodes are connected to the positive pole (+) from the B. In the

solution there are ions Cu2+ and SO42-. If the circuit connected to an electric current flow

occurs. This resulted in the dissolution of copper in the electrode A. Electrode D (which

is connected to the negative pole current source) is negative because it is rich in charged

electron from B. Electron-electrons merged with copper ions (Cu2+) in solution to form

Cu precipitate on the electrode D. The reaction can be written as follows.

Cu2+ + 2e- Cu

Reaction on the D electrode is the reduction reaction. Thus D electrode is called the

cathode. A electrode is called the anode, the place for the oxidation reaction. Oxidation

reactions that occur at the electrode A is as follows.

Cu Cu2+ + 2e-

It can be seen that the two electrons at the cathode is used to react to form Cu

and simultaneously two electrons out of the anode due to a change of Cu to form Cu2+

.

Results exemption is the transfer of two electrons in the outer circuit from anode to

cathode.

When the circuit is closed, the positive ions or cations will move to the cathode and

negative ions or anions will move to the anode which causes an electrical current can

flow. Process flow of electric current through the electrolytic conductor be accompanied

with chemical reaction called electrolysis.

From the above description of the mechanisms of electrolysis can be concluded that theelectrons in and out of the solution occurs through a chemical change in its electrodes.

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The dissociation degree of Weak Electrolytes

According to Arhenius, degree of dissociation , weak electrolyte can be

expressed by the following equation.

o

=

Where = molar conductivity of electrolytes in the concentration C and 0 =

molar conductivity at infinite dilution.

For electrolytes with dissociation reaction AB A+ + B-,

The equilibrium dissociation constant of the above reaction can be determined in

the following way. Let imagine molarity of electryte AB = a mol L-1 and its

dissociation degree , then:

AB A+ + B-

Initial a - -

React a a a

Equilibrium a(1- ) a a

Dissociation equilibrium constant,[ ][ ]

[ ] )1()1())((

2

=

==

+ a

a

aa

AB

BAK

Specific Conductivity and Molar Conductivity

a. Specific Conductivity

Specific conductivity of electrolyte solution cannot be measured directly, which

can be measured directly is the resistance of an electrolyte solution. Resistance (R) of an

electrolyte solution cannot be measured properly if used direct electrical current (DC).

This is because electrolysis will occur and cause changes in electrolyte concentration

and accumulation of electrolysis on the electrode will change the solution resistance. To

eliminate this, can be used alternating current (AC). The electrodes used is platinum

coated with platinum black. The resistance cell is stored in a bath with T fixed (constant

temperature) and placed on one side of the Wheatstone bridge.

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Figure 01. Wheatstone bridge

Based on that images, can be obtained calculation of resistance formula is as

follows.

or31

2

R

R

R

R=

1

32

R

RRR =

After R is gotten, the specific conductivity can be determined as formula bellow.

1=K ,

A

lksel =

R

kK sel=

The specific conductivity of an electrolyte is useful in determining the cell

constant where the constant cell has a value that does not depend on the type of solution

when the distance between the two electrodes in the conducting cells remain. One

example of a solution that is often used to determine the cell constant is a solution of

KCl. Value of the conducting type of KCl solution at various temperatures can bepresented in the table below.

Table 1. Value of specific conductivity of KCl

No. Concentration (mol L-1) K (ohm m-1)

0oC 18 oC 25 oC

1 1,00 0,5430 9,8200 11,1730

2 0,1 0,7154 1,1192 1,2886

3 0,01 0,0775 0,1223 0,1411

b. Molar Conductivity

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A solution with different concentrations will have a different specific

conductivity because they contain different concentrations and different number of ions.

Therefore, to obtain a measure of ability to transport electricity from a particular

electrolyte is called the molar conductivity, m.

C

Km =

Where:

K specific conductivity (ohm m-1)

m molar conductivity (S m2 mol-1)

C concentration of electrolyte (mol dm-3 or mol L-1)

Based on observations made by Kohlrausch, the relationship between molar

conductivity and specific conductivity toward concentration are as follows.

1. For a strong electrolyte, the conducting type of electrolyte will rise rapidly with

increasing concentration, whereas for a weak electrolyte conductivity type of

electrolyte will rise slowly with increasing concentration. It caused of differences in

the ionization both of electrolyte, strong electrolyte is ionized completely while

weak electrolyte will partially ionized.

2. For the strong and weak electrolytes, the molar conductivity will increase withrising dilution and would be worth a maximum at infinite dilution.

The relationship between the molar conductivity at a certain concentration (m)

and molar conductivity at infinite dilution (0) toward concentration (C) for strong

electrolytes are as follows.

Cbo =m

Graph of molar conductivity with the square root of concentration for some of the

electrolyte can be described as follows.

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Figure 02. Relationship between molar conductivity toward the rood quadrate of

electrolyte concentration

Based on the graphic above, can be explained as the follow.

1. Plot of molar conductivity against the square root of concentration are

straight lines for strong electrolytes, and the steep curve for a weak

electrolyte

2. Extrapolation data of molar conductivity to infinite dilution limit is known as

the molar conductivity (0) which is based on the mean free migration of

ions, as proposed by Kohlrausch.

Based on that law, the molar conductivity of each electrolyte at infinite dilution

(o) is the sum of the molar conductivity of ions at infinite dilution. This is due to the

infinite dilution, each ion in solution can move freely without being affected by the ion-ion opponent. If the number of positive ions and negative ions is expressed as v+ and v-

as well as the molar conductivity at infinite dilution of positive ions and negative

expressed as o+ and o

-, then it can be formulated as follows.

oo

o vvA ++ +=

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For example, v+= v- = 1 for HCl, NaCl and CuSO4, and v+ = 1, v- =2 for MgCl2. The

main application of the Kohlrausch law is to determine the value of limit molar

conductivity of the weak electrolyte. For example an electrolyte AD, molar conductivity

at infinite dilution (limit of molar conductivity) determined from the determination of

the molar conductivity solution of strong electrolyte AB, CD, CB by using the

following equation.

(CB)A(CD)A(AB)A(AD)A oooo +=

o

D

o

A

o

B

o

C

o

D

o

C

o

B

o

Ao-(AD)A

+=+++=

++++

This simple result, which can be understood on the grounds that the ion migrate

independently in the limit of zero concentration. It can uses to predict the limiting molar

conductivity of any strong electrolyte from the data bellow.

Table 02. Limiting ionic conductivity in water at 298 K (in /mol-1S cm2)

H+ 396.6 OH- 199.1

Na+ 50.1 Cl- 76.3

K+ 73.5 Br - 78.1

Zn2+ 105.6 SO42- 160.0

Ba2+

127.2

For example, the limiting molar conductivity of BaCl2 in water at 298 K, as follow;

0m = (127.2 + 2+ 73.6) mol-1S cm2

Dissociation degree of weak electrolyte

Based on Arhenius, the dissociation degree, is

o

=

In which = molar electrolyte conductivity on C and 0= molar conductivity on

unlimited dilution.

for electrolyte with reaction AB A+ + B-, the equilibrium constant of

dissociation can be determined as this way: example the molarity of electrolyte AB=a

mol L, so the dissociation degree is

AB A+ + B-

Initial a - -

Reaction a a a

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Equilibrium (1- ) a a

The dissociation constant,[ ][ ]

[ ] )1()1())(( 2

=

==

+ a

a

aa

AB

BAK

Electrolysis and Electron Mobility

Electrolysis has been studied quantitatively by Michael Faraday in 1820. He

stated that the amount of charge that passes through the electrolyte solution is

quantitatively related to the amount of products produced at the electrodes. This

quantitation was introduced as the Faraday unit charge of one mole of electrons.

1F= NA x electron charges =6.02.1023 x 1.6.10-9 Cmol-1

In 1914, Bate and Vinal doing research on a number of products produced by

passing electrical current through an electrochemical cell. Circuit containing a silver

coulometer, where silver is deposited as a result of the reaction on the electrode: Ag + +

e- Ag and other cells that contain an inert electrode in a solution of KI I2 generated

from the electrode reaction: 2I- I2 + 2e-.

In an experiment with a charge current is passed through a circuit C 3666.4 is

4.009903 grams of Ag deposited and 4.8224 grams of I2 that is formed is determined by

titration. According to the results of this experiment, what is the value F (charge of one

mole of electrons)? (Ar Ag = 107.87; Ar I = 126.7)

Answer :

Product of silver coulometer :

Ag precipitated = = 0.038 mol

Charge to precipitate 1 mol Ag = = 96.485C.mol-1

If 1 electron nedded to precipitate 1 atom Ag from Ag -. So 1 mol electron =96.485C

In forming I2 from KI solution:

Amount I2 formed = = 0.019 mol

Charge 1 mol I2 = = 192.970Cmol-1

To produce 1 molecule of I2 from I- released 2 elecctrons, so1 mol electron = 192.970/2

= 96.485C

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In electrolysis, direct electric current is passed through the solution produces a

chemical reaction on the electrode, where the positive ions move toward the negatively

charged electrode (cathode), while negative ions move toward the positively charged

electrode or anode. Due to the movement of these ions causes of differences in

concentration in certain areas in the electrolysis cell. Movement / mobility of ions can

be determined by measuring the distance traveled each ion in a given time that can be

formulated as follows:

is the ionic mobility (m2v-1s-1), x is distance (m), t is time in second (s), dE/dx is the

strenghtness of the field (v m-1).

Field strenght can be formulated as follows:

t is the electric current (A), A is the cross-sectional area (m2) and K = conductivity type

(mho m-1). Ion mobility of each ion can also be specified if known ion conductivity at acertain concentration which is formulated as follows:

is the molar ionic conductivity at a certain concentration, F is the Faraday constant,

and Z is the valence of ion

Mobility of ions can be linked with transport numbers through the following equation:

t+ = or t-=

t + is the positive ion transport number and t-is the number of negative ion transport.

t- + t+ =1

Transport Number

It is the fraction of the current carried by each ion that is present in solution

=

=

1i

i

I

It

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The sum of transport numbers in a solution should give one,

1=++tt

Where + tt and are the sum of the transport number for cations and anions

respectively.

When a solution contains c1 concentration of NaCl and c2 concentration of KNO3 the

transport number of Na+ can be given as

( )2211

1

333czuczuczuczu

lFA

lFAczu

t

NONONOKKKClClClNaNaNa

NaNaNa

Na

+++

=

We used equation for substituting the current caused by different ions. This long

equation is simplified by the following facts:

Na = 1,zNa = 1, K= 1,zK= 1, Cl = 1,zCl = 1, NO3 = 1,zNO3 = 1

( ) ( )321

1

NOKClNa

Na

Na

uucuuc

cut

+++

= 11.

When we have a singlez: zelectrolyte with mobilities u+and u-, the transport number

of anion is

+

+= uu

ut

This equation can also be given in molar conductivity representation

+

+

=t

The electrical current delivered by the solution through the displacement of

positive ions and negative in the solution. The fraction of electric current carried by

each ion is not the same. For example (i) in solution MgSO4, ion currents carried by

Mg2+ as much as 0.38 and 0.62 parts of currents carried by SO 42-. (ii) in HNO3 solution

is only 0.16 which is transported by the flow of NO 3- ions and the remaining 0.84 parts

must be transported by the H+ ions. Based on the examples can be seen that the SO42-

ions and H+ can carry more current than the ion Mg2+ and NO3-. This is due to ion-ion

SO42-and H+ can move faster in solution. If the two ions in solution move with the same

speed, then each ion will carry the same amount of electric current at a given time. But

if the speed of the ions is not equal, then the period of time, the faster ions will carry a

larger fraction of the flow. The fraction of total current carried by each ion in solution is

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known as carrier numbers. Numbers of transport is also called the number delivery /

transport numbers.

Q

Qt i

i

=

In this case ti is transport number for ions i. and Qi is a charge for ions I and Q is

the total charge.

The quantitative relationship between current fractions that carried out by ions

and its velocity can be described follow. Reviewed is conducted on two parallel plat that

separated as far as I that filled by electrolyte solution.

On both electrode is applied the different potential V. If the total of cations N

with velocity v+, the charge is z-e but the total of anions N by velocity v - and the charge

is z.e in which e is the total electricity that depend on the charge unit, and on dt, cations

move as far as v+ dt and all cations that place on volume ABCDEFGH will achieve

negative electrode on dt. The amount of cations on its volume is a fractions from all

cations that occur (v.dt/1). Thus, the positive charge dQ- that through parallel plate to

negative electrode on dt is l

Nevz+++

and the strong current that taken by cations is as

follow:

l

Nevzl ++++=

Similar to the cation, anion contribution to the strong currents expressed by the

following equationl

Nevzl =

Thus the total strong currents carried by both anions are as follows

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+ += II1

Requirement electronity solution must be satisfied that the total charge equal to the total

charge of cations anions

++ = zNzN (this applied if the charge is different)

Therefore, the total current can be expressed as follows

( )

111

++++++++

=+

=+

=vveNzevNzevNzevNzevz

I (this equation is used

if the charge is same)

Based on the relationship of dQ=1dt, the transport number of t i that define as

fraction from total current that taken on each ions.

Q

Qt 1=

Its also can expressed on the form ofI

Itt

1= because the dt is equal. So, the total

fraction that taken by cation is:

I

It ++=

The fraction of total current that taken by anion is:

I

It =

By each t- and t+ is the transport number or conductivity number from cations and

anions. Based on the equation above, the formula that obtained is:

+

+=v

v

I

I

So the transport number of ions is proportional with the velocity. If there are

similar velocity between v- and v+, so t- and t+ give similar contribution to electrical

current transfer. If v- and v+ is different, so the t- and t+ will be not same and both ion

will take portion from total current that different.

Transport numbers of ions in solution can be determined by two methods is the

method of Hittorf and moving boundary method.

A. Hittorf Method

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To determine the transport with Hittorf based on the changing the amount of

electrolyte (the concentration of electrolyte) in the electrode surrounding. The

changing of concentration has relationship with the mobility or the velocity of

motion from the concerned ion. Cell that used on Hittorf method is like follow:

In hittorf cell, the electrolyte tube inserted in A and electrolysed at a certain time.

Amount of electrical current that flows in total can be determined from the

coulometer readings. The amount of electric current carried by cations can be

determined based on analysis of the anode room solutions, while the amount of

electric current carried by anions can be determined based on analysis of the

solution in the cathode chamber. Transport numbers of cations and anions can be

determined using the following equation

coulometeronedprecipitatthateqivalentmetaloftotalthe

anodeinlostanion thateqivalentoftotalthet

coulometeronedprecipitatthateqivalentmetaloftotalthe

anodeinlosttcation thaeqivalentoftotalthet

=

=

+

To determine the transport number of experimental data, can be explained

as follows. For example, AB is an electrolyte with equivalent weights are Be, then:

1) If the silver coulometer used in experiments to measure the amount of current

that flows, equivalent silver precipitate in the same coulometer

zw==

108

'

AgBe

w', where w is the amount of silver deposited and Be is the

equivalent weight. If the copper coulometer used in experiments to measure the

current flowing and w'' is the amount of copper deposited in the coulometer, the

equivalent of copper deposited in the coulometer with

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zw

==8,31

''

CuBe

'w'

2) Let a grams of anode solution containing b g AB (before electrolysis). Weight of

water = (a-b) g, ie (a-b) grams of water containing b g AB, so the equivalent AB

prior to electrolysis = b / Be = y

3) Let c grams anode solution containing d grams of AB (after electrolysis). Weight

of water = (c-d) g, ie (c-d) containing gram gramb)(adc

dAB

= so that

water will contain an equivalent AB after electrolysis = xBedc

bad=

)(

)(

Case A : If the electrodes are used in cell hittorf is an inert electrode

concentration decreased around the anode = (y-x)

z

xylA

=

Case B : If that is not an inert electrode is used, the concentration of A- would

rise before the B-

attack the electrode, so that x> y. The increase in theconcentration around the anode = (x-y) equivalent. But the increase is

actually supposed to be equivalent if there is no z- ion A moves out.

Then ion A-kumlah moving out = z-(x-y) equivalent.

z

yxzlA

)( =

Example :

1. In the Hittorf cell which is used silver electrode and AgNO3 as electrolyte,

the amount of current are flowed. Before electrolysis, the anode space

contains 0.2157 grams of AgNO3 in 24.96 grams water and after electrolysis

27.5 mL (27.8538 gram) of solution which contain AgNO3 required 49.8 mL

of KCNS solution 0.031 N. Coulometer which is connected serially in

Hittrof cell precipitating 0.014 grams of copper by the amount of same

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electric current. Calculate transport number of Ag+ (Mr AgNO3 = 170, Ar

Cu=63.6)

Solution:

The eqivalent amount of Cu which is precipitated =

eqivalentgekiv

g 41

104.48.31

014.0

= that is increasing Ag in anode.

Before electrolysis : 24.96 g of water contain 0.2157 grams of AgNO3

After electrolysis : N AgNO3N

mL

NmL0599.0

)5.27(

)0331.0)(8.49(=

The amount of AgNO3 in 27.5 mL =

gmL

gNmLN28.0

1000

)170)(5.27)(0599.0( 1=

The amount of water in 27.8538 grams of electrolyte = 27.8538 0.28 =

27.5738 g

So that: 27.5738 g of water before electrolysis will be containing

AgNO3 = gx 238.05738.2796.24

2157.0=

The amount differences of AgNO3 = 0.28 0.238 = 0.042 g = 1.170

042.0eqivg

g

=2.47x10-4 eqiv

The increasement theoritically in anode space from Ag- have to 4.4 x 10-4

eqivalent, but actually just 2.47 x 10 -4 eqivalent, so that (4.4 2.47) x 10-4 = 1.93

x 10-4 eqiv have migrated.

44.0104.1

1093.14

4

==

eqivalentx

eqivalentxtag

2. For 0.2 % of NaOH solution was electrolysed by using Pt electrode in Hittrof

cell. Electric current about 120 mA is flowed for 30 minutes. After

electrolysis, 25.64 grams of cathode solution contain 0.0560 grams of

NaOH. Calculate transport number of Na- ion and OH- ion!

Solution:

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The amount of current which is flowed =

)96500(

)6030)(10120(

)96500(

))((1

3

1

=

Ceqiv

sxAx

Ceqiv

tsiA

Before electrolysis: 100 grams of cathode solution contain 0.2 grams of NaOH

After electrolysis : 25.64 grams of cathode solution = 0.056 grams

100 g cathode solution = ggxg

g2188.0100

6.25

056.0=

The increasement of concentration in cathode = 0.2188 0.2 = 0.0188 g

=

eqivalentxeqivg

g 41 107.4.40

0188.0 =

B. Moving Boundary Method

Determination of transport numbers by the method of boundary motion can be

described in the following figure

In the moving boundary method, the boundary between the two movements are

ionized electrolyte solution was observed as the flow of current.

MX is a salt that will be observed and NX is a more concentrated salt

(functioning as an indicator). Indicator solution as a lower layer and a solution of

79.021.01

21.010238.2

107.44

4

==

==

OH

Na

t

eqivalentx

eqivalentxt

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MX on the top layer with clear boundaries. When the current I is passed during the

time t the limit of the two solutions will move from ab to cd. When the volume

abcd is V liters of solution and the concentration = c mol / L, then the number of

ions MZ = NA.cV and the amount of M- charge is z.e.NA.c.V=z.F.c.V, where z =

positive charge. Analysis of boundary movement shown that z.F.c.V is the current

which carried by positive ion at t. The comparison of z.F.c.V to 1 t called as

positive transport ion.

I=It

VcFz ..+

Example:

1. In boundary movement experiment, KCl solution 0.1 M move in top layer ofCdCl2 on bottom layer. The surface area length of tube is 0.276 cm

2. Cell is

defended at 250C. If the current is 6.25x10-3 A is flowed through solution about

4000 second of boundary movement 4.6 cm. How the currenr which carried by

K- ion?

Solution:

49.0)sec4000)(1025.6(

)276.06.4)(1.0)(485.96)(1(..3

111

==+

=

ondAx

LxmolLCmol

It

VcFzI

2. In boundary movement methode, electric current about 25 Ma is flowed in HCl

solution 0.05 M for 10 minutes. If the capillary radii is 3 mm. Calculate the

distance of H+ if the transport number of H+ ion 0.892.

Solution:

33

11

3

10773.2)05.0)(485.96)(1(

sec)6010)(1025)(892.0(..... dmx

molLcmolxAx

FczIttV

ItVcFzI

===+=

cmmmmm

dmmmdmx

r

VI 8.914.98

)3)(14.3(

)10)(10773.2(22

3333

2====

Conductivity In Electrolyte

A. Ionic Conductivity

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Inverydilutesolution, the solution behavior is close to ideal. In this dilute

condition the molar conductivity state is the sum of the ions conductivity contained in

the electrolyte. Molar conductivity for each ion at indefinitely dilution is given by:

In this case, t+ and t- are transference number from positively and negatively ions

for the indefinitely dilution. v+ and v- are the amount of positively and negatively ions

that associate in solution.

To calculate the conductivity of ions molar at certain concentration, the above

equation can be enlarged, so obtained the following equation:

In this equation, Kohlrausch law about average free migration from ions is not used.

Therefore, the conductivity of ions molar can be calculated by this way depends on the

electrolyte properties, not the properties of the ions itself.

Example:

1. Molar conductivity of 0.05 M HCl is 399.09 x 10-4 m2 ohm-1mol-1and the

transference number of H+ ion is 0.829. Determine the molar conductivity of H+

and Cl- ions.

Solution:

. Because the V- of HCl is equals to one, so

85 x 10-4 m2 ohm-1 mol-1

= x 10-4 m2 ohm-1 mol-1

2. Determine the molar conductivity of CaCl2 at indifenitely if known the ion

mobility of Ca2+ and Cl- are 6.16 x 10-8 m2 v-1 s-1 and 7.91 x m2 v-1 s-1 .

Solution:

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B. Concentration Effect to the Electrolyte/Ionic Conductivity

Electrolytic conductivity is influenced by the concentration of electrolyte ions. If

the electrolyte concentration increases, the molar conductivity electrolyte is reduced.

According to Arrhenius, the decrease in molar conductivity is due to the increased of

concentration caused by the small ionization degree of electrolytes ( ). Ionization

degree of electrolytes can be determined by using the following equation:

At infinite dilution the degree of ionization of the electrolyte is equal to one. The

equation above is only valid for weak electrolytes where the influence of the ion

interactions is minimums.

According to the Debye-Huckel, the decrease in conductivity or the increase

electrolyte concentration are caused by the existence of an attractive force between the

ions. Where, the ions are always surrounded by ions and the shape is speris. But, with

the electric field, the shape of the ion atmosphere will be distorted and no longer speris.Distortion of ion atmosphere tends to counteract the electric field.

Atmospheric ions are effectively formed at high electrolyte concentrations. The

movement of ions will be slowed by the presence of atmospheric ions. Thus the

conducting of electrolyte will be reduced at higher concentrations.

C. Conductivity Measurement Application

Electrolytic conductivity measurements can be applied in a variety of purposes

which is to determine: 1) Dissosiaton Degree and Weak Electrolyte Equilibrium

Constant, 2) Solubility Product of Difficult Soluble Salt, and 3) Conductometry

Titration.

a. Dissociation Degree and Weak Electrolyte Equilibrium Constant

By using the equation can be determined the degree of dissociation of Arhenius

and equilibrium constants of weak electrolyte.

Example:

1) Acetic acid solution0.0185M has conductivity2.35x 102ohm-1. If

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thecellconstantis 105m-1andaceticacidmolarconductivityatinfinite dilution is

391x 10-4ohm-1m2mol-1. Calculate the degree of ionization and the equilibrium

constant of acetic acid!

Solution:

CH3COOH CH3COO- + H+

Initial: c= 1.85 x 10-2 M 0 0

Eq: c-c=c(1-) c c

2) Dissociation constant of n-butyric acid at 25C is1.515xl0-5M and molar

conductivity at infinite dilution 382x10-4

ohm-1

mol-1

m2. Calculate the degree

of dissociation and conductivity type of butyric acid with a concentration of

0.01M

C3H7COOH C3H7COO- + H+

(Because is smal)

so:

From the Arhenius equation, obtained:

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b. Solubility Product of Difficult Soluble Salt

Salt Solubility Product which is insoluble like BaS04 and AgCl are can be

determined by measuring the solution conductivity.

Example:

Determine the solubility product of BaSO4 if the molar conductivity of ions

in indefinitely dilution at 25C from Ba2+ is 127.28 x 10-4 ohm-1mol-1m2 and

SO42- is 160x10-4 ohm-1mol-1m2. The density conductivity of BaSO4 is 3.01 x

10-4 ohm-1m.

Solution:

M

Ksp BaSO4= [Ba2+][SO4

2-]

= (1.048 x 10-5) (1.048 x 10-5)

= 1.1 x 10-10 mol2 dm-6

c. Conductometry Titration

Conductivity measurements can be used to determine the titration end point.

For example is titration of strong acid with strong base.

HCl + NaOH--- NaCl + H2O

In the titration of strong acid by strong base, delivery will decline until the

equivalence point because H+ion is neutralized by OH- ions to form H2O.

After the equivalence point is reached, the excess of OH- ions will increase

the electrolyte conductivity. Curve is the relationship between the

conductivity to amounts of base which added is as follows:

Conductivity

Equivalent Point

mL base

total

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Figure 2.6. Titration Curve of strong acid by strong base

For the titration of weak acid with strong base is such as acetic acid with

sodium hydroxide. At the beginning of the titration, the conductivity will

increase slowly because of the formation of ionized sodium acetate

completely from neutralization of acetic acid which partially ionized by

NaOH. After the equivalence point is reached, conductivity rises quickly due

to an increase in OH- ions. Conductivity curve against the number of bases

which added is as follows:

Figure 2.7. Titration curve of weak acid by strong base

Conductivity

Equivalent Point

mL base

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REFERENCES

Atkins, Peter, Julio De Paula. 2006. Physical Chemistry, Eighth Edition. Oxford

University Press

Suardana, I Nyoman, I Nyoman Retug. 2003. Buku Ajar Kimia Fisika III. IKIP Negeri

Singaraja

Sukardjo.1997. Kimia Fisika.Jakarta: Rineka Cipta