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ELECTROCHEMISTRY
CONDUCTIVITY
Conductivity (electrolytic conductivity) is the reciprocal of resistivity. It may also be
defined as the conductance of a solution of 1cm length and cross sectional area of 1
square cm. It is represented by the symbol Kappa (K).
Mathematically we can write;
ĸ=1⁄ƿ where ƿ is the resistivity, the units of K are ohm-1 cm-1or 𝑠𝑐𝑚−1
The conductivity, K, of an electrolytic solution depends on the concentration of the
electrolyte, nature of the solvent and temperature;
𝐿
Ƿ = 𝑅 𝐴⁄𝐿 OR R = ƿ𝑅 𝐴
1 𝐿
ĸ = ƿ = 𝑅𝐴

Measurement of Conductivity
The electrolytic conductivity of an electrolyte is measured by determining the resistance
of a solution between two electrodes of cross sectional area (A), separated by a fixed
distance (l).
An alternating voltage is used to avoid electrolysis. The resistance is measured by a
conductivity meter. The resistance (R) of the solution in the conductivity cell is
measured by connecting the electrolyte in the Wheatstone bridge.

It is difficult to make a cell to exact dimensions so that conductivity is measured directly.

Usually a standard solution (which is 0.1M Potassium Chloride) of known conductivity is
1
put in the cell and the conductance ( ) of the cell is measured. This gives the factor
𝑅

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𝐿
(the cell constant ( ))by which the conductance of the solution in the cell must be
𝐴

multiplied to give the conductivity of the solution

𝐿1 𝑅
= R is the resistance of the solution in the conductivity cell.
𝐿2 𝑟

The position balance X may be indicated by a minimum signal on an oscilloscope or

minimum buzzing sound in earphones.
𝑙1 𝑅
= Where R is the resistance of the solution in the conductivity cell.
𝑙2 𝑟
𝑙1
R= ×r
𝑙2
1 𝐿
K= 𝑅 × 𝐴

Since R and are known K can be calculated.

Variation of conductivity (K) with concentration (C)
The variation is different for strong and weak electrolytes
For strong electrolytes, conductivity increases with increase in concentration at first
because the number of ions per unit volume carrying the current increases. At very high
concentrations, conductivity decreases because of increase in ionic interferences
resulting in decrease in ionic freedom.

For weak electrolytes, conductivity decreases with increase in concentration because

the number of ions in the solution increases. But at very high concentration it decreases
because of decrease in degree of ionization.

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Exercise;
The conductivity of a 0.1 MKCL solution at 250C is 1.280 what is the conductivity
and resistance of the solution for which the electrodes have an effective area of
2.037cm and are separated by a distance of 0.53cm.

Molar conductivity is defined as the conductivity of an electrolyte solution divided by

the molar concentration of the solution.
It may also be defined as the conductance of a volume of a solution containing one
mole of an electrolyte kept between two electrodes with unit cross sectional area and
unit length apart.
It is represented by the symbol Lambda (ƛ)
Mathematically, we can write; ƛ= ĸ⁄𝐶 or ƛ=ĸµ where µ is the dilution. The units for ƛ are
𝑜ℎ𝑚−1 𝑐𝑚2 𝑚𝑜𝑙 −1 or 5𝑐𝑚2 𝑚𝑜𝑙 −1 .
Question
A 0.01M solution of sodium chloride is placed in a conductance cell whose electrodes
are 1𝑐𝑚2 in area and 0.5cm apart. The resistance of the solution was found to be
421.93Ω. If the conductivity of water is 8×10−5 𝛺 −1 𝑚−1. Calculate
i. Conductivity of the solute ( sodium chloride)
ii. The molar conductivity of sodium chloride
Solution

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Variation of conductivity with concentration for potassium chloride and ethanoic

acid

The figure above compares the molar conductivity with concentration for a weak
electrolyte (ethanoic acid), and a strong electrolyte (potassium chloride).
For the strong electrolyte molar conductivity increases with increase in dilution because
ions get further apart and ionic interferences get reduced and ionic mobility increases.
At infinite dilution it reaches a limiting value because ions are further apart and all
interferences have been eliminated so that any further dilution results in no change in
molar conductivity. Molar conductivity at infinite dilution (the limiting value, ƛ∞) can be
obtained by extrapolation.
For the weak electrolyte, the molecules are slightly or partly dissociated into ions. The
dissociation increases with increase in dilution, and so the molar conductivity increases
as the electrolyte is diluted. The weak electrolyte does not reach a limiting value as the
electrolyte continues to dissociate with dilution.
Kohlrausch established an empirical relationship between ƛc and √𝑐
ƛc =ƛ∞ - ĸ√𝑐
For the strong electrolyte the plot of ƛc against √𝑐 is approximately linear and again ƛc
can be obtained by extrapolation. At low concentration there are few ions per unit
volume, ionic interferences are low and mobility of ions is high hence high molar
conductivity. As concentration increases, number of ions per unit volume increases and

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ionic interferences increases so the drag on the mobility of the ions increase hence the
decrease in molar conductivity with increase in concentration as seen in the figure
below;

For the weak electrolytes molar conductivity is high at low concentration due to high
degree of dissociation, but decreases as concentration increases due to decrease in
degree of dissociation and hence number of ions per unit volume decreases.

KOHLRAUSCH’S LAW OF INDEPENDENT IONIC MIGRATIONS

It states that the molar conductivity of an electrolyte at infinite dilute is equal to the sum
of the molar conductivities of the ions produced by the electrolyte.
ƛ0 = ƛ+ 0 + ƛ− 0 where ƛ+ 0 and ƛ− 0 are the molar conductivities of cation and anion
respectively in the case sodium chloride,
ƛ0NACL = ƛ0𝑁𝐴+ + ƛ0𝐶𝐿_
For Barium Chloride,
ƛ0𝐵𝑎𝐶𝐿 = ƛ0𝐵𝑎+ + 2ƛ0𝐶𝐿_
The law holds for all types of electrolytes. Therefore for ethanoic acid;
ƛ0𝐶𝐻3 𝑂𝑂𝐻 = ƛ0𝐶𝐻3 𝑂𝑂− + ƛ0𝑂𝐻 +
At infinite dilution, the dissociation of the electrolyte is complete and hence each ion
makes a definite contribution to the molar conductivity irrespective of the nature of the
other ions associated with it.

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Experimental basis and theoretical explanation of Kohlrausch’s law.

Kohlrausch observed that at infinite dilution the difference between the molar
conductivities of sodium and potassium salts is constant dependent of the associated
anions.
For example,
ƛ0KCL – ƛ0NaCL = 21.1 ohm𝑐𝑚2 𝑚𝑜𝑙 −1
130 108.9
ƛ0𝐾𝑁𝑂3 - ƛ0𝑁𝑎𝑁𝑂3 = 21.1𝑜ℎ𝑚𝑐𝑚2 𝑚𝑜𝑙 −1
126.3 105.2
It also true to say that the difference is constant independent of the cation associated
with the anions.
For example,
ƛ0KCL - ƛ0𝐾𝑁𝑂3 = 3.7 𝑜ℎ𝑚−1 𝑐𝑚2 𝑚𝑜𝑙 −1
130 126.3
ƛ0NaCL - ƛ0Na𝑁𝑂3 = 3.7 𝑜ℎ𝑚 −1 𝑐𝑚2 𝑚𝑜𝑙 −1
108.9 105.2

APPLICATIONS OF KOHLRAUSCH’S LAW

1. Calculating the molar conductivities of weak electrolytes.
E.g. the calculation of molar conductivity of methanoic acid, a weak acid is illustrated
below
The molar conductivities of HCOONa, NaCL and HCL are 113, 101.2, and 397.8
𝑜ℎ𝑚−1 𝑐𝑚2 𝑚𝑜𝑙 −1 respectively.
ƛ0HCOOH = ƛ0𝐻𝐶𝑂𝑂− + ƛ0𝑂𝐻 +
This comes out of the expression;
ƛ0HCOOH = ƛ0HCOONa + ƛ0HCL – ƛNaCL
=101.2 + 397.8 – 113
=
DETERMINATION OF THE DEGREE OF IONISATION (ᾳ)
Since molar conductance is proportional to the number of ions in the solution, the
degree of ionization is equal to the conductance ratio as given below

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ƛ𝑐 ƛ𝑐
ᾳ= = where ƛc = molar conductivity at a given concentration
ƛ0 ƛ+ 0+ƛ− 0

ƛ0 = molar conductivity at infinite dilution

Question
The molar conductivity of𝑁𝐻4 𝐶𝐿, NaOH and NaCL are 134.1, 125.1 and
113𝑜ℎ𝑚−1 𝑐𝑚2 𝑚𝑜𝑙 −1. Calculate the molar conductivity of ammonium hydroxide (𝑁𝐻4 OH)

Ionic mobility
This is the average velocity of an ion in an 𝑠 −1 under a potential gradient of 1volt 𝑐𝑚−1
The factors which affect ionic mobility are;
1. The potential gradient between the electrodes i.e. Potential difference between
the electrodes and their distance apart
2. The concentration of the solution
3. The ionic radius
4. The amount of charge on the ion
5. Viscosity of the solvent
6. Temperature
Ionic mobility of some ions is given below.
Cation Mobility Anions mobility
𝐻 + (aq) 36.3×10−4 OH(aq) 20.5×10−4
𝐾 + (aq) 7.62×10−4 𝑆𝑂4−2 (aq) 8.27×10−4
𝑁𝑎+ (aq) 5.19×10−4 B𝑟 − (aq) 8.12×10−4
𝐿𝑖 + (aq) 4.0×10−4 𝐶𝐿− (aq) 7.91×10−4
𝐶𝑎+ (aq) 6.16×10−4 𝑁𝑂3− (aq) 7.40×10−4

The relatively high mobilities of the hydrogen and hydroxide ions contribute to the high
conductivities of aqueous solutions of strong acids and strong alkalis. Multiple charged
ions tend to move faster than singly charged ions because they are attracted more
strongly by appositively charged electrodes.
Note that ions in solution exist in solvated (hydrated) form and not as bare ions. The
extent of hydration is greater the smaller the radius of the bare ion.
Thus the ionic radii of the alkali metal ions increase in the order;

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𝐿𝑖 + ˂ 𝑁𝑎+ ˂𝐾 + ˂𝑅𝑏 + ˂𝐶𝑠 + but the ionic mobilities are the order;
𝐶𝑠 + ˃𝑅𝑏 + ˃𝐾 + ˃𝑁𝑎+ ˃𝐿𝑖 +
Ionic mobility increases with increase in ionic radius. The smaller the ion the more it
attracts the water molecules around it so the drag on the mobility increases thus
lowering the conductivity.

A small ion has a high charge density and attracts more water around it and become
more heavily hydrated and more bulky than the larger ion. This reduces its mobility and
conductivity compared to a larger ion.

Temperature - the higher the temperature the higher the mobility of ions, hence the
conductivity. Also increase in temperature decreases the viscosity of the solvent
resulting into increase in mobility of the ions.

Applications of conductivity measurements

1. Determination of molar conductivity at infinite dilution (ƛ0 or ƛ∞) for strong
electrolyte. This is done by extrapolation of the graph of ƛc versus √𝑐 or ƛc
1
versus dilution (𝑐 )

2. Determination of the solubility and solubility product of a sparing soluble salt.

Since very little of such a salt dissolves in water, ƛc may be taken as
approximately equal to its ƛ0 or ƛ∞
ĸ
From ƛ0 = 𝑐
ĸ ĸ
C= =
ƛ0 (ƛ+ 0+ƛ− 0)

Conductivity for water should be subtracted because the small number of ions produced
by the electrolyte gives a conductivity of the same order of magnitude as water. ƛc
values can be deduced from measurements of strong electrolytes as in (1) above.
Example:
At 25, the molar conductivity of silver nitrate, potassium nitrate and potassium chloride
are 133.5, 145.0 and 150𝛺 −1 𝑐𝑚2 𝑚𝑜𝑙 −1 respectively. At the same temperature the
conductivity of a solution of silver chloride is 3.40×10−6 𝛺 −1 𝑐𝑚−1 while that of water is
1.6×10−6 𝛺 −1 𝑐𝑚−1

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Calculate;
I. The solubility of silver chloride in 𝑚𝑜𝑙𝑑𝑚−3 at 25oC
II. The solubility product of silver chloride at 250C
Solution
I. ĸ𝑠𝑜𝑙𝑢𝑡𝑒 = ĸ𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 - ĸ𝑠𝑜𝑙𝑣𝑒𝑛𝑡
ĸ𝐴𝑔𝐶𝐿 = 3.4×10−6 - 1.6 × 10−6
=1.8×10−6 𝛺 −1 𝑐𝑚−1
ƛ0 AgCL = ƛ0 Ag𝑁𝑂3 + ƛ0KCL – ƛ0K𝑁𝑂3
=133.5 + 150 – 145.0
=138.5𝛺 −1 𝑐𝑚2 𝑚𝑜𝑙 −1
ĸ×1000
ƛ= 𝐶
ĸ𝐴𝑔𝐶𝐿 ×1000
Solubility (c) = ƛ0𝐴𝑔𝐶𝐿

1.8×10−6 ×1000
= 138.5

=1.3×10−5 𝑚𝑜𝑙𝑑𝑚−3
II. Equation of solubility of silver chloride is;
+
𝐴𝑔𝐶𝐿(𝑆) 𝐴𝑔(𝑎𝑞) + 𝐶𝐿−(𝑎𝑞)
1−𝑐 𝑐 𝑐
Ksp = [𝐴𝑔+ ][𝐶𝐿− ]
=𝐶×𝐶
= (1.3×10−5) 2
= 1.69× 10−10 𝑚𝑜𝑙 2 𝑑𝑚−6

3. Determining the ionic product of water (kw). Water ionizes according to the
following equation;
H2O (aq) H+ (aq) + OH (aq)
1-c c c
C is the concentration of water that increases
ĸ
C= ƛ0(𝐻20)

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ƛ0 (H2O) = ƛ0H+ + ƛ0OH- which is got using selected strong electrolyte i.e.
ƛ0H2O = ƛ0HCL + ƛ0NaOH – ƛ0NaCL
If the conductivity (K) for water and H2O are known c can be calculated;
Kw = [𝐻 + ][𝑂𝐻 − ] = c × c = c2
4. Determination of dissociation constant (K) for a weak electrolyte
ᾳ2𝑐
K = 1−ᾳ
ƛ𝑐
But ᾳ = ƛ0
ƛ𝐶
( )2𝐶
ƛ0
Therefore K = ƛ𝐶
1−
ƛ0

If c, ƛc and ƛ0 are known K can be calculated.

 Calculating the dissociation constant, Ka, of ethanoic acid
The molar conductivity of hydrochloric acid, sodium chloride and sodium ethanoate at
infinite dilution at 298K are 426.1, 126.4 and 91 𝛺 −1 𝑐𝑚2 𝑚𝑜𝑙−1 respectively.
Calculate the;
a) Molar conductivity of ethanoic acid at infinite dilution
ƛ0𝐶𝐻3 𝐶𝑂𝑂𝐻 = ƛ0𝐶𝐻3 𝐶𝑂𝑂− + ƛ0𝐻 +
From; ƛ0𝐶𝐻3 COOH = ƛ0𝐶𝐻3 𝐶𝑂𝑂𝑁𝑎 + ƛ0HCL - ƛ0NaCL
= 91 + 426.1 – 126.4
= 390.7 𝛺 −1 𝑐𝑚2 𝑚𝑜𝑙 −1
b) Dissociation constant, Ka, of a 0.1M ethanoic acid solution ( the electrolytic
conductivity of ethanoic acid at 298K is 1.65×10−4 𝑜ℎ𝑚−1 𝑐𝑚−1 )
1000ĸ 1.65×10−4 ×1000
ƛc = =
𝑐 0.1

= 16.5 𝑜ℎ𝑚 𝑐𝑚2 𝑚𝑜𝑙 −1 −1

ƛ𝑐 16.5
ᾳ= = 390.7 = 0.042
ƛ0
ᾳ2 𝑐 (0.042)2 ×0.01
Ka = 1− ᾳ = 1−0.042

= 1.84×10−5 𝑚𝑜𝑙𝑑𝑚−3
5. Conductmetric titrations
Conductivity measurements can be used to follow the course of acid-base and
precipitation reactions.

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a) Strong acid vs. strong alkali ( e.g. HCL(aq) vs. NaCL(aq)

At A – conductivity is high due to high conductivity of hydrogen ions from hydrochloric

acid.
A to B – conductivity decreases because the fast moving H+ ions in the acid are
neutralized and replaced by the slow moving Na+ ions during the reaction.
At B –end point has occurred and the conductivity is due to Na+ and CL- ions after
complete neutralization.
B to C – conductivity increases due to excess hydroxyl ions from NaOH which are also
faster moving and also due to general increase in ion concentration i.e. excess Na +
ions. The rise is not as the fall from A to B because the hydroxyl ions are less mobile
than the hydrogen ions.
b) Weak acid vs. strong alkali (e.g. 𝐶𝐻3 𝐶𝑂𝑂𝐻(𝑎𝑞) vs. NaOH(aq) )

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NX – conductivity falls slightly as the small concentration of hydrogen ions from the
ionization of the weak acid is neutralized by sodium hydroxide added.
X to Y – conductivity increases due to increase in the number of ions, 𝐶𝐻3 COO- and
Na+ formed from the completely ionized salt (𝐶𝐻3 COONa) formed during neutralization.
At Y – neutralization is complete and conductivity is due to 𝐶𝐻3 COO- and Na+ ions.
Y to Z – conductivity increases due to addition of excess Na+ and OH- ions. The rise
(YZ) is steeper than that of XY because the hydroxyl ion is a better conductor than the
ethanoate ion (𝐶𝐻3 COO-).
c) Strong acid vs. weak alkali e.g. HCL vs. 𝑁𝐻4 OH(aq) or 𝑁𝐻4 (aq)

L to M – conductivity decreases because the fast moving H+ ions are neutralized and
replaced by the slow moving 𝑁𝐻4+ ions
𝑁𝐻4 OH (aq) + HCL (aq) 𝑁𝐻4+ (aq) + 𝐶𝐿− (aq)
Or 𝑁𝐻3 (aq) + HCL (aq) 𝑁𝐻4+ (aq) + CL-(aq)
At M – conductivity is due to 𝑁𝐻4+ and CL- ions after complete neutralization.
M to N – conductivity is almost constant because the ionization of ammonia is
suppressed by the 𝑁𝐻4+ ion already present in the solution from the salt (𝑁𝐻4 𝐶𝐿).
d) Weak acid vs. weak alkali (e.g. 𝐶𝐻3 COOH and 𝑁𝐻4 OH)

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R to S – conductivity falls slightly as the small concentration of H+ from the ionization of

the weak acid is neutralized by ammonia solution added.
S to T – conductivity gradually rises due to formation of other ions 𝐶𝐻3 COO- and 𝑁𝐻4+
from the
𝐶𝐻3 COONa (aq) + 𝑁𝐻3 (aq) 𝐶𝐻3 COO- (aq) + 𝑁𝐻4+ (aq) +H2O (aq)
Or 𝐶𝐻3 COOH (aq) + 𝑁𝐻3 (aq) 𝐶𝐻3 𝐶𝑂𝑂− (aq) + 𝑁𝐻4+ (aq)
At T – conductivity is due to 𝐶𝐻3 𝐶𝑂𝑂− and 𝑁𝐻4+ ions after complete neutralization
T to U – when ammonia solution is added, there is little increase in conductivity as the
ionization of ammonia solution is suppressed by the 𝑁𝐻4+ ions already present in the
solution.

𝑁𝐻3 (aq) + H2O (l) 𝑁𝐻4+ (aq) + 3H (aq)

e) Study of complexes.
Determining the ratio in which a given ion (cation) reacts with a given ligand i.e. the
coordination number in a complex ion.
A solution of known concentration of the ligand is added to a volume of a known
concentration of a particular cation and the conductivity of the solution measured.
When there is complete complexation there will be a change in conductivity and from
this the ratio in which the ligand and cation react can be calculated.
EXERCISE;
1a). Define the term molar conductivity.

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b). a solution was made by dissolving 0.095g of anhydrous magnesium chloride in water
to make a litre of the solution. Calculate the molar conductivity of the solution. (the
electrolytic conductivity of magnesium chloride is 2.58×10−4 𝛺 −1 𝑐𝑚−1)
2b). Define electrolytic conductivity.
b. the molar ionic conductivities at infinite dilution of Na+(aq), OH-, H+(aq) and CL-(aq)
are 50.1, 198.6, 349.8 and 76.4 𝛺 −1 𝑐𝑚2 𝑚𝑜𝑙 −1 .
Calculate the electrolytic conductivities for,
a. 0.01M Sodium hydroxide solution
b. A solution made by mixing 50cm3 of 0.01M sodium hydroxide and 50cm3of
0.02M hydrochloric acid.
c. State two applications of conductivity measurements.
3a). The electric conductivity of 0.032M ethanoic acid at 25 0c is 0.6392𝛺 −1 𝑚−1. The
molar conductivities at infinite dilution of the H+ (aq) and 𝐶𝐻3 𝐶𝑂𝑂− (aq) ions at 250c are
3.15×10−2 and 3.5×10−3 𝛺 −1 𝑚2 𝑚𝑜𝑙 −1 respectively. Calculate the degree of dissociation
of ethanoic acid.
4. The electrolytic conductivity of a saturated solution of silver bromide at 25 0c is
5.42×105 𝛺 −1 𝑐𝑚−1. The molar conductivities of Ag+ and Br- ions at infinite dilution are
56 and 70 𝛺 −1 𝑐𝑚2 𝑚𝑜𝑙 −1 respectively. Calculate the solubility product of the salt.
5. A solution containing 1.15 of methanoic acid per dm3 has an electrolytic conductivity
of 1.15×10−3 𝛺 −1 𝑐𝑚−1 at 250c. The molar conductivity of the acid at infinite dilution at
this temperature is 404.4 Ω𝑐𝑚2 𝑚𝑜𝑙 −1. Calculate the dissociation constant of methanoic
acid.
6. The molar conductivities of hydrochloric acid, sodium fluoride and sodium chloride
are 426.2, 105.8 and 126.5𝛺 −1 𝑐𝑚2 𝑚𝑜𝑙 −1 respectively at infinite dilution. Calculate the:
a. molar conductivity of hydrofluoric acid at infinite dilution
b. dissociation constant, Ka, of a 0.1M hydrofluoric acid solution. (The electrolytic
conductivity of hydro fluoride acid is 3.15×10−5 𝛺 −1 𝑐𝑚−1).

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7. The molar conductivity of sodium hydroxide solutions of different concentrations are

shown in table below.
Concentration/ mol 𝑑𝑚−3 0.01 0.04 0.09 0.16 0.25 0.36
Molar conductivity 𝛺 −1 𝑐𝑚2 𝑚𝑜𝑙−1 238 230 224 217 210 202

a. Draw a graph of molar conductivity against the square root of the concentration.
b. Explain the shape of the graph.
c. Determine the value of the molar conductivity at infinite dilution of sodium
hydroxide and indicate its unit.

8.Using the same conductivity cell, the resistance of a 0.1M Potassium chloride
solution and 0.1M bromoethanoic solution were found to be 24.96 and 66.50 ohms
respectively at 250c (the conductivity of potassium chloride at 250c is 0.01164𝛺 −1 𝑐𝑚−1
and the molar conductivity of bromoethanoic acid at infinite dilution is 389𝛺 −1 𝑐𝑚2 𝑚𝑜𝑙−1.
a) Calculate the cell constant.
b) Calculate the molar conductivity if the 0.1M bromoethanoic acid
c) Determine the PH of 0.1M bromoethanoic acid.

ELECTROLYSIS (ELECTROLYTIC CELLS)

Electrolysis – is the decomposition of an electrolyte by passage of an electric current
through it. An electrolyte – is a compound which conducts electricity when dissolved in
water or in molten state.
Electrolytes conduct electricity by movement of ions.
Some compounds conduct a current both when they are molten and when dissolved in
water e.g. sodium chloride, and calcium chloride, sodium hydroxide. Other compounds
form conducting solutions when dissolved in water, but are non-conductors in molten
state e.g. pure liquid hydrochloric acid and liquid ammonia.
Also there are compounds which are non-conductors both in molten state and in
aqueous solution e.g. sugar (glucose), alcohols, and petrol.
A non-electrolyte - is a solution or molten compound which cannot be decomposed by
an electric current.

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Strong and weak electrolytes.

A strong electrolyte- is a compound which completely ionizes in aqueous solution and
in molten state. For example;
 Strong acids- hydrochloric acid
Sulphuric acid
Nitric acid
 Strong alkalis- sodium hydroxide
Potassium hydroxide
 Salt s- NaCl(aq) Na+(aq) + Cl-(aq)
CuSO4 (aq) Cu+ (aq) + 𝑆𝑂42− (aq)
Weak electrolyte – is a compound which is only slightly ionized in aqueous solution and
in molten state e.g.
Water- H2O (l) H+ (aq) + OH-(aq)
Carbonic acid – H2CO3 (aq) H+ (aq) + HCO3- (aq)
Ammonia – NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
Ionic theory
The theory states that electrolytes of ions, which are positively and negatively charged
atoms or groups of atoms.
An anion is a negatively charged ion that that moves to the anode and a cation is a
positively charged ion that moves to the cathode.
The flow of current in an electrolyte is due to movement of ions in both directions.
Compound Formula Cations Anions
Sodium chloride NaCl Na+ Cl-
Sodium hydroxide NaOH Na+ OH-
Copper (II) chloride CuCl2 Cu2+ Cl-
Sulphuric acid H2SO4 2H+ SO42-
Copper (II) sulphate CuSO4 Cu2+ SO42-
Lead (II) bromide PbBr2 Pb2+ Br-

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Movement of ions and electrons during electrolysis.

During electrolysis a battery or other direct current (d.c) supply moves electrons through
the wire towards or away from the electrodes.
The electrons move to the cathode (-ve) and cations receive these electrons. In the
electrolyte, cations move towards the cathode and the anions move to the anode (+ve).
Thus flow of ions is the electric current inside the electrolyte.
The anions transfer electrons to the anode, and the electrons then move from the anode
to the d.c source (battery).

The electrodes are two pieces of metal or graphite by which electrons enter and leave
the electrolyte.
The cathode is the negative electrode (electrons enter the electrolyte here). The anode
is the positive electrode (electrons leave the electrolyte here).

Electrolysis of molten lead (II) bromide.

Ions are formed by the reaction;
heat
PbBr2 (l) Pb2+ (aq) + 2Br-(aq)
At the cathode; lead (II) ions move to the cathode, receive electrons and become lead
atoms (lead (II) ions are discharged by electron gain)
Pb2+ (aq) + 2e Pb (l)
Observation; a grey solid id formed.
At the anode; bromide ions move to the anode and are discharged by electron loss
forming bromide atoms.

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Br-(l) Br (l) + e
Overall equation at the anode; 2Br- (l) Br2 (g) + 2e
Observation; brown gas is evolved.

N.B. Electrolytes do not conduct electricity in solid form because the ions are held by
strong electrostatic forces of attraction and are not free to move.
However in molten or aqueous solution they conduct because the electrostatic forces of
attraction are weakened and broken hence the ions are free and mobile to conduct
electricity.
To investigate movement of ions during electrolysis.

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A small crystal of potassium manganate (VII) (KMnO4) solution is placed midway

between the ends of the moistened filter paper. The side and contents are connected to
a d.c supply (batter) and observation of filter paper about 10 to 15 minutes made.
KMnO4 (aq) K+ (aq) + MnO-4 (aq)
Observation; negatively charged ions migrate (move) to the electrode of opposite
charge (i.e. To the anode)
Note; manganate (VII) ion (MnO4-) is purple.
Selective discharge of ions.
When two or more ions of similar charge are present in solution e.g. H +, Na+ or OH- and
SO42- one is preferentially discharged and the selection of the ion discharged depends
on the following factors.

1) Position of the metal or group in the electrochemical series.

Cations go to the cathode; Anions go to the anode
I. K+ SO42-
II. Na+ NO3-
III. Ca2+ Cl-
IV. Mg2+ Br-
V. Zn2+ I-
VI. Fe2+ OH-
VII. Pb2+
VIII. H+
IX. Cu2+
X. Ag+
In dilute solution, if all factors are constant an ion is discharged in preference to those
above it in the E.C.S.
E.g. In sodium chloride solution containing H+ (from water) and Na +, H+ is discharged in
preference to Na+ and hydroxide ions are discharged in preference to chloride (Cl-) ions.
2) Concentration.

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Increase in concentration of an ion promotes its discharge. E.g. concentrated

hydrochloric acid containing OH- (from water) of Cl- is greater than that of OH- therefore
Cl- is discharged in preference.
3) Nature of the electrodes.
This factor may influence the choice of ion for discharge. E.g. in electrolysis of sodium
chloride solution. If a mercury cathode is used Na+ is discharged to form sodium
amalgam in the mercury.
This requires less energy than the discharge of H+.
Na+ (aq) + e Na (s)
Na (s) + Hg (l) NaHg (l)
Electrolysis of dilute sulphuric acid (so called the electrolysis of water)
At the cathode; the H+ ions are discharged to form hydrogen atoms.
H+ + e H2 (g)
The hydrogen atoms combine to form H2 molecules
H+H H2 (g)
Observation; bubbles of a colourless gas
Both the sulphate and hydroxyl ions migrate to the anode. The hydroxyl ions are
discharged in preference to the sulphate ions because they are lower than sulphate ions
in the E.C.S forming hydroxide radical.
OH-(aq) OH (g) + e
The hydroxyl radicles combine to form water and oxygen atoms.
OH (g) + OH (g) H2O (l) + O (g)
The oxygen atoms combine to form oxygen molecules (gas) seen as bubbles of a
colourless gas.
O (g) + O (g) O2 (g)
Discharge of OH- disturbs the ionic equilibrium of water. Therefore more water ionizes
to restore the equilibrium.
Excess H+ produced at the anode with sulphate ions is equivalent to increasing
concentration of sulphuric acid.
The total acidity at the anode and cathode together remains constant. Two volumes of
H2 at cathode and one volume of oxygen at anode is equivalent to electrolysis of water.

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Consider the flow of electrons;

4H+ (aq) + 4e 2H2 (g)
4OH- (aq) 2H2O (l) + O2 (g) + 4e
2 moles of H2 molecules are released for every one mole of O2 molecules released.
Set up of apparatus.

Electrolysis of concentrated sodium chloride solution (using platinum cathode

and graphite anode)
NaCl (aq) Na+ (aq) + Cl- (aq)
H2O (l) H+ (aq) + OH- (aq)
At the cathode (Pt or Carbon); H+ and Na+ move there. H+ ions being lower than Na+
in the electrochemical series are preferably discharged.
H+ (g) + e H2 (g)
H (g) + H (g) H2 (g)
Overall equation; 2H+ (aq) + 2e H2 (g)
Observation; bubbles of a colorless gas.
Discharge of hydrogen ions disturbs the equilibrium of water. More water ionises to
restore the equilibrium. Excess hydroxyl ions produced with incoming Na + is equivalent
to formation of NaOH therefore the solution at the anode becomes alkaline.
At the anode (carbon graphite not Pt because chlorine attacks Pt); Cl- and OH- move
there. Cl- ion is discharged in preference to OH- ion because it is higher in the
concentration than OH-
Cl- (aq) Cl (g) + e

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Chlorine atoms combine to form chlorine molecules;

Cl (g) + Cl (g) Cl2 (g)
Overall equation; 2Cl- (aq) Cl2 (g) + 2e
Observation; bubbles of a greenish yellow gas.

Application of electrolysis.
1. In extraction of metals e.g. Al, Na, etc.
2. In the manufacture of sodium hydroxide, chlorine and refining metals.
Faradays laws of electrolysis
1. During electrolysis, the mass of substance liberated at the electrodes is
proportional to the quantity of electricity passed.
Or Mαit (M= mass, I = current, t= time)
2. The quantity of electricity required to liberate one mole of any substance is
proportional to the charge number of its ions.
Or It α Z (Z = charge number on the ion)
Example;
A current of 2A was passed for 30 minutes through a cell containing dilute sulphuric
acid and the hydrogen produced at the cathode collected. Calculate the volume of the
hydrogen in cm3 that was produced at 230c and 100kPa. (IF = 96487)
Solution
Quantity of electricity = It
= 2×30×60
= 3600 coulombs
2H+ (aq) + 2e H2 (g)
2F 1mole
2×96487 coulombs liberate 1 mole of H2
1
3600C liberate 2×96487×3600 moles of H2

1.866×10-2 moles of H2
PV = nRT
𝑛𝑅𝑇
V= 𝑃

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1.866×10−2 ×8.31×296
= 100×103

= 4.58×10-4m3
=4.58×10-4×106cm3
=458cm3
Exercise
1. What mass of copper will be deposited from a solution of copper (II) chloride
when a current of 3A is passed for 3 hours?
2. During electrolysis of dilute sulphuric acid a current of 2A was passed for 2
hours. Calculate the volume of hydrogen gas evolved at the cathode at room
temperature. (1 mole of a gas at room temperature is 24 litres.
3. An electric current was passed through an aqueous solution of nickel (II)
sulphate. After sometime 600cm3 of oxygen were liberated at 298K. Calculate
the mass of nickel deposited at the cathode. (Ni = 59, 1mole of a gas at 298K is
24litres.

ELECTRODE PROCESSES.
Electrode potential – is the equilibrium potential difference set up by an electrode and a
solution of its ions.
Example; if a metal is placed in a solution of one of its salts or in water the metal tends
to send off positive ions into solution.
E.g. zinc in zinc sulphate solution or in water.

Zinc goes into solution as zinc ions and electrons are left on the zinc rod which
becomes negatively charged with respect to the solution.
Zn (s) Zn2+ (aq) + 2e
The formation of an electric double layer is accompanied by a potential different called
electrode potential between the surface and the solution.

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For a copper rod dipped in copper (II) ions are deposited on the copper rod and give it a
positive charge and electrons are sent into the solution and the atom that loses the
electrons remain as the positive charge on the surface.
Cu2+ (aq) Cu (s) – 2e
Or Cu2+ (aq) + 2e Cu (s)

Single electrode potential

The absolute value of a single electrode cannot be measured experimentally because a
half cell reaction cannot take place independently (cannot by itself). We can measure
only the different between the electrode potentials of any two half cells which makes the
measured Potential difference relative not absolute (i.e. it would be for the two
electrodes since we need two electrodes to measure the Potential difference between
two points).
It is possible to obtain the Potential difference of an electrode if the given electrode is
coupled (connected) with another electrode having a Potential of zero volts. The
electrode whose Potential can be taken as zero volts is called Primary Reference
Electrode (P.R.E). The Standard Hydrogen Electrode (S.H.E) has been assigned an
electrode potential of 0.00 volts at all temperatures and is used as P.R.E.
The electrode potential of any other electrode is obtained by coupling it with the S.H.E.

STANDARD ELECTRODE POTENTIAL.

The reduction potential of an electrode relative to the standard hydrogen electrode at
298K and when the concentration of the ion taking part in the electrode reaction is
1mol𝑙 −1 is called the standard electrode potential.

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Standard hydrogen electrode consists of a 1molar solution of hydrogen ions, having

platinized platinum around which pure hydrogen gas at 1 atmosphere and 298K is
bubbled.

Measurement of standard hydrogen potential of a metal.

The metal is placed in a 1 molar solution of its ions and connected to the standard
hydrogen electrode (S.H.E). The two electrodes are then joined by a salt bridge. The
voltmeter will show the e.m.f of the cell which will be the standard electrode potential of
the metal since the standard electrode potential of the S.H.E is taken as 0.00volts.

Cell diagram.

Notation; Pt. /H2 (g) /H+ (aq) // M+ (aq)/M (aq)

Note; A salt bridge is a glass tubing containing saturated KCl, KNO 3, or NH4NO3 OR
Folded filter paper saturated with these salts.
A salt bridge allows conduction of current between the two half cells.
E.M.F Sign Convection.
If the electron accepting tendency of the electrode is more than that of S.H.E, its
reduction potential get a positive sign and incase the electron accepting tendency of the
metal is less than that of S.H.E, its standard reduction potential gets a negative sign.
The electrode at which reduction occurs with respect to S.H.E has a positive (+ve)
reduction potential.

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The electrode at which oxidation occurs with respect to S.H.E has a –ve reduction
potential. According to the latest convection, standard electrode potentials are taken as
reduction potentials.

Connecting two half cells.

When connecting two half cells and one of which is not a hydrogen electrode;
 The electrode with a more negativity Potential is always written on the left as the
negative pole of the cell (and is anode)
 The electrode with a more positive electrode potential is written on the right as
the positive pole of the cell (and is the cathode)
A simple cell (e.g. Daniel’s cell)
Zinc is dipped in zinc sulphate solution and is connected to copper dipped in copper (II)
sulphate solution.
Zn2+ (aq)/ Zn (s) = -0.76V
Cu2+ (aq)/ Cu (s) = +0.34V
Cell diagram

Cell notation; Zn (s)/Zn2+ (aq) // Cu2+ (aq)/ Cu (s)

Solid line – represents boundary between two phases.
Double line – represents salt bridge
Dotted line – represents porous partition

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Half-cell reactions;
Cell reaction at left hand electrode (-ve) anode.
Cu2+ (aq) + 2e Cu(s)
Overall cell reaction;
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
E cell = ER - EL
=+0.34 – (-0.76)
=+1.10V
Since e.m.f of the cell is +ve, it indicates that the reaction is feasible.

The anode and cathode in a cell.

Oxidation occurs when a chemical loses electrons. At the negative electrode in the
Daniel cell, the Zinc loses electrons therefore oxidation takes place at this electrode.
Conversely reduction takes place at the positive electrode (copper) where copper (II)
ions are converted to copper atoms. Reduction occurs when a chemical gains electron
(s).

The electrode at which oxidation takes place is called the anode. The electrode at which
reduction takes place is called the cathode.
In the Daniel’s cell, the Zinc electrode is the anode and the copper electrode is the
cathode. However in electrolysis the electrode at which chemicals are oxidized happens
to be the positively charged electrode and chemicals are reduced at the negative
electrode. In electrolysis therefore, the cathode is negatively charged and the anode is
positively charged.

Factors that affect the standard electrode potential.

1. Atomization energy.
If it is high, the value of Eθ for the electrode is low (i.e. less negative or more positive).
This is because it becomes difficult to convert the solid metal into its gaseous ions.
2. Ionization energy (I.E)
If I.E is high the value of Eθ is low because it becomes difficult to covert the gaseous
atoms into ions.

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3. Hydration energy.
When the hydration energy is high, the value of the negative E θ is also high. This is
because it becomes easier for the gaseous ions to become hydrated in aqueous
solution.
M (s) M (g) atomization
M (g) M+ (g) + e ionization energy
M+ (g) + aq M+ (aq) hydration
The amount of charge and size of the ion affects the value of the hydration energy so
they can also have effect on the standard electrode potential.
Predicting feasibility of a reaction
Given;
i. Cu2+(aq) + e Cu+(aq) Eθ = +0.15V
ii. Cu+(aq) + e Cu(s) Eθ = +0.52V

Is 2Cu+ (aq) Cu2+(aq) + Cu(s) a feasible reaction

On reversing equation (i) and adding it to equation (ii)
Cu+(aq) Cu2+(aq) + e Eθ = -0.15V
Cu+(aq) + e Cu(s) Eθ = +0.52V
Overall equation is;
2Cu+(aq) Cu2+(aq) + Cu(s) Eθ = +0.37V
Since Eθ for the reaction is positive, the reaction is feasible.
Copper (I) compounds in aqueous solution disproportionate (i.e. undergo simultaneous
oxidation and reduction).
Is Fe2+ (aq) expected to disproportionate in the same way as Cu+ (aq)?
Consider the following data;
i. Fe2+(aq) + 2e Fe(s) Eθ = -0.44V
ii. Fe3+(aq) + e Fe2+(aq) Eθ = +0.77V
Multiplying equation (ii) by 2 and reserve it then add it to equation (i)
Note; on multiplying the equation by a factor the potential difference remains the same.
2Fe2+(aq) 2Fe3+(aq) + 3e Eθ = -0.77V
Fe2+(aq) + 2e Fe(aq) Eθ = -0.44V

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Overall equation is;

3Fe2+(aq) 2Fe3+(aq) + Fe(s) Eθ = -1.2
Since Eθ for the reaction is negative the reaction is not feasible.
Fe2+(aq) does not disproportionate like Cu+(aq). Indeed Fe can reduce Fe3+(aq) to Fe2+(aq).
Eθ for the change is +1.21V.
THE STRENGTH OF OXIDISING AND REDUCING AGENTS
Eθ/V
Li+(aq) + e Li(aq) -3.04
Na+(aq) + e Na(s) -2.71
Zn2+(aq) + e Zn(s) -0.76
1
H+(aq) + e H2(g) 0.00
2

Cu2+(aq) + 2e Cu(s) +0.34

1
I2 (aq) + e I-(aq) +0.54
2

Fe3+ (aq) + e Fe2+(aq) +0.77

1
Br2 (aq) + e Br-(aq) +1.07
2

Cr2O72-(aq) + 14H+ (aq) + 6e 2Cr3+ (aq) + H2O (l) +1.33

1
Cl2 (aq) + e Cl-(aq) +1.36
2

MnO4-(aq) + 8H+ (aq) 5e Mn2+ (aq) + 4H2O (l) +1.52

1
F2 (aq) + e F- (aq) +2.80
2

The more negative the electrode potential the stronger the reducing agent. The more
positive the electrode potential, the stronger the oxidizing agent.
Lithium is the strongest reducing agent in the set above since it has the largest
negative. Fluorine is the strongest oxidizing agent since it has the largest positive
electrode potential.
Lithium has a small atomic radius and forms a small cation (Li+) which is heavily
hydrated in aqueous solution and a lot of energy is given out.
Fluorine atom is also small and has a high electronegative so it has the greatest
tendency to form a negatively charged hydrated ion.
1
2
F2 (aq) + e F- (aq) ∆H = E.P.

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STANDARD ELECTRODE POTENTIAL AND STANDARD FREE ENERGY CHANGE

(∆Gθ)
∆Gθ = -z F Eθ where z = no. of electrons taking part in process
F = Faradays constant (approximately 96500 c)
For the reason;
Cu2+ (aq) + Zn (aq) Cu(s) + Zn2+(aq) Eθ = +1.1V
∆Gθ = -2×96500×1.1 J
= -212300 J or -212.3 Kj
Since ∆Gθ is negative, it indicates that the reaction is feasible or possible.
Are the following reactions feasible;
a. 2Fe3+(aq) + 2I-(aq) 2Fe2+(aq) + I2(aq)
b. 2MnO4-(aq) + 10Cl- (aq) + 16H+ 2Mn2+(aq) + 8H2O(l) + 5Cl2(g)
c. Cr2O72-(aq) + 14H+(aq) + 6Cl-(aq) 2Cr3+(aq) + 2Cl-(aq)
d. 2Fe2+(aq) + Cl2(g) 2Fe3+(aq) + 2Cl-(aq)
e. F2(g) + 2Cl-(aq) 2F-(aq) + Cl2(g)

Types of electrodes
The electrode including the reagents that are involved with it is called a half-cell and the
reaction that occurs in the half-cell is a half reaction or electrode reaction
1. Gas electrode.

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A gas is introduced to participate in an electrochemical reaction. The gas is in contact

with the surface of an inert- metal electrode e.g. H2 electrode.
Conventionally a reduction reaction is written as the electrode reaction.
H+ (aq) + e ½H2 (g)
Electrode representation
Pt. / H2 (g) / H+
2. Oxidation – reduction electrode.
This is for electrodes consisting of an inert metal dipping into a solution containing two
different oxidation states of a species e.g. Pt dipping in a solution of Fe2+(aq) and Fe3+(aq)
Electrode representation;
Pt / Fe2+ (aq), Fe3+ (aq)
The comma is used to separate the two chemical species which are in the same
solution.
3. Metal-metal ion electrodes.
A metal electrode dips into a solution containing ions of the metal e.g. silver dipped in
Silver nitrate solution representation.
Ag(aq) / Ag+(aq)
Very active metals react with water and cannot be used for such an electrode as Na, K,
Ca
4. Amalgam electrode.

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A metal is in form of an amalgam i.e. a metal is dissolved in mercury rather than

remaining in pure form. Electric contact is done by Pt-wire dipping into the amalgam
pool. Active metals such as sodium are used in such electrodes.
Electrode representation;
Na in Hg / Na+(aq)
Na(s) /Na+(aq)
5. Metal – insoluble salt electrode.
Here, a metal is in contact with an insoluble salt of the metal which in turn is in contact
with a solution containing the anion of the salt.
E.g. Ag(s) / AgCl(s) / Cl-(aq)
Electrode reaction is AgCl (s) + e Ag(s) +Cl-(aq)
Also common is the calomel electrode represented as;
Hg (l) / Ag2Cl2 (g) / Cl-(aq)
Electrode reaction; ½Hg2Cl2(s) + e Hg (l) + Cl-(aq)
Differences between an E.m.f cell and an electrolytic cell.
Electrolytic cell
1. Electric power is supplied from outside.
2. There is transfer of material
3. Oxidation occurs at the positive electrode and reduction at the negative
electrode.
E.m.f cell
1. Generate their power.
2. There is no transfer of material.
3. Oxidation occurs at the negative electrode.
Application of electrode potential.
1. Predicting feasibility (spontaneity) of a reaction (i.e. direction of a reaction)
2. Selection of oxidizing or reducing agent.
Electrode reaction and corresponding notation.
MnO4-(aq) + 8H+ (aq) + 5e Mn2+(aq) + 4H2O(aq) Pt/ Mn2+(aq), H+(aq), MnO4-
(aq)

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H2O2 (aq) + 2H+ (aq) + 2e H2O (l) Pt/H2O2 (l), H2O2 (aq)/ H+
(aq)

Fe3+ (aq) + e Fe2+ (aq) Pt/ Fe2+ (aq), Fe3+ (aq)

I2 (aq) + 2e I-(aq) Pt/ I-(aq), I2 (aq)
O2 (g) + 2H+ (aq) + 2e H2O2 (aq) Pt/ H2O2 (aq), H+ (aq) /
O2(g)
Exercise
The standard electrode potentials for some half-cells are shown below;
Fe3+ (aq)/Fe2+ (aq) +0.76V
I2 (aq)/I-(aq) +0.54V
a. Write;
I) The cell concentration for the combined cell
II) The equation for the overall cell reaction.
b. Calculate the overall electrode potential for the cell.
c. State whether the reaction is feasible or not.
d. Give a reason for your answer.
Answer;
a) Pt/I-(aq), I2(aq)// Fe3+(aq), Fe2+(aq)/Pt
ii) 2Fe3+ (aq) + 2I- (aq) 2Fe2+ (aq) + I2 (aq)
b) Ecell = Eright - Eleft
= +0.76 – (+0.54)
=+0.22V.

A BATTERY.
A battery is a device that stores chemical energy and converts it to electrical energy.
The chemical reaction in a battery involves flow of electrons from one electrode (anode)
to another (cathode) through an external circuit. The flow of electrons provides an
electric current that can be used to do work.

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Uses of a battery.
1. Backup power in event to power grid outage
2. Energy management.
3. Renewable integration – batteries can store energy from solar and wind and
discharge it when it is needed.

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