Applied Cemistry Book Final 1 PDF
Applied Cemistry Book Final 1 PDF
Applied Cemistry Book Final 1 PDF
WATER TECHNOLOGY
1.1 INTRODUCTION
Water is one of the most abundant commodities in nature. Water is essential for
survival of all living organisms. About 80% of the earth surface is covered by water. The
main sources of water are
i) Rain
ii) River and lakes (surface water)
iii) Wells and springs(ground water)
Among the above (sources of water) rain water is the purest form of water whereas sea
water is the most impure form. For drinking and industrial purposes we need water free
from undesirable impurities. So first we should understand about the nature and types of
impurities present in water so that the water can be treated suitably to remove the
undesirable impurities.
1
1.3 HARD AND SOFT WATER
Water which does not produce lather with soap solution but forms a white precipitate
is called hard water.
Water which lathers easily with soap solution is called soft water.
2
1.3.2 Expression of hardness as equivalents of CaCO3 (or) CaCO3 standard
The concentration of hardness causing ions are usually expressed in terms of an
equivalent amount of CaCO3. The choice of CaCO3 is standard because its molecular
weight is 100 (equivalent weight =50) and also it is the most insoluble salt that can be
precipitated in water treatment.
If the concentration of hardness producing constituent is x mg/lit., then
Equivalent amount of CaCO3 = x × 100
Molecular weight of hardness producing substance
2. A water sample contains 200 mgs of CaSO4 and 75 mgs of Mg (HCO3)2 per litre.
What is the total hardness interms of CaCO3 equivalent?
Solution:
Name of the hardness Amount in Molecular Amount equivalent
producing salts mgs/lit weight to CaCO3
CaSO4 200 136 200 X 100 / 136 =
147 mgs/lit
Mg(HCO3)2 75 146 75 x 100 / 146 =
51.4 mgs/lit 4
Temporary Hardness = Mg (HCO3)2 = 51.4 mgs/lit
Permanent hardness = CaSO4 =147 mgs/lit.
Total hardness = Mg (HCO3)2 + CaSO4
= 51.4 + 147
= 198.4mgs/lit
3. A sample of water contains 25mgs of Ca2+ ions per litre, Calculate its hardness
interms of CaCO3 equivalent?
Solution
Given:
The amount of Ca2+ ions = 25 mgs/lit
We know that, the atomic weight of calcium = 40
Amount of CaCO3 = 25 X 100 / 40
= 62.5 mgs/lit
6. A sample of water is found to contain 16.8 mg/lit Mg(HCO3)2, 12 mg/lit MgCl2, 29.6
mg/lit MgSO4, and 5.0 mg/lit NaCl. Calculate the permanent and temporary hardness of
water and express it in ppm. (Atomic weight of Mg=24; H=1; C=12; O=16; Cl=35.5;
Na=23; S=32)
1.5ALKALINITY
Alkalinity in water is due to the presence of soluble (i) hydroxide (OH-) ions, (ii)
carbonate (CO32-) ions and (iii) bicarbonate (HCO3-) ions. These can be determined by
titrimetry using standard acid and phenolphthalein and methyl orange as indicators. The
determination is based on the following reactions.
6
Depending upon the anion that is present in water alkalinity is classified into three types.
1. Hydroxide alkalinity - due to (OH-)
2. Carbonate alkalinity - due to (CO3 2-)
3. Bicarbonate alkalinity - due to (HCO3-)
Titration 2:
Similarly the same amount of water sample is titrated against a standard acid using
methyl orange indicator, the end point indicates the completion of reaction (i), (ii) and (iii).
The amount of acid used after the phenolphthalein end point corresponds to one-half of
normal carbonate {+} all the bicarbonates.
i.e., neutralization of OH-, CO32-, and HCO3- ions.
The total amount of acid used (amount of acid upto phenolphthalein end point {+}
amount of acid upto methyl orange end point) represents the total alkalinity (due to
hydroxide, carbonate and bicarbonates).
Note:
OH- and HCO3- ions cannot exist together in water, because they combine
instantaneously to form CO32- ions.
i.e., OH- + HCO3- CO32- + H2O
e.g. NaOH + NaHCO3 Na2CO3 + H2O
Thus in a water all the three ions (OH-, CO32-, HCO3-) cannot exist together.
7
1.5.2 Determination of various types of alkalinity
Various types involved in the determination of various types of alkalinity are as
follows.
V1 50 100
= ppm
100 50
P = 10V1
M = 10(V1 + V2)
Conclusions
(i) When P = 0, both OH- and CO32-are absent and alkalinity is only due to HCO3-.
(ii) When P = 1/2 M, only CO32- is present, half of the carbonate neutralization reaction
takes place (i.e CO32- + H+ HCO3-) with phenolphthalein indicator complete
carbonate neutralization reaction (i.e CO32- + H+ HCO3-. HCO3-+ H+ H2O + CO2)
occurs when methyl orange indicator is used. Thus alkalinity due to CO32- = 2P.
9
(iii) When P = M, only OH— is present, alkalinity due to OH-= P = M
(iv) When P > 1/2M, besides CO32-, OH- ions are also present. Now half of CO32-
(i.e,HCO3- + H+ CO2 + H2O) equal to (M-P) so, alkalinity is due to complete CO32- =
2(M-P) alkalinity due to OH- = M-2(M-P) = 2P-M
(v) When P < 1/2 M, Besides CO32-, HCO3- ions are also present.
Now alkalinity due to CO32- = 2P
Alkalinity due to HCO3- = (M- 2P)
The data of the above conclusions are tabulated as follows.
Hydroxide and
p>1/2 M (2P-M) 2(M-P) 0
carbonate
10
Problems based on Alkalinity
1. 100 ml of a raw water sample on titration with N/50 H2SO4 required 10.0 ml of the
acid to phenolphthalein end-point and 14.0 ml of the acid to methyl orange end-
point. Determine the type and extent of alkalinity present in the water sample.
Solution:
Strength of HCl = 0.02 N
phenolphthalein end-point =P = 10.0 ml
methyl orange end-point =M = 14.0 ml
Since P > M,
the water sample must contain only OH- and CO32- alkalinities and there cannot be any
HCO3- alkalinity.
i) Volume of std .HCl required for OH- alkalinity = 2P - M
= (2 x 10.0) ml - 14 ml
= 20 ml- 14.0 ml
= 6.0 ml
Volume of acid consumed to OH- alkalinity V1 = 6.0 ml
ii) Volume of std .HCl required for CO32- alkalinity = 2M – 2P
= 2 x 14.0 ml – 2 x 10.0 ml
= 24 – 20
Volume of acid consumed to CO32- alkalinity V1 = 4.0 ml
11
Strength of water sample due to OH- alkalinity N2 = ?
V1 N1 = V2 N2
V1N 1
N2 =
V2
6 0.02
N2 =
100
= 0.0012 N 50
= 0.06 gm x 1000
Amount of OH- alkalinity = 60 ppm
4 0.02
N2 =
100
= 0.04 gm x1000
= 40 ppm
Amount of CO32- alkalinity = 40 ppm
Total Alkalinity
Total alkalinity = Alkalinity due to OH— +
Alkalinity due to CO32-
= 60 ppm + 40 ppm
= 100 ppm
2. A water sample is not alkaline to phenolphthalein but, 100 ml of the sample on
titration with N/10 HCl, required 15 ml to methyl orange end point. What are the types
and amounts of alkalinity present in the sample.
Solution:
Strength of HCl = 0.1 N
phenolphthalein end point p = 0
methyl orange end point M = 15 ml
Since P = 0
The water sample contain only HCO3- alkalinity ,
Volume of HCl required to HCO3- alkalinity = M
M = 15 ml
1) Calculate the HCO3- alkalinity
Volume of HCl V1 = 15.0 ml
Strength of HCl N1 = 0.1 N
Volume of water sample V2 = 100 ml
13
Strength of water sample due to HCO3- alkalinity N2 = ?
V1 N1= V2 N2
V1N 1
N2 =
V2
15 0.1
N2 =
100
= 0.015 N
Strength of water sample due to HCO3- alkalinity = 0.015 N
Amount of HCO3- alkalinity present in
1 litre in terms Of CaCO3 equivalent = Strength of HCO3- alkalinity
Eq. wt of CaCO3
= 0.015 N 50
= 0.75 gm x1000
= 750 ppm
Amount of HCO3- alkalinity = 750 ppm
3. 100 ml of a water sample on titration with 0.02 N H2SO4 gave a titre value of 7.8 ml
to phenolphthalein end-point and 15.6m l to methyl orange end-point. Calculate the
alkalinity of the water sample interms of CaCO3 equivalent and comment the type of
alkalinity present.
Solution:
Given:
Strength of HCl = 0.02 N
Volume of the water sample = 100 ml
phenolphthalein end point P = 7.8 ml
methyl orange end point M = 15.6 ml
Given data satisfy the condition P = 1/2 M, Therefore water sample contains only
CO 32 alkalinity not OH - and HCO3- alkalinity,
14
Volume of HCl required to CO32- alkalinity = 2P
= 2 7.8
= 15.6 ml
Calculation for CO32- Alkalinity.
Volume of HCl V1 = 15.6 ml
Strength of HCl N1 = 0.02 N
Volume of water sample V2 = 100 ml
Strength of water sample due to CO32- alkalinity N2 = ?
V1 N1 = V2 N2
V1N 1
N2 =
V2
15.6 0.02
N2 =
100
= 0.00312 N 50
= 0.156 gm x1000
= 156 ppm
Amount of CO32- alkalinity = 156 ppm
15
4. 100 ml of a raw water sample on titration with N/50 H2SO4 required 7.5 ml of the
acid to phenolphthalein end-point and 18.0 ml of the acid to methyl orange end-point.
Determine the type and extent of alkalinity present in the water sample.
Solution.
Strength of HCl = 0.02 N
phenolphthalein end-point P = 7.5 ml
methyl orange end-point M = 18.0 ml
the water sample must contain both CO32- and HCO3- alkalinities and there cannot be any
OH- alkalinity.
i) Volume of std .HCl required for CO32- alkalinity = 2P
= 2 x 7.5 ml
= 15.0 ml
Volume of acid consumed to CO32- alkalinity V1 = 15.0 ml
ii) Volume of std .HCl required for HCO3- alkalinity = M – 2P
= 18.0 ml – 2 x 7.5.0 ml
= 3.0 ml
Volume of acid consumed to HCO3- alkalinity V1 = 3.0 ml
Calculate the CO32- alkalinity
Volume of HCl V1 = 15.0 ml
Strength of HCl N1 = 0.02 N
Volume of water sample V2 = 100 ml
Strength of water sample due to
CO32- alkalinity N2 = ?
16
V1 N1 = V2 N2
V1N 1
N2 =
V2
15 0.02
N2 =
100
= 0.003 N
Strength of water sample due to CO32- alkalinity = 0.003N
Amount of CO32- alkalinity present in
1 litre in terms Of CaCO3 equivalent = Strength of CO32- alkalinity
Eq. wt of CaCO3
= 0.003 N 50
= 0.15 gm x 1000
Amount of CO32- alkalinity = 150ppm
Calculate the HCO3- alkalinity
Volume of HCl V1 = 3.0 ml
Strength of HCl N1 = 0.02 N
Volume of water sample V2 = 100 ml
Strength of water sample due to HCO3- alkalinity N2 = ?
V1 N1 = V2 N2
V1N 1
N2 =
V2
3 0.02
N2 =
100
N2 = 0.0006 N
Strength of water sample due to HCO3- alkalinity = 0.0006 N
Amount of HCO3- alkalinity present in
1 litre in terms Of CaCO3 equivalent = Strength of of HCO3- alkalinity
Eq. wt of CaCO3
17
= 0.0006 N x 50
= 0.03 gm x1000
= 30 ppm
Amount of HCO3- alkalinity = 30 ppm
Total Alkalinity
Total alkalinity = Alkalinity due to CO32- +
Alkalinity due to HCO3-
= 160 ppm + 30 ppm
= 190 ppm
HOOCH2C CH2COOH
N-CH2-CH2-N
HOOCH2C CH2COOH
Since, EDTA is insoluble in water, its disodium salt is used as a sequestering or
complexing agent.
Principle:
The amount of hardness causing ions (Ca2+and Mg2+) can be estimated by titrating the
water sample against EDTA using Eriochrome-Black-T indicator (EBT) at a pH of 8-10. In
order to maintain the pH, buffer solution (NH4Cl-NH4OH mixture) is added. Only at this
pH such a complexation is possible.
When the EBT indicator is added to the water sample, it forms wine red coloured
weak complex with Ca2+and Mg2+ ions.
18
[ Ca2+ & Mg2+] + EBT pH = 8-10 Ca
EBT complex
Mg
When this solution is titrated against EDTA, it replaces the indicator and form stable
EDTA complex. When all the hardness causing ions are complexed by EDTA, the
indicator is set free. The colour of the free indicator is steel blue. Thus the end point is
change of colour from wine red to steel blue.
Ca
EBT complex + EDTA pH = 8-10 Ca
Mg EDTA + EBT
Mg
c) EBT Indicator
0.5 gms of EBT is dissolved in 100 ml of alcohol.
d) Buffer solution
67.5 gms of NH4Cl and 570 ml of NH3 are dissolved and the solution is made upto
1000 ml using distilled water.
19
1.6.2 Experimental procedure:
(i) Standardisation of EDTA
Pipette out 50 ml of standard hard water into a clean conical flask. Add 10 ml of buffer
solution and 4-5 drops of EBT indicator and titrate it against EDTA solution taken in the
burette. The end point is the change of colour from wine red to steel blue
Let the volume of EDTA consumed be V1 ml
Calculation:
(i) Standardization of EDTA using standard hardwater
1 ml of std. hard water = 1 mg of CaCO3
50 ml of std. hard water = 50 mgs of CaCO3
50 ml of std. hard water consumes = V1 ml of EDTA
V1 ml of EDTA 50 mgs of CaCO3 equivalent of hardness
(Or)
50
1 ml of EDTA mgs of CaCo3 equivalent of hardness
V1
20
(ii) Estimation of total hardness of water sample
50 ml of the given hard water sample consumes} = V2 ml of EDTA
50
= V2 mgs of CaCO3 equivalent of hardness
V1
50
[ 1 ml of EDTA = mgs of CaCO3]
V1
50 1000
= V2
V1 50
V2
= 1000 {mgs of CaCO3 equivalent of hardness}
V1
V2
Total Hardness = 1000 ppm
V1
V3
= 1000 {mgs of CaCO3 equivalent of hardness}
V1
V3
Permanent Hardness = 1000 ppm
V1
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(iii) Temporary hardness
Temporary hardness = Total hardness - Permanent hardness
V2 V
= 1000 1000 3
V1 V1
1000
Temporary hardness = V 2 V1 ppm
V1
15 0.02
M1 =
50
M2 = 0.006 M
3. 100 ml of a water sample of consumed 19.5 ml of 0.01 M EDTA for titration using
Erio-chrome Black-T indicator. In another experiment, 100 ml of the same water sample
was boiled to remove the carbonate hardness, the precipitate was discarded and then it
is allowed to reach the room temperature. The filtrate CaCO3 solution was then titrated
against 0.01 M EDTA which consumed 10 ml of EDTA, in the presence of EBT
indicator. Calculate (i) the total hardness (ii) permanent hardness of non carbonate
hardness (iii) carbonate hardness, in terms of mg/lit of CaCO3.
Titration: I
Strength of EDTA = 0.01M
Estimation of total hardness.
Volume of water sample V1 = 100 ml
Strength of water sample M1 = ?
Volume of std EDTA V2 = 19.5 ml
Strength of b std EDTA M2 = 0.01 M
23
V1M1 = V2M2
V2 M 2
M1 =
V1
19.5 0.01
M1 =
100
M1 = 0.00195 M
Titration: II
Estimation of permanent hardness.
Volume of boiled water V1 = 100 ml
Strength of boiled water M1 = ?
Volume of std EDTA V2 = 10 ml
Strength of b std EDTA M2 = 0.01 M
V1M1 = V2M2
V2 M 2
M1 =
V1
10 0.01
M1 =
100
M1 = 0.001 M
4. Calculate permanent and temporary hardness from the following. 250 ml of a water
sample is boiled for 1 hr. It is then cooled nd filtered. The filtrate is made upto 250 ml
again with the addition of distilled water. 20 ml of this solution requires 10 ml of 1/100
EDTA with EBT-indicator and NH4Cl - NH4OH buffer.
Titration: I
Estimation of permanent hardness.
Volume of boiled water V1 = 20 ml
Strength of boiled water M1 = ?
Volume of std EDTA V2 = 10 ml
Strength of std EDTA M2 = 0.01 M
V1M1 = V2M2
V2 M 2
M1 =
V1
10 0.01
M1 =
20
M1 = 0.005 M
Strength of boiled water = 0.005 M
Amount of permanent hardness present in per liter in
Terms of CaCO3 equivalent = Strength of boiled water x M .W of CaCO3
= 0.005 M X 100
= 0.5 gm x1000
Amount of permanent hardness = 500 ppm 25
5. 100 ml of a water sample consumed 25.0 ml of 0.01 M EDTA for the titration using
EriochromeBlack-T indicator. Calculate the total hardness.
Estimation of hardness.
Volume of water sample V1 = 100 ml
Strength of water sample M1 = ?
Volume of std EDTA V2 = 25 ml
Strength of std EDTA M2 = 0.01 M
V1M1 = V2M2
V2 M 2
M1 =
V1
25 0.01
M1 =
100
M1 = 0.0025 M
26
6. 0.25 gm of CaCO3 was dissolved in HCl and the solution made upto one litre with
distilled water. 100 ml of the above solution consumed 25 ml of EDTA solution on
titration.100 ml of hard water sample required 30 ml of same EDTA solution on
titration. 100 ml of this water, after boiling cooling and filtering required 11 ml of
EDTA solution on titration. Calculate the temporary permanent and total hardness of
water.
Calculate the strength of given std water,
Amount / Lit
Strength of Std Water = _________________________
Molecular Weight
0.25 gm
Strength of std Water =
100
Titration: I
Standardization of EDTA :
Volume of std water V1 = 100 ml
Strength of std water M1 = 0.0025M
Volume of EDTA V2 = 25 ml
Strength of EDTA M2 = ?
V1M1 = V2M2
V2 M 2
M1 =
V1
100 0.0025
M1 =
25
M2 = 0.01 M
Strength of EDTA = 0.01 M
27
Titration: II
Estimation of total hardness.
Volume of water sample V1 = 100 ml
Strength of water sample M1 = ?
Volume of std EDTA V2 = 30 ml
Strength of b std EDTA M2 = 0.01 M
V1M1 = V2M2
V2 M 2
M1 =
V1
30 0.01
M1 =
100
M2 = 0.003 M
Strength of hard water sample = 0.003 M
Amount of total hardness present in per liter in
terms of CaCO3 equivalent = Strength of hard water x M .W OF CaCO3
= 0.003M X 100
= 0.3 gm x1000
Amount total hardness = 300 ppm
Titration: III
Estimation of permanent hardness.
Volume of boiled water V1 = 100 ml
Strength of boiled water M1 = ?
Volume of std EDTA V2 = 11 ml
Strength of b std EDTA M2 = 0.01 M
V1M1 = V2M2
V2 M 2
M1 =
V1
11 0.01
M1 =
100
M2 = 0.0011 M
Strength of boiled water = 0.0011 M 28
Amount of permanent hardness present in per liter in
terms of CaCO3 equivalent = Strength of boiled water x M .W of CaCO3
= 0.0011M X 100
= 0.11 gm x1000
Amount of permanent hardness = 110 ppm
Temporary hardness = Total hardness - Permanent hardness
Temporary hardness = 300 ppm - 110 ppm
Temporary hardness = 190 ppm
Molecular Weight
1gm
Strength of std Water =
100
Titration: I
Standardization of EDTA :
Volume of std water V1 = 100 ml
Strength of std water M1 = 0.01M
Volume of EDTA V2 = 20 ml
Strength of EDTA M2 = ?
V1M1 = V2M2
V1M 1
M2 =
V2 29
100 0.01
M2 =
20
M2 = 0.05 M
Strength of EDTA = 0.05 M
Titration : II
Estimation of hardness.
Volume of water sample V1 = 100 ml
Strength of water sample M1 = ?
Volume of std EDTA V2 = 18.5 ml
Strength of b std EDTA M2 = 0.05 M
V1 M1 = V2 M2
V1M1 = V2M2
V2 M 2
M1 =
V1
18.5 0.05
M2 =
100
M2 = 0.00925 M
Strength of hard water sample = 0.00925 M
Amount of hardness present in per liter in
terms of CaCO3 equivalent = Strength of hard water x M .W of CaCO3
= 0.00925 M X 100
= 0.925 gm x1000
Amount of hardness = 925 ppm
30
1.7 BOILER FEED WATER
The water fed into the boiler for the production of steam is called boiler feed water. Water
used in steam making should be free from dissolved salts and gases, suspended
impurities, silica and oil. If it is used in boilers, these impurities lead to following
problems.
(i) Scale and sludge formation
(ii) Boiler Corrosion
(iii) Priming and foaming
(iv) Caustic embrittlement
a) Dissolved oxygen: The concentration of oxygen in water used in all types of boilers
should be minimum (0.01 ppm 0.05 ppm). Dissolved oxygen in water attacks the boiler
material at high temperature.
4Fe + 2H2O + O2 4Fe (OH)3
b) Dissolved carbon dioxide: Dissolved carbon dioxide in water produces carbonic
acidic, which is acidic and corrosive in nature.
Ca (HCO3)2 CaCO3 + H2O + CO2
CO2 + H2O H2CO3
c) Dissolved MgCl2: When water containing dissolved MgCl2 is used in the boiler, HCl is
produced which attacks boiler in a chain like reaction producing HCl again and again
which corrodes boiler severely.
MgCl2 + H2O Mg (OH)2 + 2HCl
Fe + 2HCl FeCl2 + H2
FeCl2 + 2H2O Fe (OH)2 + 2HCl
1.7.2.1 Removal of dissolved oxygen and carbon dioxide
i) Chemical method
Sodium sulphite, hydrazine are some of the chemicals used for removing oxygen.
2Na2SO3 + O2 2Na2SO4
N2H4 + O2 N2 + 2H2O
Hydrazine is found to be an ideal compound for removing dissolved oxygen in the
water, since the products are water and inert N2 gas.
Carbon dioxide can be removed from water by adding of calculated amount of
NH4OH in to water
2NH4OH + CO2 (NH4)2CO3 + H2O
Corrosion by acids can be avoided by the addition of alkali to the boiler water. 33
ii) Mechanical de-aeration method
Dissolved oxygen along with CO2 can be removed by mechanical deaeration.
34
called wet steam. These droplets of liuid water carry with them some dissolved salts and
suspended impurities. The phenomenon is called carry over. It occurs due to priming and
foaming.
Priming:
Priming is the process of formation of wet steam. It is caused by
(i) very high water level
(ii) high steam velocity
(iii) sudden steam demands leading to sudden boiling.
(iv) improper boiler design
Priming can be controlled by
(i) keeping the water level lower
(ii) good boiler design providing mechanical steam purifier
(iii) avoiding rapid changes in the steaming rate, caused by sudden steam
demands.
Foaming:
The formation of stable bubbles above the surface of water is called foaming. These
bubbles are carried along with steam leading to excessive priming.
It may be caused by presence of oil, grease in water and finely divided sludge
particles.
Foaming can be prevented by
(i) adding antifoaming agents like synthetic polyamides
(ii) finely divided sludge particles, oil and grease can be removed by the addition
of coagulants such as sodium aluminate, ferrous sulphate etc.
35
1.7.4 Caustic embrittlement (Inter-crystalline cracking)
As water evaporates in the boiler, the concentration of sodium carbonate increases in
the boiler. Sodium carbonate is used in softening of water by lime soda process, due to
this some sodium carbonate maybe left behind in the water. As the concentration of
sodium carbonate increases, it undergoes hydrolysis to form sodium hydroxide.
Na2CO3 + H2O → 2NaOH + CO2
This NaOH flows into the minute hair cracks and crevices usually present on the boiler
material by capillary action and dissolves the surrounding area of iron as sodium ferroate.
Fe + 2NaOH Na2FeO2 + H2
This causes brittlement of boiler parts, Particularly stressed parts like bends, joints,
rivets etc, causing even failure of the boiler.
Prevention :
Caustic embrittlement can be prevented by using sodium phosphate as softening
agent instead of sodium carbonate
By adding tanning lignin to the boiler water, which blocks the minute hair cracks
37
Cation exchanger:
Resins containing acidic functional groups (-COOH,-SO3H) are capable of exchanging
their H+ ions with other cations of hard water. Cation exchange resin is represented as
RH2.
Examples
i. Sulphonated coals.
ii. Sulphonate polystyrene.
R-SO3H: R-COOH = RH2
Anion exchanger:
Resin containing basic functional groups (-NH2, -OH) are capable of exchanging their
anions with other anions of hard water. Anion exchange resin is represented as R (OH)2.
Examples
i) Cross-linked quaternary ammonium salts.
ii) Urea-formaldehyde resin
R-NR3OH; R-OH; R-NH2 = R (OH)2
Process
The hard water first passed through a cation exchange column,
Which absorbs all the cations like Ca2+,Mg2+,Na+,K+,etc.,present in the hard water.
RH2 + CaCl2 RCa + 2HCl
RH2 + MgSO4 RMg + H2SO4
RH + NaCl RNa + HCl
The cation free water is then passed through a anion exchange column, which absorbs
all the anions like Cl-, SO42-, HCO3-, etc., Present in the water.
R’ (OH)2 + 2HCl R’Cl2 + 2H2O
R’ (OH)2 + H2SO4 R’SO4 + 2H2O
38
The water coming out of the anion exchanger is completely free from cations and
anions. This water is known as determineralised or deionized water.
Regeneration
When the cation exchange resin is exhausted, it can be regenerated by passing a
solution of dil. HCl or H2SO4
RCa + 2HCl RH2 + CaCl2
RNa + HCl RH + NaCl
Similarly, when the anion exchange resin exhausted, it can be regenerated by passing a
solution of dil.NaOH.
R’Cl2 + 2NaOH R’(OH)2 + 2NaCl.
39
1.8.2.1 Carbonate Conditioning
Scale formation can be avoided by adding Na2CO3 to the boiler water. It is used only
in low pressure boilers. The scale forming salt like CaSO4 is converted into CaCO3, which
can be removed easily.
CaSO4 + Na2CO3 CaCO3 + Na2SO4
40
1.9 TREATMENT OF WATER FOR DOMESTIC SUPPLY
Rivers and lakes are the most common source of water used by municipalities. These
water should be free from colloidal impurities, domestic sewages, industrial effluents and
disease producing bacteria. Hence domestic supply of water involves the following stages
in the purification process.
Sterilisation
(or)
Disinfection
i. Screening
It is a process of removing the floating materials like leaves, wood pieces, etc. from
water. The raw water is allowed to pass through a screen, having large no of holes, which
retains the floating materials and allow the water to pass.
ii. Aeration
The process of mixing water with air is known as aeration. The main purpose of
aeration is to remove gasses like CO2, H2S, and other volatile impurities causing bad taste
and odour to water and remove ferrous and manganous salts as insoluble ferric and
manganic salts.
41
iii. Sedimentation
It is a process of removing suspended impurities by allowing the water to stand
undisturbed for 2-6 hours in a big tank. Most of the suspended particles settle down at the
bottom, due to forces of gravity, and they are removed. Sedimentation removes only 75%
of the suspended impurities.
iv. Coagulation
Finely divided clay, silica, etc do not settle down easily and hence cannot be removed
by sedimentation. Such impurities are removed by coagulation method.
In this method certain chemicals, called coagulants like alum, Al2 (SO4)3 etc., are added
to water. When the Al2 (SO4)3 is added to water, it gets hydrolyzed to form a gelatinous
precipitate of Al (OH)3. The gelatinous precipitate of Al (OH)3 entraps the finely divided
clay and colloidal impurities, settles to the bottom and can be removed easily.
v. Filtration
It is the process of removing bacteria, colour, taste, odour and suspend particles by
passing the water through filter beds containing fine sand, coarse sand and gravel.
The sand filter consists of a tank containing a thick top layer of fine sand followed by
coarse sand, fine gravel and coarse gravel. When the water passes through the filtering
medium, it flows through the various beds slowly. The rate of filtration decreases slowly
due to clogging of filtration becomes very slow, the filtration is stopped and the thick top
layer of fine sand is scrapped off and replaced with clean sand. Bacteria are also partly
removed by this process.
42
Water
Fine sand
Coarse sand
Fine gravel
Coarse gravel
Water outlet
1.By Boiling
When water is boiled for 10-15 minutes all the harmful bacteria are killed and water
becomes safe for use
Disadvantages
1. Boiling alters the taste of drinking water
2. It is impossible to employ it in municipal water works.
43
2. By Using Ozone
Ozone is a powerful disinfectant and is readily absorbed by water. Ozone is highly
unstable and break down to give nascent oxygen.
O3 O2 + [O]
The nascent oxygen is a powerful oxidizing agent and kills the bacteria.
Disadvantages
1. This process is costly and cannot be used in large scale
2. Ozone is unstable and cannot be stored for long time
Disadvantages
1. It is costly
2. Turbid water cannot be treated.
4. By Chlorination
The process of adding chlorine to water is called chlorination. Chlorination can be
done by the following methods.
44
b) By Adding Chloramine
When chlorine and ammonia are mixed in the ratio 2:1, a compound chloramine is
formed.
Cl2 + NH3 ClNH2 + HCl
When chloramine is added to water it decomposes slowly to give chlorine. It is a better
disinfectant than chlorine.
45
It is seen from the graph that initially the applied chlorine is used to kill the bacteria
and it oxidises all the reducing substances present in the water and there is no free
residual chlorine.
As the amount of applied chlorine increases, the amount of combined residual
chlorine also increases. This is due to the formation of chloramines and other
chlorocompounds.
At one point, on further chlorination the oxidation of chloramines and other impurities
starts and there is a fall in the combined chlorine content. Thus the combined residual
chlorine decreases to a minimum point at which oxidation of chloramines and other
impurities complete and free residual chlorine begins to appear, this minimum point is
known as ”break point chlorination”.
Formation of Destruction
chloramines& of Free residual
chloro chloramine chlorine
Kill the bacteria, and oxidation of
reducing compounds by chlorine
compounds &chloro
compounds
Residual chlorine
Applied chlorine
Figure 1.5 Break point chlorination
Thus the break point chlorination eliminates the bacterias, reducing substances, organic
substances responsible for the bad taste and odour, from the water. 46
1.10. Reverse osmosis
When two solutions of different concentrations are separated by a semi – permeable
membrane, solvent (water) flows from a region of lower concentration to higher
concentration. This process is called Osmosis. The driving force in this phenomenon is
called osmotic pressure.
Advantages:
1. The life time of the membrane is high, and it can be replaced within few minutes.
2. It removes ionic as well as non – ionic, colloidal impurities.
3. Due to low capital cost, simplicity, low operating, this process is used for converting
sea water into drinking water.
47
IMPORTANT QUESTIONS
1. Explain how sterilization of water carried out using chlorine? Give the mechanism.
2. (i) What is meant by carbonate and non-carbonate hardness of water ?Explain with
example
(ii) What is break-point chlorination? State its significance
3. Write a detailed procedure for the determination of various forms of alkalinity.
4. (i) Define the term desalination with a neat diagram, describe desalination by
reserve osmosis’ method.
(ii) What are coagulants? Write the mechanism of coagulation process with suitable
example.
5. What are the boiler troubles? How are they caused? Suggest steps to minimize the
boiler troubles.
6. What is desalination? How is this achieved by reverse osmosis?
Explain break point chlorination, State its significant.
7. Compare the zeolite process with ion-exchange process in water softening. How will
you regenerate the used up reagents.
8. (i) How is the hardness of water determined by EDTA method
(ii) Describe briefly the different steps in the purification of water for drinking
purposes. What is the usage of breakpoint chlorination? (or) Out line the various
stages of domestic water treatment in sequence.
9. What is meant by sterilization of water? What are the chemicals that are normally
used for this purpose? Explain break-point chlorination.
(i) Distinguish between softwater and demineralized water.
(ii) How do you estate the total hardness of water by EDTA method? Explain.
10. Explain how sterilization of water carried out using chlorine? Give the mechanism.
48
11. Explain with a neat sketch the various steps in the treatment of water for municipal
supply.
12. What is the principle of EDTA method? Describe the estimation of hardness by
EDTA method.
(or)
Explain the principle and procedure involved in the determination of permanent and
temporary hardness by EDTA method.
13. Describe de-mineralisation process of water softening. Explain the reaction involved.
14. What is potable water? What are the steps taken to obtain pure drinking?
15. How is water disinfected by chlorine.
16. Describe the principle and method involved in the determination of different types
and amount of alkalinity of water.
17. 0.28 gm of CaCO3 was dissolved in HCl and the solution made upto one litre with
distilled water. 100 ml of the above solution required 28 ml of EDTA solution on
titration.100 ml of hardwater sample required 33 ml of same EDTA solution on
titration. 100 ml of this water, after boiling cooling and filtering required 10 ml of
EDTA solution on titration. Calculate the temporary and permanent hardness of
water.
18. Explain the various steps involved in the domestic water treatment.
19. How is the exhausted resin regenerated in an ion-exchanger? What are the merits
and demerits of ion-exchange method?
20. What are the requisites of water for municipal supply. How is raw water treated for
domestic purpose.
21. Discuss the causes and prevention of priming and foaming.
22. What is the principle of reverse osmosis. How is it used for desalination process.
23. Discuss desalination by reverse osmosis process.
24. Explain
(i) Phosphate conditioning. 49
(ii) Sedimentation with coagulation.
25. What is desalination. Name the different methods of desalination. Explain any one in
detail.
26. What is caustic embrittlement. How can it be prevented.
27. What are scales and sludges. Describe the disadvantages of scale and sludge
formation.
28. What are the problems one would face when hard water is used in boiler industries.
Question Bank
4. How is the hardness of water expressed? (or) bring out the significance of calcium
carbonate equivalents?
The concentration of hardness producing salts are usually expressed interms of an
equivalent amount of CaCO3.
Significance: Its molecular weight is a whole number and it is the most insoluble
salt.of the concentration of hardness producing salt is X mgs/lit, then
Amount equivalent to
X x molecular weight of CaCO3
CaCO3 = --------------------------------------------------------------
Molecular weight of hardness producing substances
5. Write the units of hardness (or) bring out the relationship between ppm and mg/lit.
(i) parts per million (ppm) : it is defined as the no of parts of CaCO3 equivalent
hardness per 106 parts of water.
(ii) milligrams per litre(mg/lit) : it is defined as the no of milligrams of CaCO3
equivalent hardness per 1 litre of water.
(iii) clarke’s degree (°Cl) : it is defined as the no of parts of CaCO3 equivalent hardness
per 70,000 parts of water.
(iv) French Degree (°Fr) : It is defined as the number of parts of CaCO3 equivalent
hardness per 105 parts of water.
Relationship between ppm and mg/lit:
1 mg/lit= 1 mg of CaCO3 equivalent in 106 parts of water
= 1 part of CaCO3 equivalent hardness in 106 parts of water.
51
= 1 ppm.
9. Draw the structure of EDTA. What happens when EDTA is added to hard water ?
HOOCH2C CH2COOH
N - CH2 - CH2 - N
HOOCH2C CH2COOH
When EDTA is added to hardwater, it forms stable complex with hardness producing ions
like Ca and Mg the pH range = 8 -10
[Ca++, Mg++] + EDTA ------------ [ Ca, Mg EDTA] complex
52
If hard water obtained from natural sources is fed directly to the boilers, the following
troubles may arise.
1. Scale and sludge formation
2. Priming and foaming (carry over)
3. Caustic embrittlement
4. Boiler corrosion.
2. Sludge
If the precipitate is loose and slimy it is called sludge. Sludges are formed by
substances like MgCl2, MgCO3, MgSO4 and CaCl2. They have greater solubilities in hot
water than cold water.
12. What is meant by priming and foaming? How can they be prevented?
Priming is the process of production of wet steam. Priming can keep prevented by
controlling the velocity of steam and keeping the water level lower.
Foaming is the formation of stable bubbles above the surface of water. Foaming can be
prevented by adding coagulants like sodium aluminate and antifoaming agents like
synthetic polyamides.
15. What are the requirements of drinking and boiler feed water?
(i) Boiler feed water Must have zero hardness and free from dissolved gases like
O2, CO2.
(ii) Drinking water (i) pH of water should be in the range of 7.0 – 8.5.
(ii) Total hardness and dissolved solids of water should be
less than 500 ppm
26. Write the principle involved in the desalination of water by reverse osmosis.
(Or)
What is meant by ‘reverse osmosis’? How is it applied in the desalination of water?
If the process in excess of osmotic pressure is applied on the higher concentration side,
the solvent flow is reversed i.e., solvent flows from higher concentration to lower
concentration side. This process is called reverse osmosis.
Salt water is taken as higher concentration and water is taken as solvent of pressure is
applied on the salt water, the water flows from salt water to water side.
57
UNIT-II
ELECTROCHEMISTRY
Introduction
Electrochemistry deals with chemical changes produced by an electric current and
the production of electricity by chemical reactions. All electrochemical reactions involve
transfer of electrons and are redox reactions. Electrochemistry is a branch of physical
chemistry, which deals with the relationship between chemical energy and electrical
energy. Electrochemistry is the basis for batteries, cells, corrosion and its prevention,
metal coatings, etc. And holds a central position in the study of chemistry for a number of
reasons:
(i) It acts as a bridge between thermodynamics and the rest of the chemistry because
it provides techniques both for measuring the thermo dynamical state functions
and for predicting equilibrium concentrations of dissolving and reacting ions.
(ii) It is of enormous commercial importance because of the costly destruction caused
by corrosion and the exciting possibilities of fuel cells which generate electricity
directly from fuel.
From the above valid reasons, the subject electrochemistry consists of major
technological importance.
58
2.1. Basic Definitions:
2.1.1 Conductors
Conductors are the substances or materials, which allow electric current to pass through
them.
Examples: all metals, graphite, fused salts, aqueous solutions of acids, bases and salts.
59
Table 2.1 Differences between metallic conduction and electrolytic conduction
S.No Metallic conduction Electrolytic conduction
1 It involves the flow of electrons in It involves the movement of ions in a
a conductor solution.
2 It does not involve any transfer of It involves transfer of electrolyte in the
matter. form of ions.
3 Conduction decreases with Conduction increases with increase in
increase in temperature. temperature.
4 No change in chemical properties Chemical reactions occur at the two
of the conductor. electrodes.
Ohm’s law:
This law can be stated as, at constant temperature, the strength of the current
flowing through a conductor is directly proportional to the potential difference and
inversely proportional to the resistance(R) of the conductor.
V
I , V RI , V = volt I ampere , R = ohm
R
Specific resistance:
The resistance (R) ohms offered by the material of the conductor to the flow of
current through it is directly proportional to its length (l) and inversely proportional to the
area of cross section (a) of the conductor. Thus,
60
l
R
a
ρ called the specific resistance and it is resistance in ohms which one meter cube of
material offers to the passage of electricity through it, Unit of specific resistance is ohm-
meter.
Specific conductance:
The reciprocal of specific resistance is called as specific conductance (or) specific
conductivity (k) [k is called ‘kappa’].k is defined as the conductance of one meter cube of
an electrolyte Solution
1 1 l 1 l
k . k .
R a R a
k = Ohm-1.cm-1
Also, 1 siemen = 1 mho.
k is also expressed as S.m-1.
Equivalent conductance:
Equivalent conductance ( C) is defined as the conductance of an electrolyte
solution containing one gram equivalent of the electrolyte. It is equal to the product of
specific conductance (k) of the electrolytic solution and the volume (V) of the solution that
contains one gram equivalent of the electrolyte.
c k V
temperature.
and are the cationic and anionic equivalent conductance at infinite
dilutions and n+ and m– correspond the valency of cations and anions furnished by each
molecule of the electrolyte
, CH3COOH = H+ + – CH3COO–
62
Application of Kohlraush’s law:
appropriate manner.
For example
HCl, and CH3COONa in such a manner that of CH3COOH is obtained. Sodium acetate
(CH3COONa) is a strong electrolyte and it ionises to acetate (CH3COO–) and sodium (Na+)
ions at all concentrations in water. Applying Kohlraush’s law,
Example 1 :The equivalent conductances at infinite dilution of HCl, CH3 COONa and
NaCl are 426.16,91.0 and 126.45 ohm–1 cm2 gm.equiv-1 respectively. Calculate the of
acetic acid.
63
2.3 Conduct metric Titration
This method of estimation is applicable to any titration in which there is a sharp
change in conductance at the end point .The principle involved in these titrations is that
electrical conductance depends upon the number and mobility of the ions. In these
titrations it is a necessary to observe the following.
1. To keep the temperature constant throughout the experiment.
2. The titrant solution should be 10 times stronger than the solution to be titrated so
that the volume change is as little as possible.
Conductance is followed during the course of titration and the values are plotted
against volume of the titrant added. Conductrometric curves should be straight line or
nearly straight lines and only few readings on each sides of the end point are required
.The end point of titration is the point of intersection of the two straight lines.
At the beginning of the titration the conductance of the HCl acid solution is due to
the H+ and Cl - .As alkali is added gradually from the burette, fast moving H+ ions are
replaced by slow moving sodium ions and the reaction is.
H+ + Cl – + [Na++OH-] Na+ + Cl- + H2O
64
points will lie on two straight lines as in figure (A).The point of intersection of these two
straight lines AB and CD gives the volume of alkali required for the neutralization
Conductance (ohm-1)
D
B
C
| l
Volume of NaOH
After the completion of neutralization of the acid, any further addition of alkali will
show a sharp increase in conductance due to the sodium and the fast moving hydroxide
ions. The plot of conductivity against volume of alkali added is shown in the figure (B).the
point of intersection straight lines gives the end point
65
D
Conductance (ohm-1)
C
l
B
l
A
l
Volume of NaOH
Fig 2.2
2.3.3 Titration of a mixture of acids (CH3COOH + HCl) against a strong base (NaOH)
(CH3COOH + HCl ) Vs NaOH
When a mixture of acetic acid and hydrochloric acid is titrated against sodium
hydroxide, usually a combination of figures (2.1 &2.2) will be obtained. It is as shown in
the figure 2.3. In these titrations the strong acid (HCl) will get titrated first and the titration
of the weak acid (CH3COOH) will commence only after the complete neutralization of the
strong acid. It is evident from the figure 2.3 , that the first end point corresponds to the
neutralization of HCI, the second end point corresponds to the neutralization of acetic
acid.
Conductance (ohm-1)
A
D
C l
l l
l l
Volume of NaOH
Fig 2.3 66
2.3.4 Precipitation titration (BaCl2 Vs Na2SO4)
Let us now see how the conductance varies during the course of titration of barium
chloride against sodium sulphate. Here BaSO4 gets precipitated; while chloride ions
remain unchanged.
A D
Conductance (mho-1)
Volume of Na2SO4
Fig 2.4
68
viii) Half-cell
Half-cell is a part of cell, containing an electrode dipped in an electrolyte solution.
If oxidation occurs at the electrode that is called oxidation half cell. If reduction occurs at
the electrode, that is called reduction half cell.
ix) The salt bridge
It allows charge transfer through the solutions but prevents mixing of the
solutions.
Mn+ + ne- M
An electrode potential forms between the metal and the solution of its own ions of
valency ‘n’.The electrical energy required to transfer 1g equivalent of metal ion from the
metal to the solution would be ‘nFE’, where F is 1 Faraday and E is the electrode potential.
The decrease in free energy namely
-G would be equal to the electrical energy released namely nFE.
-G = nFE ---------- (1)
According to Van’t Hoff reaction isotherm, at equilibrium this free energy change
in terms of concentrations is given by, 69
[Pr oduct ]
G = G 0 2.303 RT log ---------- (2)
[Re ac tan t ]
Substituting equation (1) in equation (2), we get,
[M ]
nFE = nFE 0 2.303 RT log ---------- (3)
[ Mn ]
Dividing Equation (3) both sides by – nF,
E = E0
2.302 RT
log
M
nF [ Mn ]
2.302 RT 1
E = E0 log
nF [ Mn ]
(as activity or concentration of metal=1 at all temperatures) [M]= 1
E = E0
2.302 RT
nF
log Mn
This expression is known as Nernst’s equation for electrode potential. Since R=8.314
J/K/mole; F= 96,500 Coulombs; T= 25 0 C+273 = 298K ; then, Nernst equation at 250C is
E = E0
0.0591
n
log Mn
Nernst equation for reversible electrochemical cell.
E = E0
0.0591
log
Pr oduct
nF [Re ac tan t ]
70
Problems based on Nernst equation.
1) Calculate the reduction potential of the Cu/Cu2+(0.5M) at 25 0C . E0 Cu2+ / Cu =
0.34V
Given:
E 0 0.34V ; [Cu2+] =0.5M; n =2;
Solution
Electrode reaction Cu2+ +2e Cu
E red = E0
0.0591
log
Pr oduct
n Re ac tan t
E0
0.0591 Cu
E red =
n
log
Cu 2
E red = 0.34
0.0591
log
1
n 0.5
0.0591
E red = 0.34 log 2
n
0.0591 0.3010
E red = 0.34
2
E red = 0.34 0.0089
E red = 0.3311V
E oxd = E0
0.0591
log
Pr oduct
n Re ac tan t
0.0591 0.2
E oxd = 0.76 log
n 1 71
0.0591 0. 2
E oxd = 0.76 log
2 1
0.0591 1
E oxd = 0.76 log
2 0.2
0.0591
E oxd = 0.76 log 5
2
E oxd = 0.76 0.02955 0.6990V
E oxd = 0.7806V
3) Calculate the electrode potential of Zinc electrode dipped in 0.1 M ZnSO4 solution
at 250C .
Given:
Concentration of ZnSO4 = 0.1M ; n = 2 ; E 0Zn / Zn2+ = - 0.76V
Solution
E = E0
0.0591
log
Pr oduct
n Re ac tan t
0.0591 Zn
E = 0.76
n
log
Zn 2
0.0591 1
E = 0.76 log
n 0.1
0.0591
E = 0.76 log 0.1
2
E = 0.76 0.02955 1
E = 0.76 0.02955 V
E = 0.7896V
72
4) Calculate the standard e.m.f. of the cell : Cd, Cd2+||Cu2+,Cu and determine the
cell reaction. The standard reduction potentials ofCu2+, Cu and Cd2+, Cd are
0.34V and –0.40 volts respectively. Write the cell reaction and predict the
feasibility of the cell reaction.
Solution.
0
E Cell = E0right – E0 left
0
E Cell = [Std. reduction potential of Cu2+, Cu]
5) Calculate the standard free energy of the cell reaction is the following cell at
250C Zn, Zn2+ Ni2+, Ni. The standard reduction potentials of Zn2+, Zn and
73
G 0 = – nFE0cell
n = 2 electrons
G 0 = –2 x 96495 x 0.51
= –97460 Joules
G 0 = – 97.46 kJ.
8) Calculate the equilibrium constant for the reaction between silver nitrate and
metallic zinc.
Solution:
Cell construction: Zn /Zn2+ / / Ag+ / Ag
Step 1 : Write the equation for the reaction
2Ag+ + Zn → Zn2+ + 2Ag E0cell = 1.56 V
Step 2 : Substitute values in the Nernst equation at equilibrium
0.0591
E Cell = E0 log K
n
At the Eqm, E Cell =0, Where K = Equilibrium Constant
0 0.0591
0 = E Cell log K
n
0.0591 0
log K = E Cell
n
0
nE Cell
log K =
0.0591
2 1.56
log K =
0.0591
74
log K = 53
K = 1053
n=2
0.0591
E Cell = E0 log K
n
E Cell = E0
0.0591
log
Ag 2 Zn 2 [Zn]=1, [Ag]=1
n Ag 2 Zn
E Cell = 1.56
0.0591
log
1 0.1
2
2 10 2 1
0.0591 0. 1
E Cell = 1.56 log
2 [ 10 ] 2
E Cell = 1.6487V
75
10) Calculate the potential of the following cell at 298 K Zn/Zn2+ (a = 0.1) // Cu2+ (a
= 0.01) / Cu, E 0Zn 2 Zn
0
= – 0.762 V E Cu 2
Cu
= + 0.337 V Compare the free energy
change for this cell with the free energy of the cell in the standard state.
Solution:
The overall cell reaction is
Zn + Cu2+ (a = 0.01) Zn2+ (a = 0.1) + Cu
E0cell = ERight - ELeft
E0cell = 0.337 – (- 0.76)
E0cell = 1.099 V
The cell potential given by nernst equation
Ecell = E0
0.0591
Cu Zn 2
n
log
Cu 2 Zn
0.0591 1 0.1
Ecell = 1.099 log
2 0.01 1
0.0591 0.1
Ecell = 1.099 log
2 0.01
Ecell = 1.0694V
76
2.6 Types of Electrodes:
Some common types of reversible electrodes
1) Metal-Metal ion electrodes
An electrode of this type consist of a metal rod dipping in a solution containing its
own ions, Example, zinc rod dipping in zinc sulphate solution.
Representation: Zn/Zn2+
Reaction: Zn2+ + 2e- Zn
77
2.6.1 Metal –metal ion electrodes
A metal (in the form of a rod ,strip or wire) in contact with it own ions in an
aqueous salt solution constitutes a metal –metal ion electrode represented M / Mn+ .
Reactivity of metal should be intermediate otherwise the reactive metal will reactive with
water and cannot function as electrodes.
Example of metal-metal ion electrodes : Cu/Cu2+ and Zn/Zn2+
Cell construction:
Hydrogen electrode consists of a small platinum foil(0.5cm x 0.5cm) .that is
connected to a platinum wire and sealed in a glass tube. Hydrogen gas passed through the
side arm of the glass tube. This electrode, when dipped in a1N HCl and hydrogen gas at 1
atmoshspheric pressure is maintained ,forms a normal or standard hydrogen electrode.
0
( SHE) . The emf of this cell has been arbitrarily fixed at zero. (i.e. E SHE E 0HE 0 )
0 0
E oxd HE E oxd HE 0
78
Fig 2.4
E oxd = E0
0.0591
log
H
n H 2 12
0 0.0591 H
log 2
1
2
E red = E
1 H
H2 2
1
= - 0.0591 log
H
1
H 2
1
2 1 2 1
1
= - 0.0591 log
H
= 0.0591 log H
E HE = 0.0591 p H p H log( H )
79
Limitations
i) The potential of the electrode can be altered by changing the barometric pressure.
ii) It requires considerable volume of test solution.
iii) It requires hydrogen gas and is difficult to set up and transport.
Fig 2.5
It is a mercury-mercurous chloride electrode. It is made up of a glass tube containing
mercury at the bottom above which mercurous chloride is placed. The rest of the vessel is
filled with KCl solution. The glass tube consists of two side-tubes, one is used to fill the
tube with solutions and the other side-tube is used for making electrical contact with a salt
bridge. The electrode potential depending upon the concentration of KCl solution, the
80
potential of calomel electrode varies. The activity of Cl ion is inversely proportional to
electrode potential The reduction potentials for different KCl solution concentrations at
25 0C are given below:
If the concentration of KCl solution are 0.1N, 1N and saturated, They are called as
decinormal, normal (NCE) and saturated calomel electrodes (SCE).
The reaction taking place in this electrode:
If the electrode acts as anode, the reaction is,
2 Hg(l) + 2 Cl Hg2 Cl 2 ( s ) +2 e
E = E0
0.0591
log
Pr oduct
n Re ac tan t
E = 0.2422
0.0591
2
log Cl
2
81
The cell constructed is : Zn Zn 2 SCE
E Cell = E SCE E Zn , Zn 2
Since E SCE is known, measuring the Emf of the cell & Ecell, We can calculate the potential of the
Zn electrode.
82
Cell representation: Pt, HCl (0.1M) / Glass
83
Fig 2.8
E Cell = E SCE E GE
E Cell =
E SCE E G0 0.0591 pH
E Cell = 0.2422 E a0 0.0591 pH
0.2422 E a0 E cell
pH =
0.0591
Limitations
Since the resistance is quite high, special electronic potentiometer is employed for
measurement
84
The glass electrode can be used in solutions only with pH range of 0-10. However
above pH 12 (high alkalinity), cations of the solution affect the glass and make the
electrode useless.
Zn + CuSO4 ZnSO4 + Cu
Cu + ZnSO4 No reaction
i.e. Cu²+ has a greater tendency to acquire Cu form, than Zn2+ has for
acquiring Zn form.
86
4. Determination of equilibrium constant for the reaction
Standard electrode potential can also be used to determine the equilibrium constant
(K) for an equal reaction. We know that,
G 0 = RT ln K
G 0 = 2.303 RT log K
G 0
log K =
2.303 RT
nFE 0
log K =
2.303 RT
From the value of E°, the equilibrium constant for the cell reaction can be
calculated.
87
In general, an element having lower reduction potential can displace another metal
having higher potential from its salt solution spontaneously.
Galvanic series:
Oxidation potential of various metals and alloys are measured by immersing them
in sea water using standard calomel electrode as the reference electrode. These values are
arranged in decreasing order of activity and this series is known as the galvanic series.
88
Difference between Electrochemical series and Galvanic series:
Electrolytic Cell
Electrolytic cells are cells in which electrical energy is used to bring about the
chemical change. Electrolysis is very good example for the electrolytic cell . An electrolytic
cell is an electrochemical cell in which the energy from an applied voltage is used to drive
an otherwise nonspontaneous reaction. Such a cell could be produced by applying a
reverse voltage to a voltaic cell like the Daniell cell.
89
Fig 2.9
If a voltage greater than 1.10 volts is applied as illustrated to a cell under standard
conditions, then the reaction
Cu(s) + Zn2+(aq) Zn(s) + Cu2+(aq)
will be driven by removing Cu from the copper electrode and plating zinc on the zinc
electrode. Electrolytic processes are very important for the preparation of pure substances
like aluminum and chlorine.
90
2.11.1 Galvanic cell
Galvanic cell is used to convert chemical energy released in a redox reaction into
electrical energy. The electrons transferred in the redox reaction (oxidation and reduction)
are utilized as a source of producing electrical energy.
91
The electrode reactions in the cell are
Thus the number of electrons liberated in one half-reaction is equal to the number
of electrons consumed in the other half-reaction. The overall cell reaction is called as redox
reaction.
Applying the rules for the representation of a cell, the Daniel cell can be
represented as follows;
Zn | ZnSO4 || CuSO4 | Cu
Or
Zn | Zn2+ || Cu2+ | Cu
92
6) A salt bridge is indicated by double vertical lines ( ) separating the two half-cells.
7) The symbol for an inert electrode, like the platinum electrode is often enclosed in a
bracket. For example,
Mg | Mg2+ H+ | H2(Pt)
(8) The value of emf of a cell is written on the right of the cell diagram.
Thus a zinc-copper cell has emf 1.1 V and is represented as
Zn | ZnSO4 CuSO4 | Cu E = + 1.1 V
Direction of electron flow
If the emf acts in the opposite direction through the cell circuit,it is denoted as a
negative value.
Cu | CuSO4 ZnSO4 | Zn E = – 1.1 V
Direction of electron flow
The negative sign also indicates that the cell is not feasible
93
5 The emf of the cell depends on the The extent of chemical reaction
concentration of the electrolytes and taking place at the electrodes is the
chemical nature of electrodes governed by Faradays law of
electrolysis.
6 The emf produced by the cell is Amount of electricity passed during
measured using potentiometer electrolysis is measured by
coulometer
7 In ordinary galvanic cell, two different In this type of cell ,only one type of
electrods and two different electrolytes electrolyte and two electrodes of the
are employed usually same element are employed mostly
8 Eg .Zn/Zn //Cu /Cu
2+ 2+ Eg. Zn/ZnSO4 /Zn
CONSTRUCTION: 94
The zinc electrode dipped in diluted ZnSO4 solution is the anode. The zinc
electrode dipped in concentrated ZnSO4 solution is the cathode. These two solutions are
connected through a salt bridge. The chemical reaction taking at anode and cathode can be
written as follows:
At anode (left hand electrode) oxidation takes place
Zn Zn2+(C1) + 2e-
At cathode (Right hand electrode) reduction occurs
Zn2+ (C2) + 2e- Zn
The net reaction is the sum of these two electrode reactions.
Zn2+ (C2) Zn2+ (C1)
From this, it is clear that there is no net chemical reaction and this involves only the
transfer of Zn2+ ions from a solution of higher concentration (C2) to a solution of lower
concentration (C1).
We can write the Nernst equation at 25˚c obtain the emf of the concentration cell
as follows
0.0591
At anode: ER = E0 log C1
n
0.0591
At cathode: ER = E0 log C 2
n
0.0591 C
Ecell = log 2
n C1
At anode (LHE)
Ag Ag+ (C1) + e-
At cathode (RHE)
Ag+ (C2) + e- Ag
The net cell reaction is
Ag+ (C2) Ag+ (C1)
From the net cell reaction it is very clear that, there is no net chemical reaction and
this process merely involves transfer of Ag+ ions from a solution of higher concentration
(C2) to a solution of lower concentration (C1)
Concentration cell
Fig 2.11
96
From the Nernst equation, the electrode potential of the cell is given as follows. We
can write the Nernst equation at 25˚c and obtain the emf of the concentration as follows
0.0591
At anode : E L = E0 log C1
n
0.0591
At cathode: ER = E0 log C 2
n
For the process to be feasible, emf should be positive. Hence, C2 > C1.
97
Where, C1 is the concentration of Ag+ ions furnished by AgCl in KCl solution.
By measuring emf of the cell, the concentration of AgCl is calculated. Multiplying this by
143.5, the equivalent weight of AgCl , we can get the solubility of AgCl in grams/litre
Solubility = C1×143.5 (Eq.wt of AgCl)
The value of n, the number of electrons involved in the cell reaction, (Called the
valency of ions) can be calculated.
Solution
0.0592 C
E Cell = log 2
n C1
0.0592 0.1
E Cell = log
2 0.01
0.0592
E Cell = log 10
2
E Cell = 0.0295 V
98
2) Calculate the emf of a Daniel cell at 250 C, when the concentration of ZnSO4 and
CuSO4 are 0.001 and 0.1 M respectively .The standard emf of the cell 1.1 volt
Concentration of the cell
Zn / ZnSO4 (0.001M) // CuSO4(0.1M) / Cu
Solution
Emf = Eright - Eleft
0 0.0591 0.0591
E = E Cu log 0.1 E 0Zn log 0.001
2 2
E = E 0
Cu
E 0Zn
0.0591
2
log
0.1
0.001
E = 1.1 0.02955 2
E = 1.1591 V
3) The emf of the cell consisting of a hydrogen electrode and the normal calomel
electrode is 0.664V at 250C .the emf of normal calomel electrode is 0.2802V.
Calculate the pH of the solution.
Given:
E Cell = 0.664V;
Solution
ECell = ECalomel - EHydrogen
0.664 = 0.2802 – 0.0591 pH
0.664 0.2802
pH =
0.0591
pH = 6.5
99
Excises :
1) Calculate the potential of the following concentration cell at 250C
/ Cu2+ (0.012M) // Cu2+(0.032M)/ Cu
101
If the value of Ecell is positive, it indicates that the cell reaction is feasible and if it
is negative then the reaction is not feasible. Every galvanic or voltaic cell is made up of
two half-cells, the oxidation half-cell (anode) and the reduction half-cell (cathode). The
potentials of these half-cells are always different. On account of this difference in electrode
potentials, the electric current moves from the electrode at higher potential to the electrode
at lower potential, i.e., from cathode to anode. The direction of the flow of electrons is
from anode to cathode.
Flow of electrons
Anode Cathode
Flow of current
(i) When reduction potentials of both electrodes are taken into account:
E0Cell = Reduction potential of cathode - Reduction potential of anode
E0Cell = ECathode - EAnode
E0Cell = Eright - Eleft
(ii) When oxidation potentials of both electrodes are taken into account:
E0Cell = Oxidation potential of anode - Oxidation potential of cathode
E0Cell = Eoxi (anode) - Ered (cathode)
E0Cell = Eoxi (Left) - Ered (right)
102
Measurement of emf of a cell
The potential difference or emf of a cell can be measured on the basis of
Poggendorff’s compensation principle. Here the emf of the cell is just opposed or balanced
by an emf of a standard cell so that no current flows in the circuit.
The potentiometer consists of a uniform wire AB . A storage battery (K) is
connected to the ends A and B of the wire through a rheostat (R) .The cell of unknown
emf (x) is connected in the circuit by connecting its positive pole to A and the negative
pole to a sliding contact (D) through a galvanometer G. The sliding contact is freely
moved along the wire AB till no current flows through the galvanometer. Then the
distance AD is measured. The emf of unknown cell is directly proportional to the distance
AD.
Fig 2.12
Ex α AD
Then the unknown cell (x) is replaced by a standard cell (s) in the circuit. The
sliding contact is again moved till there is null deflection in the galvanometer. Then the
distance AD’ is measured.
The emf of unknown cell can be calculated from the following equation.
103
Emf of the unknown Cell x Length AD
=
Emf of the s tan dard cell s Length AD
EX AD
=
ES AD
AD
Emf of the unknown cell Ex = Es
AD
Factors affecting emf of a cell
1. Nature of the electrolyte and electrodes
2. Concentration and composition of the electrolytes
3. pH and temperature of the solution
104
Where, C1 is the concentration of Ag+ ions furnished by AgCl in KCl solution. By
measuring emf of the cell, the concentration of AgCl is calculated. Multiplying this by
143.5, the equivalent weight of AgCl , we can get the solubility of AgCl in grams/litre
Solubility = C1×143.5(Eq.wt of AgCl)
The value of n, the number of electrons involved in the cell reaction, (called the
valency of ions) can be calculated.
105
E =
2.303
F
log H
2.303
= pH pH log[ H ]
F
i.e at 250C
E = 0.0591 pH
From the above equation the hydrogen ion concentration or the pH of the solution
can be calculated
10) Why can glass electrode not be used for a solution of high alkalinity?
The glass electrode can be used in solution only with pH range of 0 to 10.
However when the pH is above 12(high alkalinity) ,cations of the solution will affect the
glass and make the electrode useless.
107
11) Define a reference electrode. Give any example.
Reference electrode is the one ,the potential of which is known or arbitrarily fixed as
zero. It is measure the electrode potential of another unknown electrode by combining
with it. Example: Std hydrogen electrode; Calomel electrode
13) Define terms (i) single electrode potential (ii) electrode chemical cells
(i) Single electrode potential(E)
It is the measure of the tendency of an electrode to lose or gain electrons, when it is
in contact with a solution of its own salt
(ii) Electro chemical cells
It is a device which converts chemical energy in to electrical energy or electrical
energy into chemical energy
108
15) Define emf of a cell?
Electromotive force is defined as,” the difference in potential which causes flow of
electrons from one electrode of higher potential to the other electrode of lower potential’.
Thus, the emf of an electrochemical cell can readily be calculated using the following
relationship.
E0cell = E0right – E0left
E = E0
0.0591
log
Pr oduct
nF Re ac tan t
109
20) Zinc reacts with dil. H2SO4 to give hydrogen but Cu does not react with dil.
H2SO4 Why Explain
Zn reacts with dil H2SO4 and liberates hydrogen due to it possess negative
reduction potential (E0 Zn = - 0.76 V) and also it is placed higher in the emf series.Cu
does not react with dil. H2SO4 due to it’s possess positive reduction potential (E0 Zn = -0.76
V) and also it is placed higher in the emf series. Cu does not react with dil. H2SO4 due to it
possess positive reduction potential (E0Cu = + 0.34V) and also it is placed lower in the emf
series.
110
23) Write the construction of saturated calomel electrode.
Calomel electrode consists of a glass tube containing mercury at the bottom over
which mercurous chloride is placed. The remaining portion of the tube is filled with a
saturated solution of KCL. The bottom of the tube is sealed with a platinum wire. The side
tube is used for making electrical contact with a salt bridge. The electrode potential of the
calomel electrode is + 0.2422 V
111
Limitations of Glass Electrode.
(i) Although glass membrane of electrode is very thin, yet its resistance is
extremely high, which cannot be measured by ordinary potentiometers it is,
therefore, necessary to use special electronic potentiometers for
measurement.
(ii) The glass electrode can be used in solutions only with pH range of 0 to 10.
However when the pH is above12 (high alkalinity), cations of the solution
will affect the glass and make the electrode useless.
112
30) Give any two limitations of H2 electrode.
(i) The potential of the electrode is altered by changing in barometric pressure
(ii) It requires considerable volume of test solution.
0.0591 0.024
ECell = log
2 2.4
0.0591
ECell = log 0.01
2
0.0591
ECell = log 1 10 2
2
ECell = 0.0591V
IMPORTANT QUESTIONS
1. i) How is emf of an voltaic cell (or) galvanic cell measured.
ii) Calculate the reduction potential of Cu/Cu2+ 0.5Mat 25oC;EoCu/Cu2+= -0.337 V.
2. Describe the construction and working of calomel electrode and hydrogen
electrode.
3. Define electromotive force. How is it measured by potentiometric method? 113
4. i) How is the emf of a galvanic cell measured by poggendorff’s compensation
method
ii) How is pH Of a solution determined from emf measurement?
5. Derive Nernest equation for emf of a cell.
6. Give the construction and uses of a saturated calomel electrode. Write its electrode
reaction.
7. What is emf ? How is it measured potentiometrically?
8. How is calomel electrode constructed? Discuss how this electrode may be used for
the determination of pH of a solution.
9. Describe the following electrodes giving the diagram, electrode notations and
electrode reaction. Standard hydrogen electrode and calomel electrode.
10. Calculate the emf of the cell
Zn /Zn2+ (0.1M)//Ag(10.0M) / Ag.
Write down the cell reaction given that Eo Zn2+/Zn=-0.76V and EoAg+/Ag= +0.80V
11. Calculate the potential of the following concentration cell at 25oC.
Cu(s) / Cu2+ (0.012M) // Cu2+ (0.032M) / Cu(s)
12. Writ schematic curves explain the principle involved in the different conductmetric
Acid base titrations.
13. Bring out the application of emf series.
14. Dscribe the construction and working of a galvanic cell.
16. Describe a glass electrode. How can it be used for determining the pH of a
solution?
17. Discuss the titration curves in the conductometric titration of
i) A strong acid with a strong base.
ii) A weak acid with a strong base.
18. Derive the Nernest equation for emf of a cell .write Nernest equation for emf of
Daniel cell.
114
19. Consider the cell reaction
Zn(s)+ Fe2+(0.005M) Zn2+ (0.01M) + Fe(s)
Given the std emf of the cell at298K is 0.323V
i) Construct the cell.
ii) Calculate the emf of the cell.
20. What is the principle underlying conductometric titration? Discuss the titration
Curve obtained in the case of a mixture of acid with a strong base.
21. Define single electrode potential. What is the value of electrode potential of SHE?
22. Explain acid-base titrations conductomtrically.
115
UNIT - III
3.1 INTRODUCTION
Metals and alloys are generally used as fabrication or construction materials in
engineering. If the metal or alloy structures are not properly maintained, they deteriorate
slowly by the action of atmospheric gases, moisture and other chemicals. This
phenomenon of deterioration or destruction of metals and alloys is known as corrosion.
Definition
Corrosion is the deterioration of materials by chemical interaction with their
environment. The term corrosion is sometimes also applied to the degradation of plastics,
concrete and wood, but generally refers to metals.
The extracted metal in the pure form has high energy, which is thermodynamically
unstable state. It is the natural tendency of the metal to go back to the thermodynamically
stable state. Metals do this by interacting chemically with their environment to form
surface compound and undergo corrosion.
116
Extraction environment
Metal
Corrosion
Ore
products
Although corroded metal is thermodynamically more stable than pure metal but
due to corrosion, useful properties of a metal like malleability, ductility, hardness, lustre
and electrical conductivity are lost.
117
2. Hazards or injuries to people arising from structural failure or breakdown
(e.g. bridges, cars, aircraft).
3. Loss of time in the availability of profit-making industrial equipment.
4. Reduced value of goods due to deterioration of appearance.
5. Contamination of fluids in vessels and pipes
6. Perforation of vessels and pipes allowing escape of their contents and possible
harm to the surroundings. For example a leaky domestic radiator can cause
expensive damage to carpets and decorations, while corrosive sea water may enter
the boilers of a power station if the condenser tubes perforate.
7. Loss of technically important surface properties of metallic component. These
could include frictional and bearing properties, ease of fluid flow over a pipe
surface, electrical conductivity of contacts, surface reflectivity or heat transfer
across a surface.
8. Mechanical damage to valves, pumps, etc, or blockage of pipes by solid corrosion
products.
9. Added complexity and expense of equipment which needs to be designed to
withstand a certain amount of corrosion, and to allow corroded components to be
conveniently replaced.
Mechanism
Oxidation takes place at the surface of the metal forming metal ions, (M2+)
M M2+ +2e‾
Oxygen is converted to oxide ion (O2- ) due to the transfer of electrons from the
metal.
1/2O2 + 2e- O2-
Oxide ion reacts with the metal ions to form metal oxide film
M2++O2- M2+O2- = MO (film)
Once the metal surface is converted with mono-layer of oxide, the oxide film
grows perpendicular to the metal surface which causes the formation of thick oxide film.
The nature of the oxide film formed on the metal surface plays a significant role.
120
Mechanism
This type of corrosion occurs when a metal is exposed to hydrogen environment.
Iron liberates atomic hydrogen with hydrogen sulphide as follows.
Fe + H2S FeS + 2H
Hydrogen diffuses into the metal matrix in the atomic form and gets collected in
the voids present inside the metal. Further diffusion of atomic hydrogen makes them
combine with each other and form hydrogen gas.
H+H H2(g)
Collection of these gases in the voids develops high pressure, causing cracking or
blistering of metal.
b) Decarburization
We know that the presence of carbon in steel gives sufficient strength to it. But
when steel is exposed to hydrogen environment at high temperature, atomic hydrogen is
formed.
H2 ∆ 2H
Atomic hydrogen reacts with the carbon of the steel and produces methane gas.
C + 4H CH4
Hence, the carbon content in steel is decreased. The process of decrease in carbon
content in steel is known as decarburization.
Collection of methane gas in the voids of steel develops high pressure which
causes cracking. Thus, steel loses its strength.
121
3.2.1.2 Liquid Metal Corrosion
This is due to the chemical action of flowing liquid metal at high temperature. This
may involve either dissolution of solid metal by a liquid or maybe due to the penetration
of a liquid metal into the solid metal.
This type of corrosion has been found to occur in devices used for nuclear power.
Mechanism
i. Corrosion or metal dissolution (or oxidation) always occurs at anode.
M Mn+ + ne-
ii. Reduction reaction occurs at the cathode which depends on the corrosive
environment are;
a) When the corrosive environment is acidic, hydrogen gas is evolved at the
cathode
2H++ 2e- H2
b) When the corrosive environment is neutral or slightly alkaline, hydroxide ion is
formed
1/2O2 + 2e- + H2O 2OH-
122
3.2.2.1 Hydrogen evolution type corrosion
All metals placed above hydrogen in electrochemical series have a tendency to get
dissolved in acid solution with simultaneous evolution of hydrogen gas.
This process is explained by considering the corrosion of iron metal by industrial
wastes containing non oxidizing acids like HCl.
At Anode
Iron undergoes dissolution (oxidation) to form Fe2+ with the liberation of electrons.
Fe Fe2+ + 2e-
At Cathode
The liberated electrons go from anode to cathode where H+ ions are reduced to
hydrogen gas.
2H+ +2e- H2(g)
123
Mechanism
The surface of iron is usually coated with a thin film of oxide. However, while the
rest of the metal part acts as cathode.
At Anode
Iron dissolves as Fe2+ with the liberation of electrons.
Fe Fe2+ + 2e- (oxidation)
At Cathode
The liberated electrons flow from anodic to cathodic part of the metal, where the
electrons are taken up by the dissolved oxygen to form hydroxide ions, OH-.
½ O2 + H2O + 2e- 2OH-
The resulting ions Fe2+ and OH- combine together to form ferrous hydroxide as a
precipitate.
Fe2++ 2OH- Fe(OH)2 (ppt)
If enough oxygen is present, Fe(OH)2 is easily oxidized to ferric hydroxide
4Fe(OH)2 + O2 + 2H2O 4Fe(OH)3 (Rust)
124
Corrosion products accumulate at Corrosion occurs at the anode while the
4.
the place of corrosion. products are formed elsewhere.
5. It is a self controlled process. It is a continuous process.
6. It adopts adsorption mechanism. It follows electrochemical reaction.
Example: Formation of mild scale on Example: Rusting of iron in the moist
7.
the iron surface. atmosphere.
125
-------------------------------------------------- -----------------------------------------
-------------------------------------------------
Electrolyt ----------------------------------------
Electrolyt
--------------------------------------------------- ---------------------------------------
Zn Fe Fe Cu
Less active More active
More active Less active (anode) (cathode)
(anode) (cathode) Fe → Fe2+ + 2e- E0 = 0.34
Zn→Zn2+ + 2e- E0 = -0.44V
E0= -0.76V E0= -0.44
Fig. 3.3 (a) Zn-Fe Bimetallic couple Fig. 3.3 (b) Fe-Cu Bimetallic couple
But in the case of Fe and Cu iron dissolves in preference to copper. Iron acts as
anode and undergoes corrosion. Copper acts as cathode.
126
Explanation: If a metal is partially immersed in a conducting solution, the metal part
inside the solution is less aerated and becomes anodic and suffers corrosion. The metal
part above the solution is more aerated and becomes cathodic.
At anode: Corrosion occurs (less aerated)
M M2+ + 2e -
At cathode: OH- ions are produced (more aerated)
1/2O2 + H2O +2e- 2OH-
Zn2+
The rate of corrosion will be more, when the area of cathode is larger and the area
of anode is smaller. Therefore, more and more material is removed from the spot. Thus a
small hole or pit is formed on the surface of the metal.
Clay Sand
129
(iv) Corrosion on Wire Fence
The figure 3.8 shows a wire fence in which the areas where the wire cross are less
aerated than the rest of the fence and hence corrosion occurs at the wire crossings which
are anodic.
d) Over voltage
The over voltage of a metal in a corrosive environment is inversely proportional to
corrosion rate. For example, the over voltage of hydrogen is 0.7 V, when zinc metal is
placed in 1 M sulphuric acid and the rate of corrosion is low. When we add small amount
131
of copper sulphate to the dilute sulphuric acid, the hydrogen over voltages is reduced to
0.33 V. This results in the decreased rate of corrosion of zinc metal.
b) Humidity
The rate of corrosion will be more when the relative humidity of the environment
is high. The moisture acts as a solvent for O2, CO2, SO2 etc. in the air to produce the
electrolyte which is required for setting up a corrosion cell.
132
c) Impurities in the environment
Impurities in the atmosphere from industrial exhaust gases like SO2 ,SO3 ,NO2 ,CO2
H2S,CO etc and fumes of HCl,H2SO4 etc, produce electrolytes and increase the rate of the
corrosion.
Corrosion
P
Corrosion
133
This figure3.9 clearly indicates the zones of passivity, immunity and corrosion. In
this diagram, P is the point whose pH is 7(neutral condition) and the corresponding
electrode potential is -0.4 V. Evidently this point is present in the corrosion zone. Under
this condition iron undergoes corrosion or rusting.
The rate of corrosion of iron can be changed by shifting point P into the immunity
or passivity zones. This can be achieved by changing the electrode potential as follows.
i. If the potential is changed to -0.8 V, by applying external current, iron will be
immune to corrosion.
ii. By applying positive potential, the corrosion rate can be reduces as become
passive.
iii. This diagram clearly indicates that rate of corrosion will be maximum when
the pH is less than 7 i.e., the corrosive environment is acidic. The rate of
corrosion can be reduced by increasing the pH of the solution by adding
alkalies.
2. By proper design
Some of the important rules for designing, which must be observed, are given
below.
a) Galvanic corrosion is prevented by
i. Selecting the metals as close as possible in the electrochemical series.
ii. Providing smaller area for cathode and larger area for anode.
iii. Inserting insulating material between the two metals
Insulation
Fig 3.10
135
b) Drainage affects corrosion
Tanks and other containers must be designed in such a way that, the whole of the
liquid should be draining out completely.
-------------------
-------------------
-------------------
-------------------
------------
Fig.3.11.a.Poor design of
drainage corrosion Fig.3.11.b.Good design
of drainage corrosion
136
c) Avoid crevices
Crevices allow moisture and dirt, which results in increased electrochemical
corrosion. This can be prevented by filling the crevices with filler
Filler
Example
Riveted joints produce crevice corrosion, so welded joints preferred to avoid
corrosion.
3. By cathodic protection
The principle involved in the cathodic protection is to force the metal to behave
like a cathode.The important cathodic protections are (i) sacrificial anodic protection.
(ii)Impressed current cathodic protection .
Zn/Mg
Anode
138
Zn
Zn
Zn
Zn
Backfill
Underground pipeline
This can be done by connecting negative terminal of the battery to the metallic
structure to be protected, and positive terminal of the battery is connected to an inert
anode. Inert anodes used for this purpose are graphite and platinised titanium. The
anode is buried in a “back fill” (containing mixture of gypsum, coke, breeze, sodium
sulphate). The “back fill” provides good electrical contact to anode.
Structures like tanks, pipelines, transmission line towers, underground water pipe
lines, oil pipe lines, ships, etc., can be protected by this method.
140
Table3.2 Comparison of Sacrificial anodic method and impressed current method
S. No Sacrificial anodic method Impressed current method
It requires periodical replacement of Here anodes are stable and do not
1.
sacrificial anode. disintegrate.
2 Investment low. Investment is more.
Soil and microbiological corrosion Soil and microbiological corrosion effects
3.
effects are not taken into account. are taken into account.
This is most economical method
This method is well suited for large
4. especially when short-term
structures and long term operations.
protection is required.
This method is suitable when the
But this method can be practiced even if
current requirement and the
5. the current requirement and the
resistivity of the electrolytes are
resistivity of the electrolytes are high.
relatively low.
a) Deaeration
Fresh water contains dissolved oxygen. The presence of increase amount of oxygen
is harmful and increases the corrosion rate. Deaeration involves the removal of dissolved
oxygen by increase of temperatue together with mechanical agitation. It also removes
dissolved carbon dioxide from the water.
141
b) Deactivation
It is the process of removing dissolved oxygen by adding some chemicals (known
as scavenger) in aqueous solution.
Examples: sodium sulphite (Na2SO3), hydrazine (N2H4) remove the dissolved oxygen in
the following way.
N2H4 N2 + 2H2O
2Na2SO3 +O2 2Na2SO4
c) Dehumidification
In this method, moisture from air is removed by lowering the relative humidity of
the surrounding air. This is achieved by adding silica gel or alumina which adsorbs
moisture preferentially on its surface.
e) By Using Inhibitors
Inhibitors are organic or inorganic substances which decrease the rate of corrosion.
Usually, the inhibitors are added in small quantities to the corrosive medium.
Inhibitors are classified as
i) Anodic inhibitors(chemical passivators )
ii) Cathodic inhibitors(adsorption inhibitors)
iii) Vapour phase inhibitors(volatile corrosion inhibitors)
142
i) Anodic Inhibitors
These inhibitors retard the corrosion of metals by forming a sparingly soluble
compound with newly produced metal cations. This compound will then adsorb on the
corroding metal surface forming a passive film or barrier.
Examples: chromate, phosphate, tungstate, nitrate, molybdate etc.
143
3.6 PRE-TREATMENT OF THE MATERIAL SURFACES(OR) PREPARATION OF
MATERIALS FOR COATING
Generally metal surface are covered with impurities like rust, scale, oil, grease, etc.
These substances, if present at the time of coating, will produce porous and discontinuous
coating. So, in order to get a uniform, smooth, cohesive and adherent coating, these
substances are removed by the following methods.
1. Solvent Cleaning
Organic solvents such as toluene, xylene, acetone, carbon tetrachloride, etc., are
used to remove oils, grease, and fatty substances present on the metal surfaces.
2. Alkali Cleaning
This method is particularly used for the removal of old paint coating from metal
surfaces. Solution of caustic soda, disodium phosphate with some detergents, emulsifying
and wetting agents are used for the alkali cleaning. After the alkali treatment, the metal
surface is rinsed with water and then with 0.1% solution of chromic acid.
4. Mechanical Cleaning
Oxide scales, rust and corrosion products are removed by the mechanical cleaning
such as grinding, wire brushing, polishing, etc.,
144
5. Flame Cleaning
This process is done by passing high velocity hot flame over the metal surface. It is
applied to remove moisture and loosely adhering scales.
6. Sand Cleaning
This process consists of introducing sand into an air steam under the pressure of 25
to 100 atmospheres. It is used for removing oxide scale present especially on steel surface.
Objectives of Electroplating
On Metals
1. To increase the resistance to corrosion of the coated metal
2. To improve the hardness and physical appearance of the article
3. To increase the decorative and commercial value of the article
4. To improve the properties of the surface of the article
5. To increase resistance to chemical attacks
On Non-Metals
1. To increase strength
2. To preserve and decorate the surfaces like plastics, wood, glass, etc.,
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3. To introduce the surface conductivity by utilization of light weight, non-metallic
materials.
Theory
Suppose the anode is made of coating metal itself in the electrolytic cell. During
electrolysis, the concentration of electrolytic bath remains unaltered. This is because the
metal the metal ions deposited from the solution on cathode are replenished continuously
by the reaction of free ions with the anode metal. Thus, for example, If CuSO4 solution is
used as an electrolyte, it ionises as
CuSO4 Cu2+ + SO42-
Fig.3.17. Electroplating
On passing current, Cu2+ ions go to the cathode and get deposited there
Cu2+ + 2e- Cu (at cathode)
The free sulphate ions migrate to the copper anode and dissolve an equivalent amount of
copper to form CuSO4
SO42- + Cu CuSO4+2e- (at anode)
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The copper sulphate formed gets dissolved in the electrolyte. Thus there is a continuous
replenishment of electrolytic salt during electrolysis.
Procedure
The article to be plated is first treated with dil.HCl or dil.H2SO4 to clean the
surface. The cleaned article is then made cathode of an electrolytic cell. The anode is either
the coating metal itself or an inert material of good electrical conductivity.
When direct current is passed from a battery, coating metal ions migrate to the
cathode and get deposited there. Thus a thin layer of coating-metal is obtained on the
article (at cathode)
In order to get strong, adherent and smooth deposit certain additives (glue, gelatin,
etc.,) are added to the electrolytic bath. In order to improve the brightness of deposit,
brightening agents are added in the electrolytic bath. They are at optimum temperature,
optimum current density and low metal ion concentrations.
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(a). Etching: Removal of unwanted particles by acid treatment.
(b). Electroplating: A thin layer of the metal to be plated or any other suitable metal is
coated on the surface of the object.
(c). Treating with stannous chloride, followed by dipping in palladium chloride
solution.
This treatment yields a thin layer of Pd on the treated surface. This method is
applied only for plastics and printed circuit board.
Step 3: Procedure
The object to be plated is immersed in the bath containing the salt of the metal and
reducing agent. The metal ions from the solution are reduced to the corresponding metal
and get plated over the surface of the object.
Various reactions
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At cathode: M2+ + 2e- M
At anode: 2HCOH+4OH- 2HCOO-+2H2O+H2+2e-
Net reaction: M2+ + 2HCHO +4OH- M + 2HCOO- +2H2O + H2
Example
The surface of the stainless steel is activated by dipping in hot solution of 50% dil.
H2SO4. The surface of Mg alloy is activated by thin coating of Zn and Cu over it. Metals
and alloys like Al, Cu, Fe, brass, etc., can be directly Ni-plated without activation Non
metallic articles (like plastics, glass, etc., ) are activated by dipping them in the solution
containing SnCl2 + HCl followed by dipping palladium chloride solution. On drying a
thin layer of Pd is formed on the surfaces.
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Differences between electroplating and electroless Plating
Table 3.4
S.No. Electroplating Electroless plating
The base metal to be plated can be either The base metal should be non-ferrous.
2
a pure metal or an alloy.
The base metal on which oxide coating
The base metal to be plated is made
3 produced is made anode in the
cathode in the electrolytic cell.
electrolytic cell.
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Hence, they can be recharged by passing electric current and used again and again. These
are called storage cells (or) accumulators.
Examples: lead acid storage cell, Nickel –cadmium cell
3. Flow Battery (or) Fuel cell
In these cells, the reactants, products and electrolytes are continuously passing
through the cell. In this chemical energy gets converted to electrical energy.
Examples: Hydrogen-Oxygen fuel cell.
IMPORTANT PRIMARY BATTERIES
3.11 DRY CELL (or) LECLANCHE’S CELL
It is a primary cell, which works without fluid component.
Description
A dry cell consists of a Zinc cylinder, which acts as anode. This Zinc cylinder is
filled with an electrolyte consisting of NH4Cl,ZnCl2 and MnO2 in the form of paste using
starch and water. A carbon rod (graphite), acts as cathode, is inserted in the electrolyte in
the centre of the cell. The zinc cylinder has an outer insulation of cardboard case. During
use, the zinc cylinder gets consumed and at the end, it will develop holes which are
responsible for leakages.
Mixture of
ZnCl2,MnO2, and
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Cell Reactions
Anode: Zn(s) Zn+2(aq) + 2e-
Cathode: 2NH4+(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) + H2O(l)
Cell Reactions
Ecell = 1.5 V
Anode (oxidation) : Zn(s) + 2OH- (aq) Zn(OH)2(s) + 2e-
Cathode (reduction): 2MnO2 (s) + H2O (l) + 2e- Mn2O3 (s) + 2OH-(aq)
Overall Cell reaction: Zn(s)+ 2MnO2 (s) + H2O (l)) Zn(OH)2(s) + Mn2O3 (s)
Description
A typical 12-V lead-acid battery has six cells connected in series. Each cell contains
two lead grids packed with the electrode material. The anode is spongy Pb, and the
cathode is powered PbO2. The grids are immersed in an electrolyte solution of H2SO4
(38% by mass) having a density of 1.30 gm/ml. Fiberglass sheets between the grids
prevent shorting by accidental physical contact.
The cell may be represented as Pb/ PbSO4 //H2SO4(aq) / PbO2//Pb
Anode
-
+ Cathod
PbO2 plates
Pb Plates
Aqueous
H2SO4
Cell Reactions
At Anode: Pb(s) Pb2+ +2 e-
Pb2+ (aq) + SO42-(aq) PbSO4 (s)
At Cathode:
PbO2 gains electrons ie., Pb undergoes reduction at the cathode from +4 to +2 ions
then combines with SO42- forms insoluble PbSO4.
PbO2 (s) + 4H+ + 2e- Pb2+(aq) + 2 H2O
Pb2+(aq) + SO42- PbSO4 (s)
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From the above cell reactions it is clear that, PbSO4 is precipitated at both
electrodes and H2SO4 is used up. As a result, the concentration of H2SO4 decreases and
hence the density of H2SO4 falls below 1.2 gm/ml. So the battery needs recharging.
discharging
Uses:
1. This cell is used to supply current mainly in automobiles such as cars, buses,
trucks, etc.,
2. It is used in gas engine ignition, telephone exchanges, hospitals, power stations,
etc.,
Description
Nickel-Cadmium cell consists of a cadmium anode and a metal grid containing a
paste of NiO2 acting as a cathode. The electrolyte in the cell is KOH.
It is represented as: Cd /Cd(OH)2// KOH(aq) / NiO2 /Ni
Working (Discharging)
When the Nicad battery operates, at the anode cadmium is oxidized to Cd2+ ions
and insoluble Cd(OH)2 is formed. It produces about 1.4V
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Cell reactions
At Anode: Cadmium is oxidised to Cd2+ and further it combines with OH- ions to form
Cd(OH)2.
disChargin
Anode: Cd(s) + 2OH-(aq) Cd(OH)2 (s) + 2e-
charging
At cathode: NiO2 gains electrons, ie., Ni undergoes reduction at the cathode from +4 to +2.
The Ni2+ ions then combine with OH- ions to form Ni(OH)2 .
dischargin
Cathode: NiO2 (s) + 2H2O(l) + 2e- Ni(OH)2 (s) + 2 OH-(aq)
charging
Overall reaction
dischargin
Cd(s) + NiO2 (s) + 2H2O(l) Cd(OH)2 (s) + Ni(OH)2 (s) +
charging
Energy
From the above cell reactions it is clear that, there is no formation of gaseous
products, the products Cd(OH)2 adhere well to the surfaces. This can be reconverted by
charging the cell.
Construction
The lithium battery consists of a lithium anode and a TiS2 cathode. A solid
electrolyte, generally a polymer, is packed in between the electrodes. The electrolyte
(polymer) permits the passage of ions but not that of electrons.
Solid electrolyte
Li TiS2
Li+
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At Anode: Li(s) Li+ + e-
At Cathode: TiS2(s) + e- TiS22-
Overall reaction
Li(s) + TiS2(s) Li+ + TiS22-
Li+ + Tis2- LiTiS2
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3.14.2 Lithium – Sulphur Battery
Lithium-sulphur battery is a rechargeable battery. Its anode is made of Li. Sulphur
is the electron acceptor, the electron from LI is conducted to S by a graphite cathode. ß-
alumina (NaAl11O17) is used as the solid electrolyte.
This solid electrolyte allows the Li+ ions to migrate to equalize the charge, but will
not allow the big poly sulphide product ions.
+ -
Cathode Anode
Li
ß- alumina
S (l)
Cell Reactions
At Anode: 2Li 2Li+ + 2e-
At Cathode: S + 2e- S2-
Ov Net reaction: 2Li + S 2Li+ + S2-
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The S2-ions formed, react with elemental sulphur to form the polysulphide ion.
S2-+ nS [Sn+1]2-
The direct reaction between Li and S is prevented by the alumina present in the
cell.
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Fig 3.23 Solar cell
Solar battery: When a large number of solar cells are connected in series it form a solar
battery. It produces more electricity which is enough to run water pump, to run street
light,etc
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Applications
1. Solar cells are used in boilers to produce hot water for domestic and industrial
uses.
2. Solar cells, if properly designed, can be used for lighting purposes.
3. These are superior to other types of cells as they are non polluting and eco-
friendly.
4. It can be used to derive vehicles.
5. The solar cells made of silicon have been used as a source of power in space craft
and satellites.
6. Solar cells can be used to produce hydrogen by electrolysis of water. The liberated
hydrogen can be used in H2-O2 fuel cells.
7. They are used in remote areas where conventional electricity supply is a problem.
The huge capital cost is the major limitations for the large scale use of solar cells.
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Fig 3.25 Fuel Cells
Anode (Oxidation):
H2(g) + CO32-(l) H2O+ CO2(g) +2e-
Cathode (Reduction):
1/2O2(g) + CO2(g) +2e- CO32-(l)
Overall reaction
H2(g) + 1/2O2(g) H2O(l)
Uses
Provides electricity and pure water during space lights.
Advantage
1. Clean, portable and product is water. Many fuel cells produce no pollutants.
2. Fuel cells are very efficient, converting about 75% of fuel’s bond energy into
electricity. In contrast, an electric power plant converts about 35% to 40% of bond
energy into electricity, and a car engine converts about 25% of gaseous bond
energy into moving the car.
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Disadvantage
1. It cannot store electrical energy, needs continuous flow of reactant.
2. Electrodes are short lived and expensive.
Important Questions
1. What is corrosion?
2. Define Hydrogen embrittlement.
3. State Pilling- Bedworth rule
4. What is meant by decarburization?
5. What is meant by liquid metal corrosion?
6. Mention the conditions for wet corrosion to take place.
7. Distinguish between dry corrosion and wet corrosion.
8. What is pitting corrosion?
9. How steel screws in brass marine hardware corrode?
10. Why Zinc is more readily corroded when coupled with copper than with lead?
11. Mention the various factors influencing the rate of corrosion.
12. What are corrosion inhibitors? Give examples.
13. What are vapour phase inhibitors?
14. An iron pipe does not rust when connected to Zinc but corrodes rapidly when it is
in contact with copper – Why?
15. Stainless steel is a corrosion resistant alloy – Why?
16. Write the difference between electro plating and electroless plating.
17. What are the advantages of electroless plating over electroplating?
18. Explain oxygen adsorption type corrosion.
19. How is galvanic corrosion occur?
20. Explain the nature of the metal influence the rate of corrosion.
21. Substantiate the statement that nature of the environment affects corrosion.
22. Write an essay on corrosion control methods. 167
23. Mention few guidelines for proper designing of metallic structures.
24. Explain the types of electro chemical corrosion with examples.
25. What is sacrificial anode? Mention its role in prevention of corrosion.
26. Write short notes on corrosion control by impressed current method.
27. Explain anodic protection.
28. How will you control corrosion by modifying the environment?
29. What are corrosion inhibitors? How do they function?
30. How is corrosion prevented by cathodic protection method.
31. Explain electroplating with an example.
32. Explain electroless platting with an example.
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