FE EXAMINATION, May 2015 ENGG CHEMISTRY [4756]-102(2012 PATTERN)

Engineering Chemistry, May 2015 Page 1

F.E.EXAMINATION, MAY 2015

ENGINEERING CHEMISRTY

(2012 PATTERN)

Prepared By

Mr. Lalge Kiran M

(Lecturer in Chemistry,

DYPSOET, Pune)

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Engineering Chemistry, May 2015 Page 2

Q 1 a) What is zeolites? Explain zeolite process of softening of water. Give regeneration

reactions, advantages and disadvantages of the process. [6]

Answer:

Defination:

Zeolites are microporous, aluminosilicate minerals commonly used as

commercial adsorbents and catalysts. Its mineral formula is: Na2Al2Si3O10·2H2O

Diagram:

Zeolite process of softening of water:

i) The other method commonly used for water softening is ion exchange softening, also known

as zeolite softening.

ii) Ion exchange softening exchanges calcium and magnesium ions in water for sodium ions as

the hard water passes through a softener.

iii) During treatment, water enters the softener and the water passes through a bed of resin

underlain by a bed of gravel, and then is collected by an underdrain and piped out of the softener.

iv) In a softener, the bed is made up of zeolite resins, which are insoluble solids with attached

cations or anions capable of reversible exchange with mobile ions of the opposite sign in the

solutions in which they are brought in contact.

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v) In the case of the ion exchange resins used in softening, sodium ions are attached to the

insoluble solids of the resins.

vi) When water passes through the softener, the sodium ions are exchanged for calcium and

magnesium ions in the water.

vii) The calcium and magnesium ions are retained on the resin grains

viii) The water leaving the softener has sodium ions in the place of calcium and magnesium ions

in its compounds, as shown in fig.

ix) Since sodium ions do not cause hardness, the treated water is no longer hard.

Regeneration:

i)This is the part of the process in which the magnesium and calcium ions on the resin become

replaced with sodium so that the softener can be used to treat more hard water. In order to

regenerate the resin, a salt solution, known as brine, is allowed to flow through the softener for

about an hour.

ii) The salt used to regenerate the resin is ordinary table salt (sodium chloride, NaCl), so it is

easy to handle. When dissolved in water, the salt dissociates into its constituent ions - Na+ and

Cl.- The sodium ions replace the calcium and magnesium ions on the resin in the following manner (where "R" preceding an ion means that the ion is bound to the resin.)

RCa + NaCl → RNa + CaCl2

RMg + NaCl → RNa + MgCl2

iii) During regeneration, calcium chloride, magnesium chloride, and excess sodium chloride flow

to waste.

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Advantages of Zeolite Process:

i) It removes the hardness almost completely (about 10 ppm hardness only).

ii) The process automatically adjusts itself for variation in hardness of incoming water.

iii) This process does not involve any type of precipitation, thus, no problem of sludge formation

occurs.

Disadvantages of Zeolite Process:

i) The outgoing water (treated water) contains more sodium salts.

ii) This method only replaces Ca+2 and Mg+2 ions by Na+ ions.

iii) High turbidity water cannot be softened efficiently by zeolite process.

iv) More acidic and alkaline water destroy the zeolite bed.

v) It is difficult to regenerate zeolite bed when water contains Mn+2 and Fe+2 ions in hard water. vi) Zeolite process leaves all acidic ions like HCO3

- and CO3-2

vii) We cannot use hot water which dissolves zeolite bed

Q.1.b) Explain titration curve of conductometric titration in case of strong acid and weak

base. [3]

Answer:

Conductometric titration in case of strong acid and weak base:

i) Consider titration of HCl and NH4OH

ii) Take in HCl burette and NH4OH in conical flask

iii) Dip the conductivity cell in conical flask and the conductivity.

iv) As flask contain NH4OH which is weak electrolyte, conductivity is low because of less

mobile ions (NH4+)

v) After addition of strong electrolyte (HCl), conductivity increase due to small, mobile ion H+

and also Cl-.

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vi) In this case, HCl and NH4OH reacts and gives NH4Cl and water molecules. Reaction is as

follows:

HCl + NH4OH ------- NH4Cl + H2O

H+ + Cl- + NH4+ +OH- ------- NH4

+ + Cl- + H2O

vii) As excess of HCl added to burette sharp increase in in conductance will observed, which

gives end point of titration

viii) Graph:

Due to H+

Conductance End point

Due to NH4+

ml of HCl

Q 1 C) Explain the follow terms with suitable examples [3]

i)Chromophore:

a) Chromophore is that part of the molecule which when exposed to visible light will absorb

and reflect a certain color, absorbs certain wavelengths of visible light

b) For example, Carbonyls – unsaturated systems incorporating N or O can undergo n - p*

transitions (~285 nm) in addition to p -p*

ii)Auxochrome

a) An auxochrome is a functional group of atoms with nonbonded electrons which, when

attached to a chromophore, alters both the wavelength and intensity of absorption

b) For example, benzene does not display color as it does not have a chromophore;

but nitrobenzene is pale yellow color because of the presence of a nitro group (-NO2) which acts

as a chromophore. But Para-hydroxynitrobenzene exhibits a deep yellow color, in which -OH

group acts as an auxochrome. Here the auxochrome (-OH) is conjugated with the chromophore -

NO2. Similar behavior is seen in azobenzene which has a red color, but para-hydroxy

azobenzene is dark red in color.

Q 2 a) Explain the pH metric titration mixture of H3PO4 and HCl against std. NaOH,

giving chemical reaction, procedure, titration curve and calculations. [6]

Answer:

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i)Alkalimetric Titration of an Acid Mixture In this experiment the quantitative composition of a

solution, which is a mixture of a monoprotic strong acid (HCl) and a weaker triprotic acid

(H3PO4) will be determined by pH methods.

ii) This experiment will introduce you to semi-automatic volumetric analysis and potentiometric

titrations. Phosphoric Acid, H3PO4 (pK1= 2.16, pK2 = 7.16, pK3=12.3).

Phosphoric acid can be titrated as a monobasic acid: H3PO4 + NaOH → NaH2PO4 + H2O

The pH of a solution of resulting dihydrogen phosphate is: The change of the pH near

equivalence point is not very pronounced.

pH=(pK1 + Pk2)/2 =4.66

iii) Phosphoric acid can also be titrated as a dibasic acid, with the second step

NaH2PO4 + NaOH → Na2HPO4 + H2O

The pH of solution of monohydrogen phosphate is

pH=(pK1 + Pk2)/2 =9.2

At this pH, phenolphthalein (transition range 8-9.6) exists in its alkaline form; thymolphtalein

(pH transition range 8.3-10.5) can be a better choice. Nevertheless, direct measurement of pH

values with pH sensitive electrode is more reliable in both cases.

iv) Hydrochloric acid HCl is very strong and it is completely titrated at the pH of the first

equivalence point of phosphoric acid.

Procedure:

i) The unknown solution containing a mixture of HCl and H3PO4 is placed in a 100 or 200 mL

volumetric flask and made to the mark with deionized water.

ii) The burette is filled with the standard NaOH solution of known concentration.

iii) An exact aliquot of 25 mL is placed in a titration beaker, and appropriate electrode is

inserted. The stirrer is turned on. Using pH-meter as an indicator, you titrate first to a first

equivalent point (pH 4.66) and record the volume V1.

iv) At this point , all HCl is titrated and phosphoric acid is completely converted in dihydrogen

phosphate: H3PO4 + NaOH → NaH2PO4 + H2O

v) Continue titration to the second equivalent point (pH 9.7) and record the volume V2.

vi) At this point, all dihydrogen phosphate is converted in monohydrogen phosphate ion:

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NaH2PO4 + NaOH → Na2HPO4 + H2O

Calculation:

i) Amount of phosphoric acid in the aliquot is calculated using the formula:

m H3PO4 =

g

ii) Amount of hydrochloric acid in aliquot is calculated as:

m HCl =

g

iii) Total amount of phosphoric acid and of hydrochloric acid in your unknown (your starting

flask) is

m total H3PO4 =

g

m total HCl =

g

Where V total is volume of your flask and V aliquot is volume of your pipette.

iv) The burette is filled with the standard NaOH solution of known concentration. An exact

aliquot of 25 mL is placed in a titration beaker, and appropriate electrode is inserted. The stirrer

is turned on. Record the readings.

v) Determine the exact volumes of NaOH used to achieve first and second ending point.

vi) Two breaks will occur in the titration curves, the first corresponding to the titration of

hydrogen ions from the HCl and the first hydrogen ion from the H3PO4.

vii) The second break corresponds to the titration of H2PO4 - that resulted from the H3PO4.

vii) Report total amount of HCl and total amount of H3PO4 in your unknown.

viii) Recall that you titrate 25-mL portions of a 200-mL (or 100-mL) total sample. Attach a

spreadsheet plot of one full titration curve and the corresponding plots.

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Graph:

∆pH/∆V

First equivalance point

second equivalance point

Mean V

Q 2 b) Explain any three principles of Green Chemistry [3]

Answer:

Principles of Green Chemistry:

1. Prevention

It’s better to prevent waste than to treat or clean up waste afterwards.

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2. Atom Economy Design synthetic methods to maximize the incorporation of all materials used in the

process into the final product. 3. Less Hazardous Chemical Syntheses

Design synthetic methods to use and generate substances that minimize toxicity to human health and the environment.

4. Designing Safer Chemicals

Design chemical products to affect their desired function while minimizing their toxicity.

5. Safer Solvents and Auxiliaries Minimize the use of auxiliary substances wherever possible make them innocuous when used.

6. Design for Energy Efficiency Minimize the energy requirements of chemical processes and conduct synthetic

methods at ambient temperature and pressure if possible. 7. Use of Renewable Feedstocks

Use renewable raw material or feedstock rather whenever practicable.

8. Reduce Derivatives Minimize or avoid unnecessary derivatization if possible, which requires additional

reagents and generate waste. 9. Catalysis

Catalytic reagents are superior to stoichiometric reagents.

10. Design for Degradation Design chemical products so they break down into innocuous products that do not

persist in the environment. 11. Real-time Analysis for Pollution Prevention

Develop analytical methodologies needed to allow for real-time, in-process monitoring

and control prior to the formation of hazardous substances. 12. Inherently Safer Chemistry for Accident Prevention Choose substances and the

form of a substance used in a chemical process to minimize the potential for chemical accidents, including releases, explosions, and fires.

Q 2 c) 50 ml of water sample requires 18 ml of 0.05 M EDTA during titration, whereas

50 ml of boiled water sample requires 12.5 ml of same EDTA in the titration. Calculate

total, temporary and permanent hardness of water sample. [3]

Answer:

50 ml water sample requires 18 ml 0.05 M EDTA Since, 1 ml of 0.05 M EDTA reacts with 100 mg of CaCO3

Therefore, 1 ml of 0.05 M EDTA reacts with =0.05 X 100 = 5 mg of CaCO3 Therefore, 18 ml of 0.05 M EDTA reacts with =5 X 18 = 90 mg of CaCO3 Therefore, 50 ml of water reacts with = 90 mg of CaCO3

Therefore, 1000 ml of water reacts with = 90 X 1000/50= 1800 mg of CaCO3 eq Hardness Therefore, total hardness of water sample =1800 ppm

Also given, 50 ml boiled water sample requires 12.5 ml 0.05 M EDTA

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Since, 1 ml of 0.05 M EDTA reacts with 100 mg of CaCO3 Since, 1 ml of 0.05 M EDTA reacts with 5 mg of CaCO3

Therefore, 12.5 ml 0.05 M EDTA reacts with = 5 X 12.5 = 62.5 mg of CaCO3 Therefore, 50 ml boiled water sample contains =62.5 mg of CaCO3 eq hardness

Therefore,1000 ml boiled water sample contains =62.5 X 1000/50=1250 mg of CaCO3 eq H Therefore, permanent hardness of water sample =1250 ppm

Temporary hardness of water sample = Total hardness - permanent hardness

= 1800-1250

= 550 ppm

Q 3 a) What is vulcanization of rubber? Explain chemical reaction involved in

vulcanization process. Compare natural rubber with vulcanized rubber. [6]

Answer:

Vulcanization is a chemical process for converting natural rubber or related polymers into more

durable materials via the addition of sulfur or other equivalent curatives or accelerators. These

additives modify the polymer by forming cross- links (bridges) between individual polymer

chains.

Process:

i) There are various vulcanization methods. The economically most important method (the

vulcanization of tires) uses increased pressure and temperature.

ii) A typical vulcanization temperature for a passenger tire is 10 minutes at 170 °C. This type of

vulcanization is an example of the general vulcanization method named compression molding.

iii) The rubber article is intended to adopt the shape of the mold. Other methods for instance

those used to make door profiles for cars use hot air vulcanization or microwave heated

vulcanization (both continuous processes).

iv) Four types of curing systems are in common use. They are:

1. Sulfur systems

2. Peroxides

3. Urethane crosslinkers

4. Metallic oxides

v) By far the most common vulcanizing methods are those dependent on sulfur.

Sulfur, by itself, is a slow vulcanizing agent. Large amounts of sulfur, as well as high

temperatures and long heating periods are necessary and one obtains unsatisfactory crosslinking

efficiency with unsatisfactory strength and aging properties.

vi) Only with vulcanization accelerators can the quality corresponding to today's level of

technology be achieved. The multiplicity of vulcanization effects demanded cannot be achieved

with one universal substance, a large number of diverse materials is necessary.

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Reactions:

Comparison natural rubber with vulcanized rubber:

Natural rubber Vulcanized rubber

(1) Natural rubber is soft and sticky Vulcanized rubber is hard and non-sticky.

(2) It has low tensile strength. It has high tensile strength.

(3) It has low elasticity. It has high elasticity.

(4) It can be used over a narrow range of

temperature (from 10° to 60°C).

It can be used over a wide range of temperature

(–40° to 100°C).

(5) It has low wear and tear resistance. It has high wear and tear resistance.

(6) It is soluble in solvents like ether,

carbon, tetrachloride, petrol, etc.

It is insoluble in all the common solvents.

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Q 3 b) Define: [3]

i) Cetane number:

A measure of the quality of a diesel fuel expressed as the percentageof cetane in a mixture of cet

ane and 1-methylnapthalene of the samequality as the given fuel.

ii) Power alcohol:

When ethanol (5-20%) is blended with petrol to act as a fuel for internal combustion of engines,

is called "power alcohol"

iii) N.C.V. :

Net calorific value of a fuel portion is defined as the amount of heat evolved when a unit weight (or

volume in the case of gaseous fuels) of the fuel is completely burnt and water vapor leaves with the

combustion products without being condensed.

Q 3 c) Calculate carbon hydrogen and Sulphur percentage present in the coal sample from

the following data: [3]

i) 0.15 gm coal sample on burning in combustion on burning in combustion chamber in

current of pure O2 was found to increase weight of CaCl2 U-tube by 0.08 gm and KOH U-

tube by 0.49 gm

Answer:

%C =

=

= 89.09 %

% H =

=

= 5.92 %

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ii) 0.65 gm coal was combustion in bomb calorimeter solution from bomb on treatment

with BaCl2 solution, forms 0.031 gm BaSO4 dry ppt

Answer:

% S =

=

= 0.65%

Q 4 a) What are fuel cells? Explain working of phosphoric acid fuel cell (PAFC) with figure

and cell reactions. State its advantages. [6]

Answer: Definition:

A fuel cell is a device that converts the chemical energy from a fuel into electricity through a

chemical reaction of positively charged hydrogen ions with oxygen or another oxidizing agent

a)Diagram of PAFC:

b) Explaination of PAFC:

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i) Phosphoric acid fuel cells (PAFC) are a type of fuel cell that uses liquid phosphoric acid as

an electrolyte.

ii) Electrolyte is highly concentrated or pure liquid phosphoric acid (H3PO4) saturated in a silicon

carbide matrix (SiC).

iii) Operating range is about 150 to 210 °C.

iv) The electrodes are made of carbon paper coated with a finely dispersed platinum catalyst.

v) It consists of an anode, a cathode and an electrolyte that allow positively charged hydrogen

ions (or protons) to move between the two sides of the fuel cell.

vi) The anode and cathode contain catalysts that cause the fuel to undergo oxidation reactions

that generate positive hydrogen ions and electrons.

vii) The hydrogen ions are drawn through the electrolyte after the reaction.

viii) At the same time, electrons are drawn from the anode to the cathode through an external

circuit, producing direct current electricity. At the cathode, hydrogen ions, electrons, and oxygen

react to form water.

c) Cell Reactions:

Anode reaction: 2H₂ → 4H+ + 4e‾

Cathode reaction: O₂(g) + 4H+ + 4e‾ → 2H₂O

Overall cell reaction: 2 H₂ + O₂ → 2H₂O

Advantages of PAFC:

i) It used in hospitals, nursing homes and for all commercial purposes

ii) Electrical efficiency of 40%

iii) Fuel cells provide high quality DC power.

iv) The power densities are high values.

V) Cogeneration Capability.

vi) Fuel cells can be responsive to changing electrical loads.

vii) Use a variety of fuels, renewable energy and clean fossil fuels.

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Q 4 b) Explain with suitable diagram bulk polymerization technique to bring about

addition polymerization [3]

Answer:

Bulk polymerization technique:

i) Bulk polymerization or mass polymerization is carried out by adding a soluble initiator to

pure monomer in liquid state.

ii) The initiator should dissolve in the monomer.

iii) The reaction is initiated by heating or exposing to radiation. Constant agitation is required.

iv) As the reaction proceeds the mixture becomes more viscous. The reaction is exothermic and a

wide range of molecular masses are produced.

v) Bulk polymerization is carried out in the absence of any solvent or dispersant.

vi) In bulk polymerization, heat evolved may cause the reaction to become too vigorous and

difficult to control.

Diagram:

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Q 4 c) Give structure, properties and applications of polyphenylenevinylene (PPV) [3]

Answer:

a) Structure of polyphenylenevinylene (PPV):

i) Highly oriented PPV films obtained by the soluble polymeric precursor route usually have P21

symmetry with a monoclinic unit cell containing two monomer units: c (chain axis) = 0.658, a =

0.790, b = 0.605 nm, and α (monoclinic angle) = 123o

ii) The structural organization of PPV chains resembles that found in other highly oriented rigid-

rod polymers, where the molecules are oriented along the fiber axis (often the stretching

direction) but with partial axial translational disorder.

Fig PPV

b) Properties of polyphenylenevinylene (PPV):

i) Poly(p-phenylene vinylene) (PPV, or polyphenylene vinylene) is a conducting polymer of

the rigid-rod polymer family.

ii) PPV is a diamagnetic material and has a very low intrinsic electrical conductivity, on the

order of 10-13 S/cm.

iii) The electrical conductivity increases upon doping with iodine, ferric chloride, alkali metals,

or acids. However, the stability of these doped materials is relatively low.

iv) In general, unaligned, unsubstituted PPV presents only moderate conductivity with doping,

ranging from <<10-3 S/cm (I2 doped) to 100 S/cm (H2SO4-doped).

c) Applications of polyphenylenevinylene (PPV):

i) The first polymer-based light emitting diode (LED) can be using PPV as the emissive layer.

ii) Polyphenylene vinylene is capable of electroluminescence, leading to applications in polymer-

based organic light emitting diodes.

iii) PPV is also used as an electron-donating material in organic solar cells

iv) PPV and PPV derivatives (especially MEH-PPV and MDMO-PPV) find frequent application

in research cells.

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Q. 5. a) What are carbon nanotubes? Give types with respective to their structure. Give

applications of CNTs[ 5]

Answer:

Definition: A Carbon Nanotube is a tube-shaped material, made of carbon, actually rolled-up

sheets of graphene, having a diameter measuring on the nanometer scale.

Types with respective to their structure:

a) Single-walled carbon nanotubes:

i) Single-walled nanotubes (SWNTs) have a diameter of close to 1 nanometer.

ii) The structure of a SWNT can be conceptualized by wrapping a one-atom-thick layer of

graphite called graphene into a seamless cylinder.

iii) These are also divvied into three types, armchair, zigzag, chiral

b) Multi-walled carbon nanotubes:

i) Multi-walled nanotubes (MWNTs) consist of multiple rolled layers (concentric tubes) of

graphene

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.

ii) The interlayer distance in multi-walled nanotubes is close to the distance between graphene

layers in graphite, approximately 3.4 Å

c) Applications of CNTs

i) Individual CNT walls can be act as a metallic or semiconducting depending material on the

orientation of the lattice with respect to the tube axis, which is called chirality.

ii) Carbon nanotube is good storage material for Hydrogen gas.

iii) Carbon nanotube can be used for automotive parts, boat hulls, sporting goods, water filters,

thin- film electronics, coatings, actuators and electromagnetic shields.

iv) Carbon nanotubes can be used to significant improvements in the mechanical properties of

biodegradable polymeric nanocomposites for applications in bone tissue engineering

v) CNTs can be used to fluorescent and photo acoustic imaging, as well as localized heating

using near- infrared radiation

vi) SWNT can be act as biosensors.

vii) MWNTs were first used as electrically conductive fillers in metals.

viii) CNT plastics are used in electrostatic-assisted painting of mirror housings, as well as fuel

lines and filters that dissipate electrostatic charge.

ix) For load-bearing applications, CNT powders are mixed with polymers or precursor resins to

increase stiffness, strength and toughness.

x) CNTs can serve as a multifunctional coating material.

xi) CNTs can be used in solar cells, in transistors etc.

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Q 5 b) What are alanates? Explain how hydrogen gas released from sodium alanates when

used for hydrogen storage. [4]

Answer:

Alanates:

i) The hydrides of sodium, aluminium are known as alanates.

ii) Sodium aluminium hydride or sodium alanate is an inorganic compound with the chemical

formula NaAlH4.

Release of hydrogen from alanates

i) First Reaction: 3 NaAlH4 → Na3AlH6 + 2Al + 3H2

Reversible capacity of NaAlH4 is 3.0 – 3.2%. Absorption rate < 10 min. Desorption at 70oC –

100oC

ii) NaAlH4 → NaH + Al + 3/2H2

Reversible capacity of NaAlH4 is 4.5 % . Absorption rate < 10 min. Desorption at 120oC – 130oC iii) By application of heat, hydrogen get release from alanate. iv) By application of pressure, hydrogen get release from alanate.

Q 5 C) Give structure, one method of preparation and application of silane [4]

Answer:

Structure of silane:

i) Central atom is Si with four hydrogen atom.

ii) SiH4 is tetrahedral.

Method of preparation of silane

i) Industrially, silane is produced from metallurgical grade silicon in a two-step process.

Step 1: Powdered silicon is reacted with hydrogen chloride at about 300 °C to

produce trichlorosilane, HSiCl3, along with hydrogen gas, according to the chemical equation:

Si + 3 HCl → HSiCl3 + H2

Step 2: The trichlorosilane is then boiled on a resinous bed containing a catalyst which promotes

the formation of silane and silicon tetrachloride according to the chemical equation:

4 HSiCl3 → SiH4 + 3 SiCl4

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ii) Another commercial production of silane involves reduction of silicon dioxide (SiO2) under

Al and H2 gas in a mixture of NaCl and aluminum chloride (AlCl3) at high pressures:[3]

3SiO2 + 6H2 + 4Al → 3SiH4 + 2Al2O3

Application of Silane:

i) Silanes are used as coupling agents to adhere fibers such as glass fibers and carbon fibers to certain polymer matrices, stabilizing the composite material.

ii) Other applications include water repellents, masonry protection, control of graffiti, applying, polycrystalline silicon layers on silicon wafers when manufacturing semiconductors, and sealants.

Q 6 a) Explain industrial production of hydrogen steam reforming of methane and coke [5]

Answer:

Industrial production of hydrogen steam reforming of methane:

i) Specifically, bulk hydrogen is usually produced by the steam reforming of methane or natural

gas. ii) At high temperatures (700–1100 °C), steam (H2O) reacts with methane (CH4) in an endothermic reaction to yield water gas. Here we have to catalyst Ni, Fe2O3, ZnO

CH4 + H2O → CO + 3 H2

iii) In a second stage, additional hydrogen is generated through the lower-temperature,

exothermic, water gas shift reaction, performed at about 360 °C and pressures of up to 50 bar.

CO + H2O → CO2 + H2

iv) Essentially, the oxygen (O) atom is stripped from the additional water (steam) to oxidize CO

to CO2. This oxidation also provides energy to maintain the reaction. Additional heat required to

drive the process is generally supplied by burning some portion of the methane.

v) Pass the the mixture of CO2 and H2 through KOH or K2CO3, it will absorb all CO2 so that we

get pure and dry H2 gas.

Industrial production of hydrogen steam reforming of coke:

i)Petroleum coke can also be converted in hydrogen rich syngas( water gas), via coal gasification.

ii) At high temperatures (700–800 °C), steam (H2O) reacts with coke in an endothermic reaction to yield water gas. Here we have to catalyst Ni, FeO

C (coke) + H2O → CO + H2

iii) In a second stage, additional hydrogen is generated through the lower-temperature,

exothermic, water gas shift reaction, performed at about 360 °C:

CO + H2O → CO2 + H2

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iv) Pass the mixture of CO2 and H2 through KOH or K2CO3, it will absorb all CO2 so that we get

pure and dry H2gas.

Q 6 b) Explain the isotopes of carbon with their applications [4]

Answer:

The isotopes of carbon:

i) Carbon (C) has 15 known isotopes, from 8C to 22C, 2 of which (12C and 13C) are stable. ii) The longest- lived radioisotope is 14C, with a half- life of 5,700 years.

iii) The most stable artificial radioisotope is 11C, which has a half- life of 20.334 minutes. iv) 12C and 13C are stable, occurring in a natural proportion of approximately 99:1.

v) 14C is produced by thermal neutrons from cosmic radiation in the upper atmosphere, and is transported down to earth to be absorbed by living biological material. vi) 12C contains 6 protons, 6 neutrons, 6 electrons.

vii) 13C contains 6 protons, 7 neutrons, 6 electrons. viii) 14C contains 6 protons, 8 neutrons, 6 electrons.

Applications the isotopes of carbon:

i) 12C is used to compare atomic weights other elements. ii) 13C is used to in NMR Spectroscopy to find out structure of organic compound.

iii) 14C is used to in carbon dating process to determine age fossils.

Q 6 c) Explain structure of diamond based on bonding. Give its applications. [4]

Answer:

Structure of diamond based on bonding:

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i) Each carbon atom is covalently bonded to four other carbon atoms.

ii) There are no free electrons or ions in diamond, so it does not conduct electricity. iii) How to draw the structure of diamond

iv) In diamonds, each carbon atom is strongly bonded to four adjacent carbon atoms located at the

apices of a tetrahedron (a three-sided pyramid).

v) The four valence electrons of each carbon atom participate in the formation of very strong covalent

bonds. These bonds have the same strength in all directions. This gives diamonds their great hardness.

Applications of Diamond:

i) Diamond used as a Gemstone in jewellery

ii) Small particles of diamond are used in cutting, drilling or grinding.

iii) Diamond paste that is used for polishing or for very fine grinding.

iv) Diamond windows are made from thin diamond membranes and used to cover openings in

lasers, x-ray machines and vacuum chambers.

v) Diamond speaker domes enhance the performance of high quality speakers.

vi) It is used to conduct heat away from the heat sensitive-parts of high performance

microelectronics.

vii) Wear-resistant parts can be produced by coating surfaces with a thin coating of diamond.

viii) Fake diamond can be identify because of conductivity of diamond

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Q 7 a) What is dry corrosion? Explain mechanism of oxidation of corrosion with suitable

figures and reactions [5]

Answer:

Dry corrosion:

The destruction of metal due to atmospheric gases like H2S, SO2, O2, H2, CO2, CO, Cl2 and chemical attack in absence of moisture is known as dry corrosion.

Diagram:

Mechanism of oxidation corrosion:

i) When metal is exposed with air ,adsoption of O2 on the surface of metal will take place.

ii) There is interaction of O2 and metal will observed.

iii) Metal and O2 react and formation of oxide will take place called rusting.

iv) Reaction is take place at exposed part.

v) Then diffusion of oxide ions take place and these reacts with metal ions.

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vi) Oxidation first occurs at the surface of the metal and the resulting metal oxide forms a barrier which restricts further oxidation.

vii) or oxidation to continue, the metal 'must diffuse outwards through the metal oxide layer or oxygen must diffuse inwards through the scale to the underlying metal .

viii) Both the transfers occur but the outward diffusion or the metal is more rapid than inward diffusion of oxygen, since the metal ion is much smaller than oxygen molecule, thus higher mobility. (see fig. ).

Reactions:

Reactions in oxidation corrosion

2M ----- 2Mn+ + 2ne- (Loss of electrons)

n/2 O2 +2ne- -------- nO2- (Gain of electrons)

Overall reaction

2M + n/2O2 ----------- 2Mn+ + nO2- (Metal oxide)

Q 7 b) Explain how nature of metal affects the rate of corrosion. [4]

Answer:

Factors affecting on nature of metal:

1. Nature of the metal: The tendency of the metal to undergo corrosion is mainly dependent

on the nature of the metal. IN general the metal with lower electrode potential have more

reactive and more susceptible for corrosion and metal with high electrode potential are less

reactive and less susceptible for corrosion for egs: metals like K, Na, Mg, Zn etc have low

electrode potential are undergo corrosion very easily, where as noble metals like Ag, Au, Pt have

higher electrode potential, their corrosion rate are negligible but there are few exception for this

general trend as some metals show the property of passivity like Al, Cr, Ti, Ta etc.

2. Surface state of the metal or nature of the corrosion product (passivity):

The corrosion product is usually the oxide of the metal; the nature of the product determines the

rate of further corrosion process.

If the oxide layer, which forms on the surface, is stoichiometric, highly insoluble and non-porous

in nature with low ionic and electronic conductivity then that type of products laye r effectively

prevents further corrosion, which acts as a protective film. For egs: Al, Cr, Ti develop such a

layer on their surface and become passive to corrosion and some metal like Ta, Zr and Mo not

only forms such a protective layers but are capable of self repairing oxide films when it is

damaged. Hence these are extremely passive metals.

If the oxide layer forms on the metal surface is non-stochiometric, soluble, unstable and porous

in nature and have an appreciable conductivity, they cannot control corrosion on the metal

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surface for egs: oxide layer formed on metals like Zn, Fe, Mg etc.

3. Anodic and Cathodic area:

The rate of the corrosion is greatly influenced by the relative sizes of cathodic and anodic areas.

If the metal has smaller the anodic area and larger the cathodic area exposed to corrosive

atmosphere, more intense and faster is the corrosion occurring at anodic area because at anode

oxidation takes place and electrons are liberated. At the cathode these electrons are consumed.

When anode is smaller and cathode region is larger all the liberated electrons at anode are rapidly

consumed. This process makes the anodic reaction to takes place at its maximum rate thus

increasing the corrosion rate. If the cathode is smaller and reverse process takes place decrease

rate of corrosion.

For egs: If tin (Sn) coated on iron (Fe) and in that some area are not covered or some pin holes

are left, there forms smaller anodic area and larger cathodic area because tin is cathodic with

respect to iron so intense localized corrosion takes place. On the other hand if Zn coated to Fe

then if there are some pin holes are there creates larger anodic area and smaller cathodic area

because Fe is cathodic with respect to zinc so that rate of corrosion is very less.

4. Hydrogen over voltage:

A metal with low hydrogen over voltage on its surface is more susceptible for corrosion. When

the cathodic reaction is hydrogen evolution type with low hydrogen over voltage, liberation of

H2 gas is more easier so that cathodic reaction is very fast, that makes anodic reaction faster

hence overall corrosion process is very fast. If the H2 over voltage is high so cathodic reaction is

slow hence corrosion reaction also slower.

Q 7 c) What are electroless coatings? Explain with suitable example. Give its

applications. [4]

Answer:

Electroless coating:

Electroless coating (also known as autocatalytic coating) is defined as the deposition of a

metallic coating by a controlled chemical reduction that is catalyzed by the metal or alloy being

deposited.

For example, hydrated sodium hypophosphite (NaPO2H2·H2O) which reacts with the

metal ions to deposit metal. The alloys with different percentage of phosphorus, ranging from 2-

5 (low phosphorus) to up to 11-14 (high phosphorus) are possible. The metallurgical properties

of alloys depend on the percentage of phosphorus.

Applications of Electroless coating:

i) The most common form of electroless nickel plating produces a nickel phosphorus alloy

coating.

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ii) It is commonly used in engineering coating applications where wear resistance, hardness and

corrosion protection are required.

iii) Applications include coating of oil field valves, rotors, drive shafts, paper handling

equipment, fuel rails, optical surfaces for diamond turning, door knobs, kitchen

utensils, bathroom fixtures, electrical/mechanical tools and office equipment.

iv) It is also commonly used as a coating in electronics printed circuit board manufacturing, typically with an overlay of gold to prevent corrosion.

Q 8 a) Define corrosion. State the conditions under which wet corrosion occurs. Explain

oxygen absorption mechanism of wet corrosion. [5]

Answer:

Definition Corrosion:

Corrosion is defined as the damage or deterioration of a material (usually a metal) due to a

reaction or interaction with the environment.

The conditions under which wet corrosion occurs:

i) Presence of moisture

ii) Presence of aqueous medium

iii) Presence of acid, basic and neutral medium

iv) Presence of corrosive medium like soil

Oxygen absorption mechanism of wet corrosion:

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i)The general reaction that occurs at the anode is the dissolution of metal atoms as ions:

Fe -- Fe2++ 2e

ii) Electrons from the anode flow to the cathode area through the metallic circuit and force a cathodic reaction (or reactions) to occur.

iii)In alkaline and neutral aerated solutions, the predominant cathodic reaction is

O2+ 2H2O + 4e → 4OH-

iv) In aerated acids, the cathodic reaction could be

O2+ 4H++ 4e → 2H2O

v) The corrosion product formed on iron surface in the presence of oxygen is:

Fe2++ 2OH- → Fe(OH)2

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This hydrous ferrous oxide (FeO.nH2O) or ferrous oxide Fe(OH)2 composes a diffusion barrier

layer on the surface. This layer is green to greenish black in colour. In the presence of oxygen

Fe2+ is oxidised to Fe3+.

vi) Hydrous ferric oxide is orange to red-brown in colour. It exists as nonmagnetic

Fe2O3 (hematite) or as magnetic Fe2O3. Fe3O4.nH2O often forms a black intermediate layer

between hydrous Fe2O3 and FeO. Hence, rust films can consist of up to three layers of iron

oxides in different states of oxidation.

Q 8 b) Explain cementation and cladding methods of applying metallic coating on base

metal [4]

Answer:

Cementation:

i)Cementation coatings are surface alloys formed by diffusion of the coating metal into the base metal, producing little dimensional change.

ii)Parts are heated in contact with powdered coating material that diffuses into the surface to form an alloy coating, whose thickness dependson the time and the temperature of treatment.

iii) The most widely used coatings are chromium, aluminum or silicon material.

iv) Sherardizing: To coat Zn on surface of metal, we have maintained temperature nearly 3500C. Take the mixture Zn powder, Flux material like ammonium chloride in rotating drum and rotate the drum format

v)Aluminising:

To form a protective surface alloy on a metal by treatment at elevated temperature with aluminum or an aluminum compound.

vi) Chromizing: To form a protective surface of on a metal surface. Here we have to maintain temperature nearly 8000C. We have to take Chromium powder, flux like ZnCl2

Metal cladding:

i) Cladding is prevalently applied at the mill stage by the manufacturers of sheet, plate or tubing.

ii) Cladding by pressing, rolling or extrusion can produce a coating in which the thickness and distribution can be controlled over wide ranges and the coatings produced free of porosity.

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iii)Aluminum Cladding

Among the principal uses are aluminum cladding in the aircraft industry, lead and cadmium sheathing for cables, lead-sheathed sheets for architectural applications and composite extruded

tubes for heat exchangers. The thickness of the cladding is usually between 2% and 5% of the total sheet or plate thickness, and since the cladding is usually a softer and lower strength alloy, the presence of the cladding can lower the fatigue strength and abrasion resistance of the product.

Q 8 c) Compare Cathodic protection and Anodic protection. [4]

Answer:

Comparison of Anodic and Cathodic protection:

Sr No Point Anodic protection Cathodic protection

1 Defination Anodic protection (AP)

is a technique to control

the corrosion of a metal

surface by making it

the anode of

an electrochemical

cell and controlling

the electrode potential in

a zone where the metal

is passive.

Cathodic

Protection (CP) is a

technique used to

control the corrosion of

a metal surface by

making it the cathode of

an electrochemical cell

2 Applicability Active-passive

matals/alloys

All metals/alloys

3 Nature of corrosive

medium

Weak to aggressive Weak to medium

4 Cost: Installation

Maintenance

High

Very low

Low

Medium to high

5 Operating conditions Can be accurately determine

Determined by empirical testing

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Sr No Point Anodic protection Cathodic protection

6 Significance of applied current

Direct measure of protected corrosion rate

Complex to indicate corrosion rate

7 Applications Anodic protection is used for carbon steel storage tanks containing

extreme pH environments including

concentrated sulfuric acid and 50 percent caustic

soda where cathodic protection is not suitable

due to very high current requirements.

Cathodic protection systems protect a wide range of metallic

structures in various environments. Common

applications are: steel water or fuel pipelines and

steel storage tanks such as home water heaters;

steel pier piles; ship and boat hulls; offshore oil platforms and

onshoreoil well casings

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