Electrolysis of products of Potassium Iodide

Study the electrolysis of products of Potassium Iodide (KI)

 

S.No. Contents II Page No.
I. Theory 4
II. Procedure 8
III. Observation 9
IV. Precaution 10
V. Conclusion 10
VI. Bibliography 11

THEORY

Electrolysis-

It is defined as a process of decomposition of an electrolyte by the passage of electricity through its aqueous solution  or molten (fused) state.

Mechanism  Of  Electrolysis-

Whenever an electrolyte is dissolved in water or is taken in the molten state, the electrolyte dissociates to produce Positively and Negatively charged ions. On passing electric current, the positively charged ions move towards the cathode and hence are called cations, whereas the negatively charged ions move towards the anode and hence are called anions. On reaching their respective electrodes, ions lose their charge and become neutral. The cations accept electrons from the cathode to become neutral species. Thus, oxidation occurs at the anode while reduction takes place at the anode. The conversion of ions into neutral species at their respective electrodes is called Primary change. The product formed as a result of primary change may be collected as such or it may go under a Secondary change to form the final products.

Quantitative Aspects Of Electrolysis-

Michael Faraday was the first scientist who described the quantitative aspects of electrolysis.

Faraday’s Laws Of Electrolysis-

First Law:-

The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).

Second Law :-

The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (atomic mass of metal – number of electrons required to reduce the cations).

Products Of Electrolysis –

Products of electrolysis depend on the nature of material being and the type of electrodes being used .If the electrode is inert, it does not participate in the chemical reaction and acts only as source or sink for electrons. On the other hand, if the electrode is reactive, it participates in the electrode reaction. Thus, the products of electrolysis may be different for reactive and inert electrodes. The products of electrolysis depend on the different oxidizing and reducing species present in the electrolytic cell and their standard electrode potentials. Moreover, some of the electrochemical processes although feasible, are so slow kinetically that at lower voltages these do not seem to take place and extra potential (called overvoltage) has to be applied, which makes such processes more difficult to occur.

Reactions involved:-

In the electrolysis of an aqueous solution of KI, I ions are oxidized at the anode preferentially to water molecules. Possible reactions at anode are as follows:-

2 I(aq) I2 (g) + 2 e …………(1)

2 H2O (l) 4 H+ (aq) + O2 + 4e ………….(2)

Reaction (1) occurs in preference to reaction (2) due to standard electrode potential value of the following reaction.

I2 (g) + 2 e 2 I(aq)                           …………(3)

Eo/volt = + 0.53V

4 H+ (aq) + O2 (g) + 2e2 H2O (l)                …………(4)

Eo/volt = + 1.53V

Possible cathode reactions are:

K+ (aq) + e K (s)                                                 …………..(5)

Eo/volt = – 2.92V

2 H2O (l) + 2e H2 (g) + 2 OH(aq)                      …………..(6)

Eo/volt = – 0.83V

Eo value of reduction reaction (5) is much smaller than that of reaction (6). Thus, reaction (6) occurs competitively over reaction (5) at cathode .Thus, violet colour of anode is due to formation of iodine and its subsequent reaction with starch  Pink colour at cathode is due to formation of OH ions which also render the solution alkaline. OHions give pink colour with phenolphthalein.

Procedure

Prepare 0.1M solution of potassium iodide. Fix a U- shaped tube in a stand and insert two graphite electrodes into both ends of the U- tube through the corks. Assemble the apparatus as shown in the figure. Take about 30ml of 0.1M solution of potassium iodide in a 100ml beaker add five or six drops of phenolphthalein solution and five to six drops of freshly prepared starch solution. Stir the solution and transfer it into an electrolysis – tube fitted with graphite electrodes. Pass electric current through the electrolyte and observe the appearance of colour. A pink colour appears at the cathode and a violet colour appears at the anode. Bubble formation also occurs on the surface of the cathode.

Observations

TEST  SOLUTIONS OBSERVATIONS INFERENCE
Aqueous solution of potassium iodide with five drops of phenolphthalein and five drops of starch solution. At the anode, violet colour.

At the cathode:

(i )Pink colour

(ii)Formation of bubbles

Free iodine is evolved.

(i)OH ion is formed

(ii)Hydrogen is evolved

Precautions

1)    Both the electrodes should be loosely fixed into the U- tube so as to allow the escape of evolved gasses.

2)    Electrodes should be cleaned before use.

Conclusion

In the electrolysis of an aqueous solution of potassium iodide, I ions are oxidized at the anode preferentially to water molecules. Violet colour at anode is due to iodine. Pink colour at cathode is due to formation of OH ions which renders the solution alkaline. OH ions give pink colour with phenolphthalein.

BIBLIOGRAPHY

  • NCERT CHEMISTRY XII
  • ENCARTA ENCYCLOPEDIA 2009

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Effect of Metal Coupling on the Rusting of Iron – Chemistry Project

 

Study of the Effect of Metal Coupling on the Rusting of Iron

 

S.No. Contents II Page No.
I. Introduction 4
II. Aim Of The Project 6
III. Requirement 7
IV. Procedure 8
V. Observation 9
VI. Conclusion 9
VII. Bibliography 10

Introduction:

Metals and alloys undergo rusting and corrosion. The process by which some metals when exposed to atmospheric condition i.e., moist air, carbon dioxide form undesirable compounds on the surface is known as corrosion, The compounds formed are usually oxides . Rusting is also a type of corrosion but the term is restricted to iron or products made from it. Iron is easily prone to rusting making its surface rough. Chemically, rust is a hydrated ferric oxide

Titanic‘s bow exhibiting microbial corrosion damage in the form of ‘rusticles’

Rusting an Electrochemical Mechanism ;

Rusting may be explained by an electrochemical mechanism. In the presence of moist air containing dissolved oxygen or carbon dioxide, the commercial iron behave as if composed of small electrical cells. At anode of cell, iron passes into solution as ferrous ions. The electron moves towards the cathode and form hydroxyl ions. Under the influence of dissolved oxygen the ferrous ions and hydroxyl ions interact to form rust, i.e., hydrated ferric oxide.

Methods of Prevention of Corrosion and Rusting

Some of the methods used to prevent corrosion and rusting are discussed here :

1) Barrier Protection : In this method , a barrier film is introduced between Iron surface and atmospheric air. The film is obtained by painting, varnishing etc.

2) Galvanization ; The metallic iron is covered by   a layer of more reactive metal such as zinc. The active metal losses electrons in preference of iron. Thus, protecting from rusting and corrosion.

Galvanized Metals

Aim of the project

In this project the aim is to investigate effect of the metals coupling on the rusting of iron. Metal coupling affects the rusting of iron . If the nail is coupled with a more electro-positive metal like zinc, magnesium or aluminium rusting is prevented but if on the other hand , it is coupled with less electro – positive metals like copper , the rusting is facilitated.

Requirement :

1)Two Petri dishes

2) Four test – tube

3) Four iron nails

4) Beaker

5) Sand paper

6)Wire gauge

7) Gelatine

8) Copper, zinc & magnesium strips

9)Potassium ferrocyanide solution

10)Phenolphthalein

Procedure

1)At first we have to clean the surface of iron nails with the help of sand paper.

2) After that we have to wind zinc strip around one nail, a clean copper wire around the second & clean magnesium strip around the third nail. Then to put all these three and a fourth nail in Petri dishes so that they are not in contact with each other.

3) Then to fill the Petri dishes with hot agar solution in such a way that only lower half of the nails are covered with the liquids .Covered Petri dishes for one day or so.

4) The liquids set to a gel on cooling. Two types of patches are observed around the rusted nail, one is blue and the other pink. Blue patch is due to the formation of potassium ferro-ferricyanide where pink patch is due to the formation of hydroxyl ions which turns colourless phenolphthalein to pink.

Observation

S.No. Metal Pair Colour of the patch Nails rusts or not
1 Iron- Zinc
2 Iron -Magnesium
3 Iron- Copper
4 Iron – Nail

Conclusion

It is clear from the observation that coupling of iron with more electropositive metals such as zinc and magnesium resists corrosion and rusting of iron. Coupling of iron with less electropositive metals such as copper increases rusting.

BIBLIOGRAPHY

  • NCERT CHEMISTRY XII
  • ENCARTA ENCYCLOPEDIA 2009

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Study the Diffusion of Solids in Liquids – 2 – Chemistry Project

Introduction

When substances are brought in contact with each other, they intermingle with each other. This phenomenon is known as diffusion. Diffusion takes place very rapidly in case of gases, to a lesser extent in case of liquids, and not at all in the case of solids.

However, diffusion of solids in liquids does take place, albeit at a very slow rate. If a solid is kept in contact with excess of solvent in which it is soluble, some portion of the solid gets dissolved. This process is known as dissolution of a solid in liquid, and it takes place due to the diffusion of solid particles into liquid medium.

Molecules of solute are in constant random motion due to the collision between molecules of solute and that of the solvent. It is this physical interaction between solute-solvent particles that leads to diffusion.

Objective

To demonstrate that rate of diffusion depends upon the following factors:

Temperature: As temperature increases, the kinetic energy of the particles increases. Thus, the speed of particles also increases, which in turn increases the rate of diffusion.

Size of the particle: As the size of particle increases, rate of diffusion decreases. This is because the particles become less  mobile in the solvent.

Mass of the particle: As the mass of the particle increases, the rate of diffusion decreases; as the particle becomes less mobile.

Experiment 1

To study diffusion when copper sulphate is brought in contact with water (liquid).

Requirements

Copper sulphate (CuSO4) crystals, 100 mL beaker

Procedure

 Take about two grams of copper sulphate crystals in 100 mL beaker.

 Add about 50 mL of water and allow it to stand for few minutes.

 Note the development of blue colour in water.

 Allow to stand further till it is observed that all copper sulphate disappears.

 Note the blue colour change in water.

Conclusion

When solids such as copper sulphate are brought in contact with liquids such as water, intermingling of substances, i.e., diffusion takes place.

Experiment 2

To study the effect of temperature on the rate of diffusion of solids in liquids.

Requirements

Copper sulphate (CuSO4) crystals, three 100 mL beakers, watch glass, wire gauge, burner, tripod stand, thermometer, stop watch.

Procedure

 Take five gram of copper sulphate each in three beakers.

 Pour 100 mL of distilled water slowly in one of the beakers.

 Cover this beaker with a watch glass.

 Pour 100 mL of cold water in a second beaker slowly.

 Place a third beaker containing 100 mL of water on a tripod stand for heating.

 Observe the diffusion process which begins in all the beakers.

 Record of copper sulphate the time taken for the dissolution of copper sulphate in all the three cases.

Observations

S. No. Crystal Size Time Taken to Diffuse
1 Big 19 minutes
2 Medium 13 minutes
3 Small 5 minutes

Conclusion

The rate of diffusion of copper sulphate in water is in the order as given below:

Beaker 3 > Beaker 2 > Beaker 1

Thus, the rate of diffusion varies directly with temperature.

Experiment 3

To study the effect of size of particles on the rate of diffusion of solids in liquids.

Requirements

Graduated 100 mL measuring cylinders, copper sulphate (CuSO4) crystals of different sizes, stop watch.

Procedure

 Add 50 mL of water to each of the three cylinders.

 Take five gram each of big size, medium size, small size crystals of copper sulphate, and add them separately in three cylinders.

 Allow them to stand for sometime.

 Note the time taken for blue colour to reach any fixed mark in each of the cylinders and note the observations.

Observations

 

S. No. Crystal Size Time Taken to Diffuse
1 Big 19 minutes
2 Medium 13 minutes
3 Small 5 minutes

Conclusion

The rate of diffusion of copper sulphate in water is in the order as given below:

Beaker 3 > Beaker 2 > Beaker 1

Thus, smaller particles undergo diffusion more quickly than bigger particles.

Result

 When solids such as copper sulphate are brought in contact with liquids such as water, intermingling of substances, i.e., diffusion takes place.

 The rate of diffusion varies directly with temperature.

 Small particles undergo diffusion more quickly than bigger particles.

Bibliography

Chemistry (Part I) – Textbook for Class XII; National Council of Educational Research and Training

Concepts of Physics 2 by H C Verma; Bharti Bhawan (Publishers & Distributors)

Bibliography

NCERT Chemistry for Class XII

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Diffusion of Solids in Liquids – Chemistry Project

 

S.No. Contents II Page No.
I. Introduction 4
II. Objective 5
III. Experiment 1 6
IV. Experiment 2 7
V. Experiment 3 9
VI. Result 11
VII. Bibliography 12

INTRODUCTION

When substances are brought in contact with each other they intermix, this property is known as Diffusion. This property of diffusion takes place very rapidly in case of gases and to a lesser extent in case of liquids, whereas solids do not show this process of diffusion with each other. But what we can observe in case of solids is that the diffusion of solids in liquids takes place at a very slow rate.

If a solid is kept in contact with an excess of solvent in which it is soluble, some portion of the solid gets dissolved. We know that this process is known as dissolution of a solid in liquid and this process has taken place due to the diffusion of solid particles into liquid.

Molecules of solute are in constant random motion due to the collision between molecules of solute and that of the solvent.

OBJECTIVE

Rate of diffusion depends upon:-

Temperature: As temperature increases, the kinetic energy of the particles increases so the speed of particles also increases which thus increases the rate of diffusion.

Size of the particle: As the size of particle increases, rate of diffusion decreases.

Mass of the particle: As the mass of the particle increases the rate of diffusion decreases.

EXPERIMENT 1

To study diffusion when copper sulphate is brought in contact with water (liquid)

REQUIREMENTS:

Copper sulphate crystals, 100ml beaker.

PROCEDURE:

  • Ø Take about 2g of copper sulphate crystals in 100ml beaker.
  • Ø Add about 50ml of water and allow it to stand for few minutes.
  • Ø Note the development of blue colour in water.
  • Ø Allow to stand further till it is observed that all copper sulphate disappears.
  • Ø Note the blue colour change in water.

CONCLUSION:

When solids such as copper sulphate, potassium permanganate are brought in contact with liquids such as water, intermixing of substances, i.e. diffusion takes place.

EXPERIMENT 2

To study the effect of temperature on the rate of diffusion of solids in liquids

REQUIREMENTS:

Copper sulphate crystals, 200ml beaker, watch glass, wire gauge, burner, tripod stand, thermometer and stop watch.

PROCEDURE:

  • Ø Take 5g of copper sulphate each in three beakers.
  • Ø Pour 100ml of distilled water slowly in one of the beakers.
  • Ø Cover this beaker with a watch glass.
  • Ø Pour 100ml of cold water in a second beaker slowly.
  • Ø Place a third beaker containing 100ml of water on a tripod stand for heating.
  • Ø Observe the diffusion process which begins in all the beakers.
  • Ø Record the time taken for the dissolution of copper sulphate in all the three cases.

OBSERVATIONS:

S.No. Temperature of water Time Taken in Minutes
1. 25 0C 15 Min.
2. 10 0C 20 Min.
3. 70 0C 10 Min.

CONCLUSION:

The Rate of diffusion of copper sulphate in water is in the order of Beaker 3 > Beaker 1 > Beaker 2. Thus, the rate of diffusion varies directly with temperature.

EXPERIMENT 3

To study the effect of size of particles on the rate of diffusion of solids in liquids

REQUIREMENTS:

Graduated 100ml measuring cylinders, copper sulphate crystals of different sizes, stop watch

PROCEDURE:

  • Ø Add 50ml of water to each of the three cylinders.
  • Ø Take 5g each of big size, medium size, small size crystals of copper sulphate and add them separately in three cylinders.
  • Ø Allow to stand for sometime.
  • Ø Note the time taken for blue colour to reach any fixed mark in each of the cylinders and note the observations.

OBSERVATION:

S.No. Crystal size Time Taken in Minutes
1. Big 20 Min.
2. Medium 15 Min.
3. Small 10 Min.

CONCLUSION:

Small particles undergo diffusion more quickly than bigger particles.

RESULT

  • Ø When solids such as copper sulphate, potassium permanganate are brought in contact with liquid such as water, intermixing of the substances, i.e. diffusion takes place.
  • Ø The rate of diffusion varies directly with temperature.
  • Ø Small particles undergo diffusion more quickly than bigger particles.

BIBLIOGRAPHY

  • NCERT CHEMISTRY XII
  • ENCARTA ENCYCLOPEDIA 2009

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Prepare Cup of Ammonium Rayon Threads from Filter Paper

 

 

S.No. Contents II Page No.
I. Introduction 4
II. Apparatus 6
III. Procedure 7
IV. Precaution 9
V. Bibliography 9

Introduction

Cellulose is nature’s own giant molecule. It is the fibrous material that every plant from seaweed to the sequoia makes by baking glucose molecules in long chains; the chains are bound together in the fibres that give plants their shape and strength. Wood has now become the main source of cellulose. Since it contains only 40% to 50% cellulose, the substance must be extracted by ‘pulping’. The logs are flaked, and then simmered in chemicals that dissolve the tarry lignin, resins and minerals. The remaining pulp, about 93% cellulose, is dried and rolled into sheets-raw material for paper, rayon and other products.

It can be obtained in 2 ways:

  1. Viscose Process: Cellulose is soaked in 30% caustic soda solution for about 3 hrs. The alkali solution is removed and the product is treated with CSi. This gives cellulose xanthate, which is dissolved in NaOH solution to give viscous solution. This is filtered and forced through a spinneret into a dilute H2SO4 solution, both of which harden the gum like thread into rayon fibres. The process of making viscose was discovered by C.F. Cross and EJ. Bevan in 1891.
  2. Cuprammonium Rayon: Cuprammonium rayon is obtained by dissolving pieces of filter paper in a deep blue solution containing tetra-ammine cupric hydroxide. The latter is obtained from a solution of copper sulphate. To it, NH4OH solution is added to precipitate cupric hydroxide, which is then dissolved in excess of NH3.

Reactions:

CUSO4+ 2NH4OH —> Cu(OH)2+ (NH4)2S04

Pale blue ppt

Cu(OH) 2 + 4NH4OH —> [Cu(NH3)4](0H) 2 + 4H2O

[Cu(NH3) 4](OH) 2 + pieces of filter paper left for 10-15 days give a viscous solution called VISCOSE.

Apparatus Required

a) Conical flask (preferably 250 ml)

b) Funnel

c) Glass rod

d) Beaker (preferably 250 ml)

e)Water bath

f) Filter paper

Chemicals Required

a) CuSO4

b) NaOH solution

c) Liquor ammonia solution

d) Dilute H2SO4

e) Whitman Paper

f) Distilled H2O

Procedure

A. Preparation of Schweitzer’s Solution:

a)     Weight of CuSO4.5H20.

b)     Transfer this to a beaker having 100ml distilled water and add 15ml of dilute H2SO4 to prevent hydrolysis of CuSO4.

c)      Stir it with a glass rod till a clear solution is obtained. Add 11ml of liquor ammonia drop by drop with slow stirring. The precipitate of cupric hydroxide is separated out.

d)    Filter the solution containing cupric hydroxide through a funnel with filter paper.

e)     Wash the precipitate of cupric hydroxide with water until the filtrate fails to give a positive test for sulphate ions with barium chloride solution.

f)       Transfer the precipitate to a beaker that contains 50ml of liquor ammonia or wash it down the funnel. The precipitate when dissolved in liquor ammonia gives a deep blue solution of tetra-ammine cupric hydroxide. This is known as SCHWEITZER’S SOLUTION.

B. Preparation of Cellulose material

a)     After weighing 2g of filter paper divide it into very fine pieces and then transfer these pieces to the tetra-ammine cupric hydroxide solution in the beaker.

b)     Seal the flask and keep for 10 to 15 days, during this period the filter paper is dissolved completely.

C. Formation of Ravon Thread

a)      Take 50ml of distilled water in a glass container. To this add 20ml of conc H2SO4 drop by drop. Cool the solution under tap water. In a big glass container pour some of the solution.

b)     Fill the syringe with cellulose solution prepared before.

c)      Place the big glass container containing H2SO4 solution produced before in ice (the reaction being spontaneous results in excess release of energy in the form of heat which makes the fibres weak and breaks them).

d)     Immerse the tip of the syringe in the solution and press gently. Notice the fibres getting formed in the acid bath. Continue to move your hand and keep pressing the syringe to extrude more fibres into the bath.

e)      Leave the fibres in solution till they decolorize and become strong enough.

f)       Filter and wash with distilled water.

Precautions

a)     Addition of excess NH3 should be avoided.

b)     Before taking the viscose in the syringe make sure that it does not contain any particles of paper, otherwise, it would clog the needle of the syringe.

c)      Addition of NH3 should be done in a fume cupboard and with extreme care. The fumes if inhaled may cause giddiness.

d)    Use a thick needle otherwise the fibres won’t come out.

Bibliography

Chemistry (Part I) –  Textbook for Class XII

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Constituents Of An Alloy – Chemistry Project

 

S.No. Contents II Page No.
I. Introduction 4
II. Material Required 5
III. Theory 6
IV. Procedure 8
V. Conclusion 10
VI. Bibliography 10

Introduction

An alloy is a homogeneous mixture of two or more metals or a metal and non-metal.

They are generally harder than their components with reduced malleability and ductility. Alloys are prepared to enhance certain characteristics of the constituent metals, as per requirement.

In this project, we shall qualitatively analyze the chemical composition of two alloys:

MATERIALS REQUIRED

1) BRASS AND BRONZE PIECES

2) CHINA DISHES

3) FILTRATION APPARATUS

4) NITRIC ACID

5) HYDROGEN SULPHIDE GAS

7) AMMONIUM CHLORIDE

8)POTASSIUM FERROCYANIIDE

9) AMMONIUM SULPHIDE

10) DIL HYDROCHLORIC ACID

Theory

Brass

Brass contains Cu and Zn . Both dissolve in nitric acid.

4Zn+ 10 HNO3 –> 4Zn(NO)2+ N2O + 5HO

3Cu + 8 HNO3 –> 3Cu(NO3)2 + 4HO+2NO

Further analysis is carried out for respective ions.

Cu dissolves in H2S to give black ppt. of CuS. It is filtered to get the soln of Zinc Sulphide. It precipitates out in the form of ZnCl2 in an ammonical soln. of Ammonium chloride. The precipitate is dissolved in dilute HCl and then treated with Potassium ferrocyanide to get a bluish-white ppt. of Zn2[Fe(CN)6].

Bronze

Bronze contains Cu and Sn. Their nitrates are obtained by dissolving the sample in conc. Nitric acid. The nitrates are precipitated as sulphides by passing H2S through their solution in dil. HCl.

The CuS is insoluble in yellow ammonium sulphide, while SnS is soluble. The ppt. is separated by filtration.

The ppt. is dissolved in cone HNO3 and then Ammonium hydroxide solution is passed through it. Blue colouration confirms the presence of Cu.

The filtrate is treated with conc. HCl followed by Zinc dust to obtain SnCl2 . Then HgCl2 solution is added. Formation of slate-coloured ppt. indicates the presence of Sn.

Procedure

Brass:

1. A small piece of brass was placed in a china dish and dissolved in minimum quantity of 50% conc. H2SO4.

2. The soln. was heated to obtain a dry residue. The residue was dissolved in Dilute HCl gas was passed and a black.ppt. was observed. The soln. was filtered and the ppt. was dissolved in NH4OH soln. A blue coloration observed indicates the presence of Cu.

4.   The filtrate was tested for presence of Zn. Ammonium hydroxide and chloride solutions were added and then H2S gas was passed. A dull grey ppt. was separated and dissolved in dil. HCl. followed by addition of Potassium ferrocyanide. A bluish white ppt. confirms the presence of Zn.

Bronze:

1. The sample was dissolved in 50% HNO3 and then heated to obtain nitrates.

2. The nitrates were dissolved in dil. HCl and then precipitated as sulphides by passing H2S gas.

3. The precipitates were treated with yellow amm.sulphide.

4. The ppt. was tested for Cu as in the case of brass.

5. The filtrate was treated with conc. HCl followed by Fe dust.

6. Then HgCl2 soln. was added. Formation of a slate-coloured ppt. confirmed the presence of Sn.

Conclusion

Brass contains Copper and

Bronze contains Copper and Tin.

Bibliography

1. Comprehensive practical Chemistry- Class 12.

2. www.alloyanalyzer.niit.edu

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Compare Rate Of Fermentation Of Wheat Flour – Chemistry Project

 

S.No. Contents II Page No.
I. Objective 4
II. Introduction 5
III. Theory 7
IV. Material required 8
V. Procedure 9
VI. Observation 11
VII. Bibliography 11

OBJECTIVE

The purpose of the experiment is – to compare the rate of fermentation of the given samples of wheat flour, gram flour, rice flour and potatoes.

I became interested in this idea when I saw some experiments on fermentation and wanted to find out some scientific facts about fermentation. The primary benefit of fermentation is the conversion of sugars and other carbohydrates ,e.g., converting juice into wine, grains into beer, carbohydrates into carbon dioxide to leaven bread, and sugars in vegetables into preservative organic acids.

INTRODUCTION

Fermentation typically is the conversion of carbohydrates to alcohols and carbon dioxide or organic acids using yeasts, bacteria, or a combination thereof, under anaerobic conditions. A more restricted definition of fermentation is the chemical conversion of sugars into ethanol. The science of fermentation is known as zymology. Fermentation usually implies that the action of microorganisms is desirable, and the process is used to produce alcoholic beverages such as wine, beer, and cider. Fermentation is also employed in preservation techniques to create lactic acid in sour foods such as sauerkraut, dry sausages, and yoghurt, or vinegar for use in pickling foods.

Fermentation in food processing typically is the conversion of carbohydrates to alcohols and carbon dioxide or organic acids using yeasts, bacteria or a combination thereof, under anaerobic conditions. A more restricted definition of fermentation is the chemical conversion of sugars into ethanol. The science of fermentation is known as zymology.

Fermentation usually implies that the action of microorganisms is desirable, and the process is used to produce alcoholic beverages such as wine , beer, and cider. Fermentation is also employed in preservation techniques to create lactic acid in sour foods such as sauerkraut , dry sausages,  and yogurt, or vinegar (acetic acid) for use in pickling foods.

Uses

Food fermentation has been said to serve five main purposes:

#  Enrichment of the diet through development of a diversity of flavours, aromas, and textures in food substrates

#  Preservation of substantial amounts of food through lactic acid, alcohol, acetic acid and alkaline fermentations

#  Biological enrichment of food substrates with protein, essential amino acids, essential fatty acids, and vitamins

#  Elimination of ant nutrients.

#  A decrease in cooking times and fuel requirements

Theory

Wheat flour, gram flour, rice flour and potatoes contains starch as the major constituent. Starch present in these food materials is first brought into solution. In the presence of enzyme diastase, starch undergoes fermentation to give maltose.

Starch gives blue-violet colour with iodine whereas product of fermentation starch do not give any characteristic colour. When the fermentation is complete the reaction mixture stops giving blue-violet colour with iodine solution.

By comparing the time required for completion of fermentation of equal amounts of different substances containing starch the rates of fermentation can be compared. The enzyme diastase is obtained by germination of moist barley seeds in dark at 15 degree celsius. When the germination is complete the temperature is raised to 60 degree celsius to stop further growth. The seeds are crushed into water and filtered. The filtrate contains enzyme diastase and is called malt extract.

MATERIALS REQUIRED

#   Conical flask

#   Test tube

#   Funnel

#   Filter paper

#   Water bath

#   1 % Iodine solution

#   Yeast

#   Wheat flour

#   Gram flour

#   Rice flour

#   Potato

#   Aqueuos NaCl solution

PROCEDURE

#   Take 5 gms of wheat flour in 100 ml conical flask and add 30 ml of distilled water.

#   Boil the contents of the flask for about 5 minutes

#   Filter the above contents after cooling, the filtrate obtained is wheat flour extract.

#   To the wheat flour extract. taken in a conical flask.

Add 5 ml of 1% aq. NaCl solution.

#    Keep this flask in a water bath maintained at a temperature of 50-60 degree celsius. Add 2 ml  of malt extract.

#   After 2 minutes take 2 drops of the reaction mixture and add to diluted iodine solution.

#   Repeat step 6 after every 2 minutes. When no bluish colour is produced the fermentation is complete.

#   Record the total time taken for completion of fermentation.

#   Repeat the experiment with gram flour extract, rice flour extract, potato extract and record the observations

OBSERVATIONS

Time required for the fermentation—

# Wheat flour — 10 hours

#  Gram flour –  12.5 hours

#  Rice flour — 15 hours

#  Potato — 13 hours

BIBLIOGRAPHY

#   Chemistry manual

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Commercial Antacid – Chemistry Project

 

S.No. Contents II Page No.
I. Introduction 4
II. Requirements 6
III. Procedure 7
IV. Observation 10
V. Conclusion 13
VI. Bibliography 13

INTRODUCTION

An Antacid is any substance, generally a base or basic salt, which neutralizes stomach acidity. They are used to relieve acid indigestion, upset stomach, sour stomach, and heartburn.

ACTION MECHANISM

Antacids perform a neutralization reaction, i.e. they buffer gastric acid, raising the pH to reduce acidity in the stomach. When gastric hydrochloric acid reaches the nerves in the gastrointestinal mucosa, they signal pain to the central nervous system. This happens when these nerves are exposed, as in peptic ulcers. The gastric acid may also reach ulcers in the oesophagus or the duodenum.

Other mechanisms may contribute, such as the effect of aluminium ions inhibiting smooth muscle cell contraction and delaying gastric emptying.

INDICATIONS

Antacids are taken by mouth to relieve heartburn, the major symptom of gastro oesophageal reflux disease, or acid indigestion. Treatment with antacids alone is symptomatic and only justified for minor symptoms. Peptic ulcers may require H2-receptor antagonists or proton pump inhibitors.

The utility of many combinations of antacids is not clear, although the combination of magnesium and aluminium salts may prevent alteration of bowel habits.

INTERACTIONS

Altered pH or complex formation may alter the bioavailability of other drugs, such as tetracycline. Urinary excretion of certain drugs may also be affected.

PROBLEMS WITH REDUCED STOMACH ACIDITY

Reduced stomach acidity may result in an impaired ability to digest and absorb certain nutrients, such as iron and the B vitamins. Since the low pH of the stomach normally kills ingested bacteria, antacids increase the vulnerability to infection. It could also result in reduced bioavailability of some drugs. For example, the bioavailability of ketoconazole (antifungal) is reduced at high intragastric pH (low acid content).

SOME FAMOUS ANTACID BRANDS

1. Alka-Seltzer – NaHCO3 and/or KHCO3

2. Equate – Al(OH)3 and Mg(OH)2

3. Gaviscon – Al(OH)3

4. Maalox (liquid) – Al(OH)3 and Mg(OH)2

5. Maalox (tablet) – CaCO3

6. Milk of Magnesia – Mg(OH)2

7. Pepto-Bismol – HOC6H4COO

8. Pepto-Bismol Children’s – CaCO3

9. Rolaids – CaCO3 and Mg(OH)2

10. Tums – CaCO3

11. Mylanta

DRUG NAMES

Some drugs used as antacids are :

1. Aluminium hydroxide

2. Magnesium hydroxide

3. Calcium carbonate

4. Sodium bicarbonate

5. Bismuth subsalicylate

6. Histamine

7. Cimetidine

8. Ranitidine

9. Omeprazole

10. Lansoprazole

REQUIREMENTS :

  • Burettes
  • Pipettes
  • Titration flasks
  • Measuring flasks
  • Beakers
  • Weight box
  • Fractional weights
  • Sodium hydroxide
  • Sodium carbonate
  • Hydrochloric acid
  • Phenolphthalein.

PROCEDURE :

1. Prepare 1 litre of approximately  HCl solution by diluting 10 ml of the concentrated acid for one litre.

2. Similarly, make 1 litre of approximately NaOH solution by dissolving4.0g of NaOH to prepare one litre of solution.

3. Prepare  Na2CO3 solution by weighing exactly 1.325 g of anhydrous sodium carbonate and then dissolving it in water to prepare exactly 0.25 litres (250 ml) of solution.

4. Standardize the HCl solution by titrating it against the standard Na2CO3 solution using methyl orange as indicator.

5. Similarly, standardize NaOH solution by titrating it against standardized HCl solution using phenolphthalein as indicator.

6. Powder the various samples of antacid tablets and weigh 1.0 g of each.

7. Add a specific volume of standardised HCl to each of the weighed sample is taken in conical flasks. The acid should be in slight excess, so that it can neutralize all the alkaline component of the tablet.

8. Add 2 drops of phenolphthalein and warm the flask till most of powder dissolves. Filter off the insoluble material.

9. Titrate this solution against the standardised NaOH solution, till a permanent pinkish tinge is obtained. Repeat this experiment with different antacids.

OBSERVATIONS:

Standardisation of HCl solution :

Volume of Na2CO3 solution taken = 20.0 ml

S No. of obs. Burette readings

Initial       Final

Volume of acid used
1.

2.

3.

4.

5.

0 ml          15.0 ml

0 ml          15.1 ml

0 ml          15.0 ml

0 ml          15.0 ml

0 ml          15.0ml

15.0 ml

15.1 ml

15.0 ml

15.0 ml

15.0 ml

Concordant volume = 15.0 ml

Applying normality equation,

N1V1 = N2V2

N1 * 15.0 =  * 20

Normality of HCl, N1 = = 0.133 N

Standardisation of NaOH solution :

Volume of the given NaOH solution taken = 20.0 ml

S No. of obs. Burette readings

 Initial     Final

    

Volume of acid used
1.

2.

3.

4.

5.

0 ml          26.5 ml

0 ml          26.8 ml

0 ml          26.6 ml

0 ml          26.6 ml

0 ml          26.6ml

26.5 ml

26.8 ml

26.6 ml

26.6 ml

26.6 ml

Concordant volume = 26.6 ml

Applying normality equation,

N1V1 = N2V2

0.133 x26.6 = N2x20

Normality of NaOH = = 0.176 N

Analysis of antacid tablet :

Weight of antacid tablet powder = 1.0 g

Volume of HCl solution added = 20.0 ml

Antacid Vol. Of NaOH soln. Used to neutralise unused HCl Vol. Of HCl soln. Used to neutralise 1.0 g of antacid matter
1. Gelusil

2. Digene

3. Aludrox

4. Logas

5. Ranitidine

6. Ocid 20

12.1 ml

16.0 ml

19.3 ml

24.3 ml

21.4 ml

22.7 ml

12.0 ml

16.2 ml

18.9 ml

24.4 ml

21.7 ml

21.9 ml

CONCLUSION :

The antacid which has maximum volume of HCl is used for neutralizing i.e. Ocid 20 is more effective.

BIBLIOGRAPHY

1. WEBSITES :

  • www.pharmaceutical-drugmanufacturers.com

2. BOOKS :

  • Comprehensive Practical Manual for class XII
  • Pradeep’s New Course Chemistry
  • NCERT Class XII Part II

You can find other Chemistry Projects here.

Analysis Of Different Brands Of Cold Drinks – Chemistry Project

 

S.No. Contents II Page No.
I. Introduction 4
II. Theory 5
III. Apparatus 6
IV. Detection of pH 9
V. Test for Carbon Dioxide 10
VI. Test for Glucose 11
VII. Test for Alcohol 13
VIII. Test for Sucrose 15
IX. Result 17
X. Conclusion 18
XI. Bibliography 20

INTRODUCTION

The era of cold drinks began in 1952 but the indianization of industry marked its beginning with launching of limca and goldspot by parley group of companies. Since, the beginning of cold drinks was highly profitable and luring, many multinational companies launched their brands in India like pepsi and coke.

Now days, it is observed in general that majority of people viewed Sprite, Miranda, and Limca to give feeling of lightness, while Pepsi and Thumps Up to activate pulse and brain.

Theory

Cold drinks of different brands are composed of alcohol, carbohydrates, carbon dioxide, phosphate ions etc. These soft drinks give feeling of warmth, lightness and have a tangy taste which is liked by everyone. Carbon dioxide is responsible for the formation of froth on shaking the bottle.

The carbon dioxide gas is dissolved in water to form carbonic acid which is also responsible for the tangy taste. Carbohydrates are the naturally occurring organic compounds and are major source of energy to our body. General formula of carbohydrates is

CX (H2O)Y.

On the basis of their molecule size carbohydrates are classified as:-

Monosaccharide, Disaccharides and Polysaccharides. Glucose is a monosaccharide with formula C6H12O6 .It occurs in Free State in the ripen grapes in bones and also in many sweet fruits. It is also present in human blood to the extent of about 0.1%. Sucrose is one of the most useful disaccharides in our daily life. It is widely distributed in nature in juices, seeds and also in flowers of many plants. The main source of sucrose is sugar cane juice which contain 15-20 % sucrose and sugar beet which has about 10-17 % sucrose. The molecular formula of sucrose is C12H22O11. It is produced by a mixture of glucose and free dose. It is non-reducing in nature whereas glucose is reducing. Cold drinks are a bit acidic in nature and their acidity can be measured by finding their pH value. The pH values also depend upon the acidic contents such as citric acid and phosphoric acid.

APPARATUS

  • Test tube
  • Test tube holder
  • Test tube stand
  • Stop watch
  • Beaker
  • Burner
  • pH paper tripod stand
  • China dish
  • Wire gauge
  • Water bath
  • Iodine solution
  • Potassium  iodine
  • Sodium hydroxide
  • Fehling’s A & B solution
  • Lime water
  • Concentrated HNO3
  • Benedict solution
  • Ammonium molybdate

DETECTION OF PH

1-2 drops of the sample of cold drink of each brand was taken and put on the pH paper. The change in the color of pH paper was noticed and was compared with the standard pH scale.

OBSERVATION

SERIAL NO NAME OF

DRINK

COLOUR CHANGE PH VALUE
1 COCA COLA PINK 1-2
2 SPRITE ORANGE 3
3 LIMCA PINKISH 3-4
4 FANTA LIGHT DRINK 2-3

INFERENCE

Soft drinks are generally acidic because of the presence of citric acid and phosphoric acid. pH values of cold drink of different brands are different due to the variation in amount of acidic contents.

TEST FOR CARBON DIOXIDE

EXPERIMENT

As soon as the bottles were opened, one by one the sample was passed through lime water. The lime water turned milky.

OBSERVATON

SR. NO NAME OF THE DRINK TIME TAKEN (SEC.) CONCLUSION
1 COCA COLA 26.5 CO2 IS  PRESENT
2 SPRITE 21 CO2 IS  PRESENT
3 LIMCA 35 CO2 IS  PRESENT
4 FANTA 36 CO2 IS  PRESENT

INFERENCE

All the soft drinks contain dissolved carbon dioxide in water. The carbon dioxide (CO2) dissolves in water to form carbonic acid, which is responsible for its tangy taste.

CHEMICAL REACTION INVOLVED

Ca(OH)2 (s) + CO2(g) ———> CaCO3 (s) + H2O(s)

TEST FOR GLUCOSE

Glucose is a reducing sugar acid. Its presence is detected by the following test:-

1. BENIDICTS’S SOLUTION TEST:-

A small sample of cold drink of different brands was taken in a test tube and a few drops of Benedict’s reagent were added. The test tube was heated for few seconds. Formation of reddish color confirms the presence of glucose in cold drinks.

OBSERVATON

SR. NO NAME OF THE DRINK OBSERVATION CONCLUSION
1 COCA COLA REDDISH COLOUR GLUCOSE PRESENT
2 SPRITE REDDISH COLOUR GLUCOSE PRESENT
3 LIMCA REDDISH COLOUR GLUCOSE PRESENT
4 FANTA REDDISH COLOUR GLUCOSE PRESENT

INFERENCE

All the samples gave positive test for glucose with Benedict’s reagent. Hence all the drinks contain glucose.

2. FEHLING’S SOLUTION TEST

A small sample of cold drink of different brands was taken in a test tube and a few drops of Fehling’s A solution and Fehling’s B solution was added in equal amount. The test tube was heated in a water bath for 10 minutes. Appearance of brown precipitate confirms the presence of glucose in cold drinks.

OBSERVATON

SR. NO NAME OF THE DRINK OBSERVATION CONCLUSION
1 COCA COLA Reddish Brown Precipitate GLUCOSE PRESENT
2 SPRITE Reddish Brown Precipitate GLUCOSE PRESENT
3 LIMCA Reddish Brown Precipitate GLUCOSE PRESENT
4 FANTA Reddish Brown Precipitate GLUCOSE PRESENT

INFERENCE

All the samples give positive test for glucose with Fehling’s solutions (A&B).Hence all the cold drinks contain glucose.

TEST FOR PHOSPHATE

Sample of each brand of cold drink was taken in a separate test tube and ammonium molybdate followed by concentrated nitric acid (HNO3) was added to it, the solution was taken heated and the color of the precipitate confirms the presence of phosphate ions.

OBSERVATON

SR. NO NAME OF THE DRINK OBSERVATION CONCLUSION
1 COCA COLA CANARY-YELLOW PPT PHOSPHATE IS PRESENT
2 SPRITE CANARY-YELLOW PPT PHOSPHATE IS PRESENT
3 LIMCA CANARY-YELLOW PPT PHOSPHATE IS PRESENT
4 FANTA CANARY-YELLOW PPT PHOSPHATE IS PRESENT

INFERENCE

All the soft drinks contain phosphate ions which are detected by the presence of phosphate when canary yellow obtained.

CHEMICAL REACTION INVOLVED

NaHPO4 + 12 (NH4)2MoO4 + 21HNO3 +3H——>(NH4)3PO4.12MoO3 +21HN4NO3 +12H2O

TEST FOR ALCOHOL

Samples of each brand of cold drinks are taken in sample test tube and iodine followed by potassium iodide and sodium hydroxide (NaOH) solution is added to each test tube. Then the test tube are heated in hot water bath for 30 minutes yellow colored precipitate confirmed the presence of alcohol in cold drinks

OBSERVATON

SR. NO NAME OF THE DRINK OBSERVATION CONCLUSION
1 COCA COLA YELLOW PPT ALCOHOL  IS PRESENT
2 SPRITE YELLOW PPT ALCOHOL  IS  PRESENT
3 LIMCA YELLOW PPT ALCOHOL  IS PRESENT
4 FANTA YELLOW PPT ALCOHOL  IS PRESENT

INFERENCE

All the Brands of Cold Drinks Contain Alcohol.

CHEMICAL REACTION INVOLVED

CH3CH2OH +4I2+ 6NaOH——>CHI3 + HCOONa +5NaI +5H2O

TEST FOR SUCROSE

5 ml samples of each brand of cold drinks was taken in a china dish and heated very strongly until changes occur. Black colored residue left confirms the presence of sucrose in cold drinks.

OBSERVATON

SR. NO NAME OF THE DRINK OBSERVATION CONCLUSION
1 COCA COLA BLACK RESIDUE SUCROSE  IS PRESENT
2 SPRITE BLACK RESIDUE SUCROSE  IS PRESENT
3 LIMCA BLACK RESIDUE SUCROSE  IS PRESENT
4 FANTA BLACK RESIDUE SUCROSE  IS PRESENT

INFERENCE

All the brands of cold drinks contain sucrose. But amount of sucrose varies in each brand of drink. Fanta contained highest amount of sucrose.

RESULT

After conducting several tests, it was concluded that the different brands of cold drinks namely

  1. Coca cola
  2. Sprite
  3. Limca
  4. Fanta

All contains glucose, alcohol sucrose, phosphate, ions and carbon dioxide. All are acidic in nature. On comparing the pH value of different brands coca cola is most acidic and Limca is least acidic of all the four brands taken.

pH value of coca cola is nearly equal to disinfectant which is harmful for body.

CARBON DIOXIDE

AMONG THE FOUR SAMPLES OF COLD DRINKS TAKEN –SPRITE HAS MAXIMUM AMOUNT OF DISSOLVED CARBON DIOXIDE AND FANTA HAS MINIMUM AMOUNT OF DISSOLVED CARBON DIOXIDE.

CONCLUSION

DISADVANTAGES OF COLD DRINKS

  1. Soft drinks are little more harmful than sugar solution. As they contain sugar in large amount which cause “diabetes”.
  2. Soft drinks can cause weight gain as they interfere with the body’s natural ability to suppress hunger feeling.
  3. Soft drinks have ability to dissolve the calcium so they are also harmful for our bones.
  4. Soft drinks contain “phosphoric acid” which has a pH of 2.8. So they can dissolve a nail in about 4 days.
  5. For transportation of soft drinks syrup the commercial truck must use the hazardous matter place cards reserved for highly consive material.
  6. Soft drinks have also ability to remove blood so they are very harmful to our body.

USES OF COLD DRINKS

  1. Cold drinks can be used as toilet cleaners.
  2. They can remove rust spots from chrome car bumper.
  3. They clean corrosion from car battery terminals.
  4. Soft drinks are used as an excellent ‘detergent’ to remove grease from clothes.
  5. They can lose a rusted bolt.

BIBLIOGRAPHY

  • NCERT CHEMISTRY XII
  • ENCARTA ENCYCLOPEDIA 2009

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Change in EMF of a Daniel Cell – Chemistry Project

 

S.No. Contents II Page No.
I. Objective 3
II. Theory 3
III. Material and Equipment 8
IV. Procedure 9
V. Observation 10
VI. Conclusion 11
VII. Bibliography 12

Objective

The goal of this project Is to study the change in E.M.F. of a Daniel cell Due to various factors such as  Change in concentration, temperature And Area of electrodes.

Theory

The Daniell cell was invented in 1836 by John Frederic Daniell, a British chemist and meteorologist, and consisted of a copper pot filled with a copper sulphate solution, in which was immersed an unglazed earthenware container filled with sulphuric acid and a zinc electrode. He was searching for a way to eliminate the hydrogen bubble problem found in the Voltaic pile, and his solution was to use a second electrolyte to consume the hydrogen produced by the first. The Daniell cell was a great improvement over the existing technology used in the early days of battery development. A later variant of the Daniell cell called the gravity cell or crowfoot cell was invented in the 1860s by a Frenchman named Callaud and became a popular choice for electrical telegraphy.

When an external circuit is connected, the chemical equation for the zinc side (anode) half cell is:

Zn (s) Zn2+ (aq) + 2 e

For the copper sulphate side (cathode) half cell:

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

Therefore, the overall reaction of the Daniel cell is:

Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)

It is an arrangement to convert the chemical energy of the redox reaction into electric energy.

Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)

Features of Daniel Cell:-

v Zinc rod at which oxidation occurs is called the anode while the copper rod at which the reduction takes place is called cathode.

v The overall reaction occurring in electrochemical cell is due to two half-cell reaction, one occurring in each beaker.

v The half-cell reaction occurring at anode is called oxidation -half cell reaction while the occurring at cathode is called reduction.

v The two half-cell reactions always take place simultaneously i.e. . .  Half cell reaction cannot take place immediately.

v Since electrons are produced at zinc electrode, it is rich in electrons and pulls these electrons into the external circuit and hence acts as negative pole. The copper electrode on the other hand is deficient in electrons and thus pulls the electrons from the external circuit and act as positive pole.

The electrons flow from negative pole to positive  pole in the external circuit. However, conventionally the current is said to flow in opposite direction i.e. from positive pole to negative pole in the external circuit.

The concentration of copper sulphate solution decreases with passage of time as the cell operates, consequently the current fall with passage of time.

v      Salt Bridge  :-

It consists of a tube filled with semi-solid paste obtained by adding gelative or agar to the solution of strong electrolyte  such as NaCl , NH4NO3.KNO3 etc, which does not change chemically during the process.

v Function of salt bridge:-

To complete the electrical circuit by allowing the solution to flow from one solution to another without mixing the two solutions.

To maintain electrical neutrality of solution in two half-cells.

v EMF of Cells:-

When a current flows through two points a potential difference generated by a cell when the cell draws no current is called EMF.

Materials and Equipment

To do this experiment we will need the following materials and equipment:

v  Two beakers.

v  Zinc and Copper plate.

v  Filter paper.

v  Voltmeter.

v  Connecting wires.

v  Card board.

v  KNO3 solution.

v  1 M, 0.1M, 0.01 M  solution of :-

  1. a. CuSO4
  2. b. ZnSO4

Procedure

I. Take two beakers and pour the required chemicals in respective beaker and mark them for identification.

II. Take two square to slide in and connecting wire to their screw.

III. Connect negative of the voltmeter to the anode and its positive to  the cathode

IV. Take filter paper long enough to dip into both the solution.  Dip the filter paper in  KNo3 solution and put it as a salt bridge.

V. Put on the electrode voltmeter set up.  Note the reading quickly and then put of the electrode voltmeter set up.

VI. For measuring variation with temperature with change in area of electrode use the different size of electrode and then do step 5 again.

VII. For measuring variation with temperature heat the solution and then do step 5 again.

VIII. For measuring variations with change in concentration of electrolyte ,use  the electrolytes of different molarity and then do step 5 again.

Observations:-

v Electrode Potential of  Zinc =…………….V

v Electrode Potential of Copper=…………V

v     Variation with Concentration:-

Molarity  of CuSO4(M) Molarity of ZnSO4(M) Voltmeter Reading (V)

v     Variation with change in area of electrodes:-

With increase in area or decrease in area of electrode EMF of cell remains same.

v     Variation with temperature:-

Cuso4(.c) ZnSo4(.c) Voltmeter Reading(V)

Conclusions:-

v The EMF varies non-linearly with change in concentration of reactants.

v Increase in concentration of ions in anode half-cell decreases EMF and vice-verse.

v The EMF is independent of area of electrode.

v The EMF increases with increase in temperature.

BIBLIOGRAPHY

  • NCERT CHEMISTRY XII
  • ENCARTA ENCYCLOPEDIA 2009

You can find other Chemistry Projects here.