ELECTROSATATIC POTENTIAL AND CAPACITANCE [ NCERT EXERCISES QUESTIONS AND ANSWERS ]

 

CHEMICAL REACTIONS AND EQUATIONS [ ncert exercises questions and answers ]

                                                YASH   CLASSES

                    CHEMICAL REACTIONS AND EQUATIONS

              NCERT  exercises  QUESTIONS AND ANSWERS

CHEMICAL REACTIONS AND EQUATIONS

                                     YASH   CLASSES

                    CHEMICAL REACTIONS AND EQUATIONS

              NCERT  INTEXT  QUESTIONS AND ANSWERS

INTEXT QUESTIONS            [ page 6]

1. Why should magnesium ribbon be cleaned before burning in air?

Ans. Magnesium ribbon should be cleaned before burning in air so that unwanted impurities are removed  and we burn only pure magnesium metal. when Mg burn in air combines with oxygen

 give magnesium oxide [MgO].

                            2Mg (s)  + O₂ (g) → 2MgO (S)

 2. Write the balanced equation for the following chemical reactions.

  (i)Hydrogen + Chlorine → Hydrogen chloride

               Ans .   H₂ (g)+ Cl₂ (g) → 2HCI (g)

(ii)Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride

              Ans-       3 BaCl₂ (s) + Al₂(SO₄)₃ (s) → 3BaSO₄ (s)+ 2AICI₃(s)

 (iii)  Sodium Water Sodium Hydroxide Hydrogen

                Ans . 2 Na(s) +2 H₂O(l) → 2NaOH (l) + H₂ (g)

 3. Write a balanced chemical equation with state symbols for the following reactions.

(i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate

     and the solution of sodium chloride.

(ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce

      sodium chloride solution and water.

Ans. (i)         BaCl₂ (aq) + Na₂SO₄ (aq) → BaSO₄ (s) + 2NaCl(aq)

           (ii)         NaOH (aq) + HCl (aq) → NaCl(aq) + H₂O (l)

1. A solution of a substance X is used for white-washing             [ page 10]

        (i)  Name the substance 'X and write its formula.

       (ii)   Write the reaction of the substance X named in above with water.

Ans. X is quick lime, CaO.

       (ii)   CaO(s)      + H₂O (l)         →               Ca(OH)₂ (aq)

            Quick lime                               Slaked lime or Calcium hydroxide

2. Why is the amount of gas collected in one of the test tubes in Activity (1.7) (Textbook) double of

       the amount collected in the other Name this gas.

Ans. In activity (1.7) water is electrolysed to give H, gas at one electrode and O, gas at the other electrode   

           according to the given chemical reaction: 2H₂O(l) → 2H₂(g) + O₂(g)

                 Thus two molecules of water on electrolysis give 2 molecules of hydrogen gas and one molecule of oxygen gas or the amount of hydrogen gas collected would be double than that of oxygen gas.

 

 

 

 

 

  1. Why does the colour of copper sulphate solution change when an iron nail is dipped in it ?                   [ page13]

   Ans. When an iron mail is dipped in copper sulphate solution, it becomes brownish in colour and

            colour of copper sulphate fades due to the following reaction

                        Fe (s)   + CuSO₄ (aq)   →   FeS0₄ (aq) +  Cu (s)

                                    Copper sulphate

       In this reaction iron has replaced copper from its solution. Thus amount of per sulphate decreases

       and its colour fades.

 2. Give an example of a double displacement reaction other than one given in Activity 1.10.

 Ans. When a solution of hydrochloric acid is added to a solution of lend nitrate, a white precipitate of lead    

         chloride is formed.

                 Pb(NO₃ )₂  (aq) +  2HCl(aq) → PbCl₂ (s )+ 2HNO₃ (aq)

                Lead nitrate                               Lead chloride

     Pb^{2 +} ions from lead nitrate combine with 2Cl^- ions from hydrochloric acid and a precipitate of

      lead chloride  is formed. Thus there is an exchange of ions between the two reactants for example

      lead nitrate and hydrochloric  acid, and thus this reaction is a double displacement reaction.

  

LIFE PROCESS [10th] NCERT QUESTIONS AND ANSWERS

                                                         YASH CLASSES

                                                              LIFE PROCESS

NCERT INTEXT QUESTIONS (Page 95)

1. Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms like humans?

Ans. In multicellular organisms like us, all the body cells are not in direct contact with the surrounding environment. Thus, every cell of the body will not get oxygen as per need by the process of diffusion from the environment. So, complex multicellular organisms need specialised tissues, organs and organ system to take in oxygen.

2. What criteria do we use to decide whether something is alive?

Ans. Some movements either visible like locomotion from one place to another and growth-related movements or invisible movement such as movement of molecules are the criteria whether something is alive.

3. What are outside raw materials used for by an organism?

Ans. An organism needs raw materials from outside are mainly.

(a) Food: To supply energy, materials like minerals, vitamins etc., for working and maintenance of cells/tissues.

(b) Water: To provide medium in the body to live and carry out all the metabolic reactions necessary for life.

(c) Oxygen: For respiration.

4. What are processes would you consider essential for maintaining life?

Ans. Processes essential for maintaining life are:

(1) Nutrition      (ii) Respiration     (iii) Transportation   (iv) Excretory

   NCERT INTEXT QUESTIONS (Page 101)

 Q.1. What are the differences between autotrophic nutrition and heterotrophic nutrition?

Ans. Autotrophic Nutrition: In this type of nutrition, organisms like green plants take in carbon dioxide and water to convert them into carbohydrates in presence of sunlight and chlorophyll. By this process, called photosynthesis, autotrophs prepare their own food.

Heterotrophic Nutrition: In this type of nutrition, organisms such as animal’s fungi utilize complex substances and break down into simpler ones before. They are up in the body. The heterotrophs are directly or indirectly depended on autotrophs for foot

 Q.2.Where do planta  following things; materials required for photo

 Ans. Plants need the following things:

(i) Carbon dioxide: Which plants get from the environment/atmosphere Stomata.

 (ii) Water: Which plants (land plants) absorb from the soil through roots and transport stomata to leaves.

 (iii) Sunlight: Which plants get from the sun.

 (iv) Chlorophyll is present in chloroplast found in green leaves and green parts of plants.

 Q.3. What is the role of the acid in our stomach?

Ans. Role of the hydrochloric acid released by the gastric glands in the stomach

(i) to create acidic medium which is necessary for the activation of the enzyme pepsin.

 (ii) to kill germs which the food may contain.

  4. What is the function of digestive enzymes?

Ans. The function of digestive enzymes: The food we eat is complex in nature i.e., contains complex molecules. Digestive enzymes break down this complex molecule is into smaller and absorbable molecules so that they can be absorbed by the walls of the intestine.

 5. How is the small intestine designed to absorb digested food?

Ans.  The small intestine is the site where digestion of food components is almost completed. It is also the place where digested food is absorbed into the blood to transport to each and every cell of the body.

The small intestines are designed to provide maximum area for absorption of digested food and its transfer into  the blood for its circulation into the body. For this, the inner lining of the small intestine has numerous fingers like projections called villi. The villi are richly supplied with blood vessels which take the absorbed food to each and every cell of the body.

                                   

NCERT INTEXT QUESTIONS (Page 105)

 1. What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?

 Ans. The organisms that live in water use oxygen dissolved in surrounding water. Since air dissolved in water has fairly low concentration of oxygen, the aquatic organisms have much faster rate of breathing. Terrestrial organisms take oxygen from the oxygen-rich atmosphere, through respiratory organs. So, they have much less breathing rate than aquatic organisms.

 2. What are the different ways in which glucose is oxidised to provide energy in various organisms?

 Ans. First step of breakdown of glucose (6 carbon molecule) takes place in the cytoplasm of cells of all organisms. This process yields a three-carbon molecule compound called Pyruvate. Further break down of pyruvate takes place in different manners in different organisms.

  (i) Anaerobic respiration: This process takes place in absence of oxygen e.g., in yeast during fermentation. In this case pyruvate is converted into ethanol and carbon dioxide. Of carbon dioxide and water.

 (ii) Aerobic respiration: In aerobic respiration break down of pyruvate takes place in presence of oxygen to give rise The release of energy in aerobic respiration is much more than anaerobic respiration.

 

 (iii)Lack of oxygen: Sometimes, when there is lack of oxygen, especially during vigorous activity, in our muscles, pyruvate is converted into lactic acid (3 carbon molecule compound). Formation of lactic acid in muscles causes cramp.

        

Fig. Breakdown of glucose by various pathways during cellular respiration

 

3. How is oxygen and carbon dioxide transported in human beings?

Ans. (a) Transport of oxygen: The respiratory pigment present in the blood take up the oxygen from the air in the lungs. They carry the oxygen to tissues which are deficient in oxygen before releasing it. In human beings, the respiratory pigment, called hemoglobin present in the red blood corpuscles carries oxygen to different tissues of the body.

 (b) Transport of carbon dioxide: Carbon dioxide is more soluble in water. Hence is mostly transported from body tissues in the dissolved form in our blood plasma to lungs where it diffuses from blood to air in the lungs.

 

4. How are the lungs designed in human beings to maximise the area for exchange of gases?

Ans. In the lungs, the air passage (wind pipe) divide into smaller tubes, called bronchi which in turn form bronchioles. The bronchioles which terminate in balloon-like structures, called alveoli. The alveoli present in the lungs provide maximum surface for exchange of gases. The alveoli have very thin walls and contain an extensive network of blood vessels to felicitate exchange of gases.

1. What are the components of the transport system in human beings? What are the functions of these components?

 Ans. The transport system (called circulatory system) in human beings mainly consist of heart, blood and blood vessels.

 Functions of component of circulatory system:

 (1) Heart: The heart is a muscular organ which is as big as our fist. Heart is a pumping organ to push blood around the body. It receives deoxygenated blood from the body parts and pump it to lungs for enriching with oxygen. It receives purified blood from lungs and pumps it around the body.

 (ii) Blood: Blood is fluid connective tissue, it consists of a fluid medium known as plasma in which the cells are suspended, such as white blood cells and red blood cells. Plasma transports food, carbon dioxide and nitrogenous wastes (urea and uric acid) in dissolved form. Red blood cells transport oxygen. Blood also transports many other substances like hormones, salt etc. Platelets present in the blood stop bleed from injuries by forming blood clots.

 (iii) Blood vessels: Blood vessels form a closed network of tubes to reach all the tissues. Thus, the blood pushed by the heart flows in the blood vessels (arteries) and also comes back to the heart in the blood vessels (veins).

 Q. 2. Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?

 Ans. Separation of oxygenated and deoxygenated blood allows a highly efficient supply of oxygen to the body. This system is useful in animals that have high energy need. Mammals and birds constantly need oxygen to obtain energy to maintain body temperature constant.

 3. What are the components of the transport system in highly organised plants?

 Ans. The transport system in highly organised plants consists of two pathways as independently organized conducting tubes. One is xylem and other phloem.

 (i) Xylem: Consists of vessels, Tracheids and other xylem tissues. The interconnected vessels and Tracheids form a continuous system of water conducting channels reaching all parts of the plant. Xylem conducts water and dissolved minerals absorbed from the soil.

 (ii) Phloem: Consists sieve tubes and companion cells which conduct soluble products of photosynthesis from leaves to different parts of the plant body. This transportation of prepared food material through phloem is called translocation.

 4. How are water and minerals transported in plants?

 Ans. Water and dissolved minerals are transported through xylem tissues. Xylem vessels and Tracheids of roots, stems and leaves are interconnected to form a continuous system of water and dissolved minerals conducting channels reaching all the parts of the plant.

The root cells in contact of with the soil actively take up ions from the soil. This creates a difference in the concentration of these ions between the root and the soil. Therefore, there is steady movement of water into the xylem. This creates a column of water that is steadily pushed upward, called root pressure. However, this pressure is not enough to push water and dissolved minerals to leaves in the tall trees. There is continuous loss of water through stomata of the leaves in the form of water vapour. This process is called transpiration. The water lost due to transpiration is taken up from the xylem vessels and Tracheids in the leaves. This loss of water during transpiration creates a suction pressure which pulls water from the xylem cells of root. As a result, there is enhanced absorption and upward movement of water and dissolved minerals from roots to the leaves due to transpiration.

Fig. Loss of water vapour due to transpiration causing pull of water from the xylem cells of root

 The effect of root pressure in transport of water is more important at night when stomata are closed. When the stomata are open during the day, the transpiration pull is the major force in the movement of water and dissolved minerals in the xylem, especially in trees.

 5. How is food transported in plants?

Ans. Plants transport soluble product of photosynthesis like carbohydrates, amino acids and other substances through phloem to storage organs of roots, fruits and seeds and also to growing organs. This transport of soluble product of photosynthesis through phloem is known as translocation. The transport of prepared food and other substances takes place both in upward and downward directions. This movement of food materials is carried out in phloem by sieve tubes with the help of companion cells.

 Mechanism of translocation- translocation is an active process and utilizes energy. Material like sucrose is transferred from leaf cells or from the site of storage into phloem tissue. This process requires energy which is provided by ATP (adenosene triphosphate) molecules. Entry of sucrose into phloem tissue causes increase in osmotic pressure, as a result water from outside moves into phloem. The osmotic pressure moves the dissolved material in the phloem to tissues which have less pressure. Thus, material moves in phloem, mainly in sieve tubes, to place of need in the plant body. For example, sugar is translocated from its storage organ, root or stem to growing buds which need energy.

 

6. Describe the structure and functioning of nephrons.

 Ans. Structure of Nephron: Nephron is the filtration unit of kidney. It consists of a tubule which is connected with collecting duct at one end and a cup shaped structure at the other end. This cup-shaped structure is called Bowman's capsule. Every Bowman's capsule contains a cluster of capillaries, called Glomerulus, within the cup-shaped structure. The blood enters into glomerulus through afferent arteriole of renal artery and leaves it through efferent arteriole.

                               

                                                             Fig. Structure of nephron

 

(4) Functioning of nephron: Filtration of blood takes place in Bowman's capsule from the capillaries of glomerulus. The filtrate passes into the tubular part of the nephron. This filtrate contains glucose, amino acids, urea, uric acid, salts and a major amount of water.

 (ii) Reabsorption: As the filtrate flows along the tubule, useful substances such as glucose, amino acids, salts and water are selectively reabsorbed into the blood by capillaries surrounding the nephron tubule. The amount of water reabsorbed depends on the need of the body and also on the number of wastes to be excreted.

 (iii) Urine -The filtrate which remained after reabsorption is called urine. Urine contains dissolved nitrogenous waste i.e., urea and uric acid, excess salts and water. Urine is collected from nephrons by the collecting dust to carry it to the ureter.

 2. What are the methods used by plants to get rid of excretory products?

 Ans. To get rid of excretory products plants use the following ways:

(i) In many plants waste products are stored in vacuoles of the cells. Plant cells have comparatively large vacuoles.

 (ii) Some waste products are stored in the leaves. They are removed as the leaves fall off.

 (iii) Some waste products such as resins and gums are stored, especially in non- functional old xylem.

 (iv) Some waste products such as tannins, resins, gums are stored in bark, thereby removed as the peeled off them.

 (v) Plants also excrete some waste substances through roots into the soil around         

 3. How is the amount urine produced regulated?

 Ans. The amount of urine largely depends on the amount of water reabsorbed. The amount of water reabsorbed by the nephron tubule depends on:

(i) How much water is in excess in the body need to be removed?

Ans-When water is abundant in the body tissues, large quantities of dilute urine is excreted out. When water is less in quantities in the body tissues, a small quantity of concentrate urine is excreted.

 (ii) How much dissolved wastes, especially nitrogenous wastes, like urea and uric acid and salts needs to be excreted from the body?

 Ans-  When there is more quantity of dissolved wastes in the body, more quantity of water is required to excrete them. So, the amount of urine produced increases. The amount of urine produced is also regulated by certain hormones which control the movement of water and Na⁺ ions into and out of the  nephrons.

 Q. 1. The kidneys in human beings are a part of the system for

(a) nutrition    (c) excretion   (b) respiration   (d) transportation.

Ans. (c) excretion.

 Q. 2. The xylem in plants is responsible for

(a) transport of water                (b) transport of food

(c) transport of amino acids  (d) transport of oxygen.

Ans. (a) transport of water.

 Q. 3. The autotrophic mode of nutrition requires

(a) carbon dioxide and water   (c) sunlight

(b) chlorophyll    (d) all of the above.

Ans. (d) all of the above.

 Q. 4. The breakdown of pyruvate to give carbon dioxide, water and energy takes place in

(a) cytoplasm   (c) chloroplast   (b) mitochondria    (d) nucleus.

Ans. (b) mitochondria.

 Q. 5. How are fats digested in our bodies? Where does this process take place?

Ans. Digestion of fat takes place in the small intestine.

Digestion of Fat: The fats are present in the form of large globules in the small intestine. Fats digesting enzymes are not able to act upon large globules efficiently.

 Bile juice secreted by the liver is poured in the intestine along with pancreatic juice. The bile salts present in the bile juice emulsify the large globules of fats. So, by emulsification large globules break down into fine globules to provide larger surface area to act upon by the enzymes.

Lipase enzyme present in the pancreatic juice causes break down of emulsified fats. Glands present in the wall of small intestine secrete intestinal juice which contains lipase enzyme that converts fats into fatty acids and glycerol.

    Q. 6. What is the role of saliva in the digestion of food?

Ans. Role of saliva in digestion of food: The saliva contains salivary amylase enzyme that breaks down starch to sugar like    starch (complex molecule) + salivary amylase → sugar (simpler molecules)

 (ii) The saliva moisten the food that help in chewing and breaking down the big pl of food amylase can digest the food into a lore

 Q.7. What are the necessary conditions for autotrophic nutrition and what are its byproducts?

Ans. Necessary conditions for autotrophic nutrition:

(i) Presence of chlorophyll in the living cells.

 (ii) Provision of supply of water to green parts or cells of the plant either through roots or by surrounding environment.

 (iii) Availability of sufficient sunlight to provide light energy required to carry out photosynthesis.

 (iv) Sufficient supply of carbon dioxide which is one of the important components for the formation

      of carbohydrates during photosynthesis.

 Q. 8. What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration.

Ans.    Aerobic respiration

1. Aerobic respiration takes place in the presence of free oxygen.

2. In aerobic respiration complete oxidation of glucose takes place.

3. End products of aerobic respiration are CO. water and energy.

water and energy.

            C₆H₁₂O₆ + 60₂ → 6H₂O + 6CO₂ + Energy   (686 kcal)

4. Large amount of energy is released i.e., 38 molecules of ATP per glucose molecule.

5. First step of aerobic respiration (glycolysis) takes place in cytoplasm while second (Krebs's cycle) and third (electron transport chain) steps take place in mitochondria.

Anaerobic respiration

1. It takes place in the absence of free oxygen,

2. In anaerobic respiration the glucose molecule is incompletely  broken down.

3. End product of anaerobic respiration are ethyl alcohol (or lactic acid), CO₂ and a little energy.

              C₆H₁₂O₆ → 2CO₂ + 2C₂H₂OH + Energy (58 kcal),

4. Small amount of energy is released Le., 2 ATP molecules per glucose molecule.

5. Complete anaerobic respiration occurs in cytoplasm.

 Organisms which use anaerobic respiration are yeasts, bacteria and parasites like tape worm, Ascaris etc.

  Q. 9. How are the alveoli designed to maximise the exchange of gases?

 Ans. The alveoli are thin walled and richly supplied with a network of blood vessels to facilitate exchange of gases between blood and the air filled in alveoli.

 (ii) Alveoli have balloon-like structure. Thus, provides maximum surface for exchange of gases.

 10. What would be the consequences of a deficiency of hemoglobin in bodies?

 Ans. The average hemoglobin content of blood, irrespective of sex, is 14.5 g per 100 ml. If hemoglobin content of deficient a blood, its oxygen carrying capacity de 145 g per 100 person shows symptoms of deficient oxygen such as breathleness, often one of the first indications of iron deficiency, anemia.

 Q. 11. Describe double circulation in human beings. Why is it necessary?

 Ans. Double circulation: In human beings the blood goes through the heart twice during each cycle i.e., the blood passes through the human heart two times to supply once to the whole body. So, it is called double circulation of blood.

 The double circulation of blood includes

(i) Systemic circulation and

(ii) Pulmonary circulation

 (i) Systemic circulation: It supplies oxygenated blood from left auricle to left ventricle thereby pumped to various body parts. The deoxygenated blood is collected from the various body organs by the veins to pour into vena cava and finally into right atrium (auricle). Right atrium transfers the blood into the right ventricle.

                              

                               Fig. Double circulation of blood in human heart

 (ii) Pulmonary circulation: The deoxygenated blood is pushed by the right ventral into the lungs for origination. The oxygenated blood is brought back to left atrium of the human heart. From left atrium the oxygenated blood is pushed into the left ventricle The left ventricle pumps oxygenated blood into aorta for systemic circulation.

 Necessity of double circulation: The right side and the left side of the human heart is useful to keep deoxygenated and oxygenated blood from mixing. This type separation of oxygenated and deoxygenated blood ensures a highly efficient supply of organ to the body. This is useful in case of humans which constantly require energy to maintain their body temperature.

 Q. 12. What are the differences between the transport of materials in xylem and phloem?

  Ans. The differences are as follows:    Xylem

   a. Xylem conducts water and dissolved minerals from roots to leaves  and other parts.

   b. In xylem, the transport of material take place through vessels and Tracheids which are dead tissues.

   c. In xylem, upward movement of water and dissolved minerals is mainly achieved by transpiration         pull. It is caused due to suction created by evaporation of water molecules from the cells of leaves.

   d. Movement of water is achieved by simple physical forces, there is no expenditure of energy. So ATP molecules are not required.

                                                                  Phloem

   a. Phloem conducts prepared food material from leaves to other parts of plant in dissolved form.

   b. In phloem, transport of material takes place through sieve tubes with the help of companion cells,          which are living cells.

   c. In translocation, material is transferred into phloem tissue using energy from ATP. This increases             the osmotic pressure that moves the material in the phloem to tissues which have less pressure.

   d. The translocation in phloem is an active process and requires energy. This energy is taken from                 ATP.

   Q. 13. Compare the functioning of alveoli in the lungs and nephrons in the kidneys with respect to             their structure and functioning.

    Ans. Comparison between Alveoli and Nephron

                                                              Alveoli

   a. Alveoli have thin-walled balloon-like structure. Surface is fine and delicate.

   b .Alveoli are supplied with extensive network of thin-walled blood vessels i.e., capillaries for                  exchange of gases.

  c. Alveoli increase surface area for diffusion of CO, from blood to air and O, from air to blood.

  d. Alveoli only provide surface for exchange of gases in the lungs.

  e. Alveoli are very small and a large number of them are present in each lung.

                                                           Nephron

a. Nephron have thin-walled cup shaped structure attached with thin-walled tubule.

b. Bowman's capsule is supplied with a cluster of capillaries, called glomerulus for filtration A network      of blood vessels around the tubular part of nephron for reabsorption of useful substances and water.

c. Nephrons also increase surface area for filtration of blood and reabsorption of useful substances and      water from filtrate leaving behind urine.

d. Tubular part of nephron also carries the urine to collecting duct.

e. A large number of nephrons, basic filtration unit, are present in each kidney.

 

 

 

 

 

 

ELECTROCHEMISTRY 12th

 

                                                     YASH CLASSES   

                                             ELECTROCHEMISTRY

                                NCERT INTEXT UNSOLVED QUESTIONS

 Q.2.1. How would you determine the standard electrode potential of Mg²⁺ | Mg?

 Ans. We will set up a cell consisting of Mg I MgSO₄ (1 M) as one electrode (by dipping a magnesium wire in 1M MgSO₄, solution) and standard hydrogen electrode Pt, H₂ (1 atm) H1 (1 M) as the second electrode and measure the EMF of the cell and also note the direction of deflection in the voltmeter. The direction of deflection shows that electrons flow from magnesium electrode to hydrogen electrode, i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode. Hence, the cell may be represented as:    

     Mg²⁺ | Mg (1 M) || H⁺ (1 M) | H₂, (1 atm), Pt

             E⁰_cell = E⁰_H+, _1/2H2 - E⁰_Mg 2+, _Mg

    Put E⁰_H+, _1/2H = 0, therefore E⁰_Mg 2+, _Mg = -E⁰_cell

 [ where M = molarity]

Q.2.2. Can you store copper sulphate solution in a zinc pot?

 Ans. Zinc is more reactive than copper. Hence, it displaces copper from copper sulphate solution as follows:

    Zn (s) + CuSO₄ (aq)→ ZnSO₄ (aq) + Cu (s) Thus, zinc reacts with CuSO₄ solution. Hence, we cannot store copper sulphate solution in a zinc pot.

Alternatively. E _Zn2+, _Zn = - 0.76 V,

  E _Ca2+, _Ca= 0.34 V To check whether zinc reacts with CuSO₄, solution, le., whether the following reaction takes place or not Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s), find the EMF of this cell reaction.

  The cell may be represented as: Zn | Zn²⁺|| Cu²⁺ | Ca

        E⁰_cell = E _Ca2+, _Ca- E _Zn2+, _Zn = 0.34 V-(-0.76 V) = 1.10 V

        E⁰_cell is positive, the reaction takes place and we cannot store.

Q. 2.3. Consult the table of the standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions.

Ans. Oxidation of ferrous ions means that the following reaction should occur:

            Fe²⁺ → Fe³⁺ + e^-,       E⁰_ox = -0.77V

Only those substances can oxidize Fe²⁺ to Fe³ which are stronger oxidizing agents and have positive reduction potentials greater than 0.77 V so that EMF of the cell reaction is positive. This is so for elements lying below Fe³⁺/Fe²⁺ in the electrochemical series, e.g., Br₂, Cl₂ and F₂.     

Q.2.4. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

 

Ans. For hydrogen electrode, H⁺+ e^-→ [1/2]H₂ Applying Nernst equation. E_H+, _1/2H = E⁰_H+, _1/2H - (0.0591 /n) log [1/ _H+]

                  = 0 – [0.0591/1]log [1/ ₁₀-10] 

{pH= 10 means[H⁺] = 10^-10 M} = - 0.0591 x 10 = -0.0591V

 

Q. 2.5. Calculate the emf of the cell in which the reaction

takes following place

      Ni(s)+2 Ag⁺²(0.002 M) →N⁺² (0.160M) + 2Ag(s)

 

Ans. Applying to the given cell reaction,

      E _cell = E⁰_cell – [0.059/2] log [ Ni ⁺²/Ag ⁺]

              = 1.05 – [0.059/2]log [0.160/(0.002)²]

 

Q. 2.6. The cell in which following reaction occurs –

         2 Fe ⁺³(aq)+2 I^-(aq) → 2 Fe⁺² (aq) + I₂(s) , has

          E⁰_cell = 0.236V at 298 K. calculate the standard Gibbs

      Energy and equilibrium constant of the cell.

     Ans.  ∆G⁰= -nFE⁰, = -2x96500x0.236 J

            = -45548 J = -45.548 kJ.

         K = Antilog [n E ⁰ / 0.054] = Antilog [2x0.236/0.059]

             = 10 ⁸ . 

 

Q. 2.7. Why does the conductivity of a solution decrease with dilution?

Ans-

K= Antilog (7.983) =9.616 x 10⁷ dilution, the number of ions per unit volume decreases. Hence, the conductivity decreases.

 

  Q.2.8 Suggest the method to determine the ꓥ⁰_m value

            of water.

 Ans-      ꓥ⁰_m (H₂O) = λ⁰ _H⁺ + λ⁰ _{o H}^-.

                                 = ꓥ⁰_m (HCl) + ꓥ⁰_m (NaOH) + ꓥ⁰_m (NaCl)

       

  Q.2.9. The molar conductivity of 0.025 mol L^-1. Methanoic

 acid is 46.1 S cm² mol^-1. Calculate its degree of dissociation

   constant. Given λ⁰ _H⁺ = 349.6 S cm² mol^-1.   and

     λ⁰ _HCOO^- = 54.6 S cm² mol^-1.

  Sol.- ꓥ⁰_m (_Hcoo ^-) = λ⁰ _H⁺ + λ⁰_HCOO^-.

                             = 349.6 + 54.6 = 404.2

                     α = [ ꓥ^c_m /ꓥ⁰_m] = 46.1/ 404.2= 0.114

         K_α = [ C α²]/ [1 – α] = [0.025 x0.114 x0.114]/[1 – 0.114]

                            = 3.67 x 10 ^-4.

Q. 2.10. If a current of 0.5 Ampere flows through a metallic

 wire for 2hour then how many electrons would flow

 through the wire.

 Ans- Q = ne =It, n = number of electrons, e= charge of electron

  n = It /e = [0.5 x 2 x 60x 60]/ [1.6 x 10 ^-19] = 2.25 x 10 ²².

 

 Q.2.11 Suggest a list of metals that are extracted electrolytically.

  Ans.- The metals like Na, K, Mg, Al, Ca etc.

        i.e., reactive metals Can be extracted electrolytically.

 

Q.2.12. discovery what is the quantity of electricity in coulomb's needed to reduce one mole of

                  Cr₂O^2-+ 14H → 2Cr³⁺ + 8H₂O       

Ans. 1 mol Cr₂O^2- requires 6moles electron

   For reduction.

  Required charge = 6 x 96500 C

                                  = 57900 C

Q.2.13. write the chemistry of recharging the lead storage battery, highlighting all materials that are involved during the recharging?

Ans. A lead storge battery consists of anode of lead packed

with lead dioxide PbO₂ and 38% solution acid as electrolyte

when the battery is in the use, the following reactions take

 place:

    At anode:          Pb (s) + SO₄^2- → PbSO₄(s) + 2e^-.

    At cathode:     PbO₂ (s) + SO₄^2- (aq) + 4H⁺ (aq) + 2e^- →  PbSO₄(s) + 2H₂O.

Over all reaction: Pb (s) + PbO₂ (s) + 2H₂SO₄(aq) →  2PbSO₄(s) + 2H₂O

On charging the battery, the reverse reaction takes place, i.e., PbSO₄ deposited on the electrodes is converted back into Pb and PbO₂ and H₂SO₄ is regenerated. Reverse of reaction

(i) will be reduction and hence will take place at cathode. Reverse of reaction

(ii)will be oxidation and hence will take place at anode.

 

Q. 2.14 Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

Ans. Methane and methanol.

 

Q.2.15 Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

 Ans. The water layer present on the surface of iron (especially in the rainy season) dissolves acidic oxides of the air like CO₂, SO₂ etc. to form acids which dissociate to give H⁺ ions

         H₂O + CO₂→ H₂CO₃ + 2H⁺ + CO^3-

         In the presence of H⁺ ions, iron starts losing electrons at some spot to form ferrous ions, i.e., its oxidation takes place. Hence, this spot acts as the anode:

           Fe (s) → Fe²⁺ (aq) + 2 e^-.

 The electrons thus released move through the metal to reach another spot where H+ ions and the dissolved oxygen take up these electrons and reduction reaction takes place. Hence, this spot acts as the cathode:

         O₂ (g) + 4 H⁺ (aq) + 4 e^- →2 H₂O (1)

The overall reaction is: 2 Fe (s) + O₂(g) + 4H⁺ (aq) → 2Fe²⁺ (aq) + 2 H₂O (1)

Thus, an electrochemical cell is set up on the surface.

Ferrous ions are further oxidized by the atmospheric oxygen to ferric ions which combine with water molecules to form hydrated ferric oxide, Fe₂O₃.x H₂O, which is rust.