Determining the solubility of sparingly soluble salts Solubility = k x 1000 / Λ m mol L -1 where k is the conductivity

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1 Unit -3- ELECTROCHEMISTRY Points To Remember 1.. Electrochemistry may be defined as the branch of chemistry which deals with the quantitative study of inter relationship between chemical energy and electrical energy and inter-conversion of one form into another 2.A cell is of two types:- I. Galvanic Cell II. Electrolytic cell. In Galvanic cell the chemical energy of as spontaneous redox reaction is converted in to electrical work. In Electrolytic cell electrical energy is used to carry out a redox reaction. 3.Resistance R is mathematically related to length and area of cross section of a conductor as: R = ρl/a Where ρ known as resistivity and l/a as cell constant 4. The reciprocal of resistivity is called conductivity. The conductivity, k of an electrolytic solution depends on the concentration of the electrolyte, nature of solvent and temperature and it is defined as the conductance of a solution having unit volume. 5.Molar Conductivity λ m, is defined as conductivity of an electrolytic solution having 1 molar concentration placed between two electrodes in such a way that the distance between the electrodes is 1 cm and the area of cross section is I cm 2 Mathematically λ m = k/c where C is the concentration in mol lit -1 λ m = The unit of molar conductivity is Ω -1 cm 2 mol Conductivity decrease but molar conductivity increases with decrease in concentration. It increases with dilution for both strong electrolytes and weak electrolytes The increase in molar conductivity of a strong electrolyte is mainly because of increase in mobility of ions whereas the increase in weak electrolytes it is due to increase in number of ions as well as mobility of ions. 7.Kohlrausch s law of independent migration of ions: The law states that molar conductivity of an electrolyte at infinite dilution can be represented as the sum of the individual contribution of the anion and cation of the electrolyte at infinite dilution λ m = λ ( cation) + λ ( anion) Applications: Calculation of degree of Dissociation of Weak Electrolytes: c = Λ m / Λ m Λ m c = molar conductivity of solution at any concentration Λ m = limiting molar conductivity Determining the solubility of sparingly soluble salts Solubility = k x 1000 / Λ m mol L -1 where k is the conductivity Calculation of Dissociation Constant of Weak Electrolyte K = C 2 / ( 1- ) Also = Λ m c / Λ m K = C( Λ m ) 2 / Λ m (Λ m - Λ m c )

2 8.The Standard Electrode Potential for any electrode dipped in an appropriate solution is defined with respect to standard electrode potential of hydrogen electrode taken as zero volts. The standard potential of the cell can be obtained by taking the difference of the standard potentials of cathode and anode (one of that is hydrogen electrode). 9.The standard potential of the cells are related of standard Gibbs energy. rg 0 = -nf 10. The standard Gibbs energy of the cells is related to equilibrium constant. G= RT log K c 11.Nerst Equation: Concentration dependence on the potentials of the electrodes and the cells are given by Nernst equation. aa+ bb cc + dd [A] a [B] b Nernst equation can be written as E cell = log 12.Faraday s laws of Electrolysis: I. The amount of chemical substance deposited at any electrode during electrolysis by a current is proportional to the quantity of charge passed through the electrolyte. m= ZIt = g atomic wt x I x t n F where n = number of electrons gained or lost F = Faraday s constant I = current in amperes T = time in seconds II. The amount of different substances liberated by the same quantity of electricity passing through the different electrolytic solution connected in series is proportional to their chemical equivalent weights. W1/E1 =W2/E2 Where E = equivalent weight of the metal = atomic weight/valency 13.Electrochemical series: The arrangement of elements in the order of increasing reduction potential values is called electrochemical series. Applications of electrochemical series: 1. Calculation of EMF of the cell E 0 cell = E 0 cathode E 0 anode 2. Predicting feasibility of the reaction : A redox reaction is feasible if electron releasing species is present at a position lower in the series as compared to electron accepting species. 3. Comparison of relative reducing or oxidising power of the elements: The reducing character of the elements increases down the series and its oxidising character decreases. 4. To predict the reaction of a metal with dilute acids to liberate hydrogen gas :

3 Metals which lie above hydrogen in the electrochemical series can reduce H + ions to hydrogen. Corrosion of metals is an electrochemical phenomenon. In corrosion metal is oxidized by loss of electrons to oxygen and formation of oxides. Anode(Oxidation): 2Fe(s) -- 2Fe 2+ +4e - Cathode(Reduction): O 2 (g)+ 4H + (aq)+4e - 2H 2 O Atmospheric Oxidation: 2Fe 2+ (aq)+2h 2 O(l)+1/2O 2 (g) Fe 2 O 3 (s)+ 4H + (aq) 14.Products of electrolysis : 1. Electrolysis of Aqueous solution of NaCl. Reaction at anode: Cl - (aq) 1/2Cl 2 (g) + e - Reaction at cathode: H 2 O (l) + 2e - ½ H 2 (g)+ OH - (aq) Thus, Cl 2 gas is liberated at the anode whereas H 2 gas is liberated at the cathode. 2. Electrolysis of Copper Sulphate Solution Using Platinum Electrodes. Reaction at anode: H 2 O (l) 1/2O 2 (g)+2h + (g) + 2e - Reaction at cathode: Cu 2+ (aq) +2e Cu(s) Thus, copper is deposited at the cathode, and O 2 gas is liberated at the anode. 3. Electrolysis of Copper Sulphate Solution Using Copper Electrodes. Reaction at anode: Cu(s) Cu 2+ (aq) + 2e - Reaction at cathode: Cu 2+ (aq) + 2e Cu(s) Thus, copper dissolves at the anode and is deposited at thecathode 15.Fuel Cells cells where reactants are continuously supplied the energy from this reaction can be used to heat and run machines one type is the hydrogen-oxygen fuel cell hydrogen gas is pumped in at the anode while oxygen gas is pumped in at the cathode (which both have a lot of surface area) pressure is used to push the H 2 through a platinum catalyst which splits the H 2 into 2H + and 2e - (In the cell, hydrogen &oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide solution. Catalyst like finely divided platinum or palladium metal are incorporated into the electrodes for increasing the rate of electrode reaction.) the 2e - move through an external circuit towards the cathode generating electrical energy the O 2 is also pushed through the platinum catalyst forming two oxygen atoms the H + ions and oxygen atoms combine to form water.

4 CELL REACTION At cathode: O 2 +2H 2 O+4e - 4OH - At anode : 2H 2 +4OH - 4H 2 O+4e - Overall reaction: 2H 2 O +O 2 2H 2 O 16.PRIMARY CELLS: again. In the primary cells, the reaction occurs only once and after use over a period of time battery becomes dead and cannot be reused MERCURY CELL Suitable for low current devices like hearing aids, camera & watches. Anode: Zn-Hg amalgam. Cathode: A Paste of mercury(ii)oxide & carbon. Electrolyte: A moist paste of KOH-ZnO. CELL REACTION: At anode: Zn(Hg) +2OH - ZnO +H 2 O+2e - At cathode: HgO +H 2 O +2e - Hg + 2OH - Net reaction: Zn(Hg) +HgO ZnO + Hg The cell potential is approximately 1.35V & remain constant during its life as the overall reaction does not involves any ion in solution whose concentration can change during its life time. 17.SECONDARY CELL:A secondary cell after use can be recharged by passing current through it in opposite direction so that it can be used again. Eg. Lead storage cell & Ni-Cd cell LEAD STORAGE CELL ANODE: Spongy lead. CATHODE:A grid of lead packed with lead(iv) oxide Electrolyte: A 38% solution of sulphuric acid having density of 1.30gm/ml. CELL REACTION(Discharging) Anode: Pb +SO PbSO 4 + 2e - ;E 0 oxid=0.36v Cathode: PbO 2 +SO H + +2e PbSO 4 +2H 2 O E 0 oxid=1.69v Pb+PbO 2 +2H 2 SO 4 2PbSO 4 +2H 2 O ; E 0 cell=2.05v From the above equation,it is obvious that H 2 SO 4 is used up during discharging. As result of density of H 2 SO 4 falls below 1.20gm/ml,the battery needs recharging.

5 DURING RECHARGING (CHARGING) The battery reaction is reversed PbSO 4 on anode & cathode is converted in to Pb & PbO 2 respectively. Anode: PbSO 4 +2e- Pb+ SO 4 2- Cathode: PbSO 4 +2H 2 O PbO 2 +SO H + + 2e - PbSO 4 +2H 2 O Pb+PbO 2 +4H + +2SO 4 2- Very short answer type questions (1 marks) Q.1.How does the molar conductivity of strong electrolyte vary with increase in concentration? Ans 1. For strong electrolyte molar conductivity decreases with increase in concentration. Q.2.Suggest a metal which can be used for cathodic protection of iron? Ans 2. Zinc Q.3.What is the quantity of electricity in coulombs needed to reduce 1 mole of Cr 2 O 7 2-? Ans 3. Reduction of Cr 2 O 7 2- is as follows Cr 2 O H + + 6e - 2Cr H 2 O Since 6 moles of electrons are required to reduce 1 mole of Cr 2 O 2-7. Hence electricity in coulombs is = 6*96500 C = 5.79*10 5 C Q.4.What is meant by cell constant? Ans 4.Cell constant = l/a Where, l = distance between two electrode a = cross-section area of electrode Q.5.How is the cell potential related to free energy change? Ans 5. G = -nfe cell Q.6.Calculate the E 0 cell for the following electrode reaction Zn 2+ (aq) + 2e - Zn(s); E 0 = -0.76V - Cd 2+ (aq) + 2e Cd (s); E 0 = -0.40V Ans 6. E 0 cell = E 0 cathode - E 0 anode = (-0.76) = 0.36V (Note:- More E 0 metal acts as cathode) Q.7.Can you store CuSO 4 solution in a zinc pot? Ans 7. No, because zinc is more reactive than copper & it will displace copper from CuSO 4 solution. Q.8.How does concentration of sulphuric acid changes in Lead Storage Battery when current is drawn from it? Ans 8. Concentration of sulphuric acid decreases.

6 Q.9.Suggest two materials other than hydrogen that can be used as fuels in Fuel Cell? Ans 9. CO & CH 4 Q.10.HCl does not give an acidic solution in Benzene? Ans10. It does not give H + in benzene(a non-polar solvent) Q11. How can you increase the reduction potential of an electrode.? Ans For the reaction It can be increased by a. Increase in concentration of Mn+ions in solution b. By increasing the temperature Q12.. Define corrosion. Write chemical formula of rust. Ans Corrosion is a process of destruction of metal as a result of its reaction with air and water, surrounding it. It is due to formation of sulphides, oxides, carbonates, hydroxides, etc. Formula of rust-fe 2 O 3..XH 2 O Q13.. What are the products of electrolysis of molten and aqueous sodium chloride? Ans Molten sodium chloride:- Na, Cl 2, and aqueous sodium chloride:-h 2 and Cl 2. Q14. Why is it not possible to determine the molar conductivity at infinite dilution for weak electrolytes by extrapolation? Ans Because the molar conductivity at infinite dilution for weak electrolytes does not increase linearly with dilution as for strong electrolytes SHORT ANSWER TYPE QUESTION (2 marks) Q.1.Write the electrolytic product of CuSO 4 solution using Cu- electrode? Ans1. CuSO 4 Cu SO 4 At Cathode; Cu e Cu(s) At Anode; Cu(s) Cu 2+ (aq) + 2e - Q.2.In the button cell widely used in watches & in other devices ; the following reaction takes place; Zn(s) + Ag 2 O(s) + H 2 O(l) Zn 2+ (aq) + 2Ag(s) Determine E 0 cell & G o for the reaction? Given, E 0 Ag+/Ag = 0.344V & E 0 Zn2+/Zn = -0.76V. Ans2. E 0 cell = E 0 cathode - E 0 anode = ( ) = 1.104V G = -nfe cell = -2*96500*1.104 J = J Q.3.If a current of 0.5 ampere flows through a metallic wire for 2 hrs,then how many electrons would flow through the wire?

7 Ans3. Charge = Current(I) X Time(t) = 0.5X2X60X60 = 3600 C Since,1 F(96500 C) current contain 6.022X10 23 Hence,3600 C will contain 6.02X10 23 X3600/96500 = 2.246*10 22 electrons Q.4.The molar conductance at infinite dilution for sodium acetate, hydrochloric acid & sodium chloride are 91.0,425.9 & Scm 2 mol -1 respectively at 298 K. Calculate the molar conductance of acetic acid at infinite dilution. Ans4. λ m(ch 3 COOH) = λ m(ch 3 COONa) + λ m(hcl) - λ m(nacl) = = Scm 2 mol -1 Q.5.Conductivity of m acetic acid is7.896*10-5 Scm -1.Calculate it s molar conductivity & if λ m for acetic acid is Scm 2 mol -1.What is it s dissociation constant. Ans5. λ m = Κ*1000/C = 7.896*10-5 *1000/ = Scm 2 mol -1 Now, α = λ c m/ λ m Since, Κ = C α 2 /(1- α) = *(0.0839) 2 /( ) = 1.85*10-5 Q 6.Three iron sheets have been coated separately with three metals A,B and C whose standard electrode potentials are given below. Metal A B C Iron E 0 _ 0.46V V V -0.44V Identify in which case rusting will take place faster when coating is damaged. Ans. As iron (--0.44V) has lower standard reduction potential than C(--0.20V) only therefore when coating is broken, rusting will take place faster. Q 7. Tarnished silver contains Ag2S.Can this tarnish be removed by placing tarnished silver ware in an aluminium pan containing an inert electrolytic solution such as NaCl?. The standard electrode potential for the half reactions are : For Ag2S(s) + 2e Ag(s) + S2-, it is V and For Al3+ 3e Al(s),it is V Ans.Tarnish will be removed if the following reaction takes place: Al + Ag2S Al3+ + 2Ag + S2- EMF of this cell reaction = (--1.66)V =0.95V As EMF is positive, the reaction will take place and tarnish will be removed. Q.8.Is it possible to measure single electrode potential? Ans. NO,Oxidation or reduction can not take place alone. Moreover, Electrode potential is a relative tendency and can be measured with Respect to a reference electrode only. Q9. Why does a galvanic cell become dead after some time?

8 Ans. As the reaction proceeds concentration of ions in anodic half keeps on increasing while in the cathodic half it keeps on decreasing. Hence their electrode potentials also keep on changing till ultimately they become equal and then e.m.f. of cell becomes zero. Q10.In each of the following pairs which will allow greater conduction of electricity and why?. (a)silver wire at 20 0 C same silver wire 50 0 C (b)nacl solution at 20 0 C same solution at 50 0 C. (c)nh 4 OH at 20 0 C same at 50 0 C. (d)0.1 M acetic acid solution,1m acetic solution. Ans- (a) silver wire at 20 0 C because with increase in temperature, metallic conduction decreases due to vibration of kernel. (b) NaCl solution at 50 0 C because in case of strong electrolyte with increase in temperature the ionic mobilities increases. (c) NH 4 OH at 50 0 C because in case of weak electrolyte dissociation increases with increase in temperature. (d)0.1 M acetic acid because with dilution, dissociation / ionization increases. Q11. Copper is conducting as such while copper sulphate is conducting only in molten state or in aq. Solution. Explain. Ans- Copper is conducting as such because it contains free electrons. CuSO 4 is conducting in molten state or in aq. Solution because it gives ions only in the molten state or aqueous solution. SHORT ANSWER TYPE QUESTION (3marks) Q.1 State two advantages of H 2 -O 2 Fuel Cell over ordinary cell. Ans 1.High Efficiency:-The fuel cell converts the energy of fuel directly into electricity & therefore their efficiency is high. 2. Pollution Free Working:- No pollution is caused. 3.Continuous Source of Energy:- No electrode material is said to be replaced.the fuel cell can be fed continuously to produce water. Q.2.How many moles of electrons are required to i)reduce 1 mole of MnO 4 - to Mn 2+. ii)produce10.0 gm of Al from molten Al 2 O 3. Ans 2. i) MnO e - Mn 2+ - Hence,5 moles of electrons are required for conversion of MnO 4 ii) Al e - Al(s) since,1 mole or 27 gm of Al is produced by 3 mole of e - 10 gm will produce 3*10/27 = 1.11 mole of e - to Mn 2+. Q.3 Depict the Galvavanic cell in which the reaction Zn(s) + 2Ag + (aq) Zn 2+ (aq) + 2Ag(s) Takes place. State the reaction taking place at each of it s electrode & also state the carriers of current within this cell.

9 Ans. Cell representation; Zn(s)/ Zn 2+ (aq) Ag + (aq)/ag(s) Cell reaction; At anode, Zn(s) Zn e - At cathode, Ag + + e - Ag(s) The carriers of current within this cell are from silver to zinc Q.4.The resistance of a conductivity cell containing 0.001M KCl solution at 298K is 1500Ω.What is the cell constant if the conductivity of 0.001M KCl solution at 298K is 0.146*10-3 Scm -1? Ans. Cell constant = Conductivity/Conductance = Conductivity*Resistance = 0.146*10-3 *1500 = cm -1 Q.5.Calculate the equilibrium constant of the reaction ; Zn + Cd 2+ Given, E 0 cell =0.36V Ans. For the reaction at equilibrium, E cell = 0. Therefore, E cell = E 0 cell (0.059/n)log K c E 0 cell = (0.059/n)log K c 0.36=(0.059/2)log K c log K c = 0.36*2/0.059 = K c = Antilog = 1.52*10 12 Zn 2+ + Cd Q.6. The electrical resistance of a columns of 0.05M NaoH solution Of diameter 1cm & length 50cm is ohm. Calculate its. (a) resistivity. (b) Conductivity. (c) Molar conductivity. Ans. Cell Constant = l/a Area of cross section (a)= πr 2 = 3.14 (1/2) 2 = cm 2 Cell constant = 50/0.785 = cm -1 (a) Resistivity e = R 1/cell constant = / = Ωcm (b) Conductivity κ = 1/e = 1/ = s cm -1 (c) Molar conductivity (λ m ) = κ 1000/c = /0.05 = cm -2 mol -1 Q.7. The cell in which the following reactions occurs : 2Fe +3 (aq) + 2I (aq) 2Fe +2 (aq)+ I 2 (s) Has E 0 cell = 0.236v at 298 k. Calculate the ΔG 0 and Equilibrium constant of cell reaction?

10 Ans. ΔG 0 = - nfe 0 cell = = ΔG 0 = RT log k c log k c = -ΔG/2.303RT = / = K c = Antilog = Q.8. Write the cell reaction which occurs in lead Storage battery: (a). During Discharging process? Ans. (a).during discharging: At anode: Pb(s) Pb +2 +2e - At cathode: PbO 2 +SO 2-4 (aq)+4h + +2e - PbSO 4 (s)+h 2 O Pb(s)+PbO 2 (s)+4h SO 4 2PbSO 4 (s)+2h 2 O(l) Q.9. (a) Explain the following: (1). Rusting of iron is quicker in saline water than in Ordinary water (2). Presence of CO 2 in water increases rusting (3). Iron does not rust even if zinc coating is broken in a galvanized iron pipe? Ans.(a).(1). Saline water increases the electrical Conductivity of electrolytic solution formed on the metal surface.hence rusting is quicker in Saline water. (2).CO 2 dissolve in water to form H 2 CO 3 which give H + ions accelerate the process of rusting as; At anode: Fe (s) Fe e - At cathode:o 2 + 4H + +4e - 2H 2 O (3). Zinc being more reactive than iron will sacrifice Itself for the shake of iron. As result atoms of zinc Will be oxidised in preference to the iron atoms. Q 10. Explain mechanism of rusting of iron? Ans Rusting of iron is based on electrochemical theory as it is assumed to be miniature electrochemical cell: At anode : Fe(s) Fe 2+ +2e - At cathode : H 2 O +CO 2 H 2 CO 3 H 2 CO 3 2H + + CO 2-3 H 2 O H + + OH - 4H + +O 2 + 4e - 2H 2 O 2Fe(s) + 4H + + O 2 2Fe H 2 O The Fe +2 ions are oxidized to Fe +3 ion which combine with molecules of H 2 O to form rust Fe 2 O 3.XH 2 O.

11 Q 11.(a).What are secondary cells? (b). Write the cell reaction which occurs in lead Storage battery during charging.. (c) Write the Nernst equation and calculate the e.m.f. of Cell at 298 K Cu/ Cu 2+ (0.130M) Ag + ( M) /Ag(s) E 0 cell= 0.46V Ans6..(a). The cell which can be recharged & be used again and again is called secondary cell. (b). The cell reactions during charging are: At anode : Pbso 4 (s) + 2H 2 O PbO 2 +SO H + +2e - At cathode : PbSO 4 (s)+ 2e - Pb(s) + SO 4 2-2PbSO 4 (s)+2h 2 O(l) Pb(s)+ PbO 2 (s)+4h + +2SO 4 2- (c)cell reaction is an : At anode : Cu(s) Cu e - At cathode: 2Ag - + 2e - 2Ag(s) Cu(s) + 2Ag + Cu Ag(s) E cell = /2 log 0.130/( ) -2 = / = = V Value based question Q.1 In a class room, a teacher was teaching about batteries in the chapter of Electrochemistry to 12 th class. As the use of batteries is very common now days for invertors and cell phones, students were very curious to know which battery they should use for their invertors at home and for their cell phones as a number of cheap as well as costly batteries are available in the market. The teacher advised them that they should use those batteries which do not cause any pollution or health hazard though these may be a little costlier. Thus, he suggested the use of cadmium battery in place of lead storage battery for inverter and lithium ion battery in place of Ni-Cd battery for cell phones. After reading the above paragraph, answer the following questions: i. What values are expressed by the teacher? ii. How is lead storage battery harmful in the long run? iii. How is Ni-Cd battery harmful? What are other advantages of using Li-ion battery over Ni- Cd battery?

12 Answers: i. The teacher has expressed concern that we should not use those materials which cause health hazard and pollution to environment. ii. iii. Lead is poisonous. It has harmful effect on health and environment. Cd is toxic heavy metal and requires special care for disposal. Other advantages of Li-on batteries are: a. It has a voltage of V where as Ni-Cd batteries has 1.2 V. b. It does not need periodic discharge where as Ni-Cd battery is charged only after it is fully discharged. c. Li-on batteries do not lose their charge even after months of storage where as Ni-Cd batteries lose 1-5% of their charge per day. Q2. One day, Akashi s mother noticed that the water pipe line of her house was getting rusted. She asked her daughter that as she was a student of science, she should suggest some method so that further rusting stops. Now answer the following questions: i. What values are expressed in the above paragraph? ii. What do you think Akashi must have suggested to her mother and why? Answers: i. Everybody must have some basic knowledge of science and we should apply it to our day to day practical applications. ii. Akashi must have suggested that magnesium block should be attached to the water pipe line ideation in through an electric wire. It will act as anode and undergo oxidation in preference to iron and hence will protect iron-pipe from further rusting. Q3. Elecrolysis is an important industrial process used for production of metals and a no. of chemicals,for purification of metals and for elecroplating of a number of particles to make them corrosion resistance or to add beauty to those articles. Now answer the following questions: (a) Al is used for the production of a large number of articles. How elecrolysis has helped in its production? (b) How elecrolysis help in the production of pure copper from scrap copper? (c) How electrolysis is helped in the production of low cost jewellery which gives the look of good? (d) How electrolysis has contributed towards the setting up of nuclear power plant? Ans- (a) Al is obtained from bauxite ore.it is first purified to get alumina. Aluminium is obtained by electrolysis of alumina. (b) Scrap copper contains impurities. Pure copper is obtained frome scrap copper by electrolysis. Copper from scrap goes (c) Jewellery made of copper or silver is electroplate with gold. The jewellery is made the cathode, anode is of 24 carat gold chloride solution is taken in the electrolytic bath. (d) nuclear power plant use heavy water as moderator and coolant. This is obtained by repeated electrolysis of ordinary water. Q 4 We use batteries for number of purpose.e.g, torch, mobile phone,invetors, vehicles, aerospace etc. Often we do not have the right knowledge which battery to purchase for a particular requirement.generally we don t care for the last factor although people in advanced countries are very particular about the last factor.

13 (a) A lady is intrested to purchase a battery for her invertor. She has a choice b/w lead plates battery and cadmium plates battery. The former is cheaper while the latter is costly. What advice did you give to her and why? (b) Li-ion battery and Ni-Cd battery both are rechargeable. You are a society concious person and do not want that on disposal your battery should cause any environmental pollution. Which battery would you choose and why? Ans- (a) we shall advice her to buy battery with cadmium plates through it will be costly. This is because lead has harmful effects on health as well as environment. (b)we shall choose lithium ion battery. This is because it is non- hazardous whereas Ni-Cd ion battery uses Cd which is toxic heavy metal and requires special care for disposal. Other advantages of Li- ion battery are (i)it has voltage of V whereas Ni-Cd battery has 1.2V. (ii)it does not need any periodic discharge whereas Ni-Cd recharged only after it is fully discharge. Q5 In the near future,fuel cells will play an increasing role in our everyday life. Soon fuel powered cars and trucks will be seen on the road. They will find their way into cell phones and laptops computer. Your house or office will have fuel cell for heat and elecricity free from disruption. Once the stringent requirement for performance and safety are met, fuel cell system will we available in the general public. Now answer the following questions: (a) It is said that future car will be run on water.how? What will be the main benefits to the society in using these cars? What will be the risk factors? (b) Why did the first apolo moon flight in 1966 use H 2 -O 2 fuel cell? Give three important points which were the strong favour of their use. (c) Hydrogen gas on combustion produces a lot of heat. Why fossile fuels should not be replaced by hydrogen gas? Ans- (a)water on electrolysis produce H 2 gas which is a fuel. It can be used to make H 2 -O 2 fuel cell.these fuel cells can be used to produce energy require to run the car, bus, truck, etc. Benefits of using fuel cell vehicles :- (i) They have higher fuel efficiency. (ii) They run quietly and will have zero or low emission. (iii)they do not have any disposal problem as met with the batteries used in electric cars. Risk factors:- Hydrogen gas is highly combustible and required cars and safe handling. (b)(i) Because of continuosly supply of H 2 and O 2 they never become dead. (ii) they do not cause any pollution problem. (iii) they have excellent efficiency. (c) burning of hydrogen gas produces such high temperature at which N 2 and O 2 of the air combines to form nitrous oxide which pollute the atmosphere. When used in H 2- O 2 fuel cell, heat is converted into electricity and problem of pollution does not arises. Q6. Rusting is the serious problem. Every year large amount of money is spent to replace rusted iron and steel structures. A number of articles made of iron and used in every life are protected from rusting by suitable techniques and by chromate plating, galvinisation etc. Now answer the following questions:

14 (i) A mechanic wants to protect his tool from rusting. He does not want to go for painting as it is costly and not so long lasting. What alternative do you suggest to him? (ii) A person bought a bucket made of galvanized iron sheet but zinc coating got a crack and iron got exposed? Do you think that now rusting of bucket will start? Why or why not? (iii) You have a choice to purchase an article made of galvanized iron sheet coated with tin. You want that the article should last longer from rusting point of view. Which one you prefer and why? Ans- (i) He should apply oil or grease on the tools so that their surface remain protected from direct contact with air and moisture. (ii) Rusting of iron will not takes place even after crack of zinc layer because zinc is more active than iron and oxidation of zinc continues in preference to that of iron. (iii) Article made of galvanized iron sheet should be preferred because zinc is more active than iron whereas tin is less active than iron. If tin layer gets crack,iron will begin to corrode. Q 7. Due to corrosion,enormous damage occurs to building,bridges,ships,and all other objects made of metals especially iron. Crores of rupees are spent every years in replacing the rusted structures. Prevention of corrosions not only saves money but also helps in preventing accidents such as bridge collapse etc. Now answer the following questions: (i) The water pipe lines of a house has starts rusting. The lady of the house does not want exacavation of the pipe. What would you suggest her to do so that further rustinf of the pipe line stoped? (ii) How are bodies of the car protected from rusting by the manufacture before applying primer and painting them finally? (iii)how is the steel hull of ship protected from rusting? Ans- (i) Mg block should be attached to water pipe line through an electric wire. It will act as anode and undergo oxidation in preference to ironand hence will protect iron pipe from further rusting. (ii) Some manufacture first applied alkaline phosphate which from strong impervious layer before applying primer and finally paint on the body to make it rust proof. Some other manufactures galvanise the body before applying primer and finally paint on the body. (iii) Bars of titanium are attached to the steel hull of the ship since in salt water, titanium avts as the anode and is oxidized instead of stell hull.