Cu +2e Cu. 2OH H O+O+2e. Fe Fe +e. Cu +2e Cu. In the cell, Zn Zn Cu Cu, the negative terminal is. Option 2 Cu 2+

Transcription

1 Q. No. 1 In the cell, Zn Zn Cu Cu, the negative terminal is Cu Cu Zn Zn Explanation Zn has tendency to get oxidized, all has tending to get reduced. Zn at -ve terminal Zn Zn e - Cu e Cu [oxidation in anode half cell] Redox in cathode half cell Q. No. Which one of the following reaction occurs at the cathode? OH H OOe Ag Ag e 3 Fe Fe e Cu e Cu Explanation cathode reduction gain of el Cu e Cu Red. Q. No. 3 Reaction taking place at anode is Ionization Reduction Oxidation Hydrolysis Explanation At anode oxidation baseofel Q. No. 4 Which of the following reaction possible at anode? 3 - Cr 7H O Cr O 14H F F 7-1 O H H O None of these Explanation

2 Q. No. 5 In the reaction Cu s Ag aq Cu aq Ag s the reduction half cell reaction is Cue u Cu-e u Ag e Ag Ag-e Ag Explanation - - Q. No. 6 The electrochemical cell stops working after some time because Electrode potentials of both the electrode become zero Electrode potentials of both the electrode become equal One of the electrode is eaten way The reaction starts proceeding in opposite direction Correct Answer Explanation Cell stops working electrode potential at both the electrodes becomes equal. Q. No. 7 Which one of the following statement is incorrect regarding an electrochemical cell? The electrode on which oxidation takes place is called anode Anode is a negative pole The direction of current is same as that of flow of electrons The flow of current is partly due to flow of electrons and partly due to flow of ions. Explanation the direction of flow of current is opposite to flow of el. Q. No. 8 Cell reaction for the cell Zn Zn 1.0M Cd 1.0M Cd Cd e - - Zn Zn -e CdZn ZnCd ZnCd Zn Cd Explanation Anode chamber cathodic chamber Zn Zn (1.0 M) Ca (1.0 M) cal Cd is given by

3 Q. No. 9 Which one of the following is not a function of a salt bridge? To allow the flow of cations from one solution to the other To allow the flow of anions from one solution to the other To allow the electrons to flow from one solution to the other To maintain electrical neutrality of the two solutions Explanation Q. No. 10 If the salt bridge is removed suddenly from a working cell, the voltage Increases Decreases Drops to zero May increases or decrease depending upon cell reaction. Explanation Salt bridge is nuclear lising the solution in two half cell. If salt bridge is removed, wire is removed Zn which has tendency to get oxidized will experience opposition from solution and from electrode of Cu. Which is having tendency to get reduce will also exp. Opposition from solution as will as electrode so voltage will drop to zero. Q. No. 11 Which one of the following represents a standard hydrogen electrode correctly? Pt, H ( 1 atm) H (1 M ), 98 K Pt, H ( 1 atm ) H ( 0.1 M ), 98 K Pt, H ( 0.1 atm ) H ( 1 M ), 73 K Pt, H ( 0.1 atm ) H ( 0.1 M ), 73 K Explanation Standard condition concn. 1M pressure 1 atm temp. 5C 0 /98 K Electrode P, H (1 atm) H (1M) ; 98 K Q. No. 1 The reference electrode is made by using ZnCl CuSO 4 HgCl Hg Cl Explanation Hg Cl

4 Q. No. 13 The standard hydrogen electrode potential is zero, because There is no potential difference between the electrode and the solution Hydrogen ions acquire electrons from the platinum electrode It has been measured accurately It has been defined that way Explanation SHE is defined in such a way that it s potential is ±0.0V. Q. No. 14 A calomel electrode is represented as Hg, Hg Cl, KCl. If in such a half cell Reduction takes place then CI - ion concentration increases Oxidation takes place then Cl - ion concentration decreases The electrode reaction may be represented as Hg Cl - s e Hg - Cl aq Explanation l The electrode reaction taking place is - Hg Cl s Hg aq Cl aq Electrode reaction may be represented as Q. No. 15 During the working of the cell, with the passage of time Spontaniety of the cell reaction decreases, E cell decreases Q decreases, E cell increases W useful increases At equilibrium Q = K c, E cell = 0 Explanation With time l of reaction to occur decreases spontaniety vs. E cell = Tendency of reaction to occur = tendency of one metal to get oxidized other reduced with time. (Dynamic equation will setup) At equation Q = K c,e cell = 0

5 Q. No. 16 The reduction potential of two half cell reactions (occurring in an electrochemical cell) - - are PbSO e PbSO E =-0.31 V E =-0.80 V Ag aq e Ag s The feasible reaction will be - PbSO Ag aq Ag s PbSO PbSO 4Ag aq PbSO 4 Ag s - PbSO 4 Ag s Ag aq PbSO 4 - PbSO Ag s Ag aq PbSO Explanation PbSO e PbSO 0 forreaction tooccure E = -0.31V oxidation 0 A ag e Ag s E =0.80V Reduction (for reaction to be feasible) Q. No. 17 The standers potentials at 5 o C for the following half reactions are given against them - Zn e Zn, E =-0.76 V - Mg e Mg,E =-.37 V When zinc dust is added to a solution of magnesium chloride No reaction will take place Zinc chloride is formed Zinc dissolved in the solution Magnesium is precipitated. Explanation Zn e Zn E 0 = V Mg e Mg E 0 = -.37 V No reaction for reaction to take place R potential of Zn must be less than that of mg. Q. No. 18 If a strip of copper metal is placed in solution of ferrous sulphate Copper will precipitate out Iron will precipitate out Copper and iron both will be dissolved No reaction will take place Explanation No reaction will take place. for reaction to take place red. Potential of Cu should be less than Fe. Q. No. 19 To a mixture containing pieces of zinc, copper and silver, 1 M H SO 4 was added. H gas was found to be evolved. Which of the metal/metals do you think has/have reacted? Given E = 0.76V, E = 0.34V, E = 0.80V. Zn/Zn Cu /Cu Ag /Ag All the metals Only Zn Both Zn and Cu Only Ag. Correct Answer

6 Explanation 0 E =.76V tendency to get oxidised E E E Zn Zn 0 Zn Zn 0 Cu Cu 0 Ag Ag =-.76V =.34V tendency to get reduced =.80V tendency to get reduced when 1M H SO 4 was added, H gas was found to evolved. H SO H SO H e H Reduction Zn metal has reacted. Q. No. 0 In a simple electrochemical cell, which is in standard state, half cell reactions with their appropriate oxidation potentials are - Pb s -e Pb aq E = 0.13 Volt - Ag s -e Ag aq E =-0.80 Volt Which of the following reaction takes place? Pb aq. Ag s Ag aq. Pb s Pb aq. Ags Ag aq. Pbs Ag aq. Pbs Ags Pb aq. Ag aq. Pbs Ag s Pb aq. Explanation Pbs -e Pb aq E 0 =.13 V (tendency to get reduced) Ags -e Ag aq E 0 = - 80 V (tendency to get oxidized b/k of red potential tendency of reaction = tendency of Ag to get oxidized lendency to Pb to get reduce. the reaction that will take place Q. No. 1 When the cell reaction attains a state of equilibrium, the EMF of the cell is Zero Positive Negative Not definite. Explanation At equation a reverse potential will setup in both half cells. EMF = zero Q. No. I and Br are added to solution containing 1 M each of I - and Br - ions. Which of the following reaction will take place? (Given : standard reduction potentials of I and Br are 0.53 and 1.09 volts respectively) Iodine will reduce bromide ions

7 Bromine will reduce iodide ions Iodide ions will reduce bromine Bromide ions will reduce iodine. Explanation Given : R E 0 I I 0 Br Br =53 V =1.09 V We know, X X - Red.pot. F F -.87 Cl Li Br Br I I Oxidising agent Increase R.pot means will get reduce easily. For reaction in For conversion of 0 - Br Br (Reducing agent is req.) Q. No. 3 The standard reduction potential values of the three metallic cations X, Y and Z are 0.5, and -1.18V respectively. The order or reducing power of the corresponding metals is YZ X XY Z ZY X ZX Y Explanation X 0.5 V easily reduction Y V easily oxidized Z V one with R. Potential easily reduced Oxidising agent one with R. Potential easily oxidized Reducing agent Reducing power has to deal with its reactivity, that is the ease at the atom to be oxidized to form cation. So order will be, Y>Z>X Q. No. 4 Electrode potential data are given below : 3 - Fe e Fe ; E =0.77 V aq aq 3 - Al 3e Al ; E =-1.66 V aq s Br aq e - Br - aq ; E =1.08 V Based on the data, the reducing power of Fe, Al and Br - will increase in the order - Br Fe Al

8 Fe Al Br Al Br Fe - Al Fe Br Explanation Given : 3 Fe aq e Fe aq E0 = 0.77 V [easy to get oxidized reducing agent] 3 A l 3 e Al s E0 = V easy to get Br aq e Br - aq E0 = 1.08 V Reduced oxidising agent one with Red. potential get oxidized easily is good reducing agent. Reducing power is the ease at the atom to be oxidized to form cation. - So order will be Br <Fe <Al Q. No. 5 The standard reduction potentials at 98K for the following half reactions are given against each Zn e Zn ; V aq s Cr 3e Cr ; 3 aq s H e H ; aq g Fe e Fe ; 3 aq aq V 0.00 V V Which is the strongest reducing agent? Zn (s) Cs (s) H (g) Fe 3 (aq) Explanation One with Red. Potential get oxidized easily strong reducing agent. Given : Red. potential Zn e Zn s V 3 Cr 3e Cr s H e H g 0.00V 3 Fe e Fe V Zn(s) has tendency to get oxidized easily so is strongest reducing agent. Q. No. 6 Standard potentials (E o ) for some half reactions are given below : 4 (I) Sn e Sn ; E =0.15 V (I) Hg e Hg ; E =0.9 V (III) PbO 4H e Pb H O; E =1.45 V Based on the above, which one of the following statements is correct? Sn 4 is a stronger oxidising agent than Pb 4 Sn is a stronger reducing agent than Hg Pb is a stronger oxidising agent than Pb 4 Pb is a stronger reducing agent than Sn Correct Answer

9 Explanation Statement Reduction easy to get reduce oxidising agent 1) Sn 4 Pb 4 R.Potential Red.potential easy to oxidized easy to get reduced strong/good Reducing agent strong oxidising agent ) Sn Hg R.Potential R. Potential strong reducing agent strong oxidising agent 3) Pb 4 is strong oxidising agent 4) Sn oxidising potential = -.15 V, Pb oxidation potential (-1.45 V) Strong reducing agent b/c it oxidation potential is makes will be oxidized easily strong reducing agent Q. No. 7 The standard reduction potentials E o for the half reaction are as Zn Zn e ; E =0.76 V Fe Fe e ; E =0.41 V The EMF for the cell reaction will be -0.3 N 0.35 V 1.17 V V Correct Answer Explanation E 0 given 0 Zn Zn e E =0.76 V Red.Potential get reduced will act as cathode. 0 Fe Fe e E =0.41 V Red.Potential get oxidised will act on anode EMF = oxidation potential of metal at anode red. potential of metal of cathode = =.35 V Q. No. 8 The standard reduction potential for Fe /Fe and Sn /Sn electrodes are and volt respectively. For the cell reaction Fe Sn FeSn The standard emf will be 0.30 V V 0.58 V V

10 Explanation emf = oxidation of metal of anode red. potential of metal of cathode = = V Q. No. 9 The emf of the cell Ni/Ni ( 1.0M ) Au 3 ( 1.0M )/Au is [E o for Ni /Ni = V; E o for Au 3 / Au = 1.5 V] 1.5 V 1.75 V -1.5 V V Correct Answer Explanation anodic chamber cathodicchamber 3 Ni/Ni 1.0m Au 1.m A 4 emf = oxidation p.t of metal of anode red.pot of metal at cathode =.5.5 = 1.75 V Q. No. 30 Electrode potentials (E o red ) of four elements, A, B, C, D are , -0.3, 0, -1.6V respectively. The decreasing reactivity order of these elements is A, D, B and C C, B, D and A B, D, C and A C, A, D and B Explanation Metal with R. potential is most reactive So order will be metal A D B C 0 E Red Q. No. 31 An unknown metal M displaces nickel from nickel (II) sulphate solution but does not displace manganese from manganese sulphate solution. Which order represents the correct order of reducing power? Mn > Ni > M Ni > Mn > M Mn > M> Ni M> Ni > Mn Explanation For one metal to displace other its red. Potential must less than that of other metal on that metal (M) can displace Ni so its red potential will be less than that of Ni but it can t displace M n so its red. Potential win greater than that of M n. So order of reducing power will be

11 M >m>ni M has least red. potential so is having highest reducing power. n n Q. No. 3 The standard reduction potentials of four elements are given below. Which of the following will be the most suitable reducing agent? I = V II = V III = 0 V IV = 1.90 V III II I IV Explanation one with Red. potential get reduced easily oxidising agent. one with Red. Potential get oxidized easily Reducing agent. Q. No. 33 A gas X at 1 atm is bubbled through a solution containing a mixture of 1 MY - and 1 MZ - at 5 o C. if the reduction potential is ZY X, then Y will oxidise X and not Z Y will oxidise X and not X Y will oxidise both X and Z Y will reduce both X and Z Explanation As stated reduction potential follow the order Z>Y>Y So Y will oxidise X and not Z. Q. No. 34 Standard reduction electrode potentials three metals A, B and C are respectively -0.5, -3.0 V and -1. V. The reducing power these metals are BC A AB C CB A AC B Explanation order will be metal B> C> A Red.Potential -3.0 > -1.> -0.5

12 Q. No. 35 Given that E =-0.5 V, Ni /Ni Cu /Cu E =0.80 V, Ag /Ag E =0.34 V, E Zn /Zn =-0.76 V Which of the following reactions under standard conditions will not take place in the specified directions? Ni aq Cu s Ni s Cu aq Cus Ag aq Cu aq Ag s Cus H aq Cu aq H g Zns H aq Zn aq H g Explanation Given, Reduction potential (V) metal Ni Cu Ag Zn Reduction potential of Cu than that of Ni hasto so Cu Reduction So the reaction will not take place in specified direction. Cu has red. potential that hydrogen so has to undergo reduction for positive..red potential of Zn is less Zn has to undergo, oxidation.

13 Q. No. 36 For the cell :- AgPt Ag Pt E o = 0.4 volt - AgF Ag F E o =.07 volt If the potential for the reaction Pt Pt e is assigned zero. Then E = -0.4V Ag/Ag E = 0.4V Explanation Ag /Ag E = 1.67V - F / F E =.74V - F / F If pt Pt e has potential 0, then acc. to (1) equation..u = Reduction potential of pt. oxidation potential of Ag.u = 0 oxidation potential of Ag Reduction potential of Ag = = oxidation potential of Ag Red potential of F..74 =.4 Reduction potential of F = Red. potential of F 1.67 V. Q. No. 37 Assertion: We cannot add the electrode potentials in order to get the cell potential if numbers of moles of electrons exchanged are not same. Reason: Because the potentials are non thermodynamic properties. If both Assertion and Reason are true and the reason is the correct explanation of the assertion If both Assertion and Reason are true but the reason is not the correct explanation of the assertion If Assertion is true statement but Reason is false If both Assertion and Reason are false statement Explanation Both assertion of reason are correct. Standard red. potential is an intensive property since the no. of is lost must equal the no gained, the half-reaction must be multiplied by integer, as necessary to e 0 is not

14 Passage Text changed when a half reaction is multiplied by an integer. Electrochemical series is a series of elements arranged in increasing order of their reduction potential. E = 0. The metals above H have ve reduction potential, they H /H are more reactive then hydrogen whereas metals below hydrogen are less reactive than H. reduction potential of metal depends upon (i) sublimation energy (ii) ionization energy and (iii) hydration energy of ions. Q. No. 38 Which of the following is best oxidising agent? Cu Na Ag Al 3 Explanation Ag is best oxidising agent. ( Red potentialget reduced easily) Q. No. 39 Which of the following of weakest reducing agent among alkali metals in aqueous? Na K Rb Cs Explanation Na has Red. potential so is easy to get reduced is strong oxidising agent and weakest reducing agent. Q. No. 40 Which of the following cannot evolve H from dil acid? Pt Zn Mg Pb Explanation Metal which is more active than hydrogen active means less reduction potential easily oxidized strong reducing agent. So Pt can t evolve H from dil acid. Q. No. 41 The e.m.f. of a Daniell cell at 98K is E 1 ZnSO 4 CuSO 4 Zn Cu 0.01M 1.0M When the concentration of ZnSO 4 is 1.0 M and that of CuSO 4 is 0.01 M, the e.m.f. changed to E what is the relationship between E 1 and E? E1 E E 1=E E = 0 E 1 E E 1 Explanation Anodic chamber Cathodic chamber

15 .303 RT Zn 0 Daniel cell E = 0=E - log F Cu cell cell for reaction to be spontaneous E >E 1 Q. No. 4 E 0 of a cell aabb ccdd is A B a b ERT In C D c d RT E In c d C D a b C D C D A B A B a B C D nf A B RT E In nf A B RT E In nf C d Correct Answer Explanation E 0 of a cell aabb ccdd 0 RT C D E =E - ln nf A B cell cell a b RT C D E =E ln nf A B C C 0 cell a b d d Q. No. 43 The standard EMF for the cell reaction, ZnCu CuZn is 1.1 volt at 5 o C. The EMF for the cell reaction, when 0.1M Cu and 0.1 M Zn solutions used, at 5 o C is 1.10 V 0.10 V V V Explanation E cell =Ecell - log1 E cell =Ecell V = emf Q. No. 44 For the cell reaction, Cu C 1aq Zn s Zn Cu C s of an electrochemical cell, the change in free energy, function of In (C 1 ) C In C1 In (C 1 C ) G at given temperature is a

16 In (C ) Correct Answer Explanation G 0 =-n F Ecell.303 RT C and E cell = E cell = 0 E cell = log F C 0 0 Q. No. 45 The relationship between standard reduction potential of a cell and equilibrium constant is shown by n E cell = log kc E cell = log kc n E = nlog k cell log kc E cell = n Correct Answer Explanation For example in Daniel cell at equation..303 RT Zn 0 E = 0=E - log F Cu c cell cell Zn But at equation =K Cu Q. No. 46 E o for the cell, RT E cell = Kc F E = log K E cell = log K c 0 cell c c Zn Zn aq Cu Cu is 1.10 V at 5 o C. The equilibrium constant for the cell reaction aq ZnCu CuZn aq aq Correct Answer Explanation ne 10gK = c gK c = = So K =10 c is of order of

17 Q. No. 47 For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be 0.95V at 5 o C. The equilibrium constant of the reaction at 5 o C will be Explanation.95 10gK c = log K c = 10 K c = Q. No. 48 For the redox reaction : Zn s Cu 0.1M Zn 1M Cu s taking place in a cell, E is 1.10 volt. cell E cell RT cell will be.303 = F.14 volt 1.80 volt 1.07 volt 0.8 volt Explanation E cell =1.10- log =1.10- log 10 = 1.07 V For the Q. No. 49 Which graph correctly correlates E cell as a function of concentrations for the cell Zn s Ag aq Zn aq Ag s, E =1.56V Y- axis : E cell, X axis : log 10 Zn Ag cell

18 Correct Answer Explanation Graph must be a straight line with the slop. CTM ne 10gK = C Q. No. 50 Which of the following changes will increases the emf of cell? Co s CoCl M HCl M H g P Pt s 1 1 Increase volume of CoCl solution from 500 ml to 1000 ml Increase M from 0.1 to 0.5 M Decrease pressure of H (g) from 0. to 0.1atm Increase mass of cobalt electrode Explanation Anodic chamber Cathodicchamber emf for the following : Co CoCl M HCl M H g P Pt s s 1 1 emf will increases by increasing volume of CoCl decreasing concentration of HCl decreasing pressure of H (g) Q. No. 51 Assertion: If standard reduction potential for the reaction - Ag e Ag is 0.80 Volt then for the reaction - 3Ag 3e 3Ag E o =.4V Reason: If concentration is increased, reduction electrode potential is increased. If both Assertion and Reason are true and the reason is the correct explanation of the assertion If both Assertion and Reason are true but the reason is not the correct explanation of the assertion

19 If Assertion is true statement but Reason is false If both Assertion and Reason are false statement Explanation Assertion : Ag eag is E E 0 is independent of no of e change here means for 3Ag eag E But the reason : If concentration is increased, reduction electrode potential is increased. Passage Text For the reaction Zns Cu aq Cu s Zn aq Zn Reaction Quotient =, variation of E cell with Q is given by Cu ( where Q = concentration quotient ) OA = 1.10 volts, hence Q. No. 5 When E cell is volts. It implies, Cu =0.01 Zn Zn =0.01 Cu Zn =0.1 Cu Zn =1 Cu Correct Answer Explanation Given : OA = 1.10 V E cell is

20 Zn 0 E =E - log Cu cell Zn Cu =.01 Q. No. 53 The Gfor the process will be -ve if, Cu =10 Zn Zn 3 =10 Cu Zn =10 Cu Zn 5 =10 Cu Explanation 0 E G=- nf cell ve f 0 Q. No. 54 When E cell is and concentration is, Zn - =10 it implies, Cu T = 73 o C T = 98 o C T = 98 K T = 300 K Explanation When E cell is of concentration Zn ratio =10 Cu - St. means T = 5 0 C = 98 K Q. No. 55 During electrolysis of a concentrated aqueous solution of NaCl, what is the product at cathode? Na Cl O H Explanation Electrolysis of concentration equation solution of NaCl NaC l Na Cl H O - H OH At cathode

21 HH H - - At anode C l >> OH prefrencially. - Cl Cle - C l Cl Cl b/c concentration of Cl is higher so Cl will be discharged Q. No. 56 During the electrolysis of aqueous sodium chloride, the cathodic reaction is Oxidation of Cl - ion Reduction of Na ion Reduction of H O Oxidation of H O Explanation As explained above in 55 at cathode reduction of H O will be there. Q. No. 57 The electrolysis of silver nitrate solution is carried out using silver electrodes. Which of the following reaction occurs at the anode? Ag Ag e - - Ag e Ag H O 4H O 4e OH H O 4e Explanation Ag NO Ag NO H O H OH At cathode Ag e Ag (Decrease discharge potential as comp to H ions so Ag will be deposited Ag leis belowin the series than H ) At anode - NO3 Attach the Ag electrode Ag Ag e Q. No. 58 Which one of the following reactions occurs at the anode when CuSO 4 solution is electrolyzed using platinum electrodes? Cu Cu e SO H O H SO O 4e H O O 4H 4e SO SO O e Explanation CuSO Cu SO H O H OH At cathode : Cu e Cu D.P Cu <D.P of H3 O At anode :

22 O O O (Oxygen is liberated) Q. No. 59 Which one of the following reactions takes place at the anode when an aqueous solution of CuSO 4 is electrolyzed using copper electrodes? Cu Cu e - - SO4 SO O e - - SO 4 H O H SO 4O 4e - H O O 4H 4e Explanation CuSO Cu SO CTM : D.Pof oxidation of Cu is lowest, hence Cu anode dissolve. H O - H OH At cathode : Cu e Cu or Cu-e Cu - If E 0 = V So4 -e SO4 E0 = -.0 V - 4OH -4e SO4 E0 = - 1. V At anode : (When electrode is not inert rather. It Cu electrode) CuCu e Q. No. 60 The passage of current liberates H at cathode and CI at anode. The solution is Copper chloride in water NaCI in water Ferric chloride in water auci 3 in water Correct Answer Explanation NaC l Na Cl H O H OH At cathode, H eh HHH At anode if (concentration NaCl solution is used) Cl C le C l Cl Cl Q. No. 61 At an anode in an electrolytic cell where electrolysis is taking place, which of the following processes must occur? Oxidation Loss of electrons by anions Formation of cations by anode Electron density is higher Explanation At anode oxidation occurs (means lose of e ) So electron density will be higher.

23 Q. No. 6 On passing electricity through an aqueous solution of copper sulphate using copper electrodes then Copper is deposited at cathode Copper is dissolved at anode O is liberated at anode The concentration of the solution does not change Explanation - CuSO Cu SO When electrods is Cu electrode H O H OH At cathode Cu e Cu At anode CuCu e Q. No. 63 The number of coulombs required for the deposition of g of silver is Explanation Ag eag 1mol reqd 96500C is C Q. No C of electricity liberates from CuSO 4 solution 63.5 g of Cu g of Cu g of Cu 100 g of Cu Correct Answer Explanation Q CuSO Cu SO Cu e Cu 96500C 63.5gofCu C = g of Cu Q. No. 66 A current of.0 A passed for 5 hours through a molten metal salt deposits. g metal (At wt. = 177). The oxidation state of the metal in the salt is Explanation I = A

24 t = 5 hrs. Q I= =Q C = Q Let the oxidation state of metal in the metal salt is x. x C 36000=. x96500 x = 3 Q. No. 67 A 5 ampere current is passed through a solution of zinc sulphate for 40 minutes. Find the amount of zinc deposited at the cathode g g g g Correct Answer Explanation I = 5 A t=40min=40 60 sec Zn e Zn Q I= t C=Q 1000 C = Q q = g Q. No. 68 On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCI, all copper of the solution was deposited at cathode. The strength of CuCI solution was (Molar mass of Cu = 63.5, Faraday constant = C mol -1 ) M 0. N M 0.0 N Explanation I = 1 A t = 16 min 5 sec =16 605=96sec. Q I= t 965 C = Q Cu e Cu C= = sq.

25 M= =0.005m Q. No. 69 In a solution of CuSO 4 how much time will be required to precipitate g copper by 0.5 ampere current? sec 10 sec 510 sec 64 sec Explanation Cu e Cu 68.5g C g 63.5 = C Q I= t t = 0.5 = sec Q. No. 70 What is the amount of chlorine evolved when amperes of current is passed for 30 minutes in an aqueous solution of NaCI? 66 g 1.3 g 33 g 99 g Correct Answer Explanation I = A t = 30 min=30 60=1800sec - 1 Cl e Cl Q =I t = 3600 C C C =1.3 g 96500

26 Q. No. 71 When 9.65 coulombs of electricity is passed through a solution of silver nitrate (atomic mass of Ag = 108 g mol -1 ), the amount of silver deposited is 16. mg 1. mg 10.8 mg 6.4 mg Explanation Ag e Ag g = g = 10.8 mg Q. No. 7 The charge required to deposit 9 g of Al from Al 3 solution is (At. Wt. of Al = 7.0) C C 9650 C 3163 C Correct Answer Explanation 3 Al 3e Al 7g g 9 7 = C Q. No. 73 Silver is monovalent and has atomic mass of 108. Copper is divalent and has an atomic mass of The same electric current is passed for the same length of time through a silver coulometer and a copper coulometer. If 7.0 g of silver is deposited, then the corresponding amount of copper deposited is g g g 7.95 g Explanation W of metal deposited equation wt. of metal W of Cu deposited equationwtof Cu = W of Ag deposited equationwtof Ag x = x = =7.95g 108 Q. No. 74 The quantity of electricity needed to deposit g of copper is 1 Faraday 4 coulombs 4 Faraday 1 Ampere

27 Explanation Cu e Cu 63.5 F = 4 F Q. No. 75 By passing 0.1 Faraday of electricity through fused sodium chloride, the amount of chloride, the amount of chlorine liberated is g 70.9 g g g Explanation 1 Cle Cl 1F 35.5 g =3.55 g 1 Q. No. 76 Faraday s law of electrolysis are not related to the Atomic number of cation Atomic number of anion Equivalent weight of the cation as well as anion Speed of cation Explanation Faradays law =c t =c z z =electroshemical equivalent ce It is not related to atomic no of cation atomic no of anions speed cation. Q. No. 77 If 9 g of H O is electrolyzed completely with 50 current efficiency 1F of electricity will be needed F of electricity is needed 5.6 L of O at STP will be formed 11. L of O will be formed at STP Explanation - H O H OH H eh F of electrtocity is needed H H H f 5.6l or 8g O f 11.l of O Q. No. 78 No. Column A Column B Column C Id of Additional Answer 1 Electrolytic cell G = ve Galvanic cell G = -ve, salt bridge

28 3 Faraday first law m=z I t 4 Faraday s second W1 W = law E1 E Explanation Aq, ; Bp, r ; cs ; D t Q. No. 79 No. Column A Column B Column C Id of Additional Answer 1 Charge on one 1 F, C mole of electrons 108 g of silver 1 F, C deposited at electrode from 3.4L of hydrogen F at STP collected from 4 8 g of oxygen collected from 1 F, C, 5.6 L at STP Explanation Ap, ; q ; B, q,p ; cr ; D p, q, s. Q. No. 80 Find the strength of current that will liberate 5.60 L of O at NTP from acidulated water in 3 h. Correct Answer 0009 Is Integer Type Explanation If 5.6l of O or 8g. Q =it Q i = t f C = 96500AS = sec 965 i= =9A Q. No. 8 A constant current of 30 A is passed through an aqueous solution of NaCI for a time of 1.0 h. How many grams of NaOH and how many liters of CI (g) at STP are produced? Correct Answer 0013 Is Integer Type Explanation Current = 30 A Time = 1 hour Current = It = C Charge on 1 electron = C No. of electrons thus used: C=le - 1 lc = -19 electrons C= electrons

29 Moles of electrons= = 1.07moles. Thus moles of NaOH formed = 1.07 moles = g = 4.8g 1.07 Volume of Cl formed =.4l 3.968l = = 11.98l Q. No. 83 How many amperes must be passed through a Down s cell to produce Na-metal at a rate of 30 kg/hr(in 10-3 )? Correct Answer 0035 Is Integer Type Explanation Rate of formation of sodium 30 kg 1 hr. 30 moles of sodiumper hr = Mole of sodium formed moles of es liberated charge present on mole of es Q = I = A Q. No. 84 Which of the following reaction occurs at the anode during the charging of lead storage battery? - Pb e Pb - Pb SO4 PbSO 4 - Pb Pb e - - PbSO 4 H O PbO 4H SO 4 e Explanation When cell is charging PbSO S H O l Pb S Pb O s H SO aq 4 4 Q. No. 85 As lead storage battery is charged Lead dioxide dissolves Sulphuric acid is regenerated Lead electrode becomes coated with lead sulphate

30 The concentration of sulphuric acid decreases Correct Answer Explanation Sulphuric acid is regen rated Q. No. 86 The electroplating with chromium is undertaken because Electrolysis of chromium is easier Chromium can form alloys with other metals Chromium gives protective and decorative coating to the base metal Of the high reactivity of metallic chromium Explanation Chromium gives protective coating. Q. No. 87 Prevention of corrosion of iron by Zn coating is called Galvanization Cathodic protection Electrolysis Photo electrolysis Explanation Prevention of iron by Zn is called galvatation Q. No. 88 Which of the following cell is a secondary cell? Mercury cell Ni cell Dry cell Fuel cell Correct Answer Explanation Mercury cell Nonrechar geable Dry cell Pri mary cell Fuel cell can produce electrical energy as long as active material are fed to electrode Ni cell recharge cell secondary cell Q. No. 89 Which of the following material is not present in a dry cell? MnO NH 4 CI ZnCI KCI Explanation Dry cell non has Anodal Zn metal Cathode inert graphite rod surrounded by a paste of MnO carbon black Electorally moist paste of NH 4 Cl of ZnCl in starch. Q. No. 90 Which of the following material is not present in mercury cell? HgO KOH Zinc HgCI

31 Explanation Mercury cell Anode Zn-Hg amalgam cathode paste of H go of carbon electrolyte paste of KOH 40 mercury cells use KOH (potassium hydroxide) NaOH their electrolyte and cells using NaOH have nearly constant voltage at low discharge current making them Heal for hearing acids calculators, cell potential of cell is 1.35v which remain const during its life. Q. No. 91 Which cell has a constant voltage throughout its life? Leclamche cell Electrolytic cell Mercury cell Daniel cell Explanation Explained an aboved Q. No. 9 When a lead storage battery is charged, it acts as A primary cell An electrolytic cell A galvanic cell A concentration cell Correct Answer Explanation When a lead strong battery is chcoyed it act as an electrolytic cell electricity is driving force chem reaction is driven Q. No. 93 Iron tanks are protected from rusting by connecting them with magnesium wire. Which of the following statement (s) is/are correct? Mg acts as anode and iron acts as cathode Moist soil acts as electrolyte Corrosion prevention is electrochemical phenomenon Corrosion prevention is spontaneous phenomenon Explanation mg hen red pot. Less than iron so will act as anode. During the phenomenon of corrosion moisture get deposit in small cracks on surface of ion and act on electrolyte corrosion prevention is electrochemical phenomenon. Q. No. 94 Assertion : The voltage of mercury cell remains constant for longer period of time. Reason : It is because net cell reaction does not involve any ion. If both Assertion & Reason are true and the Reason is the correct explanation of the Assertion If both Assertion & Reason are true and the Reason is not the correct explanation of the Assertion If Assertion is true statement but Reason is false If both Assertion and Reason are false statement Explanation Explained as above. Q. No. 95 Conductance (unit Siemen s S) is directly proportional to area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel, then the unit of constant of proportionality is

32 S m mol -1 S m mol -1 S - m mol S m mol - Correct Answer Explanation area condensation Conductance length length =cell cast. area area Conductance =cast of proportionality concentration lenght Units of cost of proportionally = sm mole -1 given m S = cast of proportionality m 1 L = 100/M 3 Q. No. 96 The unit of specific conductivity is Ohms cm -1 Ohms cm - Ohms -1 cm Ohms -1 cm -1 Explanation Units of specific conductivity is 1 1 l l K= = V C l k q a C = conductive l = cell const. expressed in cm -1 a Units will be ohm -1 cm -1 mole Q. No. 97 The cell constant of a given cell is 0.47 cm -1. The resistance of a solution placed in this cell is measured to be 31.6 ohm. The conductivity of the solution (in Scm -1 where S has usual meaning) is Explanation 1 1 l l K = = =c & R K = =.0148 scm Q. No. 98 The specific conductivity of N/10KCI solution at 0 0 C is 0.1 ohm -1 cm -1 and the resistance of the cell containing this solution at 0 0 C is 55 ohm. The cell constant is cm cm cm -1

33 3.34 cm -1 Correct Answer Explanation 1 1 l K = = 7 k 9 l = = 11.66cm -1 Q. No. 99 The equivalent conductance at infinite dilution of a weak acid such as HF Can be determined by extrapolation of measurements on dilute solutions of HCI, HBr and HI Can be determined by measurements on very dilute HF solutions Can be determined from measurements on dilute solutions of NaF, NaCI and HCI Is an undefined quantity Explanation Kohlrausch law the equivalent conductance at infinite direction of weak acid such as HF can be determined from measurements on solution of NaF, NaCl HCl. HF= NaF HCl- NaCl. Q. No. 100 The unit of equivalent conductivity is ohm cm ohm -1 cm (g equivalent) -1 ohm cm (g equivalent) S cm - Correct Answer Explanation Equivalent conductivity=k v V = volume solution containg K = specific conductance If cg eq of electrolyte are dissolved in V cm 3 of solution V eq= K C If C = 1g eg. and V = 1000 cm eq= K C = normally of solution C Q. No. 101 The resistance of 0.1 N solution of a salt is found to be.5 10 ohm. The equivalent conductance of the solution in Scm /eq is (cell constant = 1.15 cm -1 ) Explanation N = 1N -1 cell cost = 1.15 cm 3 R =.5 10 K 1000 eq = N R K= cell cost

34 Q. No. 10 If 0.01 M solution of an electrolyte has a resistance of 40 ohms in a cell having a cell constant of 0.4 cm -1 then its molar conductance in ohm-1 cm mol -1 will be Correct Answer Explanation k1000 m= M cell constant = 0.4 cm -1 m=0.01m R=40 R k =cell cost m= 40 3 = Q. No. 103 Specific conductance of 0.1 M sodium chloride solution is ohm -1 cm -1. Its molar conductance in ohm -1 cm mol -1 is Explanation k = ohm cm M=0.1M k m= = = M Q. No. 104 Molar conductivity of a solution is cm mol -1. Its molarity is Its molarity is Its specific conductivity will be Correct Answer Explanation -1-1 m= cm mol M=0.01 K =? mm K= = = Q. No. 105 Molar ionic conductivities of a two-bivalent electrolytes x and y - are 57 and 73 respectively. The molar conductivity of the solution formed by them will be 130 S cm mol -1

35 65 S cm mol S cm mol S cm mol -1 Explanation n m sol = m 1 m -1 m 5773=130 s cm mol Q. No. 106 Equivalent conductance s of NaCI, HCI and CH 3 COONa at infinite dilution are 16.45, and 91 ohm -1 cm respectively. The equivalent conductance of CH 3 COOH at infinite dilution would be ohm -1 cm 53.6 ohm -1 cm ohm -1 cm ohm -1 cm Explanation Nacl = ohm -1 cm HCl = ohm -1 cm -1 CH3 COONa = 91 ohm cm - CH3COOH = CH3 COO H 3 = CH COONa = HCl- NaCl = = cm Q. No. 107 The conductance of 0.1 M HCI solution is greater than that of 0.1 M NaCI. This is because HCI is more ionized than NaCI HCI is an acid whereas NaCI solution is neutral H ions have greater mobility than Na ions Interionic forces in HCI are weaker than those in NaCI. Explanation H ions hane nobility than Na ions f = faraday number if = C Q. No. 108 Which of the following is wrong about molar conductivity? The solution contains Avogadro s number of molecules of the electrolyte It is the product of specific conductivity and volume of solution in cc containing 1 mole of the electrolyte Its units are ohm -1 cm mol -1 Its value for 1 M NaCI solution is same as that of 1 M glucose solution. Explanation 1 M NaCI being an electrolyte dissociates into its ions Na Ci - and hence has higher molar conductivity than 1 m glucose. Q. No m -1 is the unit of Molar conductivity

36 Specific conductivity Equivalent conductivity Molar conductivity at infinite dilution. Correct Answer Explanation l K=C =ohm cm ohm m a C=Conductance l =Cell cost. expresed cm -1. a Q. No. 110 The units of cell constant are -1-1 cm -1 cm -1 cm Explanation l cm = =cm a cm -1 Q. No. 111 The value of specific conductance is equal to the conductance of the solution when The cell constant is zero The cell constant is one The electrodes are made of copper The size of the vessel is very large Correct Answer Explanation l k =c when =cell cost =1 a Q. No. 11 The increase in the molar conductivity of HCI with dilution is due to Increase in the self ionization of water Hydrolysis of HCI Decrease in the self ionization of water Decrease in the interionic forces. Explanation Molar conductance is the conductance at solution in which 1 mol electrolyte is dissolved On ting volume ionic interference ionic mobility If weak electrolyte no at ions will above to molar conductance with dilution Q. No. 113 The increase in the value of molar conductivity of acetic acid with dilution is due to Decrease in interionic forces Increase in degree of ionization Increase in self ionization of water None of these

37 Correct Answer Explanation The increases in the value of molar conductivity of acetic acid with dilution is due to increase in degree of Ionisation CTM interfrance ions canreach to electrode easily moblity of over all Conductance for weak electrolytes is CH3 COOH degree of avocation of useable electrolyte keep on ting on dilution No of imges conduetance Q. No. 114 The variation of m of acetic acid with concentration is correctly by

38 Explanation There is a very large in conductance with detection especially near infinite dilution. Theses is b/c as the concentration of weak electrolyte is reduced more at it ionizes. Q. No. 115 According to Kohlrausch law, the limiting value of molar conductivity of an electrolyte, A B is - (A ) (B ) - - (A ) (B ) 1 - (A ) (B ) - (A ) (B ) Explanation for A B = x y x y Q. No. 116 The ionization constant of a weak electrolyte is while the equivalent conductance of its 0.01 M solution is 19.6 S cm eq -1. The equivalent conductance of the electrolyte at infinite dilution (in S cm eq -1 ) will be Explanation c K c = = 1- = = c m m

39 m -1 m cm eq = = = Q. No. 117 The limiting conductivity of NaCI, KCI and KBr are 16.5, and S cm eq -1, respectively. The limiting equivalent ionic conductance for Br- is 78Scm eq -1. The limiting equivalent ionic conductance for Na ions would be : Explanation NaCl = Na Cl = Na Cl - - KCl = Cl Cl = KBr = k Br K =73.5 NaCl = Na 76.5 Na = = 50 Q. No. 118 Assertion : Increasing the concentration increase the value of conductance. Reason : Increasing the concentration increase interionic of conductance. If both Assertion & Reason are true and the Reason is the correct explanation of the Assertion If both Assertion & Reason are true and the Reason is not the correct explanation of the Assertion If Assertion is true statement but Reason is false If both Assertion and Reason are false statement Explanation On ting concentration, the interline farces hence conductance. Q. No. 119 Assertion : The correct order of equivalent conductance at infinite dilution is KCI NaCI LiCI Reason : KCI is stronger electrolyte than NaCI which is stronger LiCI. If both Assertion & Reason are true and the Reason is the correct explanation of the Assertion If both Assertion & Reason are true and the Reason is not the correct explanation of the Assertion If Assertion is true statement but Reason is false If both Assertion and Reason are false statement Correct Answer Explanation The correct order of equivalent conductance at infinite dilution is KCl > NaCl > LiCl This is b/c of difference in mob illation of the cation K, Na, Li which is in order -K > Na > Li B/c of small size, Li is much more hydrated in H O and hence its mobility.