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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
An alloy of Pb-Ag weighing `1.08g` was dissolved in dilute `HNO_(3)` and the volume made to 100 mL.A ? Silver electrode was dipped in the solution and the emf of the cell dipped in the solution and the emf of the cell set-up as `Pt(s),H_(2)(g)|H^(+)(1M)||Ag^(+)(aq.)|Ag(s)` was `0.62 V` . If `E_("cell")^(@)` is `0.80 V`, what is the percentage of Ag in the alloy ? (At `25^(@)C, RT//F=0.06`)A. 25B. `2.50`C. 10D. 50 |
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Answer» Correct Answer - C `0.62=0.8-(0.06)/(1)"log"([H^(+)])/([Ag^(+)])` `-0.18=-0.06"log" (1)/([Ag^(+)])rArr[Ag^(+)]=10^(-3)M` w.t of `Ag=10^(-3)xx108=0.108g` w.t percent of `Ag=(0.108)/(1.08)xx100=10` |
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| 2. |
Iron rod is dipped in concentrated `HNO_(3)` After some time the iron rod dipped in `AgNO_(3)` solution. Ag is not displace by Iron. This is becauseA. SRP of silver is less than ironB. Iron and silver have same lattice structureC. Iron becomes passiveD. All the above |
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Answer» Correct Answer - C In mechanical passivity coloured oxide film is formed. |
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| 3. |
For the electrolytic production of `NaClO_(4)` from `NaClO_(3)` according to the reaction `NaClO_(3)+H_(2)O rarr NaClO_(4)+H_(2). ` How many faradays of electricity would be required to produce `0.5 mo l e` of `NaClO_(4)`?A. 1B. 2C. 3D. `1.5` |
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Answer» Correct Answer - A 1 mole of `NAClO_(4)` requires 2F |
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| 4. |
The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple mode for such a concentration cell involving a metal `M` is : `M_((s))|M_((aq.))^(@) 0.05"molar"||M_((aq.))^(@) 1"molar"|M_((s))` For the above electrolytic cell the magnitude of the cell potential `|E_(cell) | = 70mV` For the above cell :A. `E_(cell)lt0,DeltaGgt0`B. `E_(cell)gt0,Deltalt0`C. `E_(cell)lt0,DeltaH^(@)gt0`D. `E_(cell)gt0,DeltaG^(@)gt0` |
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Answer» Correct Answer - B It is an electrolyte concentration cell in which the electrode with concentrated electrolyte solution acts as cathode and second one as anode. For a concentration cell, appling Nernst s equation, `E_(cell)=(0.059)/(n)log.([M_("aq..")^(n+)]_("Cathode"))/([M_("aq..")^(n+)]_("anode"))` `E_(cell)=(0.056)/(0.1)log_(10).([1.0])/(5xx10^(-2))` `=0.076or76mV`(i.e.,positive value) Also `DeltaG=-2FE`, because E is positive so `DeltaGlt0`. |
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| 5. |
Zinc reacts with `CuSO_(4)` according to the equation `Zn+CuSO_(4)rarrZnSO_(4)+Cu` . If excees of zinc is added to `100.0` ml of `0.05` M `CuSO_(4)` the amount of copper formed in moles will beA. `0.05`B. `0.5`C. `0.005`D. `50` |
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Answer» Correct Answer - C `n=100/1000xx0.05` |
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| 6. |
Aqueous solution of `NaCl` containing a small amount of `MeOH` ( methyl orange ) is electrolysed using `Pt-` electrodes . The color of the solution after some time will .A. remains yellowB. change from yellow to colorlessC. change from yellow to redD. remain red |
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Answer» Correct Answer - 1 Remains yellow since solution is basic after electrolysis & MeOH shows red colour in acidic medium . |
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| 7. |
List-I A) Electrolysis of aq. `Na_(2)SO_(4)` using Pt electrodes B) The charge carried by `6.023xx10^(23)` electrons is C) The amount of electricity required to deposit 27 grmams of Aluminium at cathode from molten `Al_(2)O_(3)` is D) A gas in contact with an inert electrode. List-II 1) 1 Faraday 2) 3 Faradays 3) `H_(2(g))//pt` 4) `O_(2)` at anode `H_(2)` at cathodeA. `{:(A,B,C,D),(2,3,4,1):}`B. `{:(A,B,C,D),(4,1,2,3):}`C. `{:(A,B,C,D),(3,2,4,1):}`D. `{:(A,B,C,D),(4,3,2,1):}` |
| Answer» Correct Answer - B | |
| 8. |
The cell in whiCHM the following reaction occurs `:` `2Fe^(3+)(aq)+2I^(c-)(aq) rarr 2Fe^(2+)(aq)+I_(2)(s)` has `E^(c-)``_(cell)=0.2136V` at `298K`. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction. |
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Answer» `DeltaG^(0)=-nFE^(@)` `=-2 xx 96500 xx0.236J` `=-45548J=-45.548kJ` `"K=Antilog"[(nE^(@))/(0.059)]"= Antilog"(2xx0.236)/(0.059)=10^(8)` |
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| 9. |
`wedge^(@)._(m)` for `CaCl_(2)` and `MgSO_(4)` from the given data. `lambda_(Ca^(2+))^(@)=119.0S cm^(2)mol^(-1)` ltbr. `lambda_(Cl^(c-))^(@)=76.3S cm^(2)mol^(-1)` `lambda_(Mg^(2+))^(@)=106.0S cm^(2)mol^(-1)` `lambda_(SO_(4)^(2-))^(@)=160.0 cm^(2)mol^(-1)` |
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Answer» `Lambda_(m)^(0)CaCl_(2)=lambda_(Ca^(2+))^(0)+2lambda_(Cl^(-))^(0)` `=119+"2 x 76.3"=271.6S cm^(2)mol^(-1)` `Lambda_(m)^(0)MgSO_(4)=lambda_(Mg^(2+))^(0)+lambda_(SO_(4)^(2-))^(0)` `=106+160="266 S cm"^(2)mol^(-1)` |
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| 10. |
The molar conductivity of `0.25 mol L^(-1)` methanoic acid is `46.1 S cm^(2) mol^(-1)`. Calculate the degree of dissociation constant. Given `: lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) ` and `lambda_((CHM_(3)COO^(c-)))^(@)=54.6Scm^(2)mol^(-1)` |
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Answer» We know, `HCOOHhArrH^(+)+HCOO^(-)` `Lambda_(m)^(0)HCOOH=lambda_(H+)^(0)+lamda_(HCOO^(-))` `=349.6+54.6` `="404.2 S cm"^(2)mol^(-1)` Degree of dissociation ` alpha ` may be calculated as `alpha=(Lambda_(m)HCOOH)/(Lambda_(m)^(0)HCOOH)=(46.1)/(404.2)=0.114` Let us consider the ionisation of HCOOH. `HCOOHhArrHCOO^(-)+H^(+)` `{:(t=0,C,0,0),(teq.,C-Calpha,Calpha,Calpha):}` Dissociation constant of formic acid may be calculated as, `K=([HCOO^(-)][H^(+)])/([HCOOH])` |
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| 11. |
The conductivity of `0.001028M` acetic acid is `4.95xx10^(-5)Scm^(-1)`. Calculate its dissociation constant if `Lambda_(m)^(0)` for acetic acid is `390.5Scm^(2)mol^(-1)`. |
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Answer» We know, `Lambda_(m)=kxx(1000)/(M)`.....(i) Given : `"k=4.95 x 10"^(-5)"S cm"^(-1),M=0.001028` `therefore"From"(i)Lambda_(m)=4.95xx10^(-5)xx(1000)/(0.001028)` `="48.15 ohm"^(-1)cm^(2)mol^(-1)` Degree of dissociation, |
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| 12. |
Arrhenius theory is applicable only toA. weak electrolyteB. Strong electrolyteC. both 1 & 2D. non electrolyte |
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Answer» Correct Answer - A Arrhenious theory is applicable to weak electrolytes only. |
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| 13. |
What is the cell entropy change `(` in `J K ^(-1))` of the following cell `:` `Pt(s)|underset(p=1atm)(H_(2)(g))|CH_(2)underset(0.1M)(COOH,HCl)|underset(0.1M)(KCl(aq))|Hg_(2)Cl_(2)(s)|Hg` The `EMF` of the cell is found to be `0.045 V` at `298K` and temperature coefficient if `3.4xxx10^(-4)V K^(-1)` `(` Given `:K_(a(CH_(3)COOH))=10^(-5)M)`A. 60B. 65.2C. 69.2D. 63.5 |
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Answer» Correct Answer - B Considering the cell reaction `H_(2)+2erarr2H^(+)` `Hg_(2)^(2+)rarr2Hg(l)+2Cl^(-)(aq)` `DeltaS=nF((partialE)/(partialT))_(P)=2xx96500xx3.4xx10^(-4)=65.223J//K//mol e` |
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| 14. |
Agar-Agar is used in salt bridge since it isA. ElectrolyteB. Non-electrolyteC. Inert electrolyteD. A solid |
| Answer» Correct Answer - B | |
| 15. |
PREVENTION OF CORROSIONA. Painting the metal surfaceB. Alloying the metal with more anodic metalC. To prevent the contact of the metal surface with good electrical conducting mediaD. All |
| Answer» Correct Answer - D | |
| 16. |
Prevention of corrosion of iron by Zn coating is calledA. GalvanizationB. Cathodic protectionC. ElectrolysisD. Photoelectrolysis |
| Answer» Correct Answer - A | |
| 17. |
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below `(1)/(2)CL_(2)(g)overset((1)/(2)Delta_(diss)H^(Theta))(rarr)Cl(g)overset(DeltaH_(Eg)^(Theta))(rarr)` `Cl^(-)(g)overset(Delta_(hyd)H^(Theta))(rarr)Cl^(-)(aq)` The energy involved in the conversion of `(1)/(2)Cl_(2)(g)` to `Cl^(-)(aq)` (Using the data `Delta_(diss)H_(Cl_(2))^(Theta)=240KJ mol^(-1)`) `Delta_(Eg)H_(Cl)^(Theta)=-349KJmol^(-1)` , `Delta_(Eg)H_(Cl)^(Theta)=-381KJmol^(-1)`) will beA. `+"152 kJ mol"^(-1)`B. `-"610 kJ mol"^(-1)`C. `-"850 kJ mol"^(-1)`D. `+"120 kJ mol"^(-1)` |
| Answer» Correct Answer - B | |
| 18. |
The equivalent conductances of two strong electrolytes at infinite dilution in `H_(2)O` (where ions move freely through a solution) at `25^(@)C` are given below : `Lambda_(CH_(3)COONa)^(@) = 91.0 S cm^(2)//"equi v"`. `Lambda_(HCl)^(@) = 426.2 S cm^(2)//"equiv"`. What additional information//quantity one need to calculate `Lambda^(@)` of an aqueous solution of acetic acid ?A. `Lambda^(0)` of NaCLB. `Lambda^(0)` of `CH_(3)COOK`.C. The limiting equivalent conductance of `H^(+)(lambda_(H^(+))^(0))`.D. `Lambda^(0)` of cholroacetic acid `(ClCH_(2)COOH)` |
| Answer» Correct Answer - A | |
| 19. |
The equivalent conductances of two strong electrolytes at infinite dilution in `H_(2)O` (where ions move freely through a solution) at `25^(@)C` are given below : `Lambda_(CH_(3)COONa)^(@) = 91.0 S cm^(2)//"equi v"`. `Lambda_(HCl)^(@) = 426.2 S cm^(2)//"equiv"`. What additional information//quantity one need to calculate `Lambda^(@)` of an aqueous solution of acetic acid ?A. `^^ ^(@)` of chloroacetic acid `(ClCH_(2)COOH)`B. `^^ ^(@)` of NaClC. `^^ ^(@) of CH_(3)COOK`D. the limiting equivalent conductance of `H^(+)(lambda_(H^(+))^(@))` |
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Answer» Correct Answer - B `Lambda_(eq(CH_(3)COOH))=Lambda_(eq(CH_(3)COONa))^(infty)+Lambda_(eq(HCI))^(prop)-Lambda_(eq(NaCl))^(propto)` |
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| 20. |
The chemical reaction `2AgCl_("(fused)")+H_(2(g))rarr2HCl_((aq))+2Ag_((s))` taking place in a galvanic cell is represented by the notationA. `Pt_((s))|H_(2(g)).1"bar"|1MKCl_((aq))|AgCl_((s))|Ag_((s))`B. `Pt_((s))|H_(2(g)).1"bar"|"1M HCl"_((aq))|"1M Ag^(+)(aq)|Ag_((s))`C. `Pt_((s))|H_(2(g))."1 bar"|"1M HCl"_((aq))|AgCl_((fused))|Ag`D. `Pt_((s))|H_(2(g))."1 bar"|"1M HCl"_((aq))|Ag_((s))|AgCl_((s))` |
| Answer» Correct Answer - C | |
| 21. |
A cell, `Ag|Ag^(o+)||Cu^(2+)|Cu` , initially contains `1 M Ag^(o+)` and `1M Cu^(2+)` ions. Calculate the change in the cell the potential after the passage of `9.65A` of current for `1h.` |
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Answer» Note that given cell will not work as electrochemical cell since `E_(OP_(Cu))^(@)gtE_(OP_(Ag))^(@)`. The equation electrochemical cell will be: `CurarrCu^(2+)+2e` `2Ag^(+)+2erarr2Ag` Thus, e.m.f of cell `Cu|Cu^(2+)"||"Ag^(+)|Ag` will be `E_(cell)=E_(OP_(Cu))^(@)+E_(RP_(Ag))^(@)+(0.059)/(2)log_(10).([Ag^(+)]^(2))/([Cu^(2+)])` `because[Ag^(+)]="1M and"[Cu^(2+)]=1M` `thereforeE_(cell)=E_(cell)^(@)+(0.059)/(2)log_(10).(1)/(1)` `E_(cell)=E_(cell)^(@)("where E"_(cell)^(@)=E_(OP_(Cu))^(@)+E_(RP_(Ag))^(@))` After the passage of `9.65` ampere for 1 hr, i.e., `9.65xx60xx60` coulomb charge, during which the cell reaction it reversed thus, `Cu^(2+)` are discharged from solution and AG metal passes to ionic state. The reaction during passage of current are: `Cu^(2+)+2erarrCu` `2Agrarr2Ag^(+)+2e` `Ag^(+)` ions formed `=(9.65xx60xx60)/(96500)eq.=0.36eq. =0.18"mole"` Thus, `[Ag^(+)]_("left")=1+0.36=1.36M` `[Cu^(2+)]_("left")=1-0.18=0.82M` Thus, new cell is `Cu|{:(Cu^(2+)),(0.82M):}||{:(Ag^(+)),(1.36M):}|Ag` Thus, `E_(cell)=E_(cell)^(@)+(0.059)/(2)log_(10).((1.36)^(2))/((0.82))` `=E_(cell)^(@)+0.010"volt"` Thus, `E_(cell)` increases by `0.010V` |
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| 22. |
The molar conductance of NaCl vauies with the concentration as shown in the following table. And all values follows the equation `lambda_(m)^(C)=lambda_(m)^(oo)-bsqrtC` Where `lambda_(m)^(C)`= molar specific conductance `lambda_(m)^(oo)`=molar specific conductance at infinite dilution C = molar concentration `{:("Molar concentration","Molar conductance of NaCl in ohm"^(-1)"cm"^(2)"mole"^(-1)),(4xx10^(-4),107),(9xx10^(-4),97),(16xx10^(-4),87):}` When a certain conductivity cell (C) was filled with `25xx10^(-4)(M)` NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of `CI^(-)` and `SO_(4)^(-2)` are 80 `ohm^(-1)cm^(2)"mole"^(-1)` and 160 `ohm^(-1)cm^(2)"mole"^(-1)` respectively. It the cell (C) is filled with `5xx10^(-3)(N)Na_(2)SO_(4)` the obserbed resistance was 400 ohm. What is the molar conductance of `Na_(2)SO_(4)`.A. `19.25ohm^(-1)cm^(2)"mole"^(-1)`B. `96.25ohm^(-1)cm^(2)"mole"^(-1)`C. `385ohm^(-1)cm^(2)"mole"^(-1)`D. `192.5ohm^(-1)cm^(2)"mole"^(-1)` |
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Answer» Correct Answer - D `lambda=(Kxx1000)/(m)` `=(G^(**)xx1000)/("R m")=(0.1925xx1000)/(400xx[(5xx10^(-3))/(2)])=(192.5xx2)/(2000xx10^(-3))=192.5` |
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| 23. |
Can you store `CuSO_(4)` solution in `Zn` pot ? |
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Answer» Since the standard reduction potential of Zn is less than that of copper, hence zinc metal will reduce `Cu^(2+)` ions present in aqueous solution of copper sulphate. Thus, we can not store copper sulphate solution in a vessel of zinc. `Zn(s)+CuSO_(4)(aq)hArrZnSO_(4)(aq)+Cu` `E_("Redox process")^(0)=E_("Reducedspecies")^(0)-E_("Oxidisedspecies")^(0)` `=+0.34-(-0.76)=+1.10V` Positive value of `E^(@)` confirms that above redox process is spontaneous and hence we cannot store copper sulphate solution in a vessel of zinc. |
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| 24. |
Match the reaction quotients (Q) listed in coloumn II with the cell reaction listed in column I: |
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Answer» Correct Answer - A-R, B-S, C-P, D-Q (A) `2Al+3Cu^(2+)rarr 2Al^(3+)+3Cu`, `Q=([Al^(+3)]^(2))/([Cu^(2+)]^(3))=(0.1xx0.1)/(0.2xx0.2xx0.2)=1.25` (B) `Cd_((s))+2Fe_((aq))^(3+)rarr Cd_((aq))^(2+)+2Fe_((aq))^(2+)`, `Q=([Cd^(2+)][Fe^(2+)]^(2))/([Fe^(3+)]^(2))=(0.1xx0.1^(2))/(0.01^(2))=10` (C ) `Cd_((s))+2Ag_((aq))^(+)rarr Cd^(2+)+2Ag_((s))` , `Q=([Cd^(2+)])/([Ag^(+)]^(2))=(0.01)/(0.1^(2))=1` (D) `H_(2)(10atm)+2H^(+)(1M)rarr 2H_((aq))^(+)(0.1M)+H_(2)(1atm)` |
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| 25. |
The elctrical resistance of a column of `0.05 M N aOH ` solution of diameter `1 cm` and length `50 cm` is `5.55 xx 10^(3)ohm`. Calculate its resisteivity , conductivity, and molar conductivity. |
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Answer» Area of cross-section `=pir^(2)=3.14xx(0.5)^(2)cm^(2)=0.785cm^(2)` We know, `R=rhoxx(1)/(A),5.55xx10^(3)=rhoxx(50)/(0.785)` `"Resistivity",rho=87.135"ohm cm"` `"Conductivity k"=(1)/(rho)=(1)/(87.135)-0.01148"S cm"^(-1)` Molar conductivity `Lambda_(m)` can be calculated as, `Lambda_(m)=kxx(1000)/(M)=0.01148xx(1000)/(0.05)` `=229.6"S cm"^(2)"mol"^(-1)` |
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| 26. |
The conductivity of `0.1`m KCl solution is `1.29sm^(-1)`. If the resistance of the cell filled with `0.1` M KCl is 100 ohm. Calculate the cell constant. |
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Answer» `k=GxxG^(**)rArrk=(1)/(R)xxG^(**)` `G^(**)=kxxR=1.29xx100=129cm^(-1)` |
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| 27. |
Conductance of `0.1` M KCl (conductiviy = `X ohm^(-1) cm^(-1)`) filled in a conductivity cell is Y `ohm^(-1)` If the conductance of 0.1 M NaOH filled in the same cell is Z `ohm^(-1)` the molar conductance of NaOH will beA. `10^(3)(XZ)/Y`B. `10^(4) (XZ)/Y`C. `10(XZ)/Y`D. `0.1(XZ)/Y` |
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Answer» Correct Answer - B Conductivity (x)=conductance (c) x cell constant Cell constant `=x/y` `lambda=(kxx1000)/M=((x/yxxz)xx1000)/(0.1)=(xz)/yxx10^(4)` |
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| 28. |
We have taken a saturated solution of `AgBr,K_(sp)` of AgBr is `12xx10^(-14)`. If `10^(-7)` "mole" of `AgNO_(3)` are added to 1 litre of this solution then the conductivity of this solution in terms of `10^(-7) Sm^(-1)` units will be: [Given: `lambda_((Ag^(+)))^(@)=4xx10^(-3)Sm^(2)"mol"^(-1),lambda_((Br^(-)))^(@)=6xx10^(-3) S m^(2) "mol"^(-1),lambda_((NO_(3)^(-)))^(@)=5xx10^(-3)Sm^(2) "mol"^(-1)`] |
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Answer» Suppose the solubility `AgBr` in `10^(-7) AgNO_(3)` is `S moL^(-1)` `{:(therefore" "underset("S mol L"^(-1))(AgBr) underset(larr)rarr Aunderset(S)g^(+) +Bunderset(S)r^(-)),(therefore" "underset(10^(-7)M)(AgNO_(3))underset(larr)rarrunderset(10^(-7)M)(Ag^(+))+underset(10^(-7)M)(NO_(3)^(-))):}` Taking `[Ag^(+)] = (S+10^(-7))M` `K_(sp)` of `AgBr = [Ag^(+)[ [Br^(-)]` `12 xx 10^(-14) = (S +10^(-7)) (S) = S^(2) +10^(-7)S` `S^(2) + 10^(-7)S - 12 xx 10^(-14) = 0` On solving, `S = 3 xx 10^(-7)M` `[Br^(-)] = 3 xx 10^(-7) M = 3 xx 10^(-7) xx 10^(3) mol//m^(3)` `= 3 xx 10^(-4) mol//m^(3)` Hence, `[Ag^(+)] = (3 xx 10^(-7) +10^(-7)) M = 4 xx 10^(-4) mol//m^(3)` `[NO_(3)^(-)] = 10^(-7)M = 10^(-7) xx 10^(+3) = 1 xx 10^(=4)m^(3)` `: lambda = (k)/(c)` or `k = lambda xx c` `:. k_(Br^(-)) = 3 xx 10^(-4) xx 8 xx 10^(-3) Sm^(-1) = 24 xx 10^(-7) S m^(-1)` `k_(Ag^(+)) = 4 xx 10^(-4) xx 6 xx 10^(-3) Sm^(-1) = 24 xx 10^(-7) S m^(-1)` `k_(NO_(3)^(-)) = 1 xx 10^(-4) xx 7 xx 10^(-3) = 7 xx 10^(-7) S m^(-1)` `K_("total") = k_(Br^(-)) +K_(Ag^(+)) +K_(NO_(3)^(-))` Specific conductivity of solution `= (24 xx 10^(-7) +24 xx 10^(-7) +7 xx 10^(-7)) Sm^(-1)` `= 55 xx 10^(-7) Sm^(-1)` |
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| 29. |
During the discharge of a lead storage battery, density of `H_(2)SO_(4)` fall from 1.3 to `1.14g//mL` Sulphuric acid of density `1.3g//ml` is 40W % and that of `1.14g//mL` is 20W% The battery holds two litre of the acid and volume remains practically constant during dicharging. The number of ampere- sec used from the battery is.A. `3xx96,500`B. `6xx96,500`C. `9xx96,500`D. `12xx96,500` |
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Answer» Correct Answer - B `M_(1)-M_(2)=10/98[1.3xx40-1.14xx20]` `=3 "Mole"//"Litre"=6"Moles Per" 2 " litre"` |
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| 30. |
An aqueous solution containing `1M` each of `Au^(3+),Cu^(2+),Ag^(+),Li^(+)` is being electrolysed by using inert electrodes. The value of standard potentials are `:` `E_(Ag^(+)//Ag)^(@)=0.80 V,E_(Cu^(+)//Cu)^(@)=0.34V` and `E_(Ag^(+3)//Au)^(@)=1.50,E_(Li^(+)//Li)^(@)=-3.03V` will increasing voltage, the sequence of deposition of metals on the cathode will be `:`A. `Li,Cu,Ag,Au`B. `Cu,Ag,Au`C. `Au,Ag,Cu`D. `Au,Ag,Cu,Li` |
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Answer» Correct Answer - 3 Only `Au^(+3),Ag^(+)` and `Cu^(2+)` will deposit at cathode. Li will not deposit at cathode because SRP of water is `-0.8274V` so after `Cu^(2+),H_(2)` will evolve at cathode. |
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| 31. |
The passage of current through a solution of certain electrolyte results in the evolution of `H_2`(g) at cathode and `Cl_2`(g) at anode. The electrolytic solution is :A. WaterB. `H_(2)SO_(4)`C. Aqueous NaClD. Aqueous `CuCl_(2)` |
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Answer» Correct Answer - C Aqueous NaCl undergoes electrolysis and liberates `H_(2)` at cathode and chlorine at anode. |
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| 32. |
While charging the lead storage battery:A. `PbSO_(4)` anode is reduced to PbB. `PbSO_(4)` cathode is reduced to PbC. `PbSO_(4)` cathods is oxidised to PbD. `PbSO_(4)` anode is oxidised to `PbO_(2)` |
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Answer» Correct Answer - A While charging the lead storage battery the reaction occurring on cell is reversed and `PbSO_(4)` (s) on anode and cathode is converted into Pb and `PbO_(2)` respectively Hence, option (a) is the correct choice The electrode reactions are as follows At cathode `PbSO_(4)(s)+2e^(-)rarrPb(s)+SO_(4)^(2-)(aq)` (Reduction) At anode `PbSO_(4)(s)+2H_(2)Orarr PbO_(2)(s)+SO_(4)^(2-)+4H^(+)+2e^(-)` (Oxidation) Overall reaction `2PbSO_(4)(s)+2H_(2)O rarr Pb(s)+PbO_(2)(s)+4H^(+)(aq.)+2SO_(4)^(2-)(aq.)` |
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| 33. |
The variation equivalent conductance of stronge electrolyte with `sqrt(Concentration)` is correctly shown in the figure.A. B. C. D. |
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Answer» Correct Answer - A The equivalent conductance of strong electrolyte increase with dilution. |
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| 34. |
The voltage of a cell whose half-cells are given below is `Mg^(2+)+2e^(-)rarrMg(s),E^(@)=-2.37V` `Cu^(2+)+2e^(-)rarrCu(s),E^(@)=+0.34V` standard EMF of the cell isA. `-2.03V`B. `1.36V`C. `2.7V`D. `2.03V` |
| Answer» Correct Answer - C | |
| 35. |
On electrolysing `K_(2)SO_(4)` solution using inert electrodes, `1.68L(STP)` of gases was obtained. How many moles of `MnO_(4)^(-)` could be reduced to `Mn^(2+)` by the same quantity of electricity?A. `0.02`B. `0.15`C. `0.20`D. `0.10` |
| Answer» Correct Answer - A | |
| 36. |
The standard reduction potentials of `Cu^(2+)//Cu` and `Cu^(2+)//Cu^(+)` are 0.337 V and 0.153V respectively. The standard electrode potential of `Cu^(+)//Cu` half-cell isA. `0.184V`B. `0.827V`C. `0.521V`D. `0.490V` |
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Answer» Correct Answer - C `DeltaG_(3)^(@)=DeltaG_(1)^(@)-DeltaG_(2)^(@)` |
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| 37. |
EMF of an `H_(2)-O_(2)` fuel cellA. Is indepenent of partial pressures of `H_(2)` and `O_(2)`B. Decreases on increasing `P_(H_(2))` and `P_(O_(2))`C. Increases on increasing `P_(H_(2))` and `P_(O_(2))`D. Varies with the concentration of `OH^(-)` ions in the cathodic and anodic compartments. |
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Answer» Correct Answer - C `E_(cell)=E_(cell)^(@)+(0.059)/(2)logp_(H_(2)).p_(o_(2))^(1//2)` |
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| 38. |
For the redox change , `Zn_(s) + underset (0.1M)FuCu^2+ rarr underset(Zn^(2+))(1M) + Cu_((s))`, Taking place in a cell ` E_(cell)^@ ` is ` 1.10` volt . `E_(cell) ` for the cell would be :A. `1.07V`B. `0.82V`C. `2.14V`D. 180V |
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Answer» Correct Answer - A `E=E_(cell)^(0)+(0.059)/(2)log.([Cu^(2+)])/([Zn^(2+)])` `=1.10+(0.059)/(2)log.(0.1)/(1)=1.07V` |
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| 39. |
The standard reduction potentials of Ag, Cu, Co and Zn are `0.799,0.337,-0.277,-0.762V` respectively. Which of the following cell will have maximum cell e.m.f?A. `Zn|Zn^(+2)(1M)"||"Cu^(2+)(1M)|Cu`B. `Zn|Zn^(2+)(1M)"||"Ag^(+)(1M)|Ag`C. `Cu|Cu^(2+)(1M)"||"Ag^(+)(1M)|Ag`D. `Zn|Zn^(2+)(1M)"||"Co^(2+)(1M)Co` |
| Answer» Correct Answer - B | |
| 40. |
Given: `Fe_((s))rarrFe^(2+)+2e^(-), E^(@)=+0.44V` `Pb_((s))rarrPb^(2+)+2e^(-),E^(@)=+0.13V` `Ag^(+)+e^(-)rarrAg,E^(@)=+0.8V` `Cu^(2+)+2e^(-)rarrCu,E^(@)=+0.34V` Which of the following metal ion will oxidise iron?A. `Ag^(+) "only"`B. `Cu^(2+) " only"`C. `Pb^(+2) " only"`D. All |
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Answer» Correct Answer - D `Fe^(2+)+2e^(-)rarrFe, E^(@)=-0.44V` `pb^(2+)+2e^(-)rarrpb , E^(@)=-0.13V` `Ag^(+)+e^(-)rarrAg,E^(@)=+0.8V` `Cu^(2+)+2e^(-)rarrCu, E^(@)=+0.34V` SRP values of pb, Ag, Cu are higher than Fe. So, pb, Ag, Cu oxidises Fe |
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| 41. |
For the Daniel Cell involving the cell reaction `Zn_((s))+Cu_((aq))^(+2)hArrZn_((aq))^(+2)+Cu_((s))` the standard free energies of formation of `Zn_((s)),Cu_((s)),Cu_((aq))^(+2)` and `Zn_((aq))^(+2)` are 0, 0, `64.4` KJ/Mole and `-154.0` KJ/Mole, respectively. Calculate the standard EMF of the cellA. `2.13` VoltsB. `1.13` VoltsC. `2.26` VoltsD. `3.42` Volts |
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Answer» Correct Answer - D `DeltaG^(@)=-nFE^(@)` |
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| 42. |
The time required (approx) to remove electrolytically one half from `0.2` litres of 1M `AgNO_(3)` solution by a current of 1 amp isA. 320minB. 80minC. 160minD. 40min |
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Answer» Correct Answer - C Find the amount of Ag in the solution. Then use the formula m = ect and find t |
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| 43. |
The standard reducution potentials of `Zn^(2+)|Zn,Cu^(2+)|Cu and Ag^(+)|Ag` are respectively `-0.76,0.34` and `0.8V`. The following cells were constructed. `Zn|Zn^(2+)"||"Cu^(2+)|Cu` `Zn|Zn^(2+)"||"Ag^(+)|Ag` `Cu|Cu^(2+)"||"Ag^(+)|Ag` What is the correct order `E_("cell")^(0)` of these cell?A. `bgtcgta`B. `bgtagtc`C. `agtbgtc`D. `cgtagtb` |
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Answer» Correct Answer - D `E_(cell)^(@)=E_("cathode")-E_("anode")^(@)` |
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| 44. |
A solution of `CuSO_(4)` is electrolyzed for `10 ` min with a current of `1.5A`. What is the mass of `Cu` deposited at the cathode? `[` Atomic mass of `Cu=63g]` |
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Answer» `CuSO_(4)rarrCu^(+2)+SO_(4)^(-2)` `m=(Ect)/(96500)=(31.75xx1.5xx10xx60)/(96500)` `=0.269"grams"` |
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| 45. |
Assertion`(A):` Whne acidified `ZnSO_(4)` solution is electrolyzed between `Zn` electrodes, it is `Zn` that is deposited at the cathode and `H_(2)(g)` is not evolved. Reason `(R):` The electrode potential of `Zn` is more negative than hydrogen as the overpotential for hydrogen evolution in `Zn` is quite large.A. S is correct but E is wrongB. S is wrong but E is correntC. Both S and E are correct and E is corrent explanation of S.D. Both S and E are correct but E is not correct explanation of S |
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Answer» Correct Answer - D These are facts |
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| 46. |
Pure water does not conduct electricity because it :A. NeutralB. Readily decomposedC. Almost unionisedD. Completely ionised |
| Answer» Correct Answer - C | |
| 47. |
Cathodic standard reduction potential minus anodic standard reduction potential is equl toA. FaradayB. CoulombC. Cell potentialD. Ampere |
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Answer» Correct Answer - C `E=E_("cathode")-E_("anode")` (EMF is in SRP) |
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| 48. |
In a galvanic cell, the reactions taking place in the anodic half cell the cathodic half cell will beA. ReductionB. OxidationC. Oxidation and reductionD. Reduction and oxidation |
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Answer» Correct Answer - C at anode oxidation takes place while at cathode reduction takes place. |
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| 49. |
The chlorate ion can disproportinate in basic solution according to reaction, `2ClO_3^(-)iffClO_2^(-)+ClO_4^-)` what is the equilibrium concentration of perchlorate ions from a solution initially at 0.1 M in chlorate ions at 298 K? Given: `E_(Cl_4^(-)|ClO_3^(-))^(@)=0.36V "and" E_(Cl_3^(-)|ClO_2^(-))^(@)=0.33V "at"298 K`A. `0.019` MB. `0.024` MC. `0.1` MD. `0.19` M |
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Answer» Correct Answer - A `E_(cell)^(@)=0.33-0.36=-0.03E_(cell)^(@)=(RT)/(2F)"In K"` `-0.03=(0.06)/(2)"log K or K"=0.1` `{:(2ClO_(3)^(-)hArrClO_(4)^(-)+ClO_(2)^(-)(x^(2))/((0.1-2x)^(2))=(1)/(10),),(0.1-2x" x x , "3.16x=0.1-2x):}` `5.16x=0.1rArrx=0.1//5.16=0.0193~=1.9xx10^(-2)` |
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| 50. |
Represent the cell in which following reaction takes place `:` `Mg(s)+2Ag^(o+)(0.0001M)rarr Mg^(2+)(0.130M)+2Ag(s)` calculate its `E_(cell)` if `E^(c-)._(cell)=3.17V. ` |
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Answer» The cell may be represented as `Mg(s)|Mg^(2+)(0.13M)"||"Ag^(+)(0.0001M)|Ag(s)` The cell reaction may be given as, `Mg(s)rarr Mg^(2+)(0.13M)+2e^(-)` `2Ag^(+)(10^(-4)M)+2e^(-)rarr2Ag(s)` |
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