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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Which has the maximum conductivity ?A. `[Cr(NH_(3))_(3)Cl_(3)]`B. `[Cr(NH_(3))_(4)Cl_(2)]Cl`C. `[Cr(NH_(3))_(5)Cl]Cl_(2)`D. `[Cr(NH_(3))_(5)]Cl_(3)` |
| Answer» Correct Answer - D | |
| 2. |
Consider the cell potentials `E_(Mg^(2+)|Mg)^(0)=-2.37V` and `E_(Fe^(3+)|Fe)^(0)=-0.04V` The best reducing agent would beA. `Mg^(2+)`B. `Fe^(3+)`C. `Mg`D. `Fe` |
| Answer» Correct Answer - C | |
| 3. |
Choose the correct statement(s)A. At the anode, the species havig minimum reduction potential is formed from the oxidation of correcsponding oxidizable species.B. In highly alkaline medium, the anodic process during the electrolytic process is `4OH^(-)toO_(2)+2H_(2)O+4e^(-)`C. The standard potential of `Cl^(-)|AgCl|Ag` half-cell is related to that of `Ag^(+)"Ag` through the expression `E_(Ag^(+)|Ag)^(@)=E_(Cl^(-)|AgCl|Ag)^(@)+(RT)/(F)lnK_(sp)(AgCl)`D. Compounds of active metals `(Zn,Na,Mg)` are reducible by `H_(2)` whereas those of noble metals `(Cu,Ag,Au)` are not reducible. |
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Answer» Correct Answer - A::B In highly alkaline medium `OH^(-)` ion get oxidise in preference of `H_(2)O` `4OH^(-)toO_(2)+2H_(2)O+4e^(-)` `E_(Cl^(-)//AgCl//Ag)^(@)=E_(At^(+)//Ag)^(@)-(0.059)/(1)log((1)/(K_(sp))` `E_(Red)^(@)` of active metal is lower than `E_(Red)^(@)` of hydrogen so not reduced by hydrogen. |
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| 4. |
`M(s)rarrM^(n+)(aq)+n e^(-)` Choose the correct statement(s).A. `E_(M|M^(n+))` decrease with increase in `[M^(n+)]`B. `E_(M^(n+)|M)` increases on increasing temeperature.C. `E_(M^(n+)|M)` increases on increasing `[M^(n+)]`D. `E_(M|M^(n+))` increases on increasing temperature. |
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Answer» Correct Answer - A::B::C `E_(M//M^(+n))=E_(M//M^(+n))^(0)-(0.059)/(n)logM^(n+)` ltrgt `[M^(n+)]uarr,E_(M//M^(n+))darr,E_(M^(n)//M)uarr` |
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| 5. |
A current of 1.0 A was passed for 2 hr through a solution of cuprocyanide and 0.3745 g of copper was deposited on the cathode. Calculate the current efficiency for the copper deposition. (`Cu-63.5`)A. `79%`B. `39.5%`C. `63.25%`D. `63.5%` |
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Answer» Correct Answer - A m(theoretical)`=(63.5xx0.1xx7200)/(96500)=0.4738g` `therefore%` efficiency `=(0.3745)/(0.4738)xx100=79%` |
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| 6. |
A solution containing one mole per litre each of `Cu(NO_(3))_(2),AgNO_(3),Hg(NO_(3))_(2)` and `Mg(NO_(3))_(2)` is being electrolysed by using inerty electrodes. The values of the standard oxidation potentials in vlts are `Ag//Ag^(+)=-0.8V,Ag//Hg^(2+)=-79V,Cu//Cu^(2+)=-0.34V,Mg//Mg^(2+)=2.37V`. The order in which metals will be formed at cathode will be-A. `Ag,Cu,Ag,Mg`B. `Ag,Hg,Cu,Mg`C. `Ag,Hg,Cu`D. `Cu,Hg,Ag` |
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Answer» Correct Answer - C Metal having more value of SRP will deposit first, but Mg does not deposition aqueous solution. |
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| 7. |
A hydrogen electrode is immersed in a solution with pH = 0 (HCl). By how much will the potential (reduction) change if an equivalent amount of NaOH is added to this solution? (Take `p_(H_(2) = 1atm)` T=298KA. increase by 0.41 VB. increase by 59 mVC. decrease by 0.41 VD. decrease by 59 mV |
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Answer» Correct Answer - C pH changes from 0 to 7 `therfore[H^(+)]` changes from 1 to `10^(-7)` M. accordingly `E_(red)` decreases by `0.059log10^(-7)` i.e., `0.059xx(-7)=-0.41"volt"` |
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| 8. |
Using the standard electrode potential values given below, decide which of the statements, `I,II,III` and `IV` are correct. Choose the right answer from `(1) (2)` and `(4)` `Fe^(2+)+2e^(-) hArr Fe " "," "E^(0)=-0.44V` `Cu^(2+)+2e^(-) hArr Cu" "," "E^(0)=+0.34V` `Ag^(+)+e^(-)hArr Fe" "," "E^(0)=+0.80V` `I. ` Copper can displace iron from `FeSO_(4)` solution. `II.` Iron can displace copper from `CuSO_(4)` solution `III.` Silver can displace copper from `CuSO_(4)` solution `IV.` Iron can displace silver from `AgNO_(3)` solution.A. I and IIB. II and IIIC. II and IVD. I and IV |
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Answer» Correct Answer - C Lower standard reduction potential related metal ions can displace higher standard reduction potential related metals ions. |
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| 9. |
Calculate `E_(cell)` of `Pt(s)|{:(cl_(2)(g)),(0.1atm):}|{:(Cl^(-)(aq)),(10^(-2)M):}||{:(Cr_(2)O_(7)^(2-),Cr^(+3)("in" H_(2)SO_(4))=0.05M),(0.01M,0.1M):}|` given that `E_(Cr_(2)O_(7)^(2-)//Cr^(+3))^(0)=1.33V` `E^(0)Cl^(-)|Cl_(2)=-1.36V` |
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Answer» `6e^(-)+14H^(+)+Cr_(2)O_(7)^(2-)to2Cr^(+3)+7H_(2)O` `[2Cl^(-)toCl_(2)+2e^(-)]xx3` `14H^(+)+6Cl^(-)+Cr_(2)O_(7)^(2-)to3Cl_(2)+2Cr^(+3)+7H_(2)O` `E_(cell)^(0)=1.33-(+1.36)=-0.03` `E_(cell)=-0.03-(0.059)/(6)log(([Cr^(3+)]^(2)[P_(Cl_(2))]^(3))/([H^(+)]^(14)[Cl^(-)]^(6)[Cr_(2)O_(6)^(2-)]))=-0.03-(0.059xx23)/(6)` `E_(cell)=-0.26V` |
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| 10. |
The conductivity of a solution of AgCl at 298 K is found to be `1.382xx10^(-6)Omega^(-1)cm^(-1)` the ionic conductance of `Ag^(+)` and `Cl^(-)` at infinite dilution are `61.9Omega^(-1)cm^(2)col^(-1)` ad `76.3Omega^(-1)cm^(2)mol^(-1)` respectively the solubility of AgCl isA. `1.4xx10^(-5)molL^(-1)`B. `1xx10^(-2)molL^(-1)`C. `1xx10^(-5)molL^(-1)`D. `1.9xx10^(-5)molL^(-1)` |
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Answer» Correct Answer - C `K=1.382xx10^(-6)scm^(-1)` `^^_(AgCl)=61.9+76.3=138.2=(1000xx1.382xx10^(-6))/(S)` `S=10^(-5)M`. |
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| 11. |
The equivalent conductance of an infinitely dilute solution of `NH_(4)Cl` is 150 and te ionic conductance of `OH^(-)` ad `Cl^(-)` ions are 198 and 76 respectively. What will be the equivalent conductance of the solution of `NH_(4)OH` at infinite dilution. if the equivalent conductance of a 0.01 N solution `NH_(4)OH` is 9.6, what will be its degree of dissociation? |
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Answer» Correct Answer - `272, 0.0353` `^^_(eq)^(@)(NH_(4)Cl)=150=^^_(eq)^(@)(NH_(4)^(+))+^^_(eq)^(@)(Cl^(-))` `^^_(OH^(-))^(@)=198,^^_(Cl^(-))^(@)=76,^^_(eq)(NH_(4)OH)=9.6,C=0.01` `^^_(eq)^(@)(NH_(4)OH)=150-76+198=272` `alpha=(^^_(eq)(NH_(4)OH))/(^^_(eq)^(@)(NH_(4)OH))=(9.6)/(272)=0.0353` |
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| 12. |
Rusting of iron is quicker in salline water than in ordinary water. Why is it so? |
| Answer» In saline water, NaCl helps water to dissociate into `H^(+)` and `OH^(-)` greater the number of `H^(+)`, ions quicker will be rusting. | |
| 13. |
Corrosion is essentially an electrochemical phenomenon. Explain the reactions occurring during corrosion of iron kept in an open atmosphere. (b) Calculate the equilibrium constant for the equilibrium reaction `Fe_((s)) + Cd_((aq))^(2+) hArr Fe_((aq))^(2+) + Cd_((s))` (Given : `E_(Cd^(2+)|Cd)^(@) = -0.40 V , E_(Fe^(2+)|Fe)^(@) = -0.44V`). |
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Answer» (a). At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode. Electrons released at anodic spot move through the metal and go to another spot of metal which behaves as cathode. The half reaction are At anode: `2Fe(s)to2Fe^(2+)+4e^(-)` At cathode: `O_(2)(g)+4H^(+)(aq)+4e^(-)to2H_(2)O(l)` The overally reaction is `2Fe(s)+O_(2)(g)+4H^(+)(aq)to2Fe^(2+)(aq)+2H_(2)O(l)` (b). `E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)` `=-0.40-(-0.44)=0.04V` Since `E_(cell)^(@)=(0.059)/(n)logK_(c)` `0.04=(0.059)/(2)logK_(C)` `logK_(c)=(2xx0.4)/(0.059)=1.356` `K_(C)="anti-log"(1.356)=22.70` |
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| 14. |
Express the relationship between degree of dissociation of an electrolyte and its molar conductivities. |
| Answer» `prop=(overset(C)^^_(mol))/(overset(infty)(^^)_(mol))` | |
| 15. |
The `E^(@)` value in respect of the electrodes `(Z=24)`, manganese `(Z=25)` and iron `(Z=26)` are : `Cr^(3+)//Cr^(2+)=-0.4V,Mn^(3+)//Mn^(2+)=+1.5V`,`Fe^(3+)//Fe^(2)=+0.8V`. On the basic of the above information compare the feasibilities of further oxidation of their `+2` oxidation states. |
| Answer» `+2` oxidation state of Mn is mores stable than `+2` oxidation state of Fe which, in turn, is more stable that `+2` oxidation state of `Cr`. So the order of feasibility of further oxidation of `+2` oxidation state is `Cr^(2+)gtFe^(2+)gtMn^(2+)` | |
| 16. |
Chemical reaction involve interaction of atoms and molecules. A large number of atoms/molecules (approximately `6.022xx10^(23)`)are present in a few grams of any chemical compound varying with their atomic/molrcular mass. To handle such a large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/ electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of teh electrodes (atomic mass: Na=23, Hg=200, 1F=96500 coulombs) If cathode is a Hg electrode, the maximum weight(g) of amalgam formed from the solution isA. 200B. 225C. 400D. 446 |
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Answer» Correct Answer - D Taking the `1:1` molar combination of `Na-Hg` amalgam. Weight `=2xx23+2xx200=446gm`. |
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| 17. |
Chemical reaction involve interaction of atoms and molecules. A large number of atoms/molecules (approximately `6.022xx10^(23)`)are present in a few grams of any chemical compound varying with their atomic/molrcular mass. To handle such a large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/ electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of teh electrodes (atomic mass: Na=23, Hg=200, 1F=96500 coulombs) The total number of moles of chlorine gas evolved isA. 0.5B. 1C. 2D. 3 |
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Answer» Correct Answer - B Mol of `NaCl=4xx0.5=2mol` No. of mole of `Cl_(2)` evolved `=(1)/(2)xxmol` of NaCl `=(1)/(2)xx2=1` mol |
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| 18. |
An electric current is passed through three cells connected in series containing `ZnSO_(4)`, acidulated water and `CuSO_(4)` respectively. What amount of Zn and `H_(2)` are liberated when 6.25 g of Cu is deposited? Eq. wt. of `Cu` and `Zn` are 31.70 and 32.6 respectively. |
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Answer» `because` Eq. of `Cu=Eq`. Of `Zn=Eq`. Of `H_(2)` `(6.25)/(31.70)=(W_(Zn))/(32.6)=(W_(H_(2)))/(1)` |
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| 19. |
The limiting molar conductivities `Lambda^@` for `NaCL, KBr` and `KCI` are ` 126, 152` and `150 S cm^2 ,ol^(-1)` respectively . The `Lambda^@` fro `NaBr S cm^2 "mol"^(-1)` is :A. `128Scm^(2)mol^(-1)`B. `176Scm^(2)mol^(-1)`C. `278Scm^(2)mol^(-1)`D. `302Scm^(2)mol^(-1)` |
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Answer» Correct Answer - A `overset(0)(^^)_(NaBr)=126+152-150=128Scm^(2)mol^(-1)` |
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| 20. |
Specific conductance of pure water at `25^(@)C` is `0.58 xx 10^(-7)` mho `cm^(-1)`. Calculate ionic product of wter `(K_(w))` if ionic conductances of `H^(+)` and `OH^(-)`ions at infinite are `350` and `198` mho `cm^(2)` respectively at `25^(@)C` |
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Answer» Correct Answer - `1xx10^(-14)("mole"//"litre"^(2))` `(^^)_(m)^(infty)=350+198=(0.58xx10^(-7)xx1000)/(M)impliesM=1.06xx10^(-7)` and `K_(w)=M^(2)=1.12xx10^(-14)` |
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| 21. |
Two half cell reactions of an electrochemical cell are given below : `MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-),toMn^(2+)(aq)+4H_(2)O(l),E^(@)=+1.51V` `Sn^(2+)(aq)toSn^(4+)(aq)+2e^(-),E^(@)=+0.51V` Construct the redox equation from the two half cell reactions and predict if theis reaction favours formation of reaction or product shown in the equation. |
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Answer» over all redox equation is `5Sn^(2+)(aq)+2MnO_(4)^(-)(aq)+16H^(+)(aq)to5Sn^(4+)(aq)+2Mn^(2+)(aq)+8H_(2)O(l)` `E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)=1.51-(-0.15)=+1.66V` The reaction favours formation of products. |
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| 22. |
At equimolar concentration of `Fe^(2+)` and `Fe^(3+)`, what must `[Ag^(+)]` be so that the voltage of the galvanic cell made from the `(Ag^(+) | Ag)` and `(Fe^(3+)|Fe^(2+)` electrodes equals zero? `Fe^(2+) + Ag^(+) rightarrow Fe^(3+) + Ag` `E_(Ag^(+), Ag)^(@)`= 0.7991, `E_(Fe^(3+)//Fe^(2+))^(@) = 0.771`A. 0.34B. 0.44C. 0.47D. 0.61 |
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Answer» Correct Answer - A `0=(0.771+0.7991)-(0.0591)/(1)log((1)/(x))implies0=0.0281+0.051logX` `logX=-(0.0281)/(0.0591)impliesX=0.335M` |
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| 23. |
Molar conductances of `BaCl_(2), H_(2)SO_(4)` and `HCl` at infinite dilutions are `X_(1), x_(2)` and `X_(3)` respectively. Equivalent conductance of `BaSO_(4)` at infinite dilution is :A. `([x_(1)+x_(2)-x_(3)])/(2)`B. `[(x_(1)-x_(2)-x_(3)])/(2)`C. `2(x_(1)+x_(2)-2x_(3))`D. `([x_(1)+x_(2)-2x_(3)])/(2)` |
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Answer» Correct Answer - D `^^_(m_(i)BaSO_(4)=(x_(1)+x_(2)x-2x_(3))implies^^_(eq.BaSO_(4))=(^^_(eq.BaSO_(4))/(n-"factor")` `^^_(eq.BaSO_(4))=((x_(1)+x_(2)-2x_(3)))/(2)` |
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| 24. |
`MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H_(2)O`, `E^(@) = 1.51V` `MnO_(2) + 4H^(+) + 2e^(-) righarrow Mn^(2+) + 2H_(2)O` `E^(@) = 1.23V` `E_(MnO_(4)^(-)|MnO_(2)`A. 1.70 VB. 0.91 VC. 1.37 VD. 0.548 V |
| Answer» Correct Answer - A | |
| 25. |
`MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H-(2)O`, If `H^(+)` concentration is decreased from 1 M to `10^(-4)` M at `25^(@)C`, whereas concentration of `Mn^(2+)` and `MnO_(4)^(-)` remains 1M, then:A. The potential decreases by 0.38 V with decrease in oxidising powerB. The potential increases by 0.38 V with increase in oxidising powerC. The potential decreases by 0.25 V with decrease inn oxidising powerD. the potential decreases by 0.38 V without affecting oxidising power |
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Answer» Correct Answer - A `MnO_(4)^(-)+8H^(+)+5e^(-)toMn^(2+)+4H_(2)O` `E_(1)=E^(@)-(0.0591)/(5)log(([Mn^(2+)])/([MnO_(4)^(-)]xx1^(8)))` `E_(2)=E^(@)-(0.0591)/(g)log(([Mn^(2+)])/([MnO_(4)^(-)]xx(10^(-4))^(8)))=-(0.591)/(5)xx32=-0.37824` `E_(1)-E_(2)=0.38"volt"` |
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| 26. |
Standard electrode potential are `E_(Fe^(2+)|Fe)^(@)=-0.44V,E_(Fe^(3+)|Fe^(2+))^(@)=0.77V`. If `Fe,Fe^(2+)` and `Fe^(3+)` are kept together, thenA. `Fe^(3+)` increasesB. `Fe^(3+)` decreasesC. `Fe^(2+)` increasesD. `Fe^(2+)` decreases |
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Answer» Correct Answer - B::C `Fe^(2+)+Fe^(2+)toFe^(3+)+Fe` (non spontaneous) |
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| 27. |
`lamda_(m)^(0)Na^(+)=150Omega^(-1)cm^(2)"mole"^(-1)` : `lamda_(eq)^(0)Ba^(2+)=100Omega^(-1)cm^(2)aq^(-1)` `lamda_(eq)^(0)SO_(4)^(2-)=125Omega^(-1)cm^(2)eq^(-1) : `lamda_(m)^(0)Al^(3+)=300Omega^(-1)cm^(2)"mole"^(-1)` `lamda_(m)^(0)NH_(4)^(+)=200Omega^(-1)cm^(2)"mole"^(-1)` : `lamda_(m)^(0),Cl^(-)=150Omega^(-1)cm^(2)"mole"^(-1)` then calculate (a). `lamda_(eq)^(0),Al^(3+)` (b). `lamda_(eq)^(0)Al_(2)(SO_(4))^(3)` (c). `lamda_(m)^(0)(NH_(4))NaCl` (d). `lamda_(0)^(0)NaCl,BaCl_(2).6H_(2)O` (e). `lamda_(m)^(0),(NH_(4))_(2)SO_(4)Al_(2)(SO_(4))_(3).24H_(2)O` ltbr. (f). `lamda_(eq)^(0)NaCl` |
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Answer» (a) `lamda_(eq)^(0)Al^(3+)=(300)/(3)=100` (b). `lamda_(eq)^(0)Al_(2)(SO_(4))_(3)` (c). `lamda_(m)^(0)(NH_(4))_(2)SO_(4)=2xx200+2xx125=650` (d). `lamda_(m)^(0)NaCl.BaCl_(2).6H_(2)O=150+200+3xx150=800r^(-1)` (e). `lamda_(m)^(0),(NH_(4))SO_(4)Al_(2)(SO_(4))_(3).24H_(2)Olamda_(m)^(0)(NH_(4))_(2)=400=600+4xx250=2000` (f).`lamda_(eq)^(0)NaCl=300Omega^(-1)cm^(2)eq^(-1)` |
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| 28. |
An electric current is passed through electrolytic cells in series one containing `Ag(NO_(3))`(eq) and other `H_(2)SO_(4)` (aq). What volume of `O_(2)` measured at `25^(@)C` and `750`mm Hg pressure would be liberated form `H_(2)SO_(4)` if (a) one mole of `Ag^(+)` is deposited from `AgNO_(3)` solution (b) `8 xx 10^(22)` ions of `Ag^(+)` are deposited from `AgNO_(3)` solution. |
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Answer» Correct Answer - (a) `V(O_(2))=6.2L` , (b). `V(O_(2))=0.823L` (i). Equivalents of `Ag^(+)=` Equivalent of `O_(2)` `1=` equivalent of `O_(2)` `therefore` mass of `O_(2)=(1)/(4)=(PV)/(RT)` `V=(1)/(4)xx(0.081xx298xx760)/(750)=6.5lit`. (ii). Moles of `O_(2)=("equivalent of "O_(2))/(4)=(8x10^(22)//6.023xx10^(23))/(4)=0.0332`. `V=(0.0332xx0.0821xx298xx760)/(750)=0.823lit` |
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| 29. |
For the cell (at 298K) `Ag(s) | AgCl(s) | Cl^(-)(aq)|| AgNO_(3)(aq) | Ag(s)` Which of the following is correct?A. The cell emf will be zero when `[Ag^(+)]_(a)=[Ag^(+)]_(c)([Ag^(+)]` in anodic compartment`=[Ag^(+)]` in cathode compartment)B. The amount of `AgCl(s)` precipitate in anodic compartment will decrease with the working of the cell.C. The concentration of `[Ag^(+)]=` constant, in anodic compartment during work of cell.D. `E_(cell)=E_(Ag^(+)|Ag)^(0)-E_(Cl^(-)|AgCl|Ag)^(0)-(0.059)/(1)log((1)/(|Cl^(-)|_(a)))` |
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Answer» Correct Answer - A `E_(cell)=E_(cell)^(@)-(0.059)/(1)log((1)/([Cl^(-)]_(a)[Ag^(+)]_(c)))`, at `[Ag^(+)]_(c)=[Ag^(+)]_(a)` equilibrium will achieve and `K=K_(sp)=[Ag^(+)]_(a)[Cl^(-)]_(a)` |
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| 30. |
A certain electricity deposited `0.54g` of Ag from `AgNO_(3)` solution what volume of hydrogen will the same quantity of electricity liberate at `27^(@)C` and `728mmMg` pressure? |
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Answer» Correct Answer - `V_((H_(2)))=64.28mL` `Ag^(+)e^(-)toAg` `(0.54)/(108)=5xx10^(-3)mol` `2H_(2)O+2e^(-)toH_(2)+2OH^(-)` `n_(H_(2))=2.5xx10^(-3),V_(H_(2))=(2.5xx10^(-3)xx24.63)/((728)/(760))=64.28ml`. |
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| 31. |
The emf of the cell `Ag|AgI|CI(0.05M)||AgNO_(3)(0.05M)|Ag` is 0.79 V. Calculate the solubility product of AgI. |
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Answer» Correct Answer - `K_(SP)=1xx10^(-18)` `AgtoAg_((c))^(+)+e^(-),Cxx(0.05)K_(SP)` `Ag_((0.05M))^(+)+e^(-)toAg` if assume as concentration cell `Ag_((0.05M))^(+)toAg_((c))^(+)` `E=0.79=0-(0.0591)/(1)log((K_(sp))/(5xx10^(-2)xx5xx10^(-2)))` `(K_(SP))/(25xx10^(-4))=4xx10^(-14)impliesK_(SP)=1xx10^(-16)` |
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| 32. |
The same quantity of electrical charge deposited `0.583g` of Ag when passed through `AgNO_(3),AuCl_(3)` solution calculate the weight of gold formed. (At weighht of `Au=197gmol)` |
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Answer» Given: `W_(ag)=0.583`gm Asked: `W_(Au)=?` Formula used: `(W_(Ag))/(Eq. Wt. of Ag)=(W_(Au))/(Eq. Wt. of au)` Explanation: `W_(ag)="weight of Ag deposited", W_(Au)="weight of Au deposited"` `E_(Ag)="equivalent weight of Ag", E_(Au)="Equivalent weight of Au"` Substitution and calculation `(W_(Ag))/(Eq. Wt. of Ag)=(W_(Au))/(Eq. Wt. of Au)` `Eq. wt. of Ag=108`, `Eq. wt. of Au=(197)/(3)` `(0.583)/(109)=(W_(Au))/(197//3)),W_(Au)=(0.583xx197)/(108xx3)=0.354` g |
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| 33. |
In `H_(2) - O_(2)` fuel cell the reaction occuring at cathode is:A. `2H_(2)O+O_(2)+4e^(-)to4OH^(-)`B. `2H_(2)+O_(2)to2H_(2)O(l)`C. `H^(+)+OH^(-)toH_(2)O`D. `H^(+)+e^(-)to(1)/(2)H_(2)`. |
| Answer» Correct Answer - A | |
| 34. |
Three faradays of electricity was passed through an aqueous solution of iron (II) bromide. The mass of iron metal (at mass 56) deposited at the cathode is:A. 56 gB. 84 gC. 112 gD. 168 g |
| Answer» Correct Answer - B | |
| 35. |
In a electrolytic cell of `Ag//AgNO_(3)//Ag`, when current is passed, the concentration of `AgNO_(3)`A. increasesB. decreasesC. remains sameD. none of these |
| Answer» Correct Answer - C | |
| 36. |
During discharge of a lead storage cell the density of sulphuric acid in the cell:A. increasingB. decreasingC. remains unchangedD. initially increases but decreases subsequently. |
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Answer» Correct Answer - B `H_(2)-O_(2)` fuel cell at anode: `2OH^(-)+H_(2)to2H_(2)O+2e^(-)` At cathode: `2H_(2)O+O_(2)+4e^(-)to4OH^(-)` |
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| 37. |
A current of `2` A was passed for `1 h` through a solution of `CuSO_4. 0.237g` of `Cu^(2+)` ions was discharged at cathode . The current efficiency is .A. `42.2%`B. `26.1%`C. `10%`D. `40.01%` |
| Answer» Correct Answer - C | |
| 38. |
A lead storage cell is discharged which causes the `H_(2)SO_(4)` electrolyte to change from a concentration of `34.6%` by weight (density `1.261gml^(-1)` at `25^(@)C`) to one of `27%` by weight. The original volume of electrolyte is one litre. Calculate the total charge released at anode of the battery. Note that the water is produced by the cell reaction as `H_(2)SO_(4)` is used up. over all reaction is. `Pb(s)+PbO_(2)(s)+2H_(2)SO_(4)(l)to2PbSO_(4)(s)+2H_(2)O(l)` |
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Answer» Before the discharge of lead storage battery. mass of solution `=1000xx1.261=1261g` mass of `H_(2)SO_(4)=(1261xx34.6)/(100)=436.3g` mass of water `=1261-436.3=824.3g` after the discharge of lead storage battery. Let the mass of `H_(2)O` produce as a result of net reaction during discharge `(Pb+PbO_(2)+2H_(2)SO_(4)to2PbSO_(4)+2H_(2)O)` is `xxg` `therefore` moles of `H_(2)O` produced `=(x)/(18)=` moles of `H_(2)SO_(4)` consumed Mass of `H_(2)SO_(4)` consumed `=(x)/(18)xx98` Now, mass of solution after discharge `=1261-(98x)/(18)+x` `%` by the mass of `H_(2)SO_(4)` after discharge `=("mass of " H_(2)SO_(4)" left")/("mass of solution after discharge")xx100=27` `=(436.3-(98x)/(18))/(1261-(98x)/(18)+x)xx100=27` `thereforex=22.59g` |
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| 39. |
The molar conductivities `Lambda_(NaOAc)^@` and `Lambda_(HCI)^@` at infinite dilution is watter at `25^@C` are `91.0` and `426.2 S cm^@ //`mol respectively. To calculate `Lambda_(HOAc,)^2` the additional value required is:A. `^^_(H_(2)O)^(0)`B. `^^_(KCl)^(0)`C. `^^_(NaOH)^(0)`D. `^^_(NaCl)^(0)` |
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Answer» Correct Answer - D `CH_(3)COONa+HCltoCH_(3)COOH+NaCl`" From the reaction"` `overset(0)^^CH_(3)COONa+overset(0)(^^)_(HCl)=overset(0)(^^)_(CH_(3))COOH+overset(0)(^^)_(NaCl)oroverset(0)(^^)_(CH_(3)COOH)=overset(0)(^^)_(CH_(3)COONa)+overset(0)(^^)_(HCl)-overset(0)(^^)_(NaCl)` Thus to calculate the value of `overset(0)(^^)CH_(3)COOH` one should know know the value of `overset(0)(^^)_(NaCl)` along with `overset(0)(^^)_(CH_(3)COONa)` and `overset(0)(^^)_(HCl^(-))` |
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| 40. |
The cell , `Zn | Zn^(2+) (1M) || Cu^(2+) (1M) Cu (E_("cell")^@ = 1. 10 V)`, Was allowed to be completely discharfed at `298 K `. The relative concentration of `2+` to `Cu^(2+) [(Zn^(2=))/(Cu^(2+))]` is :A. `10^(37.3)`B. `9.65xx10^(4)`C. antilog `(24.08)`D. `37.3` |
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Answer» Correct Answer - A `0=+1.1-(0.0591)/(6)log(([Cr^(+3)])/([Cu^(2+)])),log(([Zn^(2+)])/[[Cu^(2+)]))=37.3,(([Zn^(2+)])/([Cu^(2+)]))=10^(37.3)` |
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| 41. |
At `25^(@)C, {:(Ag+I^(-)rarr,AgI+e,,E^(@) =0.152V),(Agrarr,Ag.^(+)+e,,E^(@) =- 0.80V):}` The `log K_(sp)` of `AgI` is: `((2.303RT)/(F)=0.059)`A. `-8.12`B. `+8.612`C. `-37.83`D. `-16.13` |
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Answer» Correct Answer - D `0.152=-0.8-(0.059)/(1)logK_(SP):logK_(SP)=-16.11` |
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| 42. |
The standard oxidation potential for the half-cell `NO_(2)^(-)(g)+H_(2)OtoNO_(3)^(-)(aq)+2H^(+)(aq)+2e` is `-0.78V`. Calculate the reduction in 9 molar `H^(+)` assuming all other species at unit concentration. What will be the reduction potential in neutral medium? |
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Answer» Correct Answer - `0.836"volt", 1.1937"volt"` `E_("oxidation")=-0.78-(0.0591)/(2)log9^(2)=-0.78-(0.0591)/(2)xx2xxlog9=-0.835"volt"` `E_("reduction")=-E_("oxidation")=0.836"volt"` In neutral medium `E_("Oxidation")=-0.78-(0.0591)/(2)log(10^(-7))^(2)=-1.1937"volt"` |
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| 43. |
Given,`{:(E_(Fe^(3+) //Fe)^o + 3e Cr E^o =- 0.036 V),(E_(Fe^(3+) //Fe)^o =- 0.439 V):}` The value of standard electrode ptoential for the charge,A. 0.385 VB. 0.770 VC. `-0.270 V`D. `-0.072` |
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Answer» Correct Answer - B `Fe^(3+)+3e^(-)toFe" "triangleG_(1)=-3xxFxxF_(Fe^(3+)//Fe)^(0)` `Fe^(2+)toFe" "triangleG_(2)=-2xxFxxE_(Fe^(2+)//Fe)^(0)` `Fe^(3+)+e^(-)toFe^(2+)" "triangleG=triangleG_(1)-triangleG_(2)` `triangleG=3xx0.036F-2xx0.439xxF=-1xxE_((Fe^(3+)//Fe^(+2)))^(0)xxF` `E_((Fe^(3+)//Fe^(+2)))^(0)=2xx0.439-3xx0.036` `=0.878-0.108` `=0.770V` |
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| 44. |
Electrode potential data are given below: `Fe^(3+)(aq)+e^(-)rarrFe^(2+)(aq):E^(@)=+0.77V` `Al^(3+)+3e^(-)rarrAl(s):E^(@)=-1.66V` `Br_(2)(aq)+2e^(-)rarr2Br^(-)(aq):E^(@)=+1.08V`, Based on the data, the reducing power of `Fe^(2+)` Al and `Br^(-)` will increase in the orderA. `Br^(-)ltFe^(2+)ltAl`B. `Fe^(2+)ltAlltBr^(-)`C. `Al ltBr^(-)lt Fe^(2+)`D. `AlltFe^(2+)ltBr^(-)` |
| Answer» Correct Answer - A | |
| 45. |
On the basis of the standard electroe potential values stated for acid solution, predict whether, `Ti^(4+)` species may be used to oxidise `Fe^(II) " to " Fe^(III)`. Given. `Ti^(4+) +_e^(-) to Ti^(3+),E^(ɵ)=+0.01 V,Fe^(3+)+e^(-) to Fe^(2+),E^(ɵ)=+0.77 V` |
| Answer» Bacause standard electrode potential `Ti^(4+)//Ti^(3+)` is less than of `Fe^(3+)//Fe^(2+)` so it cannot oxides `Fe^(II)` to `Fe^(III)` | |
| 46. |
The standard reduction potential of `TiO^(2+)` and `Ti^(3+)` are given by `TiO_(2+)+2H^(+)+e^(-)toTi^(3+)+H_(2)O" "E^(@)=0.10V` `Ti^(3+)+3e^(-)toTi" "E^(@)=-1.21V` |
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Answer» Correct Answer - `-0.8825"volt"` `TiO^(2+)+2H^(+)+e^(-)toTi^(3+)+H_(2)O,0.1V" "triangleG_(1)^(0)=-2xxFxx0.1` `Ti^(3+)+3e^(-)toTi-1.21V" "triangleG_(2)^(@)=-3xx(-1.21)xxF` `TiO^(2+)2H^(+)+4e^(-)toTi+H_(2)O` `-4xxE^(@)xxF=-1xx0.1xxF+-3xx(-1.21)xxF` `E^(@)=(0.1-3.63)/(4)=-0.8825"volt"` |
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| 47. |
Calculate the equilibrium concentration of all ions in an ideal solution prepared by mixing `25.00mL` of `0.100M TI^(+)` with `25.00 mL` of `0.200M Co^(3+)` `E^(@) (TI^(+)//TI^(3+)) =- 1.25 V, E^(@) (Co^(3+)//Co^(2+)) = 1.84V` |
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Answer» Correct Answer - `[Tl^(+)]=10^(-8)M;[Co^(3+)]=2xx10^(-8)M,[Tl^(+3)]=0.05M;[Co^(2+)]=0.1M` `Tl^(+)toTl^(3+)+2e^(-)` `underline(2Co^(3+)+2e^(-)to2Co^(2+))` `underline(Tl^(+)+2Co^(3+)toTl^(3+)+2Co^(2+)), " "E^(@)=-1.25+1.84=0.59"volt"` `{:(2.5,,5,,0,,0),(x,,2x,,2.5,,5):}` `=0=0.59-(0.059)/(2)log((2.5xx5^(2))/("x"xx(2x)^(2))` `(2.5xx(5)^(2))/(4x^(3))=10^(20)` `[Tl^(+)]=10^(-8)M,[Co^(3+)]=2xx10^(-8)M` |
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| 48. |
Conductance (Siemens, S) is directly proportional to the area of the vessel and the concentration of solution in it and is inversely proprtional to the length of the vessel, then the unit of constant of proportionlity is :A. `Smmol^(-1)`B. `Sm^(2)mol^(-1)`C. `S^(-2)m^(2)mol`D. `S^(2)m^(2)mol^(-1)`. |
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Answer» Correct Answer - B `K=(1)/(rho)=(1)/(R)(l)/(A)` |
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| 49. |
A silver coulometer is in series with a cell electrolyzing water. In a time of 1 minute at a constant current 1.08 g silver get deposited on the cathode of the coulometer. What total volume (in mL at 1atm, 273K) of the gases would have produced in other cell. In this cell that the anodic and cathodic efficiencies were `90%` and 80% respectively. Assume the gases collected are dry. (Ag = 108) (molar volume of any ideal gas at 1atm and 273K = 22.4L) |
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Answer» Correct Answer - 140 Charge passed `=0.01` faraday At the anode `(H_(2)Oto(1)/(2)O_(2)+2H^(+)+2e^(-))` with `90%` efficiency `0.01xx0.9` F have been used and will produce `(1)/(4)xx0.01xx0.9` mole of `O_(2)` i.e., `0.00225molO_(2)` At the cathode `2H_(2)Ooverset(+2e^(-))toH_(2)+2OH^(-)` Moles of `H_(2)` produced `=(0.01xx0.8)/(2)mol=0.004mol` Total moles produced of gases `=0.004+0.00225=0.00625mol` vol. at STP `=0.00625xx22400mL=140mL` |
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| 50. |
Calculate the standard cell potential of the galvanic cell in which the following reaction takes place: `2Cr(s)+3Cd^(2+)(aq)to2Cr^(3+)(aq)+3Cd(s)` Also calcuate the `triangle_(r)G^(ɵ)` value of the reaction (given `E_(cr^(3+)//Cr)^(ɵ)=-0.74V,E_(Cd^(3+)//Cd)^(ɵ)=-0.40V` and `F=96500Cmol^(-1)` |
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Answer» `E_(cell)^(ɵ)=E_(Cd^(2+)//Cd)^(@)-E_(Cr^(3-)//Cr)^(@)` `=-0.40-(-0.74)` `=-0.40+0.74=+0.34V` the half cell reaction are `2Cr(s)to2Cr^(3+)+6e^(-)` `underline(3Cd^(2+)+6e^(-)to3Cd(s))` `underline(2Cr(s)+3Cd^(2+)to2Cr^(3+)+3Cd(s))` `therefore` No. of electrons `n=6` `triangle_(r)G^(ɵ)=nFE^(ɵ)` `=-6xx96500xx0.34` `=-196.86kJmol^(-1)` |
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