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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. |
How many faraday are required to reduce one mol of `MnO_(4)^(-)` to `Mn^(2+)`:A. 1B. 2C. 3D. 5 |
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Answer» Correct Answer - D `8H^(+)+5e^(-)+underset((1"mole"))MnO_(4)^(-)toMn^(+2)+4H_(2)O` 5 mole `e^(-)=5` faraday. |
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| 2. |
`E_(Mn^(2+)//MnO_(4)^(-))^(0)=-1.51V` `E_(MnO_(2)//Mn^(+2))^(0)=+1.23V` `E_(MnO_(4)^(-)//MnO_(2)=?` (All in acidic medium) |
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Answer» `4H_(2)O+Mn^(2+)toMnO_(4)^(-)+8H^(+)+5e^(-)" triangleG_(1)` (i). `lArrMnO_(4)^(-)+8H^(+)+5e^(-)to4H_(2)O+Mn^(2+)" " -triangleG_(1)` `2e^(-)+MnO_(2)+4H^(+)toMn^(2+)+2H_(2)O" "triangleG_(2)` (ii). `lArr2H_(2)O+Mn^(2+)toMnO_(2)+4H^(+)+2e^(-)" "-triangleG_(2)` (iii). `lArr4H^(+)+MnO_(4)^(-)+3e^(-)toMnO_(2)+2H_(2)O" "triangleG_(3)` `(i)+(ii)=(iii)` `triangleG_(3)=-triangleG_(2)-triangleG_(2)` `-3E_(3)F=5E_(1)^(0)F+2E_(2)^(0)F` `E=(-[5E_(1)+2E_(2)])/(3)=(-[5(-1.51)+2(1.23)])/(3)=(-[-7.55+2.46])/(3)=(+5.09)/(3)=1.69V` |
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| 3. |
Consider the cell: `Mg(s)|Mg^(2+)(0.13M)||Ag^(+)(1.0xx10^(-4))M|Ag(s)` its e.m.f. is `2.96V.` calculate `E_("cell")^(@)` `(R=8.314JK^(-1),1F=96500Cmol^(-1))` |
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Answer» Given: For electro chemical cell `Mg(s)|Mg^(2+)(0.13M)||Ag^(+)(1.0xx10^(-4))M|Ag(s),E_(cell)=2.96V` Asked `E_(cell)=E_(cell)^(@)-(0.0591)/(n)log((["product"])/(["reactant"]))` Expalanation `E_(cell)^(@)="standard" e.m.f of cell `E_(cell)=e.m.f` of cell Substitution and calculation `Mg(s)toMg^(2+)(aq)+2e^(-)` underline(2Ag^(+)(aq)2e^(-)to2Ag(s))` `underline(Mg(s)+2Ag^(2+)(aq)toMg^(2+)(aq)+2Ag(s))` `E_(cell)^(@)=E_(cell)^(@)=-(0.591)/(2)log(([Mg^(2+)])/([Ag^(+)]^(2)))implies+2.96V=E_(cell)^(@)-(0.0591)/(2)log((0.13)/((10^(-4))^(2)))` `E_(cell)^(@)=3.169V` |
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| 4. |
How many faradays of charge are required to convert: (a). 1 mole of `MnO_(4)^(-)` to `Mn^(2+)` ion, (b). 1 mole of `Cr_(2)O_(7)^(2-)` to `Cr^(2+)` ion |
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Answer» (a). `MnO^(-)toMn^(2+), Mn^(7+)+5e^(-)toMn^(2+)` i.e., when 1 mole of `MnO_(4)^(-)` changes to `Mn^(2+)` 5 faradays of charge is required. (b). `Cr_(2)O_(7)^(2-)to2Cr^(3+),2Cr^(6)+.^(+)6e^(-)to2Cr^(3+` i.e., 6 faradays of charge is required. |
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| 5. |
During electrolysis of aqueous `CuBr_(2)` using `Pt` electrode,A. `Br_(2)` gas is evolved at the anodeB. `Cu(s)` is deposited at the cathodeC. `Br_(2)` gas is evolved at anode and `H_(2)` gas at cathodeD. `H_(2)` gas is evolved at anode. |
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Answer» Correct Answer - A::B (A,B) at cathode: `Cu^(2+)+2e^(-)toCu(s)` `underline(2Br^(-)+toBr_(2)2e^(-))` `Cu^(2+)+2Br^(-)toBr_(2)+Cu(s)`. |
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| 6. |
For the cell (at 1bar `H_(2)` pressure) Pt|`H_(2)`(g) HX(m_(1), NaX(m_(2), NaCl(m_(3)|AgCl|Ag|Pt is found that the value of `E-E^(@) + RTF^(-1)` in `[(m_(HX).mCl^(-))/(m_(X^(-)))]` approaches 0.2814 in the limit of zero concentration. Calcualte `K_(a)` for the acid HX at `25^(@)C` expressing your answer as `10^(7)`K_(a)` |
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Answer» Correct Answer - 174 `E=E^(@)+RTF^(-1)ln([X^(-)])/(K_(a)[HX][Cl^(-)])impliesE-E^(@)+RTF^(-1)ln([HX][Cl^(-)])/([X^(-)])RTF^(-)ln(1)/(K_(a))` `impliesl n(1)/(K_(a))=(96500)/(8.315xx298)(0.2814)impliesK_(a)=1.74xx10^(-5)` |
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| 7. |
Metallic sodium cannot be prepared from electrolysis of an aqueous solution of NaCl because the `SRPH_(2)OltSRPNa^(+)` |
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Answer» Correct Answer - F SRP `H_(2)OgtSRPNa^(+)` so `H_(2)O` will be reduced in preference to `Na^(+)` ion. |
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| 8. |
At 300 K specific conductivity of ethanol is `4xx10^(-10)mhocm^(-1)`. The ionic conductances of `H^(+),C_(2)H_(5)O^(-)` at his temperature is 300 and 100 `mhocm^(2)" equivalent"^(-1)` respectively. Then the negative logarithm of ionic product of alcohol will be 18. |
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Answer» Correct Answer - T `lamda^(infty)=lamda_(H^(+))^(infty)+lamda_(C_(2)H_(5)O^(-))^(infty)=400` `thereforelamda=kxx(1000)/(C)" "C=[H^(+)]=[.^(-)OC_(2)H_(5)]` `thereforeC=(4xx10^(-10)xx1000)/(400)=10^(-9)M` `thereforeK_("alcohol")=[H^(+)][OC_(2)H_(5)]=(10^(-9))^(2)` `thereforepK_("alcohol")=-log(10^(-18))=18` |
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| 9. |
When a solution of conductanes `1.342 mho m^(-1)` was placed in a conductivity cell with parallel electrodes the resistance was found to be `170.5` ohm. The area of the electrode is `1.86 xx 10^(-4)` sq meter. Calculate the distance between the two electrodes in meter. |
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Answer» Correct Answer - `4.25xx10^(-2)m`. `K=1.342sec^(-1)" "R=170.5Omega" "A=1.86xx10^(-4)sec.m.` `R=(deltal)/(A)=(l)/(KA)" or "l=RKA=170.5xx1.342xx186xx10^(-4)=4.25xx10^(-2)m`. |
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| 10. |
Red hot carbon will remove oxygen from the oxide AO and BO but not from MO, while B will remove oxygen from AO. The activity of metals A, B and M in decreasing order isA. `AgtBgtM`B. `BgtAgtM`C. `MgtBgtA`D. `MgtAgtB` |
| Answer» Correct Answer - C | |
| 11. |
Molar conductance of ` 0.1 M` acetic acid is ` 7 ohm^(-1) cm^2 "mol"^(-1)`. If the molar conductance to acetic acid at dinfintie dilution is `380.8 "ohm"^(-1) cm^2 "mol"^(-1)`, the value of dissociation constant will be :A. `226xx10^(-5)moldm^(-3)`B. `1.66xx10^(-3)moldm^(-1)`C. `1.66xx10^(-2)moldm^(-3)`D. `3.442xx10^(-5)moldm^(-3)` |
| Answer» Correct Answer - D | |
| 12. |
Equivalent conductance of 1M `CH_(3)COOH` is `10ohm^(-1) cm^(2) "equiv"^(-1)` and that at inifinite dilution is 200 `ohm^(-1) cm^(2) "equiv"^(-1)`. Hence, % ionisation of `CH_(3)COOJ` is:A. `5%`B. `2%`C. `4%`D. `1%` |
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Answer» Correct Answer - A `alpha=(^^_(infty)^(C))/(^^_(m)^(infty))` |
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| 13. |
How much electricity is requried in coulomb for the oxidation of 1 mol of `H_(2)O` to `O_(2)` |
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Answer» The electrode reaction for 1 mole of `H_(2)O` is `H_(2)Oto(1)(2)O_(2)+2H^(+)+2e^(-)` electricity `=2xx96500C` |
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| 14. |
Write Nernst equation for the electrode reaction : `M^(n+)(aq)+"n"e^(-)(aq) to M(s)` |
| Answer» `E_(M^(n+)//M)=E_(M^(n+)//M)^(@)+(2.303RT)/(nF)log(([Mn^(n+)])/([M]))` | |
| 15. |
How much will the reduction potential of a hydrogen electrode change when its solution initially at `pH=0` is neutralized to `pH=7` ?A. increase by `0.059V`B. decrease by `0.059V`C. increase by 0.41 VD. decrease by 0.41 V |
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Answer» Correct Answer - D `H^(+)+e^(-)to(1)/(2)H_(2),E=0-(0.0591)/(1)log_(10)((1)/([H^(+)]))=+0.591log_(10)[H^(+)]` `E_(1)=0{pH=0}` `E_(2)=+0.0591log_(10)[10^(-7)]=-0.0591xx7{atpH=7}=-0.41V` |
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| 16. |
The two aqueous solutions, `A (AgNO_3)` and `B (LiCI)`, were electrolysed using Pt electrodes. The `pH` of the resulting solutions will.A. increase in A and decrease BB. decreases in bothC. increase in bothD. decreases in A and increase B. |
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Answer» Correct Answer - D `AgNO_(3)` At cathode: `Ag^(+)+e^(-)toAg(s)` At anode: `2H_(2)OtoO_(2)+4H^(+)+4e^(-)` `therefore` at cathode pH will increase. |
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| 17. |
Three electrolytic cells A, B and C containing solutions of zinc sulphate, silver nitrate and copper sulphate, respectively are connected in series. A steady current of 1.5 ampere was passed through them until 1.45 g of silver were deposited at the cathode of cell B. How long did the current flow? What mass of copper and what mass of zinc were deposited in the concerned cells? (Atomic masses of Ag = 108, Zn = 65.4, Cu = 63.5) |
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Answer» `Ag^(+)e^(-)toAg` `because` 108 g of Ag is deposited by `1F=96500C` `therefore` 1.45g of Ag is deposited by `=(96500)/(108)xx1.45=1295.6C` `thereforet=(Q)/(I)=(1295)/(1.5)=863.75sec` Now, `Cu^(2+)+2e^(-)toCu` `therefore` Copper (Cu) deposited `=(63.5)/(2xx96500)xx1295.6=0.426g` `therefore` Zinc (Zn) deposited `=(65.4)/(2xx96500)xx1295.6=0.439g` |
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| 18. |
Mark out the correct statement(s).A. Copper metal cannot reduce iron (II) ions in acidic solution.B. sodium can be obtained by the electrolysis of aqueous solution of NaCl using Pt electrodes.C. the current carrying ions in an electrolytic cell are not necessarily discharged at the electrodes.D. Cations having more negative oxidation potential than `-0.828V` are reduced in preference to water. |
| Answer» Correct Answer - A::C::D | |
| 19. |
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S `cm^(-1)` . Calculate its molar conductivity. |
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Answer» Given `k=2.50xx10^(-2),S^(-1)cm^(-1),M=0.20molL^(-1)` `A_(m)=(kxx1000)/(M)=(1000xx2.50xx10^(-2))/(0.20)` `=125.S^(-1)cm^(2)mol^(-1)` |
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| 20. |
Express the relation between.conductivity and molar conductivity of a solution held in a cell. |
| Answer» `^^_(m)=(Kxx1000)/(M)` | |
| 21. |
Which of the following statements is/are correct?A. The conductance of one `cm^(3)` (or `1unit^(3))` of a solution is called conductivity.B. Specific conductance increases while molar conductivity decreases on progressive dilution.C. The limiting equivalent conductivity of weak electrolyte cannote bee determine exactly by extraplotation of the plot of `^^_(eq)` against `sqrt(c)`.D. The condcutance of metals is due to the movement of free electrons. |
| Answer» Correct Answer - A::C::D | |
| 22. |
In the given figure the electrolytic cell contains `1L` of an aqueous `1M` Copper `(II)` sulphate solution. If `0.4 ` mole of electrons passed through of cell, the concentration of copper ion after passage of the charge will be A. 0.4 MB. 0.8 MC. 1.0 MD. 1.2 M |
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Answer» Correct Answer - C Number of moles of `Cu^(2+)` discharged from anode `=` number of moles of `Cu^(2+)` deposited at cathode. |
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| 23. |
Mark out the correct statement(s) regarding electrolytic molar conductivity.A. it increase as temperature increases.B. It experiences resistance due to vibration of ion at the mean position.C. Increase in concentration decreases the electrolytic molar conductivity of both the strong as well as the weak electrolyte.D. |
| Answer» Correct Answer - A::C::D | |
| 24. |
The temperature coefficient of the emf i.e. `(dE)/(dT)=0.00065"volt".deg^(-1)` for the cell `Cd|CdCl_(2)(1M)||AgCl(s)` `|Ag` at `25^(@)C` calculate the entropy changes `triangleS_(298K)` for the cell reaction, `Cd+2AgCltoCd^(++)+2Cl^(-)+2Ag`A. `-105.5JK^(-1)`B. `-150.2JK^(-1)`C. `-75.7JK^(-1)`D. `-125.5JK^(-1)` |
| Answer» Correct Answer - D | |
| 25. |
The reaction : `Zn(s) + 2AgCl(g) rightarrow ZnCl_(2)(aq) + 2Ag(s)` occurs in the cell `Zn|ZnCl_(2)` (1M solution), AgCl(s) | Ag. The number of Faradays required from the external source for this reaction to occur in the cell is:A. 2B. 3C. 1D. zero |
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Answer» Correct Answer - D It is spontaneous change so no number of faraday is required. |
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| 26. |
At `Tl^(+) |Tl` couple was prepared by saturating 0.1 M KBr with TlBr and allowing the `Tl^(+)` from the relatively insoluble bromide to equilibrate. This couple was observed to have a potential of `-0443`V with respect to `Pb6(2+)|Pb` couple in which `Pb6(2+)` was 0.1 M. What is `K_(sp)` of TlBr? (Report answer in multiplication of `10^(-8)`) `(E_(Pb^(2+)//Pb)^(@) = -0.126`, `(E_(Tl^(+)//Tl)^(@) = -0.336V)` (Take antilog(0.5509) = 3.55, (2.303RT)/(F) = 0.059) [Hint: Take Pb as anode] |
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Answer» Correct Answer - 355 `Pb^(2+)+2e^(-)toPb` ltbr `E_(Pb^(2+)//Pb)^(0)=E_(Pb^(2+)//Pb)^(0)-(0.0591)/(2)log((1)/(0.1))=-0.1555"volt"` `E_(cell)=E_(Pb^(2+)//Pb)-E_(Tl^(+)//Tl)` `E_(Tl^(+)//Tl)-0.5985V` `Tl^(+)+e^(-)toTl` `E_(Tl^(+)//Tl)=E_(Tl^(+)//Tl)^(0)-0.059log((1)/((Tl^(+))))` `-0.5985=-0.336+0.059log((Tl^(+))` `Tl^(+)=3.55xx10^(-5)M` `KsP=[Tl^(+)][Br^(-)]=3.55xx10^(-5)xx0.1=3.55xx10^(-6)M^(2)=355xx10^(-8)M^(2)` |
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| 27. |
The overall formaion constant for the reaction of 6 mole of `CN^(-)` with cobalt (II) is `1xx10^(19)` the standard reduction potential constant of `[Co(CN)_(6)]^(3-)+e^(-)toCo(CN)_(6)^(4-)` is `-0.83V` Calcualte the formation constant of `[Co(CN)_(6)]^(3-)`. Given `Co^(3+)+e^(-)toCo^(2+),E^(@)=1.82V` |
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Answer» Correct Answer - `K_(f)=10^(63.915)` Anode: `[Co(CN)_(6)]^(3-)+e^(-)to[Co(CN)_(6)]^(4-)" "E_(SRP)^(0)=-0.83V` cathode: `Co^(3+)+e^(-)toCO^(2+)" "E_(SRP)^(0)=1.82V` so overall cell reaction is `Co^(3+)+[Co(CN)_(6)]^(4-)toCO^(2+)+[Co(CN)_(6)]^(3-)` `E_(cell)^(0)=E_(c)^(0)-E_(a)^(0)=1.82-(-0.83)=2.65V` `E_(cell)=E_(cell)^(0)-(0.059)/(1)log(([Co^(2+)][Co(CN)_(6)]^(3-))/([Co^(3+)][Co(CN)_(6)]^(4-)))` Now, `Co^(2+)+6CN^(-)hArr[Co(CN)_(6)]^(4-)" "K_(f_(1))=1xx10^(19)` `Co^(3+)+6CN^(-)hArr[Co(CN)_(6)]^(3-)" "K_(f_(2))` At equilibrium `E_(cell)=0` `E_(cell)^(0)=(0.059)/(1)log((k_(f_(2)))/(K_(f_(1)))`, solving we get `K_(1)=10^(63.915)` |
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| 28. |
In an electrolytic cell, the flow of electrons is formA. cathode to anode in solutionB. Cathode to anode through external supplyC. Cathode to anode through internal supplyD. Anode to cathode through internal supply |
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Answer» Correct Answer - C In electrolytic cell, flow of electron is possible from cathode to anode through internal supply. |
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| 29. |
Given: `E_(Zn^(+2)//Zn)^(@) =- 0.76V` `E_(Cu^(+2)//Cu)^(@) = +0.34V` `K_(f)[Cu(NH_(3))_(4)]^(2+) = 4 xx 10^(11)` `(2.303R)/(F) = 2 xx 10^(-4)` At what concentration of `Cu^(+2)` emf of the cell will be zero (at 298K) and concentration of `Zn^(+2)` is remains same:A. `1.19xx10^(-37)`B. `1.19xx10^(-20)`C. `3.78xx10^(-4)`D. `0.0068` |
| Answer» Correct Answer - A | |
| 30. |
For the cells in opposition, Zn(s) | ZnCl_(2)(sol).|AgCl(s)|Ag|AgCl(s)| `C-(1) = 0.02 M` , `ZnCl_(2)(sol)`| Zn(s) `C_(2) = 0.5M` Find out the emf (in millivolt) of the resultant cell. (take log 2 `=0.3,(RT)/F` at 298 K = 0.060) |
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Answer» Correct Answer - 42 As cell reaction is 1st cell: `Zn+2AgCltoZnCl_(2)+2Ag` 2nd cell: `underline(ZnCl_(2)+2Agto2AgCl+Zn)` overall `ZnCl_(2)(C_(2))toZnCl_(2)(C_(1))` `E=(RT)/(2F)ln ((0.5)/(0.02))V=[(0.059)/(2)log((0.5)/(0.02))]V=42` |
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| 31. |
A solution of sodium sulphate in qater is electrolysed using inert electrodes, The products at the cathode and anode are respectively.A. `H_(2),O_(2)`B. `O_(2),H_(2)`C. `O_(2),Na`D. `O_(2),SO_(2)` |
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Answer» Correct Answer - A At anode `2H_(2)OtoO_(2)+4H^(+)+4e^(-)` (not `SO_(4)^(2-))` At cathode: `2H_(2)O+2e^(-)toH_(2)+2OH^(-)` (not `Na^(+))` |
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| 32. |
Consider the following redox reaction: `2lrCl_(6)^(3-)+3HCOOHto2lr+3CO_(2)+12Cl^(-)+6H^(+)` Given: `CO_(2)+2H_(3)O^(+)+2etoHCOOH+2H_(2)O" "E^(@)=-0.20V` `lrCl_(6)^(3-)+3etolr+6Cl^(-)" "E^(@)=0.77V` (a). Determine standard state emf of cell. (b). Is this reaction thermodynamically spontaneous as written? Briefly explain. |
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Answer» Correct Answer - (a). `E_(cell)^(@)=0.97` (b). Reaction is thermodynamically spontaneous since `triangleG^(@)lt0`. n |
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| 33. |
Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potentials are given below: `MnO_(4)^(-)(aq) +8H^(+)(aq) +5e^(-) rarr Mn^(2+)(aq) +4H_(2)O(l) E^(@) = 1.51V` `Cr_(2)O_(7)^(2-)(aq) +14H^(+) (aq) +6e^(-) rarr 2Cr^(3+)(aq) +7H_(2)O(l), E^(@) = 1.38V` `Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq), E^(@) = 0.77V` `CI_(2)(g) +2e^(-) rarr 2CI^(-)(aq), E^(@) = 1.40V` Identify the only correct statement regarding quantitative estimation of aqueous `Fe(NO_(3))_(2)`A. `MnO_(4)^(-)` can be used in aqueous `HCl`B. `Cr_(2)O_(7)^(2-)` can be used in aqueous `HCl`C. `MnO_(4)^(-)` can be used in aqueous `H_(2)SO_(4)`D. `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_(2)SO_(4)`. |
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Answer» Correct Answer - A `MnO_(4)^(-)` ion can oxidise both `Fe^(2+)` to `Fe^(3+)` as well as `Cl^(-)` to `Cl_(2)`. So `Fe(NO_(3))_(2)` cannot be estimated quantitatively with `MnO_(4)^(-)` ion in HCl. `E_(cell)^(@)` for the cell Pt, `Cl_(2)(g)(1atm)|Cl^(-)(aq)||MnO_(4)^(-)(aq)|Mn^(2+)(aq)`. is equal to `(1.51-1.4)=0.11V` |
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| 34. |
Redox reactions play a pivotal role in chemistry and biology. The values standard redox potential `(E^(c-))` of two half cell reactions decided which way the reaction is expected to preceed. A simple example is a Daniell cell in which zinc goes into solution and copper sets deposited. Given below are a set of half cell reactions `(` acidic medium `)` along with their `E^(c-)(V` with respect to normal hydrogen electrode `)` values. Using this data, obtain correct explanations for Question. `I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.54` `Cl_(2)+2e^(-) rarr 2Cl^(c-), " "E^(c-)=1.36` `Mn^(3+)+e^(-) rarr Mn^(2+), " "E^(c-)=1.50` `Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.77` `O_(2)+4H^(o+)+4e^(-) rarr 2H_(2)O, " "E^(c-)=1.23` Among the following, identify the correct statement.A. Chloride ion is oxidised by `O_(2)`B. `Fe^(2+)` is oxidised by iodineC. Iodine ion is oxidised by chlorineD. `Mn^(2+)` is oxidised by chlorine |
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Answer» Correct Answer - C `Cl_(2)+2I^(-)toI_(2)+2Cl^(-)` `E^(@)=1.36+(-0.54)=0.82V(+ve)`. Spontaneous. |
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| 35. |
`Zn|Zn^(2+)(c_(1)) || Zn^(2+)(c_(2))|Zn`. For this cell `DeltaG` is negative if `:`A. `C_(1)=C_(2)`B. `C_(1)gtC_(2)`C. `C_(2)gtC_(1)`D. none |
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Answer» Correct Answer - C `ZntoZn_((c_(1)))^(2+)+2e^(-)` `Zn_((c_(2)))^(2+)+2e^(-)toZn` `Zn_((c_(2)))^(2+)hArrZn_((c_(1)))^(2+)` `E=0-(0.0591)/(2)log((C_(1))/(C_(2)))` `Eto+ve` when `C_(1)ltC_(2)` |
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| 36. |
Any redox reaction would occur spontaneously, if:A. the free energy change `(triangleG)` is negativeB. The `triangleG^(@)` is positiveC. The cell e.m.f. `(E^(@))` is negativeD. the cell e.m.f is positive |
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Answer» Correct Answer - A::D `triangleG^(0)=-nFE_(cell)^(0)` if `E_(cell)^(0)=+ve`. Then `triangleG^(@)=-ve` and reaction is spontaneous. |
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| 37. |
The standard emf of the cell, `Ni|Ni^(2+)(1.0M)||Ag^(+)(1.0M)|Ag [E^(@)` for `Ni^(2+)//Ni =- 0.25` volt, `E^(@)` for `Ag^(+)//Ag = 0.80` volt]A. `-0.25+0.80=0.55`VoltB. `-0.25-(+0.80)=-1.05`voltC. `0+0.80-(-0.25)=+1.05` voltD. `-0.80-(-0.25)=-0.55`volt |
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Answer» Correct Answer - C `E_(cell)=E_(Ni//Ni^(2+))^(@)+E_(Ag^(+))//Ag)^(@)` `=0.25+0.80=1.05"volt"` |
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| 38. |
Which of the followig statements about the spontaneous reaction occurring in a galvanic cell is always true?A. `E_(cell)^(@)lt0,triangleG^(@)0` and `QltK`B. `E_(cell)^(@)gt0,triangleG^(@)lt0`, and `QgtK`C. `E_(cell)^(@)gt0,triangleG^(@)gt0` and `QgtK`D. `E_(cell)gt0,triangleGlt0` and `QltK` |
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Answer» Correct Answer - D For spontaneous reaction in every condition `E_(cell)gt0,triangleGlt0` and Q (reaction quotient) `ltK` (equilibrium constant) |
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| 39. |
Calculate `E_(cell)^(0)` of (at 298K) `Zn(s)//ZnSO_(4)(aq)||CuSO_(4)(aq)//Cu(s)` given that `E_(Zn//Zn^(2+)(aq))^(0)=0.76V` `E_(Cu(s)//Cu^(2+)(aq))^(0)=-0.34V` |
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Answer» `E_(cell)^(0)=(S.R.P)_("cathode")-(S.R.P)_("anode")` `=0.34-(-0.76)=1.1V` |
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| 40. |
Write the electrode reaction and the net cell reaction for the following cells, which electrode would be the positive terminal in each cell? (a) `Zn|Zn^(2+)||Br^(-),Br_(2)|Pt` (b). `Cr|Cr^(3+)||I^(-),I_(2)|Pt`. (c). `Pt|H_(2),H^(+)||Cu^(2+)|Cu` (d). `Cd|Cd^(2+)||Cl^(-),AgCl|Ag` |
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Answer» (a). Oxidation half cell reaction `ZntoZn^(2+)+2e^(-)` Reduction half cell rection, `Br_(2)+2e^(-)to2Br^(-)` Net cell reaction `Zn+Br_(2)toZn^(2+)+2Br^(-)` (positive terminal cathode Pt) (b). Oxidation half reaction, `[CrtoCr^(3+)+3e^(-)]xx2` reduction half reaction, `[I_(2)+2e^(-)to2I^(-)]xx3` Net cell reaction `2Cr+3I_(2)to2Cr^(3+)+6I^(-)` (positive terminal cathode Pt) (c). Oxidation half reaction, `H_(2)to2H^(+)+2e^(-)` reduction half reaction, `Cu^(2+)+2e^(-)toCu` Net cell reaction `H_(2)+Cu^(2+)toCu+2H^(+)` (Positive terminal cathode Cu) (d). Oxidation half raction, `CdtoCd^(2+)+2e^(-)` reduction half reaction, `[AgCl+e^(-)toAg+Cl^(-)]xx2` Net cell reaction `Cd+2AgCltoCd^(2)+2Ag+2Cl^(-)` (positive terminal cathode Ag) |
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| 41. |
Write short hand notation for the following reaction `Sn^(2+)(aq)+2Ag^(+)(aq)toSn^(4+)(aq)+2Ag(s)`. |
| Answer» The cell consists of a platinum wire anode dipping into an `Sn^(+2)` solution and a silver cathode dipping into an `Ag^(+)` solution therefore `Pt(s)|Sn^(2+)(aq)Sn^(4+)(aq)||Ag^(+)(aq)|Ag(s)`. | |
| 42. |
A current of 9.95 amp following for 10 minutes, deposits 3 gm of a metal. Equivalent weight of the metal is:A. 12.5B. 18.5C. 21.5D. 48.5 |
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Answer» Correct Answer - D `(W)/(E)=(it)/(96500)implies(3)/(E)=(9.95xx10xx60)/(96500)impliesE=48.5` |
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| 43. |
What is the amount of chlorine evoled when `2` amperes of current is passed for `30` minumtes in an aqueous solution of `NaCI` ?A. `9.81g`B. `1.32`gC. `4.56g`D. `12.6g` |
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Answer» Correct Answer - B `K=(1)/(R)((l)/(a))=(2xx30xx60)/(96500),W=1.32gm` |
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| 44. |
A certain current liberated 0.504 g of hydrogen in 2 hours. How many gram of copper can be liberated by the same current flowing for the same time in `CuSO_(4)` solution ?A. 31.8 gB. 16.0 gC. 12.7 gD. 63.5 g |
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Answer» Correct Answer - C `(W_(1))/(E_(1))=(W_(2))/(E_(2))` `(W_(Cu))/((63.5)/(2))=(0.504)/((2)/(2))` `W_(Cu)=0.504xx63.5=31.8gm`. |
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