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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.
| 51. |
What is the potential of an electrode which originally contained `0.1 M NO_(3)^(-)` and `0.4 M H^(+)` and which has been treated by `8%` of the cadmium necessary to reduce all the `NO_(3)^(-)` to `NO(g)` at 1 bar ? Given : `NO_(3)^(-)+4H^(+)+3e^(-)rarr NO+2H_(2)O`, `E^(@)=0.96 , log 2 = 0.3`A. `0.84 V`B. `1.08 V`C. `1.23 V`D. `1.36 V` |
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Answer» Correct Answer - A After addition of Cd and it s oxidation into `Cd^(2+)` `{:(NO_(3)^(-)(aq)+4H^(+)(aq)+3e^(-)rarr NO(g)+2H_(2)O(l)),(0.1-x " " 0.4-4x " , where" x=0.08):}` `[NO_(3)^(-)]` remaining `=0.02 M, [H^(+)]` remaining `=0.08 M` `E_(NO_(3)^(-)NO)=E_(NO_(3)^(-)NO)^(@)-(0.0591)/(3)"log"(1)/([NO_(3)^(-)][H^(+)]^(4))` `=0.96-(0.0591)/(3)"log"(1)/((0.02)(0.08)^(4))` |
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| 52. |
One electrolysis 1 mole Al atoms will be deposited byA. 1 mole of electronsB. 2 moles of electronsC. 3 mole of electronsD. 6 mole of electrons |
| Answer» Correct Answer - C | |
| 53. |
When the same quantity of electricity is passed through the solution of different electrolytes in series, the amount of product obtained is proportional to theirA. Atomic weightsB. Chemical EquivalentC. Gram molecular massD. Gram atomic ions |
| Answer» Correct Answer - B | |
| 54. |
The number of faradays required to liberate 1 mole of any element indicatesA. weight of the element.B. conductance of the electrolyteC. charge on the ion of the elementD. isotopic number |
| Answer» Correct Answer - C | |
| 55. |
During electrolysis electrons flow fromA. cations to cathodeB. anode to anionsC. cathode to anodeD. anions to anode |
| Answer» Correct Answer - D | |
| 56. |
The cathode of an electrolysis and a reducing agent are similar because bothA. are metalsB. supply electronsC. remove electronsD. absorb electrons |
| Answer» Correct Answer - B | |
| 57. |
Schematic diagram of an electrolytic-cell is:A. B. C. D. None is correct presentation |
| Answer» Correct Answer - B | |
| 58. |
Consider following sets `:` Blue colour solution changes to colourless ( or fades ) inA. I,II,IIIB. I,IIC. II,IIID. I,III |
| Answer» Correct Answer - D | |
| 59. |
The dissociation constant of `[M (NH_(3))_(2)]^(+)` complex into `M^(+)` and Ammonia is `1xx10^(-17)`. The standard reduction poential of the same metal electrode `[E_(M^(+),M)^(@)]` is `1.02 V`. Hence `E^(@)` of the half cell reaction : `[M(NH_(3))_(2)]^(+)+e^(-)rarrM(s)+2NH_(3)` is found to be `chi xx 10^(-2)`. Express the value of x to the nearest integer ? (Given R = 8.3 and 2.3 log x = `log_(e )x`)__ |
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Answer» Correct Answer - 2 `Eq(A)rarr[M(NH_(3))_(2)]^(+)underset(larr)rarr M^(+)+2NH_(3)` `Delta G_(1)^(@)=-(kJ.2.303 log 10^(-17))/(96,500)=1.0F` `Eq(B)rarr M^(+)+e^(-)underset(larr)rarr M(s)` `Delta G_(2)^(@)=-1.02F` on adding the two `[M(NH_(3))_(2)]^(+)+e^(-)rarr M(S)+2NH_(3)` `Delta G_(3)^(@)=-nFE^(@)` `Delta G_(3)^(@)=Delta G_(1)^(@)+Delta G_(3)^(@)=-0.02F-1xxFxxE^(@)=-0.02F` `E^(@)=2xx10^(-2)` x = 2 |
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| 60. |
The oxidation potential of hydrogen electrode `H_(2)//H_(2)O^(+)` (aq) will be greater than zero if,A. Concentration of `H_(3)O^(+)` ions is 1MB. Concentration of `H_(3)O^(+)` ions is 2MC. Partial pressure of `H_(2)` gas is 2 atmD. E(oxidation)can never be +ve |
| Answer» Correct Answer - C | |
| 61. |
Statement : If two half reaction with electrode potential `E_(1)^(@)` and `E_(2)^(@)` gives a third reaction then, `DeltaG_(3)^(@) = DeltaG_(1)^(@) + DeltaG_(2)^(@)` Explanation : `E_(3)^(@) = E_(1)^(@) + E_(2)^(@)`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 - A In such case `E^(@)` are not additive |
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| 62. |
Water is a non-electrolyte but conducts electricity on dissovling a small amount ofA. NaClB. SugarC. AcetoneD. Oxygen |
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Answer» Correct Answer - A Because NaCl when dissolved in water produces Ions. |
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| 63. |
`0.05M` aqueous solution of NaCl is electrolysed. If a current of strength `0.5`amp is used for 193sec. The final concentration of `Na^(+)` ions in the electrolyte will be (volume of solution will be constant)A. `0.05M`B. `0.049M`C. `0.051M`D. `0.04M` |
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Answer» Correct Answer - A Aqueous solution of NaCl does not produce Na cathode. Hence `Na^(+)` concentration reamins same |
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| 64. |
Which one of the following could not be liberated from a suitable electrolyte by the passage of `0.25` faraday of electricity through that electrolyteA. `0.25` mole of AgB. 16gm of CuC. 2gm of `O_(2)` (g)D. `2.8` ltrs of `H_(2)` at STP |
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Answer» Correct Answer - B `0.25F` will deposit `0.25` equivalents. |
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| 65. |
Iron rod is immersed in Kcl solution such that half its length is exposed to air and the other half immersed in KCl solution. The part corroded isA. Part of the rod exposed to airB. Part of the rod immerced in KCl solutionC. Both 1 & 2D. None of the above |
| Answer» Correct Answer - B | |
| 66. |
The unit of electrochemical equivalent isA. GramB. Gram/AmpereC. Gram/CoulombD. Coulomb/Gram |
| Answer» Correct Answer - C | |
| 67. |
The density of copper is 8 gm/cc. Number of coulombs required to plate an area of 10 cm x 10 cm on both sides to a thickness of `10^(-2)` cm using `CuSO_(4)` solution as electrolyte is (Atomic weight of Cu - 64g)A. `48,250`B. `24,125`C. `96,500`D. `10,000` |
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Answer» Correct Answer - A `V="area"xx"thickness"=1cc` Total volume 2cc 1cc volume = 8 gms of copper wt of copper for 2cc = 16 gms 96500C = 32 gms of copper 32 gms of copper = 96500C 16gms = ? |
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| 68. |
A very thin copper plate is electro`-` plated with gold using gold chloride in `HCl`. The current was passed for `20 mi n`. And the increase in the weight of the plate was found to be `2g.[Au=197]`. The current passed was `-`A. `0.816 amp`B. `1.632 amp`C. `2.448 amp`D. `3.264 amp` |
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Answer» Correct Answer - 3 Gold chloride `AuCl_(3)` `m=(M)/(96500xxn)1xxt` `l=(2xx96500xx3)/(197xx20xx60)=2.448A`. |
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| 69. |
The electro chemical equivalent of an element is `0.0006735g//C`. Its equivalent weight isA. 65B. `67.65`C. 130D. `32.5` |
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Answer» Correct Answer - A `E=exx96500` |
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| 70. |
The electrochemical equivalent of two substanes are `E_(1)` and `E_(2)`. The current that flows to deposit their equal amount at the cathodes in the same time must be in the ratio ofA. `E_(1):E_(2)`B. `E_(2):E_(1)`C. `E_(1):E_(2)-E_(1)`D. `E_(1)XE_(2):E_(1)+E_(2)` |
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Answer» Correct Answer - B `E_(1)Q_(1)=E_(2)Q_(2)` |
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| 71. |
The same quantity of electricity is passed through `0.1` M `H_(2)SO_(4)` and `0.1` M HCl. The amounts of `H_(2)` obtained at the cathodes are in the ratioA. `1:1`B. `2:1`C. `1:2`D. `3:1` |
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Answer» Correct Answer - A W`alpha`E |
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| 72. |
The electrochemical equivalent of a metal is 'x' g `"coulomb"^(-1)` . The equivalent weight of metal isA. xB. `x xx96500`C. `x//96500`D. `1.6xx10^(-19)xx x` |
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Answer» Correct Answer - B `e(E)/(96500)` |
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| 73. |
Consult the table of standard electrode potential and suggest three substances that can oxidize `Fe^(2+)` ions under suitable conditions. |
| Answer» Only those species can oxidise ferrous ions `(Fe^(2+))` whose standard reduction potential is more positive than `0.77V`. Thus, suitable oxidising agents will be `F_(2),Cl_(2),Br_(2)`. | |
| 74. |
A cell is containing two H electrodes. The negative electrode is in contact with a solution of `10^(-6)MH^(+)` ion. The e.m.f. of the cell is `0.118` volt at `25^(@)C` . Calculate `[H^(+)]` at positive electrode. |
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Answer» Anode: `H_(2)rarr2H^(+)2e` (nagative polarity) `[H^(+)]=10^(6)M` `"Cathode: "2H^(+)+2erarrH_(2)` (positive polarity) `[H^(+)]rarraM` `thereforeE_(cell) =E_(OP_(H//H^(+)))+E_(Rp_(H^(+)//H))` `=E_(OP_(H//H^(+)))^(@) (0.059)/(2) log_(10)[H^(+)]_("anode")^(2)+E_(RP_(H^(+)//H))^(0)+(0.059)/(2)log_(10) [H^(+)]_("cathode")^(2)` `=(0.059)/(2)log_(10).([H^(+)]_("cathode")^(2))/([H^(+)]_("anode")^(-2))` `0.118=(0.059)/(2)log_(10).([H^(+)]_("cathode")^(2))/((10^(-6))^(2))` `=(0.059)/(2)log_(10).([H^(+)]_("cathode")^(2))/(10^(-6))` `therefore[H^(+)]_("cathode")=10^(-6)M` |
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| 75. |
`9.65` amp of current was passed for one hour through Daniel cell. The loss of mass of zinc anode isA. `11.76g`B. `1.176g`C. `5.88g`D. `2.94g` |
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Answer» Correct Answer - A `m=((65.4//2))/(96500)xx9.65xx1xx60xx60` |
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| 76. |
In a gydrogen - oxygen fuel cell, `67.2` litre of `H_(2)` at S.T.P is used in 5 min. What is the average current produced?A. ` 549.4` ampB. `643.33` ampC. 965 ampD. 1930 amp |
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Answer» Correct Answer - D Use Faraday s first law |
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| 77. |
`E_(Al^(3+)//Al)^(@)=-1.66 V` and `K_(SP)` of `Al(OH)_(3)=1.0xx10^(-33)`. Reduction potential of the above couple at `pH=14` is `:`A. `-2.31 V`B. `+2.31`C. `-1.01V`D. `+1.01V` |
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Answer» Correct Answer - A `pH=14" "rArr" "[CH^(-)]=1M` `[Al^(3+)]=(K_(sp))/([CH^(-)]^(3))=10^(-33)` `Al^(3+)+3e^(-)rarr Al` `E_(cell)=E_(Al^(3+)//Al)^(@)-(0.0591)/(3)log.(1)/([Al^(3+)])` |
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| 78. |
`E_(Al^(3+)//Al)^(*)=-1.66V` and `K_(SP)` of `Al(OH)_(3)=1.0xx10^(-33)` . Reduction potential of the above couple at `pH =14` is `:`A. `-2.31 V`B. `+2.31`C. `-1.01V`D. `+1.01V` |
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Answer» Correct Answer - 1 `pH=14" "rArr" "[CH^(-)]=1M` `[Al^(3+)]=(K_(sp))/([OH^(-)]^(3))=10^(-33),Al^(3+)+3e^(-)rarr Al` `E_(cell)=E_(Al^(3+)//Al)^(@)-(0.0591)/(3)log.(1)/([Al^(3+)])` |
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| 79. |
The standard electrode potential for the reactions, `Ag^(+)(aq)+e^(-) rarr Ag(s)` `Sn^(2+)(aq)+2e^(-) rarr Sn(s)` at `25^(@)C` are `0.80 ` volt and `-0.14 ` volt, respectively. The `emf`of the cell `Sn|Sn^(2+)(1M)||Ag^(+)(1M)Ag` is `:`A. `0.66 `voltB. `0.80 ` voltC. `1.08` voltD. `0.94` volt |
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Answer» Correct Answer - 4 `E_(cell)=E_(cell)^(@)-(0.0591)/(2)log.([Zn^(2+)])/([Ag^(+)]^(2))" "=0.8-(-0.14)-(0.0591)/(2)log.(1)/(1^(2))=0.94V.` |
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| 80. |
Which of the following is not true for a galvanic cell represented in IUPAC systemA. Right hand electrode is a `+ve` terminalB. Right hand electrode acts as cathodeC. Electrons are given out in the external circuit from the anodeD. Electrons are given out in the external circuit from the cathode. |
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Answer» Correct Answer - D At cathode reduction takes place |
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| 81. |
The `E^(@)` at `25^(@)` for the following reaction at the indicated concentration is 1.50 V. Calculate the `DeltaG` in kJ at `25^(@)` C: `Cr(s)+3Ag^(+)(aq,0.1M)` `rarrAg(s)+Cr^(3+)(aq,0.1M)`A. `-140.94`B. `-295`C. `-212`D. `-422.83` |
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Answer» Correct Answer - D `DeltaG=DeltaG^(@)+RT " In" Q` `rArr-nFE^(@)+2.303RT " log"([Cr^(3+)])/[Ag^(3)]^(3)` `DeltaG=-3xx96500xx1.50+8.314xx298` `2.303"log"((0.1))/((0.1)^(3))rArr-422838.3J or -422.83 kJ` |
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| 82. |
One faraday of electericity is passed separately through one litre of one molar aqueous soltion of I) `AgNO_(3)`, ii) `SnCl_(4)` and iii) `CuSO_(4)`. The number of moles of Ag, Sn and Cu deposited at cathode are respectivelyA. `1.0,0.25,0.5`B. `1.0,0.5,0.25`C. `0.5,1.0,0.25`D. `0.25,0.5,1.0` |
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Answer» Correct Answer - A `Ag^(+):Sn^(+4):Cu^(+2)therefore(1)/(1):(1)/(4):(1)/(2)` |
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| 83. |
`Zn` acts as sacrifical or cathodic protect iont to prevent rusting of iron becauseA. `E_(OP)^(@) " of " Zn lt E_(OP)^(@) " of " Fe`B. `E_(OP)^(@)" of "Zn gt E_(OP)^(@) " of "Fe`C. `E_(OP)^(@) " of " Zn = E_(OP)^(@) " of "Fe`D. Zn is cheaper than iron |
| Answer» Correct Answer - B | |
| 84. |
In the Hall process, aluminium is produced by the electrolysis of molten `Al_2O_3`. How many second would it take to produce enough aluminium by the Hall process to make a case of 24 cans of auminium soft-drink, if each can uses 5.0g of Al, a current of 9650amp is employed and the current efficiency of the cell is 90%:A. 203.2B. 148.14C. 333D. 6.17 |
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Answer» Correct Answer - B No. of equivalent of aluminu, `(W)/(E)=(Exxetaxxt)/(96500)` `(24xx5)/(27)xx3=(9650xx0.9xxt)/(96500),t=148.14sec` |
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| 85. |
The charge required for the oxidation of one mole of `Mn_(3)O_(4)` to `MnO_(4)^(2-)` in alkaliine medium is (assume `100%` current efficiency):A. `10//3F`B. 6FC. 10FD. 4F |
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Answer» Correct Answer - C Change in O.N per mole of `Mn_(3)O_(4)=10` 1 mole of `Mn_(3)O_(4)=10 "equiv"-=10F` |
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| 86. |
`Na-` amalgam is prepared by electrolysis of `NaCl` solution using liquid `Hg` as cathode . How long should the current of `10 amp`. Is passed to produce `10% Na-Hg` on a cathode of `10gm Hg`. ( atomic mass of `Na=23)`.A. `7.77 `minB. `9.44` min.C. `5.24 mi n.`D. `11.39 mi n`. |
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Answer» Correct Answer - 1 90 gm Hg has 10 gm Na 10 gm Hg `=(10)/(90)xx10=(10)/(9)gm" "Na` weight of Na`=(M)/(n)xx(ixxt)/(96500)` `(10)/(9)=(23)/(1)xx(10xxt)/(96500)" "[:.Na^(+)+erarrNa]` `t=(10xx96500)/(9xx10xx23)=7.77mi n` |
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| 87. |
The concentration of potassium ions inside a biological cell is at least 20 times higher than 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 simplel model for a concentration cell involving a metal `M` is `M(s)|M^(o+)(aq,0.05 ` molar`)||M^(o+)(aq,1` molar`)|M(s)` For the abov electrolytic cell, the magnitude of the cell potential is `|E_(cell)|=70mV.` If the `0.05` moolar solution of `M^(o+)` is replaced by a `0.0025` molar `M^(o+)` solution, then the magnitude of the cell potential would beA. 35 mVB. 70 mVC. 140 mVD. 700 mV |
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Answer» Correct Answer - C In this case `E_(cell)=(0.059)/(1)log_(10).([1.0])/(2.5xx10^(-3))` `=0.152Vor ~~140mV` |
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| 88. |
Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing `0.2` Faraday electricty through an aqueous solution of potassium succimate, the total volume of gases (all both cathode and anode) liberated at STP (1 atm and 273 K) isA. `6.72L`B. `2.24L`C. `4.48L`D. `8.96L` |
| Answer» Correct Answer - D | |
| 89. |
A student made the following observations in the laboratory, A) Clean copper metal did not react with 1mole `Pb(NO_(3))_(2)` solution B) Clean lead metal dissolved in a 1 molar `AgNO_(3)` solution and crystals of Ag metal appeared C) Clean silver metal did not react with 1 molar `Cu(NO_(3))_(2)` solution. The order of decreasing reducing character of the three metals isA. `Cu, Pb, Ag`B. `Cu, Ag, Pb`C. `Pb, Cu, Ag`D. `Pb, Ag, Cu` |
| Answer» Correct Answer - C | |
| 90. |
At any instant during the reaction `Zn+Cu^(++)rarrZn^(++)+Cu` Occurring in an open beaker at temperature TA. `DeltaG^(@)=-RT"ln"([Zn^(++)].[Cu])/([Zn].[Cu^(++)]`B. `-DeltaG`= work available from the reactionC. `DeltaG=0`D. `DeltaG gt 0` |
| Answer» Correct Answer - B | |
| 91. |
A 1 M solution of `H_2SO_4` is electrolysed. Select correct statement in respect of products obtain at anode and cathode respectively: Given : `2SO_4^(2-) toS_2O_8^(2-)+2e^-,E^(@)=-1.23V` `H_2O(l)to2H^+(aq)+1//2O_2(g)+2e^-,E^(@)=-1.23V`A. concentration of `H_(2)SO_(4)` remain constant ,`H_(2),O_(2)`B. concentration of `H_(2)SO_(4)` increases,`O_(2),H_(2)`C. concentraion of `H_(2)SO_(4)` decreases , `O_(2),H_(2)`D. concentration of `H_(2)SO_(4)` remains constant, `S_(2)O_(8)^(2-),H_(2)` |
| Answer» Correct Answer - B | |
| 92. |
The standard oxidation potential of `Ni//Ni^(2+)` electrode is `0.236 V`. If this is combined with a hydrogen electrode in acid solution, at what `pH` of this solution will be measured e.m.f. be zero at `25^(@)C`? Assume that `[Ni^(2+)] = 1 M` and `P_(H_(2)) = 1 atm`. |
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Answer» Correct Answer - D `NirarrNi^(2+)+2e,E_(OP)^(@)=0.236V` `2H^(+)+2erarrH_(2),(E_(RP)^(@)=0)/(E_(cell)^(0)=0.236V)` `E_(cell)=E_(cell)^(@)+(0.059)/(2)log_(10).([H^(+)]^(2))/([Ni^(2+)])` `0=0.236+(0.059)/(2)log_(10)[H^(+)]` `-log_(10)H^(+)=4thereforepH=4` |
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| 93. |
We observe blue colour if:A. Cu electrode is placed in the `AgNO_(3)` solutionB. Cu electrode is placed in the `ZnSO_(4)`C. Cu electrode is placed in the dil `HNO_(3)`D. Cu electrode is placed in the `NiSO_(4)` |
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Answer» Correct Answer - A::C Due to formation of `Cu^(2+)` ions in the solution |
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| 94. |
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 oxidized by `O_(2)`B. `Fe^(2+)` is oxidized by iodineC. Iodide ion is oxidized by chlorineD. `Mn^(2+)` is oxidized by chlorine |
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Answer» Correct Answer - C Reduction potential of `I_(2)` is less than `CI_(2)`. |
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| 95. |
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` While `Fe^(3+)` is stable, `Mn^(3+)` is not stable in acid solution becauseA. `O_(2)` oxidises `Mn^(2+)` to `Mn^(3+)`B. `O_(2)` oxidises both `Mn^(2+)` and `Fe^(2+)` to `Fe^(3+)`C. `Fe^(3+)` oxidizes `H_(2)O` to `O_(2)`D. `Mn^(3+)` oxidizes `H_(2)O` to `O_(2)` |
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Answer» Correct Answer - D Reaction of `Mn^(3+)` with `H_(2)O` is spontaneous. |
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| 96. |
Redox reactions play a vital role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cells reactions decide which way the reaction is expected to proceed. A simple example is a Daniell cell in which zince goes into solution and copper gets deposited. Given below are set of half-cell reactions (acidic medium ) along with their `E^(@)` in V with respect to normal hydrogen electrode values. `{:(l_(2)+2e^(-)rarr2l^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-)" ",E^(@)=1.36),(Mn^(3+)+e^(-)rarrMn^(2+),E^(2)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+)" ",E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):}` Among the following, identify the correct statementA. Chloride ion is oxidised by `O_(2)`B. `Fe^(2+)` is oxidised by chlorineC. Iodide ion is oxidised by chlorineD. `Mn^(2+)` is oxidised by chlorine |
| Answer» Correct Answer - C | |
| 97. |
Redox reactions play a vital role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cells reactions decide which way the reaction is expected to proceed. A simple example is a Daniell cell in which zince goes into solution and copper gets deposited. Given below are set of half-cell reactions (acidic medium ) along with their `E^(@)` in V with respect to normal hydrogen electrode values. `{:(l_(2)+2e^(-)rarr2l^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-)" ",E^(@)=1.36),(Mn^(3+)+e^(-)rarrMn^(2+),E^(2)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+)" ",E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):}` while `Fe^(3+)` is stable, `Mn^(3+)` is not stable in acid solution because :A. `O_(2)`oxidises `Mn^(2+)` to `Mn^(3+)`B. `O_(2)` oxidises both `MN=n^(2+)` and `Fe^(2+)`C. `Fe^(3+)` oxidizes `H_(2)O` to `O_(2)`D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)` |
| Answer» Correct Answer - D | |
| 98. |
For the feasibility of a redox reaction in a cell, the emf should be.A. negativeB. positiveC. zeroD. some times positive and some times `-ve` |
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Answer» Correct Answer - B For a cell to be spontaneous, EMF should be positive. |
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| 99. |
Arrange the following in the order of their decreasing electrode potentials:Mg, K, Ba,CaA. K,Ba,Ca, MgB. BA,Ca,K,MgC. Ca,Mg,K,BaD. Mg,Ca,Ba,K |
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Answer» Correct Answer - D `Ma gt Ca gt Ba gt K`. This is the order of SRP. |
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| 100. |
What will be the emf for the given cell ? `Pt|H_(2)(g,P_(1))|H^(+)(aq)|H_(2)(g,P_(2))|Pt`A. `(RT)/(F)log.(P_(1))/(P_(2))`B. `(RT)/(2F)log.(P_(1))/(P_(2))`C. `(RT)/(F)log.(P_(2))/(P_(1))`D. none of these |
| Answer» Correct Answer - D | |